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Proposition 1.11. Let \( K, f \) be as above. The element a of \( K \) is a multiple root of \( f \) if and only if it is a root and \( {f}^{\prime }\left( a\right) = 0 \) .
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Proof. Factoring \( f \) as above, we get\n\n\[ \n{f}^{\prime }\left( X\right) = {\left( X - a\right) }^{m}{g}^{\prime }\left( X\right) + m{\left( X - a\right) }^{m - 1}g\left( X\right) .\n\]\n\nIf \( m > 1 \), then obviously \( {f}^{\prime }\left( a\right) = 0 \) . Conversely, if \( m = 1 \) then\n\n\[ \n{f}^{\prime }\left( X\right) = \left( {X - a}\right) {g}^{\prime }\left( X\right) + g\left( X\right)\n\]\n\nwhence \( {f}^{\prime }\left( a\right) = g\left( a\right) \neq 0 \) . Hence if \( {f}^{\prime }\left( a\right) = 0 \) we must have \( m > 1 \), as desired.
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Yes
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Proposition 1.12. Let \( f \in K\left\lbrack X\right\rbrack \) . If \( K \) has characteristic 0, and \( f \) has degree \( \geqq 1 \), then \( {f}^{\prime } \neq 0 \) . Let \( K \) have characteristic \( p > 0 \) and \( f \) have degree \( \geqq 1 \) . Then \( {f}^{\prime } = 0 \) if and only if, in the expression for \( f\left( X\right) \) given by\n\n\[ f\left( X\right) = \mathop{\sum }\limits_{{v = 0}}^{n}{a}_{v}{X}^{v} \]\n\n\( p \) divides each integer \( v \) such that \( {a}_{v} \neq 0 \) .
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Proof. If \( K \) has characteristic 0, then the derivative of a monomial \( {a}_{v}{X}^{v} \) such that \( v \geqq 1 \) and \( {a}_{v} \neq 0 \) is not zero since it is \( v{a}_{v}{X}^{v - 1} \) . If \( K \) has characteristic \( p > 0 \), then the derivative of such a monomial is 0 if and only if \( p \mid v \), as contended.
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Yes
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Theorem 2.1. (Gauss Lemma). Let \( A \) be a factorial ring, and let \( K \) be its quotient field. Let \( f, g \in K\left\lbrack X\right\rbrack \) be polynomials in one variable. Then\n\n\[ \operatorname{cont}\left( {fg}\right) = \operatorname{cont}\left( f\right) \operatorname{cont}\left( g\right) \]
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Proof. Writing \( f = c{f}_{1} \) and \( g = d{g}_{1} \) where \( c = \operatorname{cont}\left( f\right) \) and \( d = \operatorname{cont}\left( g\right) \) , we see that it suffices to prove: If \( f, g \) have content 1, then \( {fg} \) also has content 1, and for this, it suffices to prove that for each prime \( p \) , \( {\operatorname{ord}}_{p}\left( {fg}\right) = 0 \) . Let\n\n\[ f\left( X\right) = {a}_{n}{X}^{n} + \cdots + {a}_{0},\;{a}_{n} \neq 0, \]\n\n\[ g\left( X\right) = {b}_{m}{X}^{m} + \cdots + {b}_{0},\;{b}_{m} \neq 0, \]\n\nbe polynomials of content 1 . Let \( p \) be a prime of \( A \) . It will suffice to prove that \( p \) does not divide all coefficients of \( {fg} \) . Let \( r \) be the largest integer such that \( 0 \leqq r \leqq n,{a}_{r} \neq 0 \), and \( p \) does not divide \( {a}_{r} \) . Similarly, let \( {b}_{s} \) be the coefficient of \( g \) farthest to the left, \( {b}_{s} \neq 0 \), such that \( p \) does not divide \( {b}_{s} \) . Consider the coefficient of \( {X}^{r + s} \) in \( f\left( X\right) g\left( X\right) \) . This coefficient is equal to\n\n\[ c = {a}_{r}{b}_{s} + {a}_{r + 1}{b}_{s - 1} + \cdots \]\n\n\[ + {a}_{r - 1}{b}_{s + 1} + \cdots \]\n\nand \( p \nmid {a}_{r}{b}_{s} \) . However, \( p \) divides every other non-zero term in this sum since in each term there will be some coefficient \( {a}_{i} \) to the left of \( {a}_{r} \) or some coefficient \( {b}_{j} \) to the left of \( {b}_{s} \) . Hence \( p \) does not divide \( c \), and our lemma is proved.
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Yes
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Corollary 2.2. Let \( f\left( X\right) \in A\left\lbrack X\right\rbrack \) have a factorization \( f\left( X\right) = g\left( X\right) h\left( X\right) \) in \( K\left\lbrack X\right\rbrack \) . If \( {c}_{g} = \operatorname{cont}\left( g\right) ,{c}_{h} = \operatorname{cont}\left( h\right) \), and \( g = {c}_{g}{g}_{1}, h = {c}_{h}{h}_{1} \), then \[ f\left( X\right) = {c}_{g}{c}_{h}{g}_{1}\left( X\right) {h}_{1}\left( X\right) \] and \( {c}_{g}{c}_{h} \) is an element of \( A \) . In particular, if \( f, g \in A\left\lbrack X\right\rbrack \) have content 1, then \( h \in A\left\lbrack X\right\rbrack \) also.
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Proof. The only thing to be proved is \( {c}_{g}{c}_{h} \in A \) . But \[ \operatorname{cont}\left( f\right) = {c}_{g}{c}_{h}\operatorname{cont}\left( {{g}_{1}{h}_{1}}\right) = {c}_{g}{c}_{h} \] whence our assertion follows.
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Yes
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Theorem 2.3. Let \( A \) be a factorial ring. Then the polynomial ring \( A\left\lbrack X\right\rbrack \) in one variable is factorial. Its prime elements are the primes of \( A \) and polynomials in \( A\left\lbrack X\right\rbrack \) which are irreducible in \( K\left\lbrack X\right\rbrack \) and have content 1 .
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Proof. Let \( f \in A\left\lbrack X\right\rbrack, f \neq 0 \) . Using the unique factorization in \( K\left\lbrack X\right\rbrack \) and the preceding corollary, we can find a factorization\n\n\[ f\left( X\right) = c \cdot {p}_{1}\left( X\right) \cdots {p}_{r}\left( X\right) \]\n\nwhere \( c \in A \), and \( {p}_{1},\ldots ,{p}_{r} \) are polynomials in \( A\left\lbrack X\right\rbrack \) which are irreducible in \( K\left\lbrack X\right\rbrack \) . Extracting their contents, we may assume without loss of generality that the content of \( {p}_{i} \) is 1 for each \( i \) . Then \( c = \operatorname{cont}\left( f\right) \) by the Gauss lemma. This gives us the existence of the factorization. It follows that each \( {p}_{i}\left( X\right) \) is irreducible in \( A\left\lbrack X\right\rbrack \) . If we have another such factorization, say\n\n\[ f\left( X\right) = d \cdot {q}_{1}\left( X\right) \cdots {q}_{s}\left( X\right) \]\n\nthen from the unique factorization in \( K\left\lbrack X\right\rbrack \) we conclude that \( r = s \), and after a permutation of the factors we have\n\n\[ {p}_{i} = {a}_{i}{q}_{i} \]\n\nwith elements \( {a}_{i} \in K \) . Since both \( {p}_{i},{q}_{i} \) are assumed to have content 1, it follows that \( {a}_{i} \) in fact lies in \( A \) and is a unit. This proves our theorem.
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Yes
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Corollary 2.4. Let \( A \) be a factorial ring. Then the ring of polynomials in \( n \) variables \( A\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is factorial. Its units are precisely the units of \( A \), and its prime elements are either primes of \( A \) or polynomials which are irreducible in \( K\left\lbrack X\right\rbrack \) and have content 1 .
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Proof. Induction.\n\nIn view of Theorem 2.3, when we deal with polynomials over a factorial ring and having content 1 , it is not necessary to specify whether such polynomials are irreducible over \( A \) or over the quotient field \( K \) . The two notions are equivalent.
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No
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Theorem 3.1. (Eisenstein’s Criterion). Let \( A \) be a factorial ring. Let \( K \) be its quotient field. Let \( f\left( X\right) = {a}_{n}{X}^{n} + \cdots + {a}_{0} \) be a polynomial of degree \( n \geqq 1 \) in \( A\left\lbrack X\right\rbrack \) . Let \( p \) be a prime of \( A \), and assume:\n\n\[ \n{a}_{n} ≢ 0\;\left( {\;\operatorname{mod}\;p}\right) ,\;{a}_{i} \equiv 0\;\left( {\;\operatorname{mod}\;p}\right) \;\text{ for all }\;i < n, \n\]\n\n\[ \n{a}_{0} ≢ 0\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) \n\]\n\nThen \( f\left( X\right) \) is irreducible in \( K\left\lbrack X\right\rbrack \) .
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Proof. Extracting a g.c.d. for the coefficients of \( f \), we may assume without loss of generality that the content of \( f \) is 1 . If there exists a factorization into factors of degree \( \geqq 1 \) in \( K\left\lbrack X\right\rbrack \), then by the corollary of Gauss’ lemma there exists a factorization in \( A\left\lbrack X\right\rbrack \), say \( f\left( X\right) = g\left( X\right) h\left( X\right) \), \n\n\[ \ng\left( X\right) = {b}_{d}{X}^{d} + \cdots + {b}_{0} \n\]\n\n\[ \nh\left( X\right) = {c}_{m}{X}^{m} + \cdots + {c}_{0} \n\]\n\nwith \( d, m \geqq 1 \) and \( {b}_{d}{c}_{m} \neq 0 \) . Since \( {b}_{0}{c}_{0} = {a}_{0} \) is divisible by \( p \) but not \( {p}^{2} \), it follows that one of \( {b}_{0},{c}_{0} \) is not divisible by \( p \), say \( {b}_{0} \) . Then \( p \mid {c}_{0} \) . Since \( {c}_{m}{b}_{d} = {a}_{n} \) is not divisible by \( p \), it follows that \( p \) does not divide \( {c}_{m} \) . Let \( {c}_{r} \) be the coefficient of \( h \) furthest to the right such that \( {c}_{r} ≢ 0\left( {\;\operatorname{mod}\;p}\right) \) . Then \n\n\[ \n{a}_{r} = {b}_{0}{c}_{r} + {b}_{1}{c}_{r - 1} + \cdots . \n\]\n\nSince \( p \nmid {b}_{0}{c}_{r} \) but \( p \) divides every other term in this sum, we conclude that \( p \nmid {a}_{r} \), a contradiction which proves our theorem.
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Yes
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Theorem 3.2. (Reduction Criterion). Let \( A, B \) be entire rings, and let\n\n\[ \n\varphi : A \rightarrow B \n\] \n\nbe a homomorphism. Let \( K, L \) be the quotient fields of \( A \) and \( B \) respectively. Let \( f \in A\left\lbrack X\right\rbrack \) be such that \( {\varphi f} \neq 0 \) and \( \deg {\varphi f} = \deg f \) . If \( {\varphi f} \) is irreducible in \( L\left\lbrack X\right\rbrack \), then \( f \) does not have a factorization \( f\left( X\right) = g\left( X\right) h\left( X\right) \) with \n\n\[ \ng, h \in A\left\lbrack X\right\rbrack \;\text{ and }\;\deg g,\deg h \geqq 1. \n\]
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Proof. Suppose \( f \) has such a factorization. Then \( {\varphi f} = \left( {\varphi g}\right) \left( {\varphi h}\right) \) . Since \( \deg {\varphi g} \leqq \deg g \) and \( \deg {\varphi h} \leqq \deg h \), our hypothesis implies that we must have equality in these degree relations. Hence from the irreducibility in \( L\left\lbrack X\right\rbrack \) we conclude that \( g \) or \( h \) is an element of \( A \), as desired.
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Yes
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Theorem 4.1. Let \( A \) be a commutative Noetherian ring. Then the polynomial ring \( A\left\lbrack X\right\rbrack \) is also Noetherian.
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Proof. Let \( \mathfrak{A} \) be an ideal of \( A\left\lbrack X\right\rbrack \) . Let \( {\mathfrak{a}}_{i} \) consist of 0 and the set of elements \( a \in A \) appearing as leading coefficient in some polynomial\n\n\[ \n{a}_{0} + {a}_{1}X + \cdots + a{X}^{i} \n\]\n\nlying in \( \mathfrak{A} \) . Then it is clear that \( {\mathfrak{a}}_{i} \) is an ideal. (If \( a, b \) are in \( {\mathfrak{a}}_{i} \), then \( a \pm b \) is in \( {a}_{i} \) as one sees by taking the sum and difference of the corresponding polynomials. If \( x \in A \), then \( {xa} \in {\mathfrak{a}}_{i} \) as one sees by multiplying the corresponding polynomial by \( x \) .) Furthermore we have\n\n\[ \n{\mathfrak{a}}_{0} \subset {\mathfrak{a}}_{1} \subset {\mathfrak{a}}_{2} \subset \cdots , \n\]\n\nin other words, our sequence of ideals \( \left\{ {a}_{i}\right\} \) is increasing. Indeed, to see this multiply the above polynomial by \( X \) to see that \( a \in {\mathfrak{a}}_{i + 1} \) .\n\nBy criterion (2) of Chapter X, \( §1 \), the sequence of ideals \( \left\{ {a}_{i}\right\} \) stops, say at \( {a}_{r} \) :\n\n\[ \n{\mathfrak{a}}_{0} \subset {\mathfrak{a}}_{1} \subset {\mathfrak{a}}_{2} \subset \cdots \subset {\mathfrak{a}}_{r} = {\mathfrak{a}}_{r + 1} = \cdots . \n\]\n\nLet\n\n\[ \n{a}_{01},\ldots ,{a}_{0{n}_{0}}\text{be generators for}{\mathfrak{a}}_{0}\text{,} \n\]\n\n..............................\n\n\( {a}_{r1},\ldots ,{a}_{r{n}_{r}} \) be generators for \( {\mathfrak{a}}_{r} \).\n\nFor each \( i = 0,\ldots, r \) and \( j = 1,\ldots ,{n}_{i} \) let \( {f}_{ij} \) be a polynomial in \( \mathfrak{A} \), of degree \( i \), with leading coefficient \( {a}_{ij} \) . We contend that the polynomials \( {f}_{ij} \) are a set of generators for \( \mathfrak{A} \).\n\nLet \( f \) be a polynomial of degree \( d \) in \( \mathfrak{A} \) . We shall prove that \( f \) is in the ideal generated by the \( {f}_{ij} \), by induction on \( d \) . Say \( d \geqq 0 \) . If \( d > r \), then we note that the leading coefficients of\n\n\[ \n{X}^{d - r}{f}_{r1},\ldots ,{X}^{d - r}{f}_{r{n}_{r}} \n\]\n\ngenerate \( {\mathfrak{a}}_{d} \) . Hence there exist elements \( {c}_{1},\ldots ,{c}_{{n}_{r}} \in A \) such that the polynomial\n\n\[ \nf - {c}_{1}{X}^{d - r}{f}_{r1} - \cdots - {c}_{{n}_{r}}{X}^{d - r}{f}_{r{n}_{r}} \n\]\n\nhas degree \( < d \), and this polynomial also lies in \( \mathfrak{A} \) . If \( d \leqq r \), we can subtract a linear combination\n\n\[ \nf - {c}_{1}{f}_{d1} - \cdots - {c}_{{n}_{d}}{f}_{d{n}_{d}} \n\]\n\nto get a polynomial of degree \( < d \), also lying in \( \mathfrak{A} \) . We note that the polynomial we have subtracted from \( f \) lies in the ideal generated by the \( {f}_{ij} \) . By induction, we can subtract a polynomial \( g \) in the ideal generated by the \( {f}_{ij} \) such that \( f - g = 0 \), thereby proving our theorem.
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Yes
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Corollary 4.2. Let \( A \) be a Noetherian commutative ring, and let \( B = \) \( A\left\lbrack {{x}_{1},\ldots ,{x}_{m}}\right\rbrack \) be a commutative ring finitely generated over \( A \) . Then \( B \) is Noetherian.
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Proof. Use Theorem 4.1 and the preceding remark, representing \( B \) as a factor ring of a polynomial ring.
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No
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Theorem 5.1. Let \( A \) be a principal entire ring, and let \( P \) be a set of representatives for its irreducible elements. Let \( K \) be the quotient field of \( A \), and let \( \alpha \in K \) . For each \( p \in P \) there exists an element \( {\alpha }_{p} \in A \) and an integer \( j\left( p\right) \geqq 0 \), such that \( j\left( p\right) = 0 \) for almost all \( p \in P,{\alpha }_{p} \) and \( {p}^{j\left( p\right) } \) are relatively prime, and \[ \alpha = \mathop{\sum }\limits_{{p \in P}}\frac{{\alpha }_{p}}{{p}^{j\left( p\right) }} \] If we have another such expression \[ \alpha = \mathop{\sum }\limits_{{p \in P}}\frac{{\beta }_{p}}{{p}^{i\left( p\right) }} \] then \( j\left( p\right) = i\left( p\right) \) for all \( p \), and \( {\alpha }_{p} \equiv {\beta }_{p}{\;\operatorname{mod}\;{p}^{j\left( p\right) }} \) for all \( p \) .
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Proof. We first prove existence, in a special case. Let \( a, b \) be relatively prime non-zero elements of \( A \) . Then there exists \( x, y \in A \) such that \( {xa} + {yb} = 1 \) . Hence \[ \frac{1}{ab} = \frac{x}{b} + \frac{y}{a} \] Hence any fraction \( c/{ab} \) with \( c \in A \) can be decomposed into a sum of two fractions (namely \( {cx}/b \) and \( {cy}/a \) ) whose denominators divide \( b \) and \( a \) respectively. By induction, it now follows that any \( \alpha \in K \) has an expression as stated in the theorem, except possibly for the fact that \( p \) may divide \( {\alpha }_{p} \) . Canceling the greatest common divisor yields an expression satisfying all the desired conditions. As for uniqueness, suppose that \( \alpha \) has two expressions as stated in the theorem. Let \( q \) be a fixed prime in \( P \) . Then \[ \frac{{\alpha }_{q}}{{q}^{j\left( q\right) }} - \frac{{\beta }_{q}}{{q}^{i\left( q\right) }} = \mathop{\sum }\limits_{{p \neq q}}\frac{{\beta }_{p}}{{p}^{i\left( p\right) }} - \frac{{\alpha }_{p}}{{p}^{j\left( p\right) }} \] If \( j\left( q\right) = i\left( q\right) = 0 \), our conditions concerning \( q \) are satisfied. Suppose one of \( j\left( q\right) \) or \( i\left( q\right) > 0 \), say \( j\left( q\right) \), and say \( j\left( q\right) \geqq i\left( q\right) \) . Let \( d \) be a least common multiple for all powers \( {p}^{j\left( p\right) } \) and \( {p}^{i\left( p\right) } \) such that \( p \neq q \) . Multiply the above equation by \( d{q}^{j\left( q\right) } \) . We get \[ d\left( {{\alpha }_{q} - {q}^{j\left( q\right) - i\left( q\right) }{\beta }_{q}}\right) = {q}^{j\left( q\right) }\beta \] for some \( \beta \in A \) . Furthermore, \( q \) does not divide \( d \) . If \( i\left( q\right) < j\left( q\right) \) then \( q \) divides \( {\alpha }_{q} \), which is impossible. Hence \( i\left( q\right) = j\left( q\right) \) . We now see that \( {q}^{j\left( q\right) } \) divides \( {\alpha }_{q} - {\beta }_{q} \), thereby proving the theorem.
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Yes
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Theorem 5.2. Let \( A = k\left\lbrack X\right\rbrack \) be the polynomial ring in one variable over a field \( k \). Let \( P \) be the set of irreducible polynomials in \( k\left\lbrack X\right\rbrack \) with leading coefficient 1. Then any element \( f \) of \( k\left( X\right) \) has a unique expression
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Proof. The existence follows at once from our previous remarks. The uniqueness follows from the fact that if we have two expressions, with elements \( {f}_{p} \) and \( {\varphi }_{p} \) respectively, and polynomials \( g, h \), then \( {p}^{j\left( p\right) } \) divides \( {f}_{p} - {\varphi }_{p} \), whence \( {f}_{p} - {\varphi }_{p} = 0 \), and therefore \( {f}_{p} = {\varphi }_{p}, g = h \) .
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Yes
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Theorem 5.3. Let \( k \) be a field and \( k\left\lbrack X\right\rbrack \) the polynomial ring in one variable. Let \( f, g \in k\left\lbrack X\right\rbrack \), and assume \( \deg g \geqq 1 \) . Then there exist unique polynomials\n\n\[ \n{f}_{0},{f}_{1},\ldots ,{f}_{d} \in k\left\lbrack X\right\rbrack \n\] \n\nsuch that \( \deg {f}_{i} < \deg g \) and such that\n\n\[ \nf = {f}_{0} + {f}_{1}g + \cdots + {f}_{d}{g}^{d} \n\]
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Proof. We first prove existence. If \( \deg g > \deg f \), then we take \( {f}_{0} = f \) and \( {f}_{i} = 0 \) for \( i > 0 \) . Suppose \( \deg g \leqq \deg f \) . We can find polynomials \( q, r \) with \( \deg r < \deg g \) such that\n\n\[ \nf = {qg} + r \n\] \n\nand since \( \deg g \geqq 1 \) we have \( \deg q < \deg f \) . Inductively, there exist polynomials \( {h}_{0},{h}_{1},\ldots ,{h}_{s} \) such that\n\n\[ \nq = {h}_{0} + {h}_{1}g + \cdots + {h}_{s}{g}^{s} \n\] \n\nand hence\n\n\[ \nf = r + {h}_{0}g + \cdots + {h}_{s}{g}^{s + 1}, \n\] \n\nthereby proving existence.\n\nAs for uniqueness, let\n\n\[ \nf = {f}_{0} + {f}_{1}g + \cdots + {f}_{d}{g}^{d} = {\varphi }_{0} + {\varphi }_{1}g + \cdots + {\varphi }_{m}{g}^{m} \n\] \n\nbe two expressions satisfying the conditions of the theorem. Adding terms equal to 0 to either side, we may assume that \( m = d \) . Subtracting, we get\n\n\[ \n0 = \left( {{f}_{0} - {\varphi }_{0}}\right) + \cdots + \left( {{f}_{d} - {\varphi }_{d}}\right) {g}^{d}. \n\] \n\nHence \( g \) divides \( {f}_{0} - {\varphi }_{0} \), and since \( \deg \left( {{f}_{0} - {\varphi }_{0}}\right) < \deg g \) we see that \( {f}_{0} = {\varphi }_{0} \) . Inductively, take the smallest integer \( i \) such that \( {f}_{i} \neq {\varphi }_{i} \) (if such \( i \) exists). Dividing the above expression by \( {g}^{i} \) we find that \( g \) divides \( {f}_{i} - {\varphi }_{i} \) and hence that such \( i \) cannot exist. This proves uniqueness.
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Yes
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Theorem 6.1. Let \( f\left( t\right) \in A\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) be symmetric of degree \( d \) . Then there exists a polynomial \( g\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) of weight \( \leqq d \) such that\n\n\[ f\left( t\right) = g\left( {{s}_{1},\ldots ,{s}_{n}}\right) \]\n\nIf \( f \) is homogeneous of degree \( d \), then every monomial occurring in \( g \) has weight \( d \) .
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Proof. By induction on \( n \) . The theorem is obvious if \( n = 1 \), because \( {s}_{1} = {t}_{1} \) . Assume the theorem proved for polynomials in \( n - 1 \) variables.\n\nIf we substitute \( {t}_{n} = 0 \) in the expression for \( F\left( X\right) \), we find\n\n\[ \left( {X - {t}_{1}}\right) \cdots \left( {X - {t}_{n - 1}}\right) X = {X}^{n} - {\left( {s}_{1}\right) }_{0}{X}^{n - 1} + \cdots + {\left( -1\right) }^{n - 1}{\left( {s}_{n - 1}\right) }_{0}X, \]\n\nwhere \( {\left( {s}_{i}\right) }_{0} \) is the expression obtained by substituting \( {t}_{n} = 0 \) in \( {s}_{i} \) . We see that \( {\left( {s}_{1}\right) }_{0},\ldots ,{\left( {s}_{n - 1}\right) }_{0} \) are precisely the elementary symmetric polynomials in \( {t}_{1},\ldots ,{t}_{n - 1} \) .\n\nWe now carry out induction on \( d \) . If \( d = 0 \), our assertion is trivial. Assume \( d > 0 \), and assume our assertion proved for polynomials of degree \( < d \) . Let \( f\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) have degree \( d \) . There exists a polynomial \( {g}_{1}\left( {{X}_{1},\ldots ,{X}_{n - 1}}\right) \) of weight \( \leqq d \) such that\n\n\[ f\left( {{t}_{1},\ldots ,{t}_{n - 1},0}\right) = {g}_{1}\left( {{\left( {s}_{1}\right) }_{0},\ldots ,{\left( {s}_{n - 1}\right) }_{0}}\right) . \]\n\nWe note that \( {g}_{1}\left( {{s}_{1},\ldots ,{s}_{n - 1}}\right) \) has degree \( \leqq d \) in \( {t}_{1},\ldots ,{t}_{n} \) . The polynomial\n\n\[ {f}_{1}\left( {{t}_{1},\ldots ,{t}_{n}}\right) = f\left( {{t}_{1},\ldots ,{t}_{n}}\right) - {g}_{1}\left( {{s}_{1},\ldots ,{s}_{n - 1}}\right) \]\n\nhas degree \( \leqq d \) (in \( {t}_{1},\ldots ,{t}_{n} \) ) and is symmetric. We have\n\n\[ {f}_{1}\left( {{t}_{1},\ldots ,{t}_{n - 1},0}\right) = 0. \]\n\nHence \( {f}_{1} \) is divisible by \( {t}_{n} \), i.e. contains \( {t}_{n} \) as a factor. Since \( {f}_{1} \) is symmetric, it contains \( {t}_{1}\cdots {t}_{n} \) as a factor. Hence\n\n\[ {f}_{1} = {s}_{n}{f}_{2}\left( {{t}_{1},\ldots ,{t}_{n}}\right) \]\n\nfor some polynomial \( {f}_{2} \), which must be symmetric, and whose degree is\n\n\( \leqq d - n < d \) . By induction, there exists a polynomial \( {g}_{2} \) in \( n \) variables and weight \( \leqq d - n \) such that\n\n\[ {f}_{2}\left( {{t}_{1},\ldots ,{t}_{n}}\right) = {g}_{2}\left( {{s}_{1},\ldots ,{s}_{n}}\right) . \]\n\nWe obtain\n\n\[ f\left( t\right) = {g}_{1}\left( {{s}_{1},\ldots ,{s}_{n - 1}}\right) + {s}_{n}{g}_{2}\left( {{s}_{1},\ldots ,{s}_{n}}\right) , \]\n\nand each term on the right has weight \( \leqq d \) . This proves our theorem, except for the last statement which will be left to the reader.
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Yes
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Theorem 7.1 (Mason-Stothers, [Mas 84], [Sto 81]). Let \( a\left( t\right), b\left( t\right), c\left( t\right) \) be relatively prime polynomials such that \( a + b = c \). Then\n\n\[ \max \deg \{ a, b, c\} \leqq {n}_{0}\left( {abc}\right) - 1. \]
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Proof. (Mason) Dividing by \( c \), and letting \( f = a/c, g = b/c \) we have\n\n\[ f + g = 1, \]\n\nwhere \( f, g \) are rational functions. Differentiating we get \( {f}^{\prime } + {g}^{\prime } = 0 \), which we rewrite as\n\n\[ \frac{{f}^{\prime }}{f}f + \frac{{g}^{\prime }}{g}g = 0 \]\n\nso that\n\n\[ \frac{b}{a} = \frac{g}{f} = - \frac{{f}^{\prime }/f}{{g}^{\prime }/g} \]\n\nLet\n\n\[ a\left( t\right) = {c}_{1}\prod {\left( t - {\alpha }_{i}\right) }^{{m}_{i}},\;b\left( t\right) = {c}_{2}\prod {\left( t - {\beta }_{j}\right) }^{{n}_{j}},\;c\left( t\right) = {c}_{3}\prod {\left( t - {\gamma }_{k}\right) }^{{r}_{k}}. \]\n\nThen by calculus algebraicized in Exercise 11(c), we get\n\n\[ \frac{b}{a} = - \frac{{f}^{\prime }/f}{{g}^{\prime }/g} = - \frac{\sum \frac{{m}_{i}}{t - {\alpha }_{i}}-\sum \frac{{r}_{k}}{t - {\gamma }_{k}}}{\sum \frac{{n}_{j}}{t - {\beta }_{j}}-\sum \frac{{r}_{k}}{t - {\gamma }_{k}}}. \]\n\nA common denominator for \( {f}^{\prime }/f \) and \( {g}^{\prime }/g \) is given by the product\n\n\[ {N}_{0} = \prod \left( {t - {\alpha }_{i}}\right) \prod \left( {t - {\beta }_{j}}\right) \prod \left( {\mathrm{t} - {\gamma }_{k}}\right) \]\n\nwhose degree is \( {n}_{0}\left( {abc}\right) \). Observe that \( {N}_{0}{f}^{\prime }/f \) and \( {N}_{0}{g}^{\prime }/g \) are both polynomials of degrees at most \( {n}_{0}\left( {abc}\right) - 1 \). From the relation\n\n\[ \frac{b}{a} = - \frac{{N}_{0}{f}^{\prime }/f}{{N}_{0}{g}^{\prime }/g} \]\n\nand the fact that \( a, b \) are assumed relatively prime, we deduce the inequality in the theorem.
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No
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Proposition 8.1. Let \( K \) be a subfield of a field \( L \), and let \( {f}_{a},{g}_{b} \) be polynomials in \( K\left\lbrack X\right\rbrack \) having a common root \( \xi \) in \( L \) . Then \( R\left( {a, b}\right) = 0 \) .
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Proof. If \( {f}_{a}\left( \xi \right) = {g}_{b}\left( \xi \right) = 0 \), then we substitute \( \xi \) for \( X \) in the expression obtained for \( R\left( {a, b}\right) \) and find \( R\left( {a, b}\right) = 0 \) .
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Yes
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Lemma 8.2. Let \( h\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) be a polynomial in \( n \) variables over the integers \( \mathbf{Z} \) . If \( h \) has the value 0 when we substitute \( {X}_{1} \) for \( {X}_{2} \) and leave the other \( {X}_{i} \) fixed \( \left( {i \neq 2}\right) \), then \( h\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) is divisible by \( {X}_{1} - {X}_{2} \) in \( \mathbf{Z}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) .
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Proof. Exercise for the reader.
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No
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Proposition 8.3. Notation being as above, we have\n\n\\[ \n\\operatorname{Res}\\left( {{f}_{v},{g}_{w}}\\right) = {v}_{0}^{m}{w}_{0}^{n}\\mathop{\\prod }\\limits_{{i = 1}}^{n}\\mathop{\\prod }\\limits_{{j = 1}}^{m}\\left( {{t}_{i} - {u}_{j}}\\right) \n\\]
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Proof. Let \\( S \\) be the expression on the right-hand side of the equality in the statement of the proposition.\n\nSince \\( R\\left( {v, w}\\right) \\) is homogeneous of degree \\( m \\) in its first variables, and homogeneous of degree \\( n \\) in its second variables, it follows that\n\n\\[ \nR = {v}_{0}^{m}{w}_{0}^{n}h\\left( {t, u}\\right) \n\\]\n\nwhere \\( h\\left( {t, u}\\right) \\in \\mathbf{Z}\\left\\lbrack {t, u}\\right\\rbrack \\) . By Proposition 8.1, the resultant vanishes when we substitute \\( {t}_{i} \\) for \\( {u}_{j}\\left( {i = 1,\\ldots, n\\text{and}j = 1,\\ldots, m}\\right) \\), whence by the lemma, viewing \\( R \\) as an element of \\( \\mathbf{Z}\\left\\lbrack {{v}_{0},{w}_{0}, t, u}\\right\\rbrack \\) it follows that \\( R \\) is divisible by \\( {t}_{i} - {u}_{j} \\) for each pair \\( \\left( {i, j}\\right) \\) . Hence \\( S \\) divides \\( R \\) in \\( \\mathbf{Z}\\left\\lbrack {{v}_{0},{w}_{0}, t, u}\\right\\rbrack \\), because \\( {t}_{i} - {u}_{j} \\) is obviously a prime in that ring, and different pairs \\( \\left( {i, j}\\right) \\) give rise to different primes.\n\nFrom the product expression for \\( S \\), namely\n\n(1)\n\n\\[ \nS = {v}_{0}^{m}{w}_{0}^{n}\\mathop{\\prod }\\limits_{{i = 1}}^{n}\\mathop{\\prod }\\limits_{{j = 1}}^{m}\\left( {{t}_{i} - {u}_{j}}\\right) \n\\]\n\nwe obtain\n\n\\[ \n\\mathop{\\prod }\\limits_{{i = 1}}^{n}g\\left( {t}_{i}\\right) = {w}_{0}^{n}\\mathop{\\prod }\\limits_{{i = 1}}^{n}\\mathop{\\prod }\\limits_{{j = 1}}^{m}\\left( {{t}_{i} - {u}_{j}}\\right) \n\\]\n\nwhence\n\n(2)\n\n\\[ \nS = {v}_{0}^{m}\\mathop{\\prod }\\limits_{{i = 1}}^{n}g\\left( {t}_{i}\\right) \n\\]\n\nSimilarly,\n\n(3)\n\n\\[ \nS = {\\left( -1\\right) }^{nm}{w}_{0}^{n}\\mathop{\\prod }\\limits_{{j = 1}}^{m}f\\left( {u}_{j}\\right) \n\\]\n\nFrom (2) we see that \\( S \\) is homogeneous and of degree \\( n \\) in (w), and from (3) we see that \\( S \\) is homogeneous and of degree \\( m \\) in \\( \\left( v\\right) \\) . Since \\( R \\) has exactly the same homogeneity properties, and is divisible by \\( S \\), it follows that \\( R = {cS} \\) for some integer \\( c \\) . Since both \\( R \\) and \\( S \\) have a monomial \\( {v}_{0}^{m}{w}_{m}^{n} \\) occurring in them with coefficient 1, it follows that \\( c = 1 \\), and our proposition is proved.
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Yes
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Corollary 8.4. Let \( {f}_{a},{g}_{b} \) be polynomials with coefficients in a field \( K \), such that \( {a}_{0}{b}_{0} \neq 0 \), and such that \( {f}_{a},{g}_{b} \) split in factors of degree 1 in \( K\left\lbrack X\right\rbrack \) . Then \( \operatorname{Res}\left( {{f}_{a},{g}_{b}}\right) = 0 \) if and only if \( {f}_{a} \) and \( {g}_{b} \) have a root in common.
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Proof. Assume that the resultant is 0 . If\n\n\[ \n{f}_{a} = {a}_{0}\left( {X - {\alpha }_{1}}\right) \cdots \left( {X - {\alpha }_{n}}\right) \n\] \n\n\[ \n{g}_{b} = {b}_{0}\left( {X - {\beta }_{1}}\right) \cdots \left( {X - {\beta }_{n}}\right) \n\] \n\nis the factorization of \( {f}_{a},{g}_{b} \), then we have a homomorphism \n\n\[ \n\mathbf{Z}\left\lbrack {{v}_{0}, t,{w}_{0}, u}\right\rbrack \rightarrow K \n\] \n\nsuch that \( {v}_{0} \mapsto {a}_{0},{w}_{0} \mapsto {b}_{0},{t}_{i} \mapsto {\alpha }_{i} \), and \( {u}_{j} \mapsto {\beta }_{j} \) for all \( i, j \) . Then \n\n\[ \n0 = \operatorname{Res}\left( {{f}_{a},{g}_{b}}\right) = {a}_{0}^{m}{b}_{0}^{n}\mathop{\prod }\limits_{i}\mathop{\prod }\limits_{j}\left( {{\alpha }_{i} - {\beta }_{j}}\right) \n\] \n\nwhence \( {f}_{a},{f}_{b} \) have a root in common. The converse has already been proved.
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Yes
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Proposition 8.5. Let \( {f}_{v} \) be as above and have algebraically independent coefficients over \( \mathbf{Z} \) . Then\n\n\[ \operatorname{Res}\left( {{f}_{v},{f}_{v}^{\prime }}\right) = {v}_{0}^{{2n} - 1}\mathop{\prod }\limits_{{i \neq j}}\left( {{t}_{i} - {t}_{j}}\right) = {\left( -1\right) }^{n\left( {n - 1}\right) /2}{v}_{0}D\left( {f}_{v}\right) . \]
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Proof. One substitutes the expression obtained for \( {f}_{v}^{\prime }\left( {t}_{i}\right) \) into the product (4). The result follows at once.
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No
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Theorem 9.1. Let \( \mathfrak{o} \) be a complete local ring with maximal ideal \( \mathfrak{m} \) . Let\n\n\[ f\left( X\right) = \mathop{\sum }\limits_{{i = 0}}^{\infty }{a}_{i}{X}^{i} \]\n\nbe a power series in \( \mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) (one variable), such that not all \( {a}_{i} \) lie in \( \mathfrak{m} \) . Say \( {a}_{0},\ldots ,{a}_{n - 1} \in \mathfrak{m} \), and \( {a}_{n} \in {\mathfrak{v}}^{ * } \) is a unit. Given \( g \in \mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) we can solve the equation\n\n\[ g = {qf} + r \]\n\nuniquely with \( q \in \mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack, r \in \mathfrak{o}\left\lbrack X\right\rbrack \), and \( \deg r \leqq n - 1 \) .
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Proof (Manin). Let \( \alpha \) and \( \tau \) be the projections on the beginning and tail end of the series, given by\n\n\[ \alpha : \sum {b}_{i}{X}^{i} \mapsto \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}{b}_{i}{X}^{i} = {b}_{0} + {b}_{1}X + \cdots + {b}_{n - 1}{X}^{n - 1}, \]\n\n\[ \tau : \sum {b}_{i}{X}^{i} \mapsto \mathop{\sum }\limits_{{i = n}}^{\infty }{b}_{i}{X}^{i - n} = {b}_{n} + {b}_{n + 1}X + {b}_{n + 2}{X}^{2} + \cdots . \]\n\nNote that \( \tau \left( {h{X}^{n}}\right) = h \) for any \( h \in \mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) ; and \( h \) is a polynomial of degree \( < n \) if and only if \( \tau \left( h\right) = 0 \) .\n\nThe existence of \( q, r \) is equivalent with the condition that there exists \( q \) such that\n\n\[ \tau \left( g\right) = \tau \left( {qf}\right) \]\n\nHence our problem is equivalent with solving\n\n\[ \tau \left( g\right) = \tau \left( {{q\alpha }\left( f\right) }\right) + \tau \left( {{q\tau }\left( f\right) {X}^{n}}\right) = \tau \left( {{q\alpha }\left( f\right) }\right) + {q\tau }\left( f\right) . \]\n\nNote that \( \tau \left( f\right) \) is invertible. Put \( Z = {q\tau }\left( f\right) \) . Then the above equation is equivalent with\n\n\[ \tau \left( g\right) = \tau \left( {Z\frac{\alpha \left( f\right) }{\tau \left( f\right) }}\right) + Z = \left( {I + \tau \circ \frac{\alpha \left( f\right) }{\tau \left( f\right) }}\right) Z. \]\n\nNote that\n\n\[ \tau \circ \frac{\alpha \left( f\right) }{\tau \left( f\right) } : \mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \rightarrow \mathfrak{{mo}}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \]\n\nbecause \( \alpha \left( f\right) /\tau \left( f\right) \in \mathfrak{m}\mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) . We can therefore invert to find \( Z \), namely\n\n\[ Z = {\left( I + \tau \circ \frac{\alpha \left( f\right) }{\tau \left( f\right) }\right) }^{-1}\tau \left( g\right) \]\n\nwhich proves both existence and uniqueness and concludes the proof.
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Yes
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Theorem 9.2. (Weierstrass Preparation). The power series \( f \) in the previous theorem can be written uniquely in the form\n\n\[ f\left( X\right) = \left( {{X}^{n} + {b}_{n - 1}{X}^{n - 1} + \cdots + {b}_{0}}\right) u \]\n\nwhere \( {b}_{i} \in \mathfrak{m} \), and \( u \) is a unit in \( \mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) .
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Proof. Write uniquely\n\n\[ {X}^{n} = {qf} + r \]\n\nby the Euclidean algorithm. Then \( q \) is invertible, because\n\n\[ q = {c}_{0} + {c}_{1}X + \cdots ,\]\n\[ f = \cdots + {a}_{n}{X}^{n} + \cdots ,\]\n\nso that\n\n\[ 1 \equiv {c}_{0}{a}_{n}\;\left( {\;\operatorname{mod}\;m}\right) \]\n\nand therefore \( {c}_{0} \) is a unit in 0 . We obtain \( {qf} = {X}^{n} - r \), and\n\n\[ f = {q}^{-1}\left( {{X}^{n} - r}\right) \]\n\nwith \( r \equiv 0\left( {\;\operatorname{mod}\;m}\right) \) . This proves the existence. Uniqueness is immediate.
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Yes
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Theorem 9.3. Let \( k \) be a field. Then \( k\left\lbrack \left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \right\rbrack \) is factorial.
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Proof. Let \( f\left( x\right) = f\left( {{X}_{1},\ldots ,{X}_{n}}\right) \in k\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) be \( \neq 0 \) . After making a sufficiently general linear change of variables (when \( k \) is infinite)\n\n\[ \n{x}_{i} = \sum {c}_{ij}{Y}_{j}\;\text{ with }\;{c}_{ij} \in k, \n\]\n\nwe may assume without loss of generality that \( f\left( {0,\ldots ,0,{x}_{n}}\right) \neq 0 \) . (When \( k \) is finite, one has to make a non-linear change, cf. Theorem 2.1 of Chapter VIII.) Indeed, if we write \( f\left( X\right) = {f}_{d}\left( X\right) + \) higher terms, where \( {f}_{d}\left( X\right) \) is a homogeneous polynomial of degree \( d \geqq 0 \), then changing the variables as above preserves the degree of each homogeneous component of \( f \), and since \( k \) is assumed infinite, the coefficients \( {c}_{ij} \) can be taken so that in fact each power \( {Y}_{i}^{d}\left( {i = 1,\ldots, n}\right) \) occurs with non-zero coefficient.\n\nWe now proceed by induction on \( n \) . Let \( {R}_{n} = k\left\lbrack \left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \right\rbrack \) be the power series in \( n \) variables, and assume by induction that \( {R}_{n - 1} \) is factorial. By Theorem 9.2, write \( f = {gu} \) where \( u \) is a unit and \( g \) is a Weierstrass polynomial in \( {R}_{n - 1}\left\lbrack {X}_{n}\right\rbrack \) . By Theorem 2.3, \( {R}_{n - 1}\left\lbrack {X}_{n}\right\rbrack \) is factorial, and so we can write \( g \) as a product of irreducible elements \( {g}_{1},\ldots ,{g}_{r} \in {R}_{n - 1}\left\lbrack {X}_{n}\right\rbrack \), so \( f = {g}_{1}\cdots {g}_{r}u \), where the factors \( {\mathbf{g}}_{i} \) are uniquely determined up to multiplication by units. This proves the existence of a factorization. As to uniqueness, suppose \( f \) is expressed as a product of irreducible elements in \( {R}_{n}, f = {f}_{1}\cdots {f}_{s} \) . Then \( {f}_{q}\left( {0,\ldots ,0,{x}_{n}}\right) \neq 0 \) for each \( q = 1,\ldots, s \), so we can write \( {f}_{q} = {h}_{q}{u}_{q}^{\prime } \) where \( {u}_{q}^{\prime } \) is a unit and \( {h}_{q} \) is a Weierstrass polynomial, necessarily irreducible in \( {R}_{n - 1}\left\lbrack {X}_{n}\right\rbrack \) . Then \( f = {gu} = \prod {h}_{q}\prod {u}_{q}^{\prime } \) with \( g \) and all \( {h}_{q} \) Weierstrass polynomials. By Theorem 9.2, we must have \( g = \prod {h}_{q} \), and since \( {R}_{n - 1}\left\lbrack {X}_{n}\right\rbrack \) is factorial, it follows that the polynomials \( {h}_{q} \) are the same as the polynomials \( {g}_{i} \), up to units. This proves uniqueness.
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Yes
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Theorem 9.4. If \( A \) is Noetherian, then \( A\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) is also Noetherian.
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Proof. Our argument will be a modification of the argument used in the proof of Hilbert's theorem for polynomials. We shall consider elements of lowest degree instead of elements of highest degree.\n\nLet \( \mathfrak{A} \) be an ideal of \( A\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) . We let \( {\mathfrak{a}}_{i} \) be the set of elements \( a \in A \) such that \( a \) is the coefficient of \( {X}^{i} \) in a power series\n\n\[ a{X}^{i} + \text{terms of higher degree} \]\n\nlying in \( \mathfrak{A} \) . Then \( {\mathfrak{a}}_{i} \) is an ideal of \( A \), and \( {\mathfrak{a}}_{i} \subset {\mathfrak{a}}_{i + 1} \) (the proof of this assertion being the same as for polynomials). The ascending chain of ideals stops:\n\n\[ {\mathfrak{a}}_{0} \subset {\mathfrak{a}}_{1} \subset {\mathfrak{a}}_{2} \subset \cdots \subset {\mathfrak{a}}_{r} = {\mathfrak{a}}_{r + 1} = \cdots \]\n\nAs before, let \( {a}_{ij}\left( {i = 0,\ldots, r}\right. \) and \( \left. {j = 1,\ldots ,{n}_{i}}\right) \) be generators for the ideals \( {a}_{i} \), and let \( {f}_{ij} \) be power series in \( A \) having \( {a}_{ij} \) as beginning coefficient. Given \( f \in \mathfrak{A} \), starting with a term of degree \( d \), say \( d \leqq r \), we can find elements \( {c}_{1},\ldots ,{c}_{{n}_{d}} \in A \) such that\n\n\[ f - {c}_{1}{f}_{d1} - \cdots - {c}_{{n}_{d}}{f}_{d{n}_{d}} \]\n\nstarts with a term of degree \( \geqq d + 1 \) . Proceeding inductively, we may assume that \( d > r \) . We then use a linear combination\n\n\[ f - {c}_{1}^{\left( d\right) }{X}^{d - r}{f}_{r1} - \cdots - {c}_{{n}_{r}}^{\left( d\right) }{X}^{d - r}{f}_{r{n}_{r}} \]\n\nto get a power series starting with a term of degree \( \geqq d + 1 \) . In this way, if we start with a power series of degree \( d > r \), then it can be expressed as a linear combination of \( {f}_{r1},\ldots ,{f}_{r{n}_{r}} \) by means of the coefficients\n\n\[ {g}_{1}\left( X\right) = \mathop{\sum }\limits_{{v = d}}^{\infty }{c}_{1}^{\left( v\right) }{X}^{v - r},\ldots ,{g}_{{n}_{r}}\left( X\right) = \mathop{\sum }\limits_{{v = d}}^{\infty }{c}_{{n}_{r}}^{\left( v\right) }{X}^{v - r}, \]\n\nand we see that the \( {f}_{ij} \) generate our ideal \( \mathfrak{A} \), as was to be shown.
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Yes
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Proposition 1.1. Let \( E \) be a finite extension of \( F \) . Then \( E \) is algebraic over \( F \) .
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Proof. Let \( \alpha \in E,\alpha \neq 0 \) . The powers of \( \alpha \) ,\n\n\[ 1,\alpha ,{\alpha }^{2},\ldots ,{\alpha }^{n}, \]\n\ncannot be linearly independent over \( F \) for all positive integers \( n \), otherwise the dimension of \( E \) over \( F \) would be infinite. A linear relation between these powers shows that \( \alpha \) is algebraic over \( F \) .
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Yes
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Proposition 1.2. Let \( k \) be a field and \( F \subset E \) extension fields of \( k \) . Then\n\n\[ \left\lbrack {E : k}\right\rbrack = \left\lbrack {E : F}\right\rbrack \left\lbrack {F : k}\right\rbrack . \]\n\nIf \( {\left\{ {x}_{i}\right\} }_{i \in I} \) is a basis for \( F \) over \( k \) and \( {\left\{ {y}_{j}\right\} }_{j \in J} \) is a basis for \( E \) over \( F \), then \( {\left\{ {x}_{i}{y}_{j}\right\} }_{\left( {i, j}\right) \in I \times J} \) is a basis for \( E \) over \( k \) .
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Proof. Let \( z \in E \) . By hypothesis there exist elements \( {\alpha }_{j} \in F \), almost all \( {\alpha }_{j} = 0 \), such that\n\n\[ z = \mathop{\sum }\limits_{{j \in J}}{\alpha }_{j}{y}_{j} \]\n\nFor each \( j \in J \) there exist elements \( {b}_{ji} \in k \), almost all of which are equal to 0, such that\n\n\[ {\alpha }_{j} = \mathop{\sum }\limits_{{i \in I}}{b}_{ji}{x}_{i} \]\n\nand hence\n\n\[ z = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{i}{b}_{ji}{x}_{i}{y}_{j} \]\n\nThis shows that \( \left\{ {{x}_{i}{y}_{j}}\right\} \) is a family of generators for \( E \) over \( k \) . We must show that it is linearly independent. Let \( \left\{ {c}_{ij}\right\} \) be a family of elements of \( k \), almost all of which are 0 , such that\n\n\[ \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{i}{c}_{ij}{x}_{i}{y}_{j} = 0 \]\n\nThen for each \( j \) ,\n\n\[ \mathop{\sum }\limits_{i}{c}_{ij}{x}_{i} = 0 \]\n\nbecause the elements \( {y}_{j} \) are linearly independent over \( F \) . Finally \( {c}_{ij} = 0 \) for each \( i \) because \( \left\{ {x}_{i}\right\} \) is a basis of \( F \) over \( k \), thereby proving our proposition.
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Yes
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Proposition 1.4. Let \( \alpha \) be algebraic over \( k \) . Then \( k\left( \alpha \right) = k\left\lbrack \alpha \right\rbrack \), and \( k\left( \alpha \right) \) is finite over \( k \) . The degree \( \left\lbrack {k\left( \alpha \right) : k}\right\rbrack \) is equal to the degree of \( \operatorname{Irr}\left( {\alpha, k, X}\right) \) .
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Proof. Let \( p\left( X\right) = \operatorname{Irr}\left( {\alpha, k, X}\right) \) . Let \( f\left( X\right) \in k\left\lbrack X\right\rbrack \) be such that \( f\left( \alpha \right) \neq 0 \) . Then \( p\left( X\right) \) does not divide \( f\left( X\right) \), and hence there exist polynomials \( g\left( X\right) \) , \( h\left( X\right) \in k\left\lbrack X\right\rbrack \) such that\n\n\[ g\left( X\right) p\left( X\right) + h\left( X\right) f\left( X\right) = 1. \]\n\nFrom this we get \( h\left( \alpha \right) f\left( \alpha \right) = 1 \), and we see that \( f\left( \alpha \right) \) is invertible in \( k\left\lbrack \alpha \right\rbrack \) . Hence \( k\left\lbrack \alpha \right\rbrack \) is not only a ring but a field, and must therefore be equal to \( k\left( \alpha \right) \) . Let \( d = \deg p\left( X\right) \) . The powers\n\n\[ 1,\alpha ,\ldots ,{\alpha }^{d - 1} \]\n\nare linearly independent over \( k \), for otherwise suppose\n\n\[ {a}_{0} + {a}_{1}\alpha + \cdots + {a}_{d - 1}{\alpha }^{d - 1} = 0 \]\n\nwith \( {a}_{i} \in k \), not all \( {a}_{i} = 0 \) . Let \( g\left( X\right) = {a}_{0} + \cdots + {a}_{d - 1}{X}^{d - 1} \) . Then \( g \neq 0 \) and \( g\left( \alpha \right) = 0 \) . Hence \( p\left( X\right) \) divides \( g\left( X\right) \), contradiction. Finally, let \( f\left( \alpha \right) \in k\left\lbrack \alpha \right\rbrack \) , where \( f\left( X\right) \in k\left\lbrack X\right\rbrack \) . There exist polynomials \( q\left( X\right), r\left( X\right) \in k\left\lbrack X\right\rbrack \) such that \( \deg r < d \) and\n\n\[ f\left( X\right) = q\left( X\right) p\left( X\right) + r\left( X\right) \]\n\nThen \( f\left( \alpha \right) = r\left( \alpha \right) \), and we see that \( 1,\alpha ,\ldots ,{\alpha }^{d - 1} \) generate \( k\left\lbrack \alpha \right\rbrack \) as a vector space over \( k \) . This proves our proposition.
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Yes
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Proposition 1.5. Let \( E \) be a finite extension of \( k \). Then \( E \) is finitely generated.
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Proof. Let \( \\left\{ {{\\alpha }_{1},\\ldots ,{\\alpha }_{n}}\\right\} \) be a basis of \( E \) as vector space over \( k \). Then certainly\n\n\[ E = k\\left( {{\\alpha }_{1},\\ldots ,{\\alpha }_{n}}\\right) \]
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Yes
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Proposition 1.6. Let \( E = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) be a finitely generated extension of a field \( k \), and assume \( {\alpha }_{i} \) algebraic over \( k \) for each \( i = 1,\ldots, n \) . Then \( E \) is finite algebraic over \( k \) .
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Proof. From the above remarks, we know that \( E \) can be obtained as the end of a tower each of whose steps is generated by one algebraic element, and is therefore finite by Proposition 1.4. We conclude that \( E \) is finite over \( k \) by Corollary 1.3, and that it is algebraic by Proposition 1.1.
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Yes
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Proposition 1.7. The class of algebraic extensions is distinguished, and so is the class of finite extensions.
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Proof. Consider first the class of finite extensions. We have already proved condition (1). As for (2), assume that \( E/k \) is finite, and let \( F \) be any extension of \( k \) . By Proposition 1.5 there exist elements \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in E \) such that \( E = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) . Then \( {EF} = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \), and hence \( {EF}/F \) is finitely generated by algebraic elements. Using Proposition 1.6 we conclude that \( {EF}/F \) is finite.\n\nConsider next the class of algebraic extensions, and let\n\n\[ k \subset F \subset E \]\n\nbe a tower. Assume that \( E \) is algebraic over \( k \) . Then a fortiori, \( F \) is algebraic over \( k \) and \( E \) is algebraic over \( F \) . Conversely, assume each step in the tower to be algebraic. Let \( \alpha \in E \) . Then \( \alpha \) satisfies an equation\n\n\[ {a}_{n}{\alpha }^{n} + \cdots + {a}_{0} = 0 \]\n\nwith \( {a}_{i} \in F \), not all \( {a}_{i} = 0 \) . Let \( {F}_{0} = k\left( {{a}_{n},\ldots ,{a}_{0}}\right) \) . Then \( {F}_{0} \) is finite over \( k \) by Proposition 1.6, and \( \alpha \) is algebraic over \( {F}_{0} \) . From the tower\n\n\[ k \subset {F}_{0} = k\left( {{a}_{n},\ldots ,{a}_{0}}\right) \subset {F}_{0}\left( \alpha \right) \]\n\nand the fact that each step in this tower is finite, we conclude that \( {F}_{0}\left( \alpha \right) \) is finite over \( k \), whence \( \alpha \) is algebraic over \( k \), thereby proving that \( E \) is algebraic over \( k \) and proving condition (1) for algebraic extensions. Condition (2) has already been observed to hold, i.e. an element remains algebraic under lifting, and hence so does an extension.
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Yes
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Lemma 2.1. Let \( E \) be an algebraic extension of \( k \), and let \( \sigma : E \rightarrow E \) be an embedding of \( E \) into itself over \( k \) . Then \( \sigma \) is an automorphism.
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Proof. Since \( \sigma \) is injective, it will suffice to prove that \( \sigma \) is surjective. Let \( \alpha \) be an element of \( E \), let \( p\left( X\right) \) be its irreducible polynomial over \( k \), and let \( {E}^{\prime } \) be the subfield of \( E \) generated by all the roots of \( p\left( X\right) \) which lie in \( E \) . Then \( {E}^{\prime } \) is finitely generated, hence is a finite extension of \( k \) . Furthermore, \( \sigma \) must map a root of \( p\left( X\right) \) on a root of \( p\left( X\right) \), and hence \( \sigma \) maps \( {E}^{\prime } \) into itself. We can view \( \sigma \) as a \( k \) -homomorphism of vector spaces because \( \sigma \) induces the identity on \( k \) . Since \( \sigma \) is injective, its image \( \sigma \left( {E}^{\prime }\right) \) is a subspace of \( {E}^{\prime } \) having the same dimension \( \left\lbrack {{E}^{\prime } : k}\right\rbrack \) . Hence \( \sigma \left( {E}^{\prime }\right) = {E}^{\prime } \) . Since \( \alpha \in {E}^{\prime } \), it follows that \( \alpha \) is in the image of \( \sigma \), and our lemma is proved.
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Yes
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Lemma 2.2. Let \( {E}_{1},{E}_{2} \) be extensions of a field \( k \), contained in some bigger field \( E \), and let \( \sigma \) be an embedding of \( E \) in some field \( L \) . Then \[ \sigma \left( {{E}_{1}{E}_{2}}\right) = \sigma \left( {E}_{1}\right) \sigma \left( {E}_{2}\right) \]
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Proof. We apply \( \sigma \) to a quotient of elements of the above type, say \[ \sigma \left( \frac{{a}_{1}{b}_{1} + \cdots + {a}_{n}{b}_{n}}{{a}_{1}^{\prime }{b}_{1}^{\prime } + \cdots + {a}_{m}^{\prime }{b}_{m}^{\prime }}\right) = \frac{{a}_{1}^{\sigma }{b}_{1}^{\sigma } + \cdots + {a}_{n}^{\sigma }{b}_{n}^{\sigma }}{{a}_{1}^{\prime \sigma }{b}_{1}^{\prime \sigma } + \cdots + {a}_{m}^{\prime \sigma }{b}_{m}^{\prime \sigma }}, \] and see that the image is an element of \( \sigma \left( {E}_{1}\right) \sigma \left( {E}_{2}\right) \) . It is clear that the image \( \sigma \left( {{E}_{1}{E}_{2}}\right) \) is \( \sigma \left( {E}_{1}\right) \sigma \left( {E}_{2}\right) \) .
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Yes
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Proposition 2.3. Let \( k \) be a field and \( f \) a polynomial in \( k\left\lbrack X\right\rbrack \) of degree \( \geqq 1 \) . Then there exists an extension \( E \) of \( k \) in which \( f \) has a root.
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Proof. We may assume that \( f = p \) is irreducible. We have shown that there exists a field \( F \) and an embedding \[ \sigma : k \rightarrow F \] such that \( {p}^{\sigma } \) has a root \( \xi \) in \( F \) . Let \( S \) be a set whose cardinality is the same as that of \( F - {\sigma k} \) ( \( = \) the complement of \( {\sigma k} \) in \( F \) ) and which is disjoint from \( k \) . Let \( E = k \cup S \) . We can extend \( \sigma : k \rightarrow F \) to a bijection of \( E \) on \( F \) . We now define a field structure on \( E \) . If \( x, y \in E \) we define \[ {xy} = {\sigma }^{-1}\left( {\sigma \left( x\right) \sigma \left( y\right) }\right) \] \[ x + y = {\sigma }^{-1}\left( {\sigma \left( x\right) + \sigma \left( y\right) }\right) \] Restricted to \( k \), our addition and multiplication coincide with the given addition and multiplication of our original field \( k \), and it is clear that \( k \) is a subfield of \( E \) . We let \( \alpha = {\sigma }^{-1}\left( \xi \right) \) . Then it is also clear that \( p\left( \alpha \right) = 0 \), as desired.
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Yes
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Corollary 2.4. Let \( k \) be a field and let \( {f}_{1},\ldots ,{f}_{n} \) be polynomials in \( k\left\lbrack X\right\rbrack \) of degrees \( \geqq 1 \) . Then there exists an extension \( E \) of \( k \) in which each \( {f}_{i} \) has a root, \( i = 1,\ldots, n \) .
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Proof. Let \( {E}_{1} \) be an extension in which \( {f}_{1} \) has a root. We may view \( {f}_{2} \) as a polynomial over \( {E}_{1} \) . Let \( {E}_{2} \) be an extension of \( {E}_{1} \) in which \( {f}_{2} \) has a root. Proceeding inductively, our corollary follows at once.
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Yes
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Corollary 2.6. Let \( k \) be a field. There exists an extension \( {k}^{\mathrm{a}} \) which is algebraic over \( k \) and algebraically closed.
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Proof. Let \( E \) be an extension of \( k \) which is algebraically closed and let \( {k}^{\mathrm{a}} \) be the union of all subextensions of \( E \), which are algebraic over \( k \) . Then \( {k}^{\mathrm{a}} \) is algebraic over \( k \) . If \( \alpha \in E \) and \( \alpha \) is algebraic over \( {k}^{\mathrm{a}} \) then \( \alpha \) is algebraic over \( k \) by Proposition 1.7. If \( f \) is a polynomial of degree \( \geqq 1 \) in \( {k}^{\mathrm{a}}\left\lbrack X\right\rbrack \), then \( f \) has a root \( \alpha \) in \( E \), and \( \alpha \) is algebraic over \( {k}^{\mathrm{a}} \) . Hence \( \alpha \) is in \( {k}^{\mathrm{a}} \) and \( {k}^{\mathrm{a}} \) is algebraically closed.
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Yes
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Theorem 2.8. Let \( k \) be a field, \( E \) an algebraic extension of \( k \), and \( \sigma : k \rightarrow L \) an embedding of \( k \) into an algebraically closed field \( L \) . Then there exists an extension of \( \sigma \) to an embedding of \( E \) in \( L \) . If \( E \) is algebraically closed and \( L \) is algebraic over \( {\sigma k} \), then any such extension of \( \sigma \) is an isomorphism of \( E \) onto \( L \) .
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Proof. Let \( S \) be the set of all pairs \( \left( {F,\tau }\right) \) where \( F \) is a subfield of \( E \) containing \( k \), and \( \tau \) is an extension of \( \sigma \) to an embedding of \( F \) in \( L \) . If \( \left( {F,\tau }\right) \) and \( \left( {{F}^{\prime },{\tau }^{\prime }}\right) \) are such pairs, we write \( \left( {F,\tau }\right) \leqq \left( {{F}^{\prime },{\tau }^{\prime }}\right) \) if \( F \subset {F}^{\prime } \) and \( {\tau }^{\prime } \mid F = \tau \) . Note that \( S \) is not empty [it contains \( \left( {k,\sigma }\right) \) ], and is inductively ordered: If \( \left\{ \left( {{F}_{i},{\tau }_{i}}\right) \right\} \) is a totally ordered subset, we let \( F = \bigcup {F}_{i} \) and define \( \tau \) on \( F \) to be equal to \( {\tau }_{i} \) on each \( {F}_{i} \) . Then \( \left( {F,\tau }\right) \) is an upper bound for the totally ordered subset. Using Zorn’s lemma, let \( \left( {K,\lambda }\right) \) be a maximal element in \( S \) . Then \( \lambda \) is an extension of \( \sigma \), and we contend that \( K = E \) . Otherwise, there exists \( \alpha \in E \) , \( \alpha \notin K \) . By what we saw above, our embedding \( \lambda \) has an extension to \( K\left( \alpha \right) \) , thereby contradicting the maximality of \( \left( {K,\lambda }\right) \) . This proves that there exists an extension of \( \sigma \) to \( E \) . We denote this extension again by \( \sigma \) .\n\nIf \( E \) is algebraically closed, and \( L \) is algebraic over \( {\sigma k} \), then \( {\sigma E} \) is algebraically closed and \( L \) is algebraic over \( {\sigma E} \), hence \( L = {\sigma E} \) .
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Yes
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Let \( k \) be a field and let \( E,{E}^{\prime } \) be algebraic extensions of \( k \) . Assume that \( E,{E}^{\prime } \) are algebraically closed. Then there exists an isomorphism\n\n\[ \tau : E \rightarrow {E}^{\prime } \]\n\nof \( E \) onto \( {E}^{\prime } \) inducing the identity on \( k \) .
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Proof. Extend the identity mapping on \( k \) to an embedding of \( E \) into \( {E}^{\prime } \) and apply the theorem.
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No
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Theorem 3.1. Let \( K \) be a splitting field of the polynomial \( f\left( X\right) \in k\left\lbrack X\right\rbrack \) . If \( E \) is another splitting field of \( f \), then there exists an isomorphism \( \sigma : E \rightarrow K \) inducing the identity on \( k \) . If \( k \subset K \subset {k}^{\mathrm{a}} \), where \( {k}^{\mathrm{a}} \) is an algebraic closure of \( k \), then any embedding of \( E \) in \( {k}^{\mathrm{a}} \) inducing the identity on \( k \) must be an isomorphism of \( E \) onto \( K \) .
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Proof. Let \( {K}^{\mathrm{a}} \) be an algebraic closure of \( K \) . Then \( {K}^{\mathrm{a}} \) is algebraic over \( k \), hence is an algebraic closure of \( k \) . By Theorem 2.8 there exists an embedding\n\n\[ \sigma : E \rightarrow {K}^{\mathrm{a}} \]\n\ninducing the identity on \( k \) . We have a factorization\n\n\[ f\left( X\right) = c\left( {X - {\beta }_{1}}\right) \cdots \left( {X - {\beta }_{n}}\right) \]\n\nwith \( {\beta }_{i} \in E, i = 1,\ldots, n \) . The leading coefficient \( c \) lies in \( k \) . We obtain\n\n\[ f\left( X\right) = {f}^{\sigma }\left( X\right) = c\left( {X - \sigma {\beta }_{1}}\right) \cdots \left( {X - \sigma {\beta }_{n}}\right) . \]\n\nWe have unique factorization in \( {K}^{\mathrm{a}}\left\lbrack X\right\rbrack \) . Since \( f \) has a factorization\n\n\[ f\left( X\right) = c\left( {X - {\alpha }_{1}}\right) \cdots \left( {X - {\alpha }_{n}}\right) \]\n\nin \( K\left\lbrack X\right\rbrack \), it follows that \( \left( {\sigma {\beta }_{1},\ldots ,\sigma {\beta }_{n}}\right) \) differs from \( \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) by a permutation. From this we conclude that \( \sigma {\beta }_{i} \in K \) for \( i = 1,\ldots, n \) and hence that \( {\sigma E} \subset K \) . But \( K = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = k\left( {\sigma {\beta }_{1},\ldots ,\sigma {\beta }_{n}}\right) \), and hence \( {\sigma E} = K \), because\n\n\[ E = k\left( {{\beta }_{1},\ldots ,{\beta }_{n}}\right) \]\n\nThis proves our theorem.
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Yes
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Corollary 3.2. Let \( K \) be a splitting field for the family \( {\left\{ {f}_{i}\right\} }_{i \in I} \) and let \( E \) be another splitting field. Any embedding of \( E \) into \( {K}^{a} \) inducing the identity on \( k \) gives an isomorphism of \( E \) onto \( K \) .
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Proof. Let the notation be as above. Note that \( E \) contains a unique splitting field \( {E}_{i} \) of \( {f}_{i} \) and \( K \) contains a unique splitting field \( {K}_{i} \) of \( {f}_{i} \) . Any embedding \( \sigma \) of \( E \) into \( {K}^{\mathrm{a}} \) must map \( {E}_{i} \) onto \( {K}_{i} \) by Theorem 3.1, and hence maps \( E \) into \( K \) . Since \( K \) is the compositum of the fields \( {K}_{i} \), our map \( \sigma \) must send \( E \) onto \( K \) and hence induces an isomorphism of \( E \) onto \( K \) .
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Yes
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Theorem 3.3. Let \( K \) be an algebraic extension of \( k \), contained in an algebraic closure \( {k}^{\mathrm{a}} \) of \( k \) . Then the following conditions are equivalent:\n\nNOR 1. Every embedding of \( K \) in \( {k}^{\mathrm{a}} \) over \( k \) induces an automorphism of \( K \) .\n\nNOR 2. \( K \) is the splitting field of a family of polynomials in \( k\left\lbrack X\right\rbrack \) .\n\nNOR 3. Every irreducible polynomial of \( k\left\lbrack X\right\rbrack \) which has a root in \( K \) splits into linear factors in \( K \) .
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Proof. Assume NOR 1. Let \( \alpha \) be an element of \( K \) and let \( {p}_{\alpha }\left( X\right) \) be its irreducible polynomial over \( k \) . Let \( \beta \) be a root of \( {p}_{\alpha } \) in \( {k}^{a} \) . There exists an isomorphism of \( k\left( \alpha \right) \) on \( k\left( \beta \right) \) over \( k \), mapping \( \alpha \) on \( \beta \) . Extend this isomorphism to an embedding of \( K \) in \( {k}^{\mathrm{a}} \) . This extension is an automorphism \( \sigma \) of \( K \) by hypothesis, hence \( {\sigma \alpha } = \beta \) lies in \( K \) . Hence every root of \( {p}_{\alpha } \) lies in \( K \) , and \( {p}_{\alpha } \) splits in linear factors in \( K\left\lbrack X\right\rbrack \) . Hence \( K \) is the splitting field of the family \( {\left\{ {p}_{\alpha }\right\} }_{\alpha \in K} \) as \( \alpha \) ranges over all elements of \( K \), and NOR 2 is satisfied.\n\nConversely, assume NOR 2, and let \( {\left\{ {f}_{i}\right\} }_{i \in I} \) be the family of polynomials of which \( K \) is the splitting field. If \( \alpha \) is a root of some \( {f}_{i} \) in \( K \), then for any embedding \( \sigma \) of \( K \) in \( {k}^{\mathrm{a}} \) over \( k \) we know that \( {\sigma \alpha } \) is a root of \( {f}_{i} \) . Since \( K \) is generated by the roots of all the polynomials \( {f}_{i} \), it follows that \( \sigma \) maps \( K \) into itself. We now apply Lemma 2.1 to conclude that \( \sigma \) is an automorphism.\n\nOur proof that NOR 1 implies NOR 2 also shows that NOR 3 is satisfied. Conversely, assume NOR 3. Let \( \sigma \) be an embedding of \( K \) in \( {k}^{\mathrm{a}} \) over \( k \) . Let \( \alpha \in K \) and let \( p\left( X\right) \) be its irreducible polynomial over \( k \) . If \( \sigma \) is an embedding of \( K \) in \( {k}^{\mathrm{a}} \) over \( k \) then \( \sigma \) maps \( \alpha \) on a root \( \beta \) of \( p\left( X\right) \), and by hypothesis \( \beta \) lies in \( K \) . Hence \( {\sigma \alpha } \) lies in \( K \), and \( \sigma \) maps \( K \) into itself. By Lemma 2.1, it follows that \( \sigma \) is an automorphism.
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Yes
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Normal extensions remain normal under lifting. If \( K \supset E \supset k \) and \( K \) is normal over \( k \), then \( K \) is normal over \( E \). If \( {K}_{1},{K}_{2} \) are normal over \( k \) and are contained in some field \( L \), then \( {K}_{1}{K}_{2} \) is normal over \( k \), and so is \( {K}_{1} \cap {K}_{2} \).
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For our first assertion, let \( K \) be normal over \( k \), let \( F \) be any extension of \( k \), and assume \( K, F \) are contained in some bigger field. Let \( \sigma \) be an embedding of \( {KF} \) over \( F \) (in \( {F}^{\mathrm{a}} \) ). Then \( \sigma \) induces the identity on \( F \), hence on \( k \), and by hypothesis its restriction to \( K \) maps \( K \) into itself. We get \( {\left( KF\right) }^{\sigma } = {K}^{\sigma }{F}^{\sigma } = {KF} \) whence \( {KF} \) is normal over \( F \) . Assume that \( K \supset E \supset k \) and that \( K \) is normal over \( k \) . Let \( \sigma \) be an embedding of \( K \) over \( E \) . Then \( \sigma \) is also an embedding of \( K \) over \( k \), and our assertion follows by definition. Finally, if \( {K}_{1},{K}_{2} \) are normal over \( k \), then for any embedding \( \sigma \) of \( {K}_{1}{K}_{2} \) over \( k \) we have \[ \sigma \left( {{K}_{1}{K}_{2}}\right) = \sigma \left( {K}_{1}\right) \sigma \left( {K}_{2}\right) \] and our assertion again follows from the hypothesis. The assertion concerning the intersection is true because \[ \sigma \left( {{K}_{1} \cap {K}_{2}}\right) = \sigma \left( {K}_{1}\right) \cap \sigma \left( {K}_{2}\right) \]
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Yes
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Theorem 4.1. Let \( E \supset F \supset k \) be a tower. Then\n\n\[{\left\lbrack E : k\right\rbrack }_{s} = {\left\lbrack E : F\right\rbrack }_{s}{\left\lbrack F : k\right\rbrack }_{s}.\]\n\nFurthermore, if \( E \) is finite over \( k \), then \( {\left\lbrack E : k\right\rbrack }_{s} \) is finite and\n\n\[{\left\lbrack E : k\right\rbrack }_{s} \leqq \left\lbrack {E : k}\right\rbrack .\]
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Proof. Let \( \sigma : k \rightarrow L \) be an embedding of \( k \) in an algebraically closed field \( L \) . Let \( {\left\{ {\sigma }_{i}\right\} }_{i \in I} \) be the family of distinct extensions of \( \sigma \) to \( F \), and for each \( i \), let \( \left\{ {\tau }_{ij}\right\} \) be the family of distinct extensions of \( {\sigma }_{i} \) to \( E \) . By what we saw before, each \( {\sigma }_{i} \) has precisely \( {\left\lbrack E : F\right\rbrack }_{s} \) extensions to embeddings of \( E \) in \( L \) . The set of embeddings \( \left\{ {\tau }_{ij}\right\} \) contains precisely\n\n\[{\left\lbrack E : F\right\rbrack }_{s}{\left\lbrack F : k\right\rbrack }_{s}\]\n\nelements. Any embedding of \( E \) into \( L \) over \( \sigma \) must be one of the \( {\tau }_{ij} \), and thus we see that the first formula holds, i.e. we have multiplicativity in towers.\n\nAs to the second, let us assume that \( E/k \) is finite. Then we can obtain \( E \) as a tower of extensions, each step being generated by one element:\n\n\[k \subset k\left( {\alpha }_{1}\right) \subset k\left( {{\alpha }_{1},{\alpha }_{2}}\right) \subset \cdots \subset k\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) = E.\]\n\nIf we define inductively \( {F}_{v + 1} = {F}_{v}\left( {\alpha }_{v + 1}\right) \) then by Proposition 2.7,\n\n\[{\left\lbrack {F}_{v}\left( {\alpha }_{v + 1}\right) : {F}_{v}\right\rbrack }_{s} \leqq \left\lbrack {{F}_{v}\left( {\alpha }_{v + 1}\right) : {F}_{v}}\right\rbrack .\]\n\nThus our inequality is true in each step of the tower. By multiplicativity, it follows that the inequality is true for the extension \( E/k \), as was to be shown.
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Yes
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Corollary 4.2. Let \( E \) be finite over \( k \), and \( E \supset F \supset k \) . The equality\n\n\[{\left\lbrack E : k\right\rbrack }_{s} = \left\lbrack {E : k}\right\rbrack\]\n\nholds if and only if the corresponding equality holds in each step of the\n\ntower, i.e. for \( E/F \) and \( F/k \) .
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Proof. Clear.
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No
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Theorem 4.3. Let \( E \) be a finite extension of \( k \) . Then \( E \) is separable over \( k \) if and only if each element of \( E \) is separable over \( k \) .
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Proof. Assume \( E \) is separable over \( k \) and let \( \alpha \in E \) . We consider the tower\n\n\[ k \subset k\left( \alpha \right) \subset E\text{.} \]\n\nBy Corollary 4.2, we must have \( \left\lbrack {k\left( \alpha \right) : k}\right\rbrack = {\left\lbrack k\left( \alpha \right) : k\right\rbrack }_{s} \) whence \( \alpha \) is separable over \( k \) . Conversely, assume that each element of \( E \) is separable over \( k \) . We can write \( E = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) where each \( {\alpha }_{i} \) is separable over \( k \) . We consider the tower\n\n\[ k \subset k\left( {\alpha }_{1}\right) \subset k\left( {{\alpha }_{1},{\alpha }_{2}}\right) \subset \cdots \subset k\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) .\n\nSince each \( {\alpha }_{i} \) is separable over \( k \), each \( {\alpha }_{i} \) is separable over \( k\left( {{\alpha }_{1},\ldots ,{\alpha }_{i - 1}}\right) \) for \( i \geqq 2 \) . Hence by the tower theorem, it follows that \( E \) is separable over \( k \) .
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Yes
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Theorem 4.4. Let \( E \) be an algebraic extension of \( k \), generated by a family of elements \( {\left\{ {\alpha }_{i}\right\} }_{i \in I} \) . If each \( {\alpha }_{i} \) is separable over \( k \) then \( E \) is separable over \( k \) .
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Proof. Every element of \( E \) lies in some finitely generated subfield\n\n\[ k\left( {{\alpha }_{{i}_{1}},\ldots ,{a}_{{i}_{n}}}\right) \]\n\nand as we remarked above, each such subfield is separable over \( k \) . Hence every element of \( E \) is separable over \( k \) by Theorem 4.3, and this concludes the proof.
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No
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Theorem 4.5. Separable extensions form a distinguished class of extensions.
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Proof. Assume that \( E \) is separable over \( k \) and let \( E \supset F \supset k \) . Every element of \( E \) is separable over \( F \), and every element of \( F \) is an element of \( E \) , so separable over \( k \) . Hence each step in the tower is separable. Conversely, assume that \( E \supset F \supset k \) is some extension such that \( E/F \) is separable and \( F/k \) is separable. If \( E \) is finite over \( k \), then we can use Corollary 4.2. Namely, we have an equality of the separable degree and the degree in each step of the tower, whence an equality for \( E \) over \( k \) by multiplicativity.\n\nIf \( E \) is infinite, let \( \alpha \in E \) . Then \( \alpha \) is a root of a separable polynomial \( f\left( X\right) \) with coefficients in \( F \) . Let these coefficients be \( {a}_{n},\ldots ,{a}_{0} \) . Let \( {F}_{0} = \) \( k\left( {{a}_{n},\ldots ,{a}_{0}}\right) \) . Then \( {F}_{0} \) is separable over \( k \), and \( \alpha \) is separable over \( {F}_{0} \) . We now deal with the finite tower\n\n\[ k \subset {F}_{0} \subset {F}_{0}\left( \alpha \right) \]\n\nand we therefore conclude that \( {F}_{0}\left( \alpha \right) \) is separable over \( k \), hence that \( \alpha \) is separable over \( k \) . This proves condition (1) in the definition of \
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Yes
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Corollary 5.2. Let \( {\mathbf{F}}_{q} \) be a finite field. Let \( n \) be an integer \( \geqq 1 \) . In a given algebraic closure \( {\mathbf{F}}_{q}^{a} \), there exists one and only one extension of \( {\mathbf{F}}_{q} \) of degree \( n \), and this extension is the field \( {\mathbf{F}}_{{q}^{n}} \) .
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Proof. Let \( q = {p}^{m} \) . Then \( {q}^{n} = {p}^{mn} \) . The splitting field of \( {X}^{{q}^{n}} - X \) is precisely \( {F}_{{p}^{mn}} \) and has degree \( {mn} \) over \( \mathbf{Z}/p\mathbf{Z} \) . Since \( {\mathbf{F}}_{q} \) has degree \( m \) over \( \mathbf{Z}/p\mathbf{Z} \), it follows that \( {\mathbf{F}}_{{q}^{n}} \) has degree \( n \) over \( {\mathbf{F}}_{q} \) . Conversely, any extension of degree \( n \) over \( {\mathbf{F}}_{q} \) has degree \( {mn} \) over \( {\mathbf{F}}_{p} \) and hence must be \( {\mathbf{F}}_{{p}^{mn}} \) . This proves our corollary.
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Yes
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Theorem 5.3. The multiplicative group of a finite field is cyclic.
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Proof. This has already been proved in Chapter IV, Theorem 1.9.
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No
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Theorem 5.4. The group of automorphisms of \( {\mathbf{F}}_{q} \) is cyclic of degree \( n \) , generated by \( \varphi \) .
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Proof. Let \( G \) be the group generated by \( \varphi \) . We note that \( {\varphi }^{n} = \mathrm{{id}} \) because \( {\varphi }^{n}\left( x\right) = {x}^{{p}^{n}} = x \) for all \( x \in {\mathbf{F}}_{q} \) . Hence \( n \) is an exponent for \( \varphi \) . Let \( d \) be the period of \( \varphi \), so \( d \geqq 1 \) . We have \( {\varphi }^{d}\left( x\right) = {x}^{{p}^{d}} \) for all \( x \in {\mathbf{F}}_{q} \) . Hence each \( x \in {\mathbf{F}}_{q} \) is a root of the equation\n\n\[ \n{X}^{{p}^{d}} - X = 0.\n\]\n\nThis equation has at most \( {p}^{d} \) roots. It follows that \( d \geqq n \), whence \( d = n \) .\n\nThere remains to be proved that \( G \) is the group of all automorphisms of \( {\mathbf{F}}_{q} \) . Any automorphism of \( {\mathbf{F}}_{q} \) must leave \( {\mathbf{F}}_{p} \) fixed. Hence it is an automorphism of \( {\mathbf{F}}_{q} \) over \( {\mathbf{F}}_{p} \) . By Theorem 4.1, the number of such auto-morphisms is \( \leqq n \) . Hence \( {\mathbf{F}}_{q} \) cannot have any other automorphisms except for those of \( G \) .
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Yes
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Theorem 5.5. Let \( m, n \) be integers \( \geqq 1 \) . Then in any algebraic closure of \( {\mathbf{F}}_{p} \), the subfield \( {\mathbf{F}}_{{p}^{n}} \) is contained in \( {\mathbf{F}}_{{p}^{m}} \) if and only if \( n \) divides \( m \) . If that is the case, let \( q = {p}^{n} \), and let \( m = {nd} \) . Then \( {\mathbf{F}}_{{p}^{m}} \) is normal and separable over \( {\mathbf{F}}_{q} \) , and the group of automorphisms of \( {\mathbf{F}}_{{p}^{m}} \) over \( {\mathbf{F}}_{q} \) is cyclic of order \( d \), generated by \( {\varphi }^{n} \).
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Proof. All the statements are trivial consequences of what has already been proved and will be left to the reader.
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No
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Proposition 6.1. Let \( \alpha \) be algebraic over \( k,\alpha \in {k}^{\mathrm{a}} \), and let\n\n\[ f\left( X\right) = \operatorname{Irr}\left( {\alpha, k, X}\right) .\n\]\nIf char \( k = 0 \), then all roots of \( f \) have multiplicity \( 1 \) ( \( f \) is separable). If\n\n\[ \text{char}k = p > 0\text{,}\n\]\nthen there exists an integer \( \mu \geqq 0 \) such that every root of \( f \) has multiplicity \( {p}^{\mu } \). We have\n\n\[ \left\lbrack {k\left( \alpha \right) : k}\right\rbrack = {p}^{\mu }{\left\lbrack k\left( \alpha \right) : k\right\rbrack }_{s},\n\]\nand \( {\alpha }^{{p}^{\mu }} \) is separable over \( k \) .
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Proof. Let \( {\alpha }_{1},\ldots ,{\alpha }_{r} \) be the distinct roots of \( f \) in \( {k}^{\mathrm{a}} \) and let \( \alpha = {\alpha }_{1} \). Let \( m \) be the multiplicity of \( \alpha \) in \( f \) . Given \( 1 \leqq i \leqq r \), there exists an isomorphism\n\n\[ \sigma : k\left( \alpha \right) \rightarrow k\left( {\alpha }_{i}\right)\n\]\nover \( k \) such that \( {\sigma \alpha } = {\alpha }_{i} \). Extend \( \sigma \) to an automorphism of \( {k}^{\mathrm{a}} \) and denote this extension also by \( \sigma \). Since \( f \) has coefficients in \( k \) we have \( {f}^{\sigma } = f \). We note that\n\n\[ f\left( X\right) = \mathop{\prod }\limits_{{j = 1}}^{r}{\left( X - \sigma {\alpha }_{j}\right) }^{{m}_{j}}\n\]\nif \( {m}_{j} \) is the multiplicity of \( {\alpha }_{j} \) in \( f \) . By unique factorization, we conclude that \( {m}_{i} = {m}_{1} \) and hence that all \( {m}_{i} \) are equal to the same integer \( m \).\n\nConsider the derivative \( {f}^{\prime }\left( X\right) \). If \( f \) and \( {f}^{\prime } \) have a root in common, then \( \alpha \) is a root of a polynomial of lower degree than \( \deg f \). This is impossible unless \( \deg {f}^{\prime } = - \infty \), in other words, \( {f}^{\prime } \) is identically 0 . If the characteristic is 0, this cannot happen. Hence if \( f \) has multiple roots, we are in characteristic \( p \), and \( f\left( X\right) = g\left( {X}^{p}\right) \) for some polynomial \( g\left( X\right) \in k\left\lbrack X\right\rbrack \). Therefore \( {\alpha }^{p} \) is a root of a polynomial \( g \) whose degree is \( < \deg f \). Proceeding inductively, we take the smallest integer \( \mu \geqq 0 \) such that \( {\alpha }^{{p}^{\mu }} \) is the root of a separable polynomial in \( k\left\lbrack X\right\rbrack \), namely the polynomial \( h \) such that\n\n\[ f\left( X\right) = h\left( {X}^{{p}^{\mu }}\right) .\n\]\nComparing the degree of \( f \) and \( g \), we conclude that\n\n\[ \left\lbrack {k\left( \alpha \right) : k\left( {\alpha }^{p}\right) }\right\rbrack = p.\n\]\nInductively, we find\n\n\[ \left\lbrack {k\left( \alpha \right) : k\left( {\alpha }^{{p}^{\mu }}\right) }\right\rbrack = {p}^{\mu }.\n\]\nSince \( h \) has roots of multiplicity 1, we know that\n\n\[ {\left\lbrack k\left( {\alpha }^{{p}^{\mu }}\right) : k\right\rbrack }_{s} = \left\lbrack {k\left( {\alpha }^{{p}^{\mu }}\right) : k}\right\rbrack\n\]\nand comparing the degree of \( f \) and the degree of \( h \), we see that the number of distinct roots of \( f \) is equal to the number of distinct roots of \( h \). Hence\n\n\[ {\left\lbrack k\left( \alpha \right) : k\right\rbrack }_{s} = {\left\lbrack k\left( {\alpha }^{{p}^{\mu }}\right) : k\right\rbrack }_{s}.\n\]\nFrom this our formula for the degree follows by multiplicativity, and our proposition is proved. We note that the roots of \( h \) are\n\n\[ {\alpha }_{1}^{{p}^{\mu }},\ldots ,{\alpha }_{r}^{{p}^{\mu }}\n\]
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Yes
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For any finite extension \( E \) of \( k \), the separable degree \( {\left\lbrack E : k\right\rbrack }_{s} \) divides the degree \( \left\lbrack {E : k}\right\rbrack \) . The quotient is 1 if the characteristic is 0, and a power of \( p \) if the characteristic is \( p > 0 \) .
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We decompose \( E/k \) into a tower, each step being generated by one element, and apply Proposition 6.1, together with the multiplicativity of our indices in towers.
|
No
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Corollary 6.3. A finite extension is separable if and only if \( {\left\lbrack E : k\right\rbrack }_{i} = 1 \) .
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Proof. By definition.
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No
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Corollary 6.4 If \( E \supset F \supset k \) are two finite extensions, then\n\n\[{\left\lbrack E : k\right\rbrack }_{i} = {\left\lbrack E : F\right\rbrack }_{i}{\left\lbrack F : k\right\rbrack }_{i}.\]
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Proof. Immediate by Theorem 4.1.
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No
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Proposition 6.5. Purely inseparable extensions form a distinguished class of extensions.
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Proof. The tower theorem is clear from Theorem 4.1, and the lifting property is clear from condition P. Ins. 4.
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No
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Proposition 6.6. Let \( E \) be an algebraic extension of \( k \) . Let \( {E}_{0} \) be the compositum of all subfields \( F \) of \( E \) such that \( F \supset k \) and \( F \) is separable over \( k \) . Then \( {E}_{0} \) is separable over \( k \), and \( E \) is purely inseparable over \( {E}_{0} \) .
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Proof. Since separable extensions form a distinguished class, we know that \( {E}_{0} \) is separable over \( k \) . In fact, \( {E}_{0} \) consists of all elements of \( E \) which are separable over \( k \) . By Proposition 6.1, given \( \alpha \in E \) there exists a power of \( p \), say \( {p}^{n} \) such that \( {\alpha }^{{p}^{n}} \) is separable over \( k \) . Hence \( E \) is purely inseparable over \( {E}_{0} \), as was to be shown.
|
Yes
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Corollary 6.7. If an algebraic extension \( E \) of \( k \) is both separable and purely inseparable, then \( E = k \) .
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Proof. Obvious.
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No
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Corollary 6.8. Let \( K \) be normal over \( k \) and let \( {K}_{0} \) be its maximal separable subextension. Then \( {K}_{0} \) is also normal over \( k \) .
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Proof. Let \( \sigma \) be an embedding of \( {K}_{0} \) in \( {K}^{\mathrm{a}} \) over \( k \) and extend \( \sigma \) to an embedding of \( K \) . Then \( \sigma \) is an automorphism of \( K \) . Furthermore, \( \sigma {K}_{0} \) is separable over \( k \), hence is contained in \( {K}_{0} \), since \( {K}_{0} \) is the maximal separable subfield. Hence \( \sigma {K}_{0} = {K}_{0} \), as contended.
|
Yes
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Corollary 6.9. Let \( E, F \) be two finite extensions of \( k \), and assume that \( E/k \) is separable, \( F/k \) is purely inseparable. Assume \( E, F \) are subfields of a common field. Then\n\n\[ \left\lbrack {{EF} : F}\right\rbrack = \left\lbrack {E : k}\right\rbrack = {\left\lbrack EF : k\right\rbrack }_{s}, \]\n\n\[ \left\lbrack {{EF} : E}\right\rbrack = \left\lbrack {F : k}\right\rbrack = {\left\lbrack EF : k\right\rbrack }_{i}. \]
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Proof. The picture is as follows:\n\n\n\nThe proof is a trivial juggling of indices, using the corollaries of Proposition 6.1. We leave it as an exercise.
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No
|
Corollary 6.10. Let \( {E}^{p} \) denote the field of all elements \( {x}^{p}, x \in E \) . Let \( E \) be a finite extension of \( k \) . If \( {E}^{p}k = E \), then \( E \) is separable over \( k \) . If \( E \) is separable over \( k \), then \( {E}^{{p}^{n}}k = E \) for all \( n \geqq 1 \) .
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Proof. Let \( {E}_{0} \) be the maximal separable subfield of \( E \) . Assume \( {E}^{p}k = E \) . Let \( E = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) . Since \( E \) is purely inseparable over \( {E}_{0} \) there exists \( m \) such that \( {\alpha }_{i}^{{p}^{m}} \in {E}_{0} \) for each \( i = 1,\ldots, n \) . Hence \( {E}^{{p}^{m}} \subset {E}_{0} \) . But \( {E}^{{p}^{m}}k = E \) whence \( E = {E}_{0} \) is separable over \( k \) . Conversely, assume that \( E \) is separable over \( k \) . Then \( E \) is separable over \( {E}^{p}k \) . Since \( E \) is also purely inseparable over \( {E}^{p}k \) we conclude that \( E = {E}^{p}k \) . Similarly we get \( E = {E}^{{p}^{n}}k \) for \( n \geqq 1 \), as was to be shown.
|
Yes
|
Proposition 6.11. Let \( K \) be normal over \( k \) . Let \( G \) be its group of automorphisms over \( k \) . Let \( {K}^{G} \) be the fixed field of \( G \) (see Chapter VI,§1). Then \( {K}^{G} \) is purely inseparable over \( k \), and \( K \) is separable over \( {K}^{G} \) . If \( {K}_{0} \) is the maximal separable subextension of \( K \), then \( K = {K}^{G}{K}_{0} \) and \( {K}_{0} \cap {K}^{G} = k \) .
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Proof. Let \( \alpha \in {K}^{G} \) . Let \( \tau \) be an embedding of \( k\left( \alpha \right) \) over \( k \) in \( {K}^{\mathrm{a}} \) and extend \( \tau \) to an embedding of \( K \), which we denote also by \( \tau \) . Then \( \tau \) is an automorphism of \( K \) because \( K \) is normal over \( k \) . By definition, \( {\tau \alpha } = \alpha \) and hence \( \tau \) is the identity on \( k\left( \alpha \right) \) . Hence \( {\left\lbrack k\left( \alpha \right) : k\right\rbrack }_{s} = 1 \) and \( \alpha \) is purely inseparable. Thus \( {K}^{G} \) is purely inseparable over \( k \) . The intersection of \( {K}_{0} \) and \( {K}^{G} \) is both separable and purely inseparable over \( k \), and hence is equal to \( k \) .\n\nTo prove that \( K \) is separable over \( {K}^{G} \), assume first that \( K \) is finite over \( k \), and hence that \( G \) is finite, by Theorem 4.1. Let \( \alpha \in K \) . Let \( {\sigma }_{1},\ldots ,{\sigma }_{r} \) be a maximal subset of elements of \( G \) such that the elements\n\n\[{\sigma }_{1}\alpha ,\ldots ,{\sigma }_{r}\alpha\]\n\nare distinct, and such that \( {\sigma }_{1} \) is the identity, and \( \alpha \) is a root of the polynomial\n\n\[f\left( X\right) = \mathop{\prod }\limits_{{i = 1}}^{r}\left( {X - {\sigma }_{i}\alpha }\right)\]\n\nFor any \( \tau \in G \) we note that \( {f}^{\tau } = f \) because \( \tau \) permutes the roots. We note that \( f \) is separable, and that its coefficients are in the fixed field \( {K}^{G} \) . Hence \( \alpha \) is separable over \( {K}^{G} \) . The reduction of the infinite case to the finite case is done by observing that every \( \alpha \in K \) is contained in some finite normal subextension of \( K \) . We leave the details to the reader.\n\nWe now have the following picture:\n\n\n\nBy Proposition \( {6.6}, K \) is purely inseparable over \( {K}_{0} \), hence purely inseparable over \( {K}_{0}{K}^{G} \) . Furthermore, \( K \) is separable over \( {K}^{G} \), hence separable over \( {K}_{0}{K}^{G} \) . Hence \( K = {K}_{0}{K}^{G} \), thereby proving our proposition.
|
No
|
Corollary 6.12. If \( k \) is perfect, then every algebraic extension of \( k \) is separable, and every algebraic extension of \( k \) is perfect.
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Proof. Every finite algebraic extension is contained in a normal extension, and we apply Proposition 6.11 to get what we want.
|
No
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Theorem 1.2. Let \( K \) be a Galois extension of \( k \) . Let \( G \) be its Galois group. Then \( k = {K}^{G} \) . If \( F \) is an intermediate field, \( k \subset F \subset K \), then \( K \) is Galois over F. The map\n\n\[ F \mapsto G\left( {K/F}\right) \]\n\nfrom the set of intermediate fields into the set of subgroups of \( G \) is injective.
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Proof. Let \( \alpha \in {K}^{G} \) . Let \( \sigma \) be any embedding of \( k\left( \alpha \right) \) in \( {K}^{\mathrm{a}} \), inducing the identity on \( k \) . Extend \( \sigma \) to an embedding of \( K \) into \( {K}^{\mathrm{a}} \), and call this extension \( \sigma \) also. Then \( \sigma \) is an automorphism of \( K \) over \( k \), hence is an element of \( G \) . By assumption, \( \sigma \) leaves \( \alpha \) fixed. Therefore\n\n\[ {\left\lbrack k\left( \alpha \right) : k\right\rbrack }_{s} = 1 \]\n\nSince \( \alpha \) is separable over \( k \), we have \( k\left( \alpha \right) = k \) and \( \alpha \) is an element of \( k \) . This proves our first assertion.\n\nLet \( F \) be an intermediate field. Then \( K \) is normal and separable over \( F \) by Theorem 3.4 and Theorem 4.5 of Chapter V. Hence \( K \) is Galois over \( F \) . If \( H = \) \( G\left( {K/F}\right) \) then by what we proved above we conclude that \( F = {K}^{H} \) . If \( F,{F}^{\prime } \) are intermediate fields, and \( H = G\left( {K/F}\right) ,{H}^{\prime } = G\left( {K/{F}^{\prime }}\right) \), then\n\n\[ F = {K}^{H}\text{ and }{F}^{\prime } = {K}^{{H}^{\prime }}. \]\n\nIf \( H = {H}^{\prime } \) we conclude that \( F = {F}^{\prime } \), whence our map\n\n\[ F \mapsto G\left( {K/F}\right) \]\n\nis injective, thereby proving our theorem.
|
Yes
|
Corollary 1.3. Let \( K/k \) be Galois with group \( G \) . Let \( F,{F}^{\prime } \) be two intermediate fields, and let \( H,{H}^{\prime } \) be the subgroups of \( G \) belonging to \( F,{F}^{\prime } \) respectively. Then \( H \cap {H}^{\prime } \) belongs to \( F{F}^{\prime } \) .
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Proof. Every element of \( H \cap {H}^{\prime } \) leaves \( F{F}^{\prime } \) fixed, and every element of \( G \) which leaves \( F{F}^{\prime } \) fixed also leaves \( F \) and \( {F}^{\prime } \) fixed and hence lies in \( H \cap {H}^{\prime } \) . This proves our assertion.
|
Yes
|
Corollary 1.4. Let the notation be as in Corollary 1.3. The fixed field of the smallest subgroup of \( G \) containing \( H,{H}^{\prime } \) is \( F \cap {F}^{\prime } \) .
|
Proof. Obvious.
|
No
|
Corollary 1.5. Let the notation be as in Corollary 1.3. Then \( F \subset {F}^{\prime } \) if and only if \( {H}^{\prime } \subset H \) .
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Proof. If \( F \subset {F}^{\prime } \) and \( \sigma \in {H}^{\prime } \) leaves \( {F}^{\prime } \) fixed then \( \sigma \) leaves \( F \) fixed, so \( \sigma \) lies in \( H \) . Conversely, if \( {H}^{\prime } \subset H \) then the fixed field of \( H \) is contained in the fixed field of \( {H}^{\prime } \), so \( F \subset {F}^{\prime } \) .
|
Yes
|
Corollary 1.6. Let \( E \) be a finite separable extension of a field \( k \) . Let \( K \) be the smallest normal extension of \( k \) containing \( E \) . Then \( K \) is finite Galois over \( k \) . There is only a finite number of intermediate fields \( F \) such that \( k \subset F \subset E \) .
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Proof. We know that \( K \) is normal and separable, and \( K \) is finite over \( k \) since we saw that it is the finite compositum of the finite number of conjugates of \( E \) . The Galois group of \( K/k \) has only a finite number of subgroups. Hence there is only a finite number of subfields of \( K \) containing \( k \), whence a fortiori a finite number of subfields of \( E \) containing \( k \) .
|
Yes
|
Lemma 1.7. Let \( E \) be an algebraic separable extension of \( k \) . Assume that there is an integer \( n \geqq 1 \) such that every element \( \alpha \) of \( E \) is of degree \( \leqq n \) over \( k \) . Then \( E \) is finite over \( k \) and \( \left\lbrack {E : k}\right\rbrack \leqq n \) .
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Proof. Let \( \alpha \) be an element of \( E \) such that the degree \( \left\lbrack {k\left( \alpha \right) : k}\right\rbrack \) is maximal, say \( m \leqq n \) . We contend that \( k\left( \alpha \right) = E \) . If this is not true, then there exists an element \( \beta \in E \) such that \( \beta \notin k\left( \alpha \right) \), and by the primitive element theorem, there exists an element \( \gamma \in k\left( {\alpha ,\beta }\right) \) such that \( k\left( {\alpha ,\beta }\right) = k\left( \gamma \right) \) . But from the tower\n\n\[ k \subset k\left( \alpha \right) \subset k\left( {\alpha ,\beta }\right) \]\n\nwe see that \( \left\lbrack {k\left( {\alpha ,\beta }\right) : k}\right\rbrack > m \) whence \( \gamma \) has degree \( > m \) over \( k \), contradiction.
|
Yes
|
Theorem 1.8. (Artin). Let \( K \) be a field and let \( G \) be a finite group of auto-morphisms of \( K \), of order \( n \) . Let \( k = {K}^{G} \) be the fixed field. Then \( K \) is a finite Galois extension of \( k \), and its Galois group is \( G \) . We have \( \left\lbrack {K : k}\right\rbrack = n \) .
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Proof. Let \( \alpha \in K \) and let \( {\sigma }_{1},\ldots ,{\sigma }_{r} \) be a maximal set of elements of \( G \) such that \( {\sigma }_{1}\alpha ,\ldots ,{\sigma }_{r}\alpha \) are distinct. If \( \tau \in G \) then \( \left( {\tau {\sigma }_{1}\alpha ,\ldots ,\tau {\sigma }_{r}\alpha }\right) \) differs from \( \left( {{\sigma }_{1}\alpha ,\ldots ,{\sigma }_{r}\alpha }\right) \) by a permutation, because \( \tau \) is injective, and every \( \tau {\sigma }_{i}\alpha \) is among the set \( \left\{ {{\sigma }_{1}\alpha ,\ldots ,{\sigma }_{r}\alpha }\right\} \) ; otherwise this set is not maximal. Hence \( \alpha \) is a root of the polynomial\n\n\[ f\left( X\right) = \mathop{\prod }\limits_{{i = 1}}^{r}\left( {X - {\sigma }_{i}\alpha }\right) \]\n\nand for any \( \tau \in G,{f}^{\tau } = f \) . Hence the coefficients of \( f \) lie in \( {K}^{G} = k \) . Furthermore, \( f \) is separable. Hence every element \( \alpha \) of \( K \) is a root of a separable polynomial of degree \( \leqq n \) with coefficients in \( k \) . Furthermore, this polynomial splits in linear factors in \( K \) . Hence \( K \) is separable over \( k \), is normal over \( k \), hence Galois over \( k \) . By Lemma 1.7, we have \( \left\lbrack {K : k}\right\rbrack \leqq n \) . The Galois group of \( K \) over \( k \) has order \( \leqq \left\lbrack {K : k}\right\rbrack \) (by Theorem 4.1 of Chapter V), and hence \( G \) must be the full Galois group. This proves all our assertions.
|
Yes
|
Corollary 1.9. Let \( K \) be a finite Galois extension of \( k \) and let \( G \) be its Galois group. Then every subgroup of \( G \) belongs to some subfield \( F \) such that \( k \subset F \subset K \) .
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Proof. Let \( H \) be a subgroup of \( G \) and let \( F = {K}^{H} \) . By Artin’s theorem we know that \( K \) is Galois over \( F \) with group \( H \) .
|
Yes
|
Theorem 1.10. Let \( K \) be a Galois extension of \( k \) with group \( G \) . Let \( F \) be a subfield, \( k \subset F \subset K \), and let \( H = G\left( {K/F}\right) \) . Then \( F \) is normal over \( k \) if and only if \( H \) is normal in \( G \) . If \( F \) is normal over \( k \), then the restriction map \( \sigma \mapsto \sigma \mid F \) is a homomorphism of \( G \) onto the Galois group of \( F \) over \( k \), whose kernel is \( H \) . We thus have \( G\left( {F/k}\right) \approx G/H \) .
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Proof. Assume \( F \) is normal over \( k \), and let \( {G}^{\prime } \) be its Galois group. The restriction map \( \sigma \rightarrow \sigma \mid F \) maps \( G \) into \( {G}^{\prime } \), and by definition, its kernel is \( H \) . Hence \( H \) is normal in \( G \) . Furthermore, any element \( \tau \in {G}^{\prime } \) extends to an embedding of \( K \) in \( {K}^{\mathrm{a}} \), which must be an automorphism of \( K \), so the restriction map is surjective. This proves the last statement. Finally, assume that \( F \) is not normal over \( k \) . Then there exists an embedding \( \lambda \) of \( F \) in \( K \) over \( k \) which is not an automorphism, i.e. \( {\lambda F} \neq F \) . Extend \( \lambda \) to an automorphism of \( K \) over \( k \) . The Galois groups \( G\left( {K/{\lambda F}}\right) \) and \( G\left( {K/F}\right) \) are conjugate, and they belong to distinct subfields, hence cannot be equal. Hence \( H \) is not normal in \( G \) .
|
Yes
|
Corollary 1.11. Let \( K/k \) be abelian (resp. cyclic). If \( F \) is an intermediate field, \( k \subset F \subset K \), then \( F \) is Galois over \( k \) and abelian (resp. cyclic).
|
Proof. This follows at once from the fact that a subgroup of an abelian group is normal, and a factor group of an abelian (resp. cyclic) group is abelian (resp. cyclic).
|
Yes
|
Theorem 1.12. Let \( K \) be a Galois extension of \( k \), let \( F \) be an arbitrary extension and assume that \( K, F \) are subfields of some other field. Then \( {KF} \) is Galois over \( F \), and \( K \) is Galois over \( K \cap F \) . Let \( H \) be the Galois group of \( {KF} \) over \( F \) , and \( G \) the Galois group of \( K \) over \( k \) . If \( \sigma \in H \) then the restriction of \( \sigma \) to \( K \) is in \( G \), and the map\n\n\[ \sigma \mapsto \sigma \mid K \]\n\n\ngives an isomorphism of \( H \) on the Galois group of \( K \) over \( K \cap F \) .
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Proof. Let \( \sigma \in H \) . The restriction of \( \sigma \) to \( K \) is an embedding of \( K \) over \( k \) , whence an element of \( G \) since \( K \) is normal over \( k \) . The map \( \sigma \mapsto \sigma \mid K \) is clearly a homomorphism. If \( \sigma \mid K \) is the identity, then \( \sigma \) must be the identity of \( {KF} \) (since every element of \( {KF} \) can be expressed as a combination of sums, products, and quotients of elements in \( K \) and \( F \) ). Hence our homomorphism \( \sigma \mapsto \sigma \mid K \) is injective. Let \( {H}^{\prime } \) be its image. Then \( {H}^{\prime } \) leaves \( K \cap F \) fixed, and conversely, if an element \( \alpha \in K \) is fixed under \( {H}^{\prime } \), we see that \( \alpha \) is also fixed under \( H \), whence \( \alpha \in F \) and \( \alpha \in K \cap F \) . Therefore \( K \cap F \) is the fixed field. If \( K \) is finite over \( k \) , or even \( {KF} \) finite over \( F \), then by Theorem 1.8, we know that \( {H}^{\prime } \) is the Galois group of \( K \) over \( K \cap F \), and the theorem is proved in that case.\n\n(In the infinite case, one must add the remark that for the Krull topology, our map \( \sigma \mapsto \sigma \mid K \) is continuous, whence its image is closed since \( H \) is compact. See Theorem 14.1; Chapter I, Theorem 10.1; and Exercise 43.)
|
Yes
|
Corollary 1.13. Let \( K \) be a finite Galois extension of \( k \) . Let \( F \) be an arbitrary extension of \( k \) . Then \( \left\lbrack {{KF} : F}\right\rbrack \) divides \( \left\lbrack {K : k}\right\rbrack \) .
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Proof. Notation being as above, we know that the order of \( H \) divides the order of \( G \), so our assertion follows.
|
No
|
Theorem 1.14. Let \( {K}_{1} \) and \( {K}_{2} \) be Galois extensions of a field \( k \), with Galois groups \( {G}_{1} \) and \( {G}_{2} \) respectively. Assume \( {K}_{1},{K}_{2} \) are subfields of some field. Then \( {K}_{1}{K}_{2} \) is Galois over \( k \) . Let \( G \) be its Galois group. Map \( G \rightarrow {G}_{1} \times {G}_{2} \) by restriction, namely\n\n\[ \n\sigma \mapsto \left( {\sigma \left| {{K}_{1},\sigma }\right| {K}_{2}}\right) \n\]\n\nThis map is injective. If \( {K}_{1} \cap {K}_{2} = k \) then the map is an isomorphism.
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Proof. Normality and separability are preserved in taking the compositum of two fields, so \( {K}_{1}{K}_{2} \) is Galois over \( k \) . Our map is obviously a homomorphism of \( G \) into \( {G}_{1} \times {G}_{2} \) . If an element \( \sigma \in G \) induces the identity on \( {K}_{1} \) and \( {K}_{2} \) then it induces the identity on their compositum, so our map is injective. Assume that \( {K}_{1} \cap {K}_{2} = k \) . According to Theorem 1.12, given an element \( {\sigma }_{1} \in {G}_{1} \) there exists an element \( \sigma \) of the Galois group of \( {K}_{1}{K}_{2} \) over \( {K}_{2} \) which induces \( {\sigma }_{1} \) on \( {K}_{1} \) . This \( \sigma \) is a fortiori in \( G \), and induces the identity on \( {K}_{2} \) . Hence \( {G}_{1} \times \left\{ {e}_{2}\right\} \) is contained in the image of our homomorphism (where \( {e}_{2} \) is the unit element of \( \left. {G}_{2}\right) \) . Similarly, \( \left\{ {e}_{1}\right\} \times {G}_{2} \) is contained in this image. Hence their product is contained in the image, and their product is precisely \( {G}_{1} \times {G}_{2} \) . This proves Theorem 1.14.
|
Yes
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Corollary 1.15. Let \( {K}_{1},\ldots ,{K}_{n} \) be Galois extensions of \( k \) with Galois groups \( {G}_{1},\ldots ,{G}_{n} \) . Assume that \( {K}_{i + 1} \cap \left( {{K}_{1}\cdots {K}_{i}}\right) = k \) for each \( i = 1,\ldots, n - 1 \) . Then the Galois group of \( {K}_{1}\cdots {K}_{n} \) is isomorphic to the product \( {G}_{1} \times \cdots \times {G}_{n} \) in the natural way.
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Proof. Induction.
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No
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Corollary 1.16. Let \( K \) be a finite Galois extension of \( k \) with group \( G \), and assume that \( G \) can be written as a direct product \( G = {G}_{1} \times \cdots \times {G}_{n} \) . Let \( {K}_{i} \) be the fixed field of\n\n\[ \n{G}_{1} \times \cdots \times \{ 1\} \times \cdots \times {G}_{n} \n\] \n\nwhere the group with 1 element occurs in the i-th place. Then \( {K}_{i} \) is Galois over \( k \), and \( {K}_{i + 1} \cap \left( {{K}_{1}\cdots {K}_{i}}\right) = k \) . Furthermore \( K = {K}_{1}\cdots {K}_{n} \) .
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Proof. By Corollary 1.3, the compositum of all \( {K}_{i} \) belongs to the intersection of their corresponding groups, which is clearly the identity. Hence the composi-tum is equal to \( K \) . Each factor of \( G \) is normal in \( G \), so \( {K}_{i} \) is Galois over \( k \) . By Corollary 1.4, the intersection of normal extensions belongs to the product of their Galois groups, and it is then clear that \( {K}_{i + 1} \cap \left( {{K}_{1}\cdots {K}_{i}}\right) = k \) .
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Yes
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Theorem 1.17. Assume all fields contained in some common field.\n\n(i) If \( K, L \) are abelian over \( k \), so is the composite \( {KL} \) .
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Proof. Immediate from Theorems 1.12 and 1.14.
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No
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Quadratic extensions. Let \( k \) be a field and \( a \in k \). If \( a \) is not a square in \( k \), then the polynomial \( {X}^{2} - a \) has no root in \( k \) and is therefore irreducible. Assume char \( k \neq 2 \). Then the polynomial is separable (because \( 2 \neq 0) \), and if \( \alpha \) is a root, then \( k\left( \alpha \right) \) is the splitting field, is Galois, and its Galois group is cyclic of order 2.
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Conversely, given an extension \( K \) of \( k \) of degree 2, there exists \( a \in k \) such that \( K = k\left( \alpha \right) \) and \( {\alpha }^{2} = a \). This comes from completing the square and the quadratic formula as in elementary school. The formula is valid as long as the characteristic of \( k \) is \( \neq 2 \).
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No
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Let \( f\left( X\right) \) be a cubic polynomial in \( k\left\lbrack X\right\rbrack \), and assume char \( k \neq 2 \) ,3 . Then: (a) \( f \) is irreducible over \( k \) if and only if \( f \) has no root in \( k \) . (b) Assume \( f \) irreducible. Then the Galois group of \( f \) is \( {S}_{3} \) if and only if the discriminant of \( f \) is not a square in \( k \) . If the discriminant is a square, then the Galois group is cyclic of order 3, equal to the alternating group \( {A}_{3} \) as a permutation of the roots of \( f \) .
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Let \( k \) be a field of characteristic \( \neq 2 \) or 3. Let \( f\left( X\right) = {X}^{3} + {aX} + b \). Any polynomial of degree 3 can be brought into this form by completing the cube. Assume that \( f \) has no root in \( k \). Then \( f \) is irreducible because any factorization must have a factor of degree 1 . Let \( \alpha \) be a root of \( f\left( X\right) \). Then \( \left\lbrack {k\left( \alpha \right) : k}\right\rbrack = 3 \). Let \( K \) be the splitting field. Since char \( k \neq 2,3, f \) is separable. Let \( G \) be the Galois group. Then \( G \) has order 3 or 6 since \( G \) is a subgroup of the symmetric group \( {S}_{3} \). In the second case, \( k\left( \alpha \right) \) is not normal over \( k \). There is an easy way to test whether the Galois group is the full symmetric group. We consider the discriminant. If \( {\alpha }_{1},{\alpha }_{2},{\alpha }_{3} \) are the distinct roots of \( f\left( X\right) \), we let \( \delta = \left( {{\alpha }_{1} - {\alpha }_{2}}\right) \left( {{\alpha }_{2} - {\alpha }_{3}}\right) \left( {{\alpha }_{1} - {\alpha }_{3}}\right) \text{ and }\Delta = {\delta }^{2} \). If \( G \) is the Galois group and \( \sigma \in G \) then \( \sigma \left( \delta \right) = \pm \delta \). Hence \( \sigma \) leaves \( \Delta \) fixed. Thus \( \Delta \) is in the ground field \( k \), and in Chapter IV, \( §6 \), we have seen that \( \Delta = - 4{a}^{3} - {27}{b}^{2} \). The set of \( \sigma \) in \( G \) which leave \( \delta \) fixed is precisely the set of even permutations. Thus \( G \) is the symmetric group if and only if \( \Delta \) is not a square in \( k \). We may summarize the above remarks as follows.
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Yes
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We consider the polynomial \( f\left( X\right) = {X}^{4} - 2 \) over the rationals \( \mathbf{Q} \). It is irreducible by Eisenstein’s criterion. Let \( \alpha \) be a real root.
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Let \( i = \sqrt{-1} \). Then \( \pm \alpha \) and \( \pm {i\alpha } \) are the four roots of \( f\left( X\right) \), and\n\n\[ \left\lbrack {\mathbf{Q}\left( \alpha \right) : \mathbf{Q}}\right\rbrack = 4\text{.}\]\n\nHence the splitting field of \( f\left( X\right) \) is\n\n\[ K = \mathbf{Q}\left( {\alpha, i}\right) \]\n\nThe field \( \mathbf{Q}\left( \alpha \right) \cap \mathbf{Q}\left( i\right) \) has degree 1 or 2 over \( \mathbf{Q} \). The degree cannot be 2 otherwise \( i \in \mathbf{Q}\left( \alpha \right) \), which is impossible since \( \alpha \) is real. Hence the degree is 1. Hence \( i \) has degree 2 over \( \mathbf{Q}\left( \alpha \right) \) and therefore \( \left\lbrack {K : \mathbf{Q}}\right\rbrack = 8 \). The Galois group of \( f\left( X\right) \) has order 8.\n\nThere exists an automorphism \( \tau \) of \( K \) leaving \( \mathbf{Q}\left( \alpha \right) \) fixed, sending \( i \) to \( - i \), because \( K \) is Galois over \( \mathbf{Q}\left( \alpha \right) \), of degree 2. Then \( {\tau }^{2} = \mathrm{{id}} \).\n\n\n\nBy the multiplicativity of degrees in towers, we see that the degrees are as indicated in the diagram. Thus \( {X}^{4} - 2 \) is irreducible over \( \mathbf{Q}\left( i\right) \). Also, \( K \) is normal over \( \mathbf{Q}\left( i\right) \). There exists an automorphism \( \sigma \) of \( K \) over \( \mathbf{Q}\left( i\right) \) mapping the root \( \alpha \) of \( {X}^{4} - 2 \) to the root \( {i\alpha } \). Then one verifies at once that \( 1,\sigma ,{\sigma }^{2},{\sigma }^{3} \) are distinct and \( {\sigma }^{4} = \) id. Thus \( \sigma \) generates a cyclic group of order 4. We denote it by \( \langle \sigma \rangle \). Since \( \tau \notin \langle \sigma \rangle \) it follows that \( G = \langle \sigma ,\tau \rangle \) is generated by \( \sigma \) and \( \tau \) because \( \langle \sigma \rangle \) has index 2. Furthermore, one verifies directly that\n\n\[ {\tau \sigma } = {\sigma }^{3}\tau \]\n\nbecause this relation is true when applied to \( \alpha \) and \( i \) which generate \( K \) over \( \mathbf{Q} \). This gives us the structure of \( G \). It is then easy to verify that the lattice of subgroups is as follows:\n\n
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Yes
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Let \( k \) be a field and let \( {t}_{1},\ldots ,{t}_{n} \) be algebraically independent over \( k \) . Let \( K = k\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) . The symmetric group \( G \) on \( n \) letters operates on \( K \) by permuting \( \left( {{t}_{1},\ldots ,{t}_{n}}\right) \) and its fixed field is the field of symmetric functions, by definition the field of those elements of \( K \) fixed under \( G \) . Let \( {s}_{1},\ldots ,{s}_{n} \) be the elementary symmetric polynomials, and let\n\n\[ f\left( X\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {X - {t}_{i}}\right) \]\n\nUp to a sign, the coefficients of \( f \) are \( {s}_{1},\ldots ,{s}_{n} \) . We let \( F = {K}^{G} \) . We contend that \( F = k\left( {{s}_{1},\ldots ,{s}_{n}}\right) \) .
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Indeed,\n\n\[ k\left( {{s}_{1},\ldots ,{s}_{n}}\right) \subset F\text{.} \]\n\nOn the other hand, \( K \) is the splitting field of \( f\left( X\right) \), and its degree over \( F \) is \( n \) !. Its degree over \( k\left( {{s}_{1},\ldots ,{s}_{n}}\right) \) is \( \leqq n \) ! and hence we have equality, \( F = k\left( {{s}_{1},\ldots ,{s}_{n}}\right) \) .
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Yes
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We shall prove that the complex numbers are algebraically closed.
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We use the following properties of the real numbers \( \mathbf{R} \) : It is an ordered field, every positive element is a square, and every polynomial of odd degree in \( \mathbf{R}\left\lbrack X\right\rbrack \) has a root in \( \mathbf{R} \) . We shall discuss ordered fields in general later, and our arguments apply to any ordered field having the above properties.\n\nLet \( i = \sqrt{-1} \) (in other words a root of \( {X}^{2} + 1 \) ). Every element in \( \mathbf{R}\left( i\right) \) has a square root. If \( a + {bi} \in \mathbf{R}\left( i\right), a, b \in \mathbf{R} \), then the square root is given by \( c + {di} \), where\n\n\[ \n{c}^{2} = \frac{a + \sqrt{{a}^{2} + {b}^{2}}}{2}\text{ and }{d}^{2} = \frac{-a + \sqrt{{a}^{2} + {b}^{2}}}{2}.\n\]\n\nEach element on the right of our equalities is positive and hence has a square root in \( \mathbf{R} \) . It is then trivial to determine the sign of \( c \) and \( d \) so that \( {\left( c + di\right) }^{2} = a + {bi} \) .\n\nSince \( \mathbf{R} \) has characteristic 0, every finite extension is separable. Every finite extension of \( \mathbf{R}\left( i\right) \) is contained in an extension \( K \) which is finite and Galois over R. We must show that \( K = \mathbf{R}\left( i\right) \) . Let \( G \) be the Galois group over \( \mathbf{R} \) and let \( H \) be a 2-Sylow subgroup of \( G \) . Let \( F \) be its fixed field. Counting degrees and orders, we find that the degree of \( F \) over \( \mathbf{R} \) is odd. By the primitive element theorem, there exists an element \( \alpha \in F \) such that \( F = \mathbf{R}\left( \alpha \right) \) . Then \( \alpha \) is the root of an irreducible polynomial in \( \mathbf{R}\left\lbrack X\right\rbrack \) of odd degree. This can happen only if this degree is 1 . Hence \( G = H \) is a 2-group.\n\nWe now see that \( K \) is Galois over \( \mathbf{R}\left( i\right) \) . Let \( {G}_{1} \) be its Galois group. Since \( {G}_{1} \) is a \( p \) -group (with \( p = 2 \) ), if \( {G}_{1} \) is not the trivial group, then \( {G}_{1} \) has a subgroup \( {G}_{2} \) of index 2. Let \( F \) be the fixed field of \( {G}_{2} \) . Then \( F \) is of degree 2 over \( \mathbf{R}\left( i\right) \) ; it is a quadratic extension. But we saw that every element of \( \mathbf{R}\left( i\right) \) has a square root, and hence that \( \mathbf{R}\left( i\right) \) has no extensions of degree 2 . It follows that \( {G}_{1} \) is the trivial group and \( K = \mathbf{R}\left( i\right) \), which is what we wanted.\n\n(The basic ideas of the above proof were already in Gauss. The variation of the ideas which we have selected, making a particularly efficient use of the Sylow group, is due to Artin.)
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Yes
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Let \( f\left( X\right) \) be an irreducible polynomial with rational coefficients and of degree \( p \) prime. If \( f \) has precisely two nonreal roots in the complex numbers, then the Galois group of \( f \) is \( {S}_{p} \) .
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Proof. The order of \( G \) is divisible by \( p \), and hence by Sylow’s theorem, \( G \) contains an element of order \( p \) . Since \( G \) is a subgroup of \( {S}_{p} \) which has order \( p \) !, it follows that an element of order \( p \) can be represented by a \( p \) -cycle \( \left\lbrack {{123}\cdots p}\right\rbrack \) after a suitable ordering of the roots, because any smaller cycle has order less than \( p \), so relatively prime to \( p \) . But the pair of complex conjugate roots shows that complex conjugation induces a transposition in \( G \) . Hence the group is all of \( {S}_{p} \) .
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Yes
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Let \( t \) be transcendental over the complex numbers \( \mathbf{C} \), and let \( k = \mathbf{C}\left( t\right) \). The values of \( t \) in \( \mathbf{C} \), or \( \infty \), correspond to the points of the Gauss sphere \( S \), viewed as a Riemann surface. Let \( {P}_{1},\ldots ,{P}_{n + 1} \) be distinct points of \( S \). The finite coverings of \( S - \left\{ {{P}_{1},\ldots ,{P}_{n + 1}}\right\} \) are in bijection with certain finite extensions of \( \mathbf{C}\left( t\right) \), those which are unramified outside \( {P}_{1},\ldots ,{P}_{n + 1} \). Let \( K \) be the union of all these extension fields corresponding to such coverings, and let \( {\pi }_{1}^{\left( n\right) } \) be the fundamental group of \( S - \left\{ {{P}_{1},\ldots ,{P}_{n + 1}}\right\} \). Then it is known that \( {\pi }_{1}^{\left( n\right) } \) is a free group on \( n \) generators, and has an embedding in the Galois group of \( K \) over \( \mathbf{C}\left( t\right) \), such that the finite subfields of \( K \) over \( \mathbf{C}\left( t\right) \) are in bijection with the subgroups of \( {\pi }_{1}^{\left( n\right) } \) which are of finite index.
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Given a finite group \( G \) generated by \( n \) elements \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) we can find a surjective homomorphism \( {\pi }_{1}^{\left( n\right) } \rightarrow G \) mapping the generators of \( {\pi }_{1}^{\left( n\right) } \) on \( {\sigma }_{1},\ldots ,{\sigma }_{n} \). Let \( H \) be the kernel. Then \( H \) belongs to a subfield \( {K}^{H} \) of \( K \) which is normal over \( \mathbf{C}\left( t\right) \) and whose Galois group is \( G \). In the language of coverings, \( H \) belongs to a finite covering of \( S - \left\{ {{P}_{1},\ldots ,{P}_{n + 1}}\right\} \).
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Yes
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Theorem 3.1. Let \( \zeta \) be a primitive \( n \) -th root of unity. Then\n\n\[ \left\lbrack {\mathbf{Q}\left( \zeta \right) : \mathbf{Q}}\right\rbrack = \varphi \left( n\right) \]\n\nwhere \( \varphi \) is the Euler function. The map \( \sigma \mapsto i\left( \sigma \right) \) gives an isomorphism\n\n\[ {G}_{\mathbf{Q}\left( \zeta \right) /\mathbf{Q}}\overset{ \approx }{ \rightarrow }{\left( \mathbf{Z}/n\mathbf{Z}\right) }^{ * } \]
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Proof. Let \( f\left( X\right) \) be the irreducible polynomial of \( \zeta \) over \( \mathbf{Q} \) . Then \( f\left( X\right) \) divides \( {X}^{n} - 1 \), say \( {X}^{n} - 1 = f\left( X\right) h\left( X\right) \), where both \( f, h \) have leading coefficient 1. By the Gauss lemma, it follows that \( f, h \) have integral coefficients. We shall now prove that if \( p \) is a prime number not dividing \( n \), then \( {\zeta }^{p} \) is also a root of \( f \) . Since \( {\zeta }^{p} \) is also a primitive \( n \) -th root of unity, and since any primitive \( n \) -th root of unity can be obtained by raising \( \zeta \) to a succession of prime powers, with primes not dividing \( n \), this will imply that all the primitive \( n \) -th roots of unity are roots of \( f \), which must therefore have degree \( \geqq \varphi \left( n\right) \), and hence precisely \( \varphi \left( n\right) \) .\n\nSuppose \( {\zeta }^{p} \) is not a root of \( f \) . Then \( {\zeta }^{p} \) is a root of \( h \), and \( \zeta \) itself is a root of \( h\left( {X}^{p}\right) \) . Hence \( f\left( X\right) \) divides \( h\left( {X}^{p}\right) \), and we can write\n\n\[ h\left( {X}^{p}\right) = f\left( X\right) g\left( X\right) \]\n\nSince \( f \) has integral coefficients and leading coefficient 1, we see that \( g \) has integral coefficients. Since \( {a}^{p} \equiv a\left( {\;\operatorname{mod}\;p}\right) \) for any integer \( a \), we conclude that\n\n\[ h\left( {X}^{p}\right) \equiv h{\left( X\right) }^{p}\;\left( {\;\operatorname{mod}\;p}\right) \]\n\nand hence\n\n\[ h{\left( X\right) }^{p} \equiv f\left( X\right) g\left( X\right) \;\left( {\;\operatorname{mod}\;p}\right) . \]\n\nIn particular, if we denote by \( \bar{f} \) and \( \bar{h} \) the polynomials in \( \mathbf{Z}/p\mathbf{Z} \) obtained by reducing \( f \) and \( h \) respectively mod \( p \), we see that \( \bar{f} \) and \( \bar{h} \) are not relatively prime, i.e. have a factor in common. But \( {X}^{n} - \overline{1} = \bar{f}\left( X\right) \bar{h}\left( X\right) \), and hence \( {X}^{n} - \overline{1} \) has multiple roots. This is impossible, as one sees by taking the derivative, and our theorem is proved.
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Yes
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Corollary 3.2. If \( n, m \) are relative prime integers \( \geqq 1 \), then\n\n\[ \mathbf{Q}\left( {\zeta }_{n}\right) \cap \mathbf{Q}\left( {\zeta }_{m}\right) = \mathbf{Q} \]
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Proof. We note that \( {\zeta }_{n} \) and \( {\zeta }_{m} \) are both contained in \( \mathbf{Q}\left( {\zeta }_{mn}\right) \) since \( {\zeta }_{mn}^{n} \) is a primitive \( m \) -th root of unity. Furthermore, \( {\zeta }_{m}{\zeta }_{n} \) is a primitive \( {mn} \) -th root of unity. Hence\n\n\[ \mathbf{Q}\left( {\zeta }_{n}\right) \mathbf{Q}\left( {\zeta }_{m}\right) = \mathbf{Q}\left( {\zeta }_{mn}\right) \]\n\nOur assertion follows from the multiplicativity \( \varphi \left( {mn}\right) = \varphi \left( m\right) \varphi \left( n\right) \) .
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Yes
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Theorem 3.3. Let \( \zeta \) be a primitive p-th root of unity, and let\n\n\[ S = \mathop{\sum }\limits_{v}\left( \frac{v}{p}\right) {\zeta }^{v} \]\n\nthe sum being taken over non-zero residue classes mod p. Then\n\n\[ {S}^{2} = \left( \frac{-1}{p}\right) p \]
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Proof. The last statement follows at once from the explicit expression of \( \pm p \) as a square in \( \mathbf{Q}\left( \zeta \right) \), because the square root of an integer is contained in the\n\nfield obtained by adjoining the square root of the prime factors in its factorization, and also \( \sqrt{-1} \) . Furthermore, for the prime 2, we have \( {\left( 1 + i\right) }^{2} = {2i} \) . We now prove our assertion concerning \( {S}^{2} \) . We have\n\n\[ {S}^{2} = \mathop{\sum }\limits_{{v,\mu }}\left( \frac{v}{p}\right) \left( \frac{\mu }{p}\right) {\zeta }^{v + \mu } = \mathop{\sum }\limits_{{v,\mu }}\left( \frac{v\mu }{p}\right) {\zeta }^{v + \mu }. \]\n\nAs \( v \) ranges over non-zero residue classes, so does \( {v\mu } \) for any fixed \( \mu \), and hence replacing \( v \) by \( {v\mu } \) yields\n\n\[ {S}^{2} = \mathop{\sum }\limits_{{v,\mu }}\left( \frac{v{\mu }^{2}}{p}\right) {\zeta }^{\mu \left( {v + 1}\right) } = \mathop{\sum }\limits_{{v,\mu }}\left( \frac{v}{p}\right) {\zeta }^{\mu \left( {v + 1}\right) } \]\n\n\[ = \mathop{\sum }\limits_{\mu }\left( \frac{-1}{p}\right) {\zeta }^{0} + \mathop{\sum }\limits_{{v \neq - 1}}\left( \frac{v}{p}\right) \mathop{\sum }\limits_{\mu }{\zeta }^{\mu \left( {v + 1}\right) }. \]\n\nBut \( 1 + \zeta + \cdots + {\zeta }^{p - 1} = 0 \), and the sum on the right over \( \mu \) consequently yields -1 . Hence\n\n\[ {S}^{2} = \left( \frac{-1}{p}\right) \left( {p - 1}\right) + \left( {-1}\right) \mathop{\sum }\limits_{{v \neq - 1}}\left( \frac{v}{p}\right) \]\n\n\[ = p\left( \frac{-1}{p}\right) - \mathop{\sum }\limits_{v}\left( \begin{array}{l} v \\ p \end{array}\right) \]\n\n\[ = p\left( \frac{-1}{p}\right) \]\n\nas desired.
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Yes
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Theorem 4.1. (Artin). Let \( G \) be a monoid and \( K \) a field. Let \( {\chi }_{1},\ldots ,{\chi }_{n} \) be distinct characters of \( G \) in \( K \). Then they are linearly independent over \( K \).
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Proof. One character is obviously linearly independent. Suppose that we have a relation\n\n\[ \n{a}_{1}{\chi }_{1} + \cdots + {a}_{n}{\chi }_{n} = 0 \n\]\n\nwith \( {a}_{i} \in K \), not all 0 . Take such a relation with \( n \) as small as possible. Then \( n \geqq 2 \), and no \( {a}_{i} \) is equal to 0 . Since \( {\chi }_{1},{\chi }_{2} \) are distinct, there exists \( z \in G \) such that \( {\chi }_{1}\left( z\right) \neq {\chi }_{2}\left( z\right) \). For all \( x \in G \) we have\n\n\[ \n{a}_{1}{\chi }_{1}\left( {xz}\right) + \cdots + {a}_{n}{\chi }_{n}\left( {xz}\right) = 0, \n\]\n\nand since \( {\chi }_{i} \) is a character,\n\n\[ \n{a}_{1}{\chi }_{1}\left( z\right) {\chi }_{1} + \cdots + {a}_{n}{\chi }_{n}\left( z\right) {\chi }_{n} = 0. \n\]\n\nDivide by \( {\chi }_{1}\left( z\right) \) and subtract from our first relation. The term \( {a}_{1}{\chi }_{1} \) cancels, and we get a relation\n\n\[ \n\left( {{a}_{2}\frac{{\chi }_{2}\left( z\right) }{{\chi }_{1}\left( z\right) } - {a}_{2}}\right) {\chi }_{2} + \cdots = 0. \n\]\n\nThe first coefficient is not 0 , and this is a relation of smaller length than our first relation, contradiction.
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Yes
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Corollary 4.2. Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be distinct non-zero elements of a field \( K \) . If \( {a}_{1},\ldots ,{a}_{n} \) are elements of \( K \) such that for all integers \( \nu \geqq 0 \) we have\n\n\[ \n{a}_{1}{\alpha }_{1}^{v} + \cdots + {a}_{n}{\alpha }_{n}^{v} = 0 \n\]\n\nthen \( {a}_{i} = 0 \) for all \( i \) .
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Proof. We apply the theorem to the distinct homomorphisms\n\n\[ \nv \mapsto {\alpha }_{i}^{v} \n\]\n\nof \( {\mathbf{Z}}_{ \geqq 0} \) into \( {K}^{ * } \) .
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Yes
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Theorem 5.2. Let \( E \) be a finite separable extension of \( k \) . Then \( \operatorname{Tr} : E \rightarrow k \) is a non-zero functional. The map \[ \left( {x, y}\right) \mapsto \operatorname{Tr}\left( {xy}\right) \] of \( E \times E \rightarrow k \) is bilinear, and identifies \( E \) with its dual space.
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Proof. That \( \mathrm{{Tr}} \) is non-zero follows from the theorem on linear independence of characters. For each \( x \in E \), the map \[ {\operatorname{Tr}}_{x} : E \rightarrow k \] such that \( {\operatorname{Tr}}_{x}\left( y\right) = \operatorname{Tr}\left( {xy}\right) \) is obviously a \( k \) -linear map, and the map \[ x \mapsto {\operatorname{Tr}}_{x} \] is a \( k \) -homomorphism of \( E \) into its dual space \( {E}^{ \vee } \) . (We don’t write \( {E}^{ * } \) for the dual space because we use the star to denote the multiplicative group of \( E \) .) If \( {\operatorname{Tr}}_{x} \) is the zero map, then \( \operatorname{Tr}\left( {xE}\right) = 0 \) . If \( x \neq 0 \) then \( {xE} = E \) . Hence the kernel of \( x \mapsto {\operatorname{Tr}}_{x} \) is 0 . Hence we get an injective homomorphism of \( E \) into the dual space \( {E}^{ \vee } \) . Since these spaces have the same finite dimension, it follows that we get an isomorphism. This proves our theorem.
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Yes
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Corollary 5.3. Let \( {\omega }_{1},\ldots ,{\omega }_{n} \) be a basis of \( E \) over \( k \) . Then there exists a basis \( {\omega }_{1}^{\prime },\ldots ,{\omega }_{n}^{\prime } \) of \( E \) over \( k \) such that \( \operatorname{Tr}\left( {{\omega }_{i}{\omega }_{j}^{\prime }}\right) = {\delta }_{ij} \) .
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Proof. The basis \( {\omega }_{1}^{\prime },\ldots ,{\omega }_{n}^{\prime } \) is none other than the dual basis which we defined when we considered the dual space of an arbitrary vector space.
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Yes
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Corollary 5.4. Let \( E \) be a finite separable extension of \( k \), and let \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) be the distinct embeddings of \( E \) into \( {k}^{\mathrm{a}} \) over \( k \) . Let \( {w}_{1},\ldots ,{w}_{n} \) be elements of E. Then the vectors\n\n\[ \n{\xi }_{1} = \left( {{\sigma }_{1}{w}_{1},\ldots ,{\sigma }_{1}{w}_{n}}\right) \]\n\n\[ \n{\xi }_{n} = \left( {{\sigma }_{n}{w}_{1},\ldots ,{\sigma }_{n}{w}_{n}}\right) \]\n\nare linearly independent over \( E \) if \( {w}_{1},\ldots ,{w}_{n} \) form a basis of \( E \) over \( k \) .
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Proof. Assume that \( {w}_{1},\ldots ,{w}_{n} \) form a basis of \( E/k \) . Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be elements of \( E \) such that\n\n\[ \n{\alpha }_{1}{\xi }_{1} + \cdots + {\alpha }_{n}{\xi }_{n} = 0. \]\n\nThen we see that\n\n\[ \n{\alpha }_{1}{\sigma }_{1} + \cdots + {\alpha }_{n}{\sigma }_{n} \]\napplied to each one of \( {w}_{1},\ldots ,{w}_{n} \) gives the value 0 . But \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) are linearly independent as characters of the multiplicative group \( {E}^{ * } \) into \( {k}^{\mathrm{a} * } \) . It follows that \( {\alpha }_{i} = 0 \) for \( i = 1,\ldots, n \), and our vectors are linearly independent.
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Yes
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Proposition 5.5. Let \( E = k\left( \alpha \right) \) be a separable extension. Let\n\n\[ f\left( X\right) = \operatorname{Irr}\left( {\alpha, k, X}\right) \]\n\nand let \( {f}^{\prime }\left( X\right) \) be its derivative. Let\n\n\[ \frac{f\left( X\right) }{\left( X - \alpha \right) } = {\beta }_{0} + {\beta }_{1}X + \cdots + {\beta }_{n - 1}{X}^{n - 1} \]\n\nwith \( {\beta }_{i} \in E \) . Then the dual basis of \( 1,\alpha ,\ldots ,{\alpha }^{n - 1} \) is\n\n\[ \frac{{\beta }_{0}}{{f}^{\prime }\left( \alpha \right) },\ldots ,\frac{{\beta }_{n - 1}}{{f}^{\prime }\left( \alpha \right) }.\]
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Proof. Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be the distinct roots of \( f \) . Then\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}\frac{f\left( X\right) }{\left( X - {\alpha }_{i}\right) }\frac{{\alpha }_{i}^{r}}{{f}^{\prime }\left( {\alpha }_{i}\right) } = {X}^{r}\;\text{ for }\;0 \leqq r \leqq n - 1. \]\n\nTo see this, let \( g\left( X\right) \) be the difference of the left- and right-hand side of this equality. Then \( g \) has degree \( \leqq n - 1 \), and has \( n \) roots \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) . Hence \( g \) is identically zero.\n\nThe polynomials\n\n\[ \frac{f\left( X\right) }{\left( X - {\alpha }_{i}\right) }\frac{{\alpha }_{i}^{r}}{{f}^{\prime }\left( {\alpha }_{i}\right) } \]\n\nare all conjugate to each other. If we define the trace of a polynomial with coefficients in \( E \) to be the polynomial obtained by applying the trace to the coefficients, then\n\n\[ \operatorname{Tr}\left\lbrack {\frac{f\left( X\right) }{\left( X - \alpha \right) }\frac{{\alpha }^{r}}{{f}^{\prime }\left( \alpha \right) }}\right\rbrack = {X}^{r} \]\n\nLooking at the coefficients of each power of \( X \) in this equation, we see that\n\n\[ \operatorname{Tr}\left( {{\alpha }^{i}\frac{{\beta }_{j}}{{f}^{\prime }\left( \alpha \right) }}\right) = {\delta }_{ij} \]\n\nthereby proving our proposition.
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Yes
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Proposition 5.6. Let \( E \) be a finite extension of \( k \) and let \( \alpha \in E \) . Then\n\n\[ \det \left( {m}_{\alpha }\right) = {N}_{E/k}\left( \alpha \right) \;\text{ and }\;\operatorname{Tr}\left( {m}_{\alpha }\right) = {\operatorname{Tr}}_{E/k}\left( \alpha \right) .
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Proof. Let \( F = k\left( \alpha \right) \) . If \( \left\lbrack {F : k}\right\rbrack = d \), then \( 1,\alpha ,\ldots ,{\alpha }^{d - 1} \) is a basis for \( F \) over \( k \) . Let \( \left\{ {{w}_{1},\ldots ,{w}_{r}}\right\} \) be a basis for \( E \) over \( F \) . Then \( \left\{ {{\alpha }^{i}{w}_{j}}\right\} \) \( \left( {i = 0,\ldots, d - 1;j = 1,\ldots, r}\right) \) is a basis for \( E \) over \( k \) . Let\n\n\[ f\left( X\right) = {X}^{d} + {a}_{d - 1}{X}^{d - 1} + \ldots + {a}_{0} \]\n\nbe the irreducible polynomial of \( \alpha \) over \( k \) . Then \( {N}_{F/k}\left( \alpha \right) = {\left( -1\right) }^{d}{a}_{0} \), and by the transitivity of the norm, we have\n\n\[ {N}_{E/k}\left( \alpha \right) = {N}_{F/k}{\left( \alpha \right) }^{r}. \]\n\nThe reader can verify directly on the above basis that \( {N}_{F/k}\left( \alpha \right) \) is the determinant of \( {m}_{\alpha } \) on \( F \), and then that \( {N}_{F/k}{\left( \alpha \right) }^{r} \) is the determinant of \( {m}_{\alpha } \) on \( E \), thus concluding the proof for the determinant. The trace is handled exactly in the same way, except that \( {\operatorname{Tr}}_{E/k}\left( \alpha \right) = r \cdot {\operatorname{Tr}}_{F/k}\left( \alpha \right) \) . The trace of the matrix for \( {m}_{\alpha } \) on \( F \) is equal to \( - {a}_{d - 1} \) . From this the statement identifying the two traces is immediate, as it was for the norm.
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Yes
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Theorem 6.1. (Hilbert’s Theorem 90). Let \( K/k \) be cyclic of degree \( n \) with Galois group \( G \). Let \( \sigma \) be a generator of \( G \). Let \( \beta \in K \). The norm \( {N}_{k}^{K}\left( \beta \right) = N\left( \beta \right) \) is equal to 1 if and only if there exists an element \( \alpha \neq 0 \) in \( K \) such that \( \beta = \alpha /{\sigma \alpha } \).
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Proof. Assume such an element \( \alpha \) exists. Taking the norm of \( \beta \) we get \( N\left( \alpha \right) /N\left( {\sigma \alpha }\right) \). But the norm is the product over all automorphisms in \( G \). Inserting \( \sigma \) just permutes these automorphisms. Hence the norm is equal to 1.\n\nIt will be convenient to use an exponential notation as follows. If \( \tau ,{\tau }^{\prime } \in G \) and \( \xi \in K \) we write\n\n\[{\xi }^{\tau + {\tau }^{\prime }} = {\xi }^{\tau }{\xi }^{{\tau }^{\prime }}\]\n\nBy Artin's theorem on characters, the map given by\n\n\[ \mathrm{{id}} + {\beta \sigma } + {\beta }^{1 + \sigma }{\sigma }^{2} + \cdots + {\beta }^{1 + \sigma + \cdots + {\sigma }^{n - 2}}{\sigma }^{n - 1} \]\n\non \( K \) is not identically zero. Hence there exists \( \theta \in K \) such that the element\n\n\[ \alpha = \theta + \beta {\theta }^{\sigma } + {\beta }^{1 + \sigma }{\theta }^{{\sigma }^{2}} + \cdots + {\beta }^{1 + \sigma + \cdots + {\sigma }^{n - 2}}{\theta }^{{\sigma }^{n - 1}} \]\n\nis not equal to 0. It is then clear that \( \beta {\alpha }^{\sigma } = \alpha \) using the fact that \( N\left( \beta \right) = 1 \), and hence that when we apply \( {\beta \sigma } \) to the last term in the sum, we obtain \( \theta \). We divide by \( {\alpha }^{\sigma } \) to conclude the proof.
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Yes
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Theorem 6.2. Let \( k \) be a field, \( n \) an integer \( > 0 \) prime to the characteristic of \( k \) (if not 0 ), and assume that there is a primitive \( n \) -th root of unity in \( k \) .\n\n(i) Let \( K \) be a cyclic extension of degree \( n \) . Then there exists \( \alpha \in K \) such that \( K = k\left( \alpha \right) \), and \( \alpha \) satisfies an equation \( {X}^{n} - a = 0 \) for some \( a \in k \).\n\n(ii) Conversely, let \( a \in k \) . Let \( \alpha \) be a root of \( {X}^{n} - a \) . Then \( k\left( \alpha \right) \) is cyclic over \( k \), of degree \( d, d \mid n \), and \( {\alpha }^{d} \) is an element of \( k \) .
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Proof. Let \( \zeta \) be a primitive \( n \) -th root of unity in \( k \), and let \( K/k \) be cyclic with group \( G \) . Let \( \sigma \) be a generator of \( G \) . We have \( N\left( {\zeta }^{-1}\right) = {\left( {\zeta }^{-1}\right) }^{n} = 1 \) . By Hilbert’s theorem 90, there exists \( \alpha \in K \) such that \( {\sigma \alpha } = {\zeta \alpha } \) . Since \( \zeta \) is in \( k \), we have \( {\sigma }^{i}\alpha = {\zeta }^{i}\alpha \) for \( i = 1,\ldots, n \) . Hence the elements \( {\zeta }^{i}\alpha \) are \( n \) distinct conjugates of \( \alpha \) over \( k \), whence \( \left\lbrack {k\left( \alpha \right) : k}\right\rbrack \) is at least equal to \( n \) . Since \( \left\lbrack {K : k}\right\rbrack = n \), it follows that \( K = k\left( \alpha \right) \) . Furthermore,\n\n\[ \sigma \left( {\alpha }^{n}\right) = \sigma {\left( \alpha \right) }^{n} = {\left( \zeta \alpha \right) }^{n} = {\alpha }^{n}. \]\n\nHence \( {\alpha }^{n} \) is fixed under \( \sigma \), hence is fixed under each power of \( \sigma \), hence is fixed under \( G \) . Therefore \( {\alpha }^{n} \) is an element of \( k \), and we let \( a = {\alpha }^{n} \) . This proves the first part of the theorem.\n\nConversely, let \( a \in k \) . Let \( \alpha \) be a root of \( {X}^{n} - a \) . Then \( \alpha {\zeta }^{i} \) is also a root for each \( i = 1,\ldots, n \), and hence all roots lie in \( k\left( \alpha \right) \) which is therefore normal over \( k \) . All the roots are distinct so \( k\left( \alpha \right) \) is Galois over \( k \) . Let \( G \) be the Galois group.\n\nIf \( \sigma \) is an automorphism of \( k\left( \alpha \right) /k \) then \( {\sigma \alpha } \) is also a root of \( {X}^{n} - a \) . Hence \( {\sigma \alpha } = {\omega }_{\sigma }\alpha \) where \( {\omega }_{\sigma } \) is an \( n \) -th root of unity, not necessarily primitive. The map \( \sigma \mapsto {\omega }_{\sigma } \) is obviously a homomorphism of \( G \) into the group of \( n \) -th roots of unity, and is injective. Since a subgroup of a cyclic group is cyclic, we conclude that \( G \) is cyclic, of order \( d \), and \( d \mid n \) . The image of \( G \) is a cyclic group of order \( d \) . If \( \sigma \) is a generator of \( G \), then \( {\omega }_{\sigma } \) is a primitive \( d \) th root of unity. Now we get\n\n\[ \sigma \left( {\alpha }^{d}\right) = {\left( \sigma \alpha \right) }^{d} = {\left( {\omega }_{\sigma }\alpha \right) }^{d} = {\alpha }^{d}. \]\n\nHence \( {\alpha }^{d} \) is fixed under \( \sigma \), and therefore fixed under \( G \) . It is an element of \( k \), and our theorem is proved.
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Yes
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Theorem 6.3. (Hilbert’s Theorem 90, Additive Form). Let \( k \) be a field and \( K/k \) a cyclic extension of degree \( n \) with group \( G \) . Let \( o \) be a generator of \( G \) . Let \( \beta \in K \) . The trace \( {\operatorname{Tr}}_{k}^{K}\left( \beta \right) \) is equal to 0 if and only if there exists an element \( \alpha \in K \) such that \( \beta = \alpha - {\sigma \alpha } \) .
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Proof. If such an element \( \alpha \) exists, then we see that the trace is 0 because the trace is equal to the sum taken over all elements of \( G \), and applying \( \sigma \) permutes these elements. Conversely, assume \( \operatorname{Tr}\left( \beta \right) = 0 \) . There exists an element \( \theta \in K \) such that \( \operatorname{Tr}\left( \theta \right) \neq 0 \) . Let \[ \alpha = \frac{1}{\operatorname{Tr}\left( \theta \right) }\left\lbrack {\beta {\theta }^{\sigma } + \left( {\beta + {\sigma \beta }}\right) {\theta }^{{\sigma }^{2}} + \cdots + \left( {\beta + {\sigma \beta } + \cdots + {\sigma }^{n - 2}\beta }\right) {\theta }^{{\sigma }^{n - 1}}}\right\rbrack . \] From this it follows at once that \( \beta = \alpha - {\sigma \alpha } \) .
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Yes
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Theorem 6.4. (Artin-Schreier) Let \( k \) be a field of characteristic \( p \) .\n\n(i) Let \( K \) be a cyclic extension of \( k \) of degree \( p \) . Then there exists \( \alpha \in K \) such that \( K = k\left( \alpha \right) \) and \( \alpha \) satisfies an equation \( {X}^{p} - X - a = 0 \) with some \( a \in k \) .
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Proof. Let \( K/k \) be cyclic of degree \( p \) . Then \( {\operatorname{Tr}}_{k}^{K}\left( {-1}\right) = 0 \) (it is just the sum of -1 with itself \( p \) times). Let \( \sigma \) be a generator of the Galois group. By the additive form of Hilbert’s theorem 90, there exists \( \alpha \in K \) such that \( {\sigma \alpha } - \alpha = 1 \) , or in other words, \( {\sigma \alpha } = \alpha + 1 \) . Hence \( {\sigma }^{i}\alpha = \alpha + i \) for all integers \( i = 1,\ldots, p \) and \( \alpha \) has \( p \) distinct conjugates. Hence \( \left\lbrack {k\left( \alpha \right) : k}\right\rbrack \geqq p \) . It follows that \( K = k\left( \alpha \right) \) . We note that\n\n\[ \sigma \left( {{\alpha }^{p} - \alpha }\right) = \sigma {\left( \alpha \right) }^{p} - \sigma \left( \alpha \right) = {\left( \alpha + 1\right) }^{p} - \left( {\alpha + 1}\right) = {\alpha }^{p} - \alpha .\n\]\n\nHence \( {\alpha }^{p} - \alpha \) is fixed under \( \sigma \), hence it is fixed under the powers of \( \sigma \), and therefore under \( G \) . It lies in the fixed field \( k \) . If we let \( a = {\alpha }^{p} - \alpha \) we see that our first assertion is proved.
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Yes
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Proposition 7.1. Solvable extensions form a distinguished class of extensions.
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Proof. Let \( E/k \) be solvable. Let \( F \) be a field containing \( k \) and assume \( E, F \) are subfields of some algebraically closed field. Let \( K \) be Galois solvable over \( k \) , and \( E \subset K \) . Then \( {KF} \) is Galois over \( F \) and \( G\left( {{KF}/F}\right) \) is a subgroup of \( G\left( {K/k}\right) \) by Theorem 1.12. Hence \( {EF}/F \) is solvable. It is clear that a subextension of a solvable extension is solvable. Let \( E \supset F \supset k \) be a tower, and assume that \( E/F \) is solvable and \( F/k \) is solvable. Let \( K \) be a finite solvable Galois extension of \( k \) containing \( F \) . We just saw that \( {EK}/K \) is solvable. Let \( L \) be a solvable Galois extension of \( K \) containing \( {EK} \) . If \( \sigma \) is any embedding of \( L \) over \( k \) in a given algebraic closure, then \( {\sigma K} = K \) and hence \( {\sigma L} \) is a solvable extension of \( K \) . We let \( M \) be the compositum of all extensions \( {\sigma L} \) for all embeddings \( \sigma \) of \( L \) over \( k \) .\n\nThen \( M \) is Galois over \( k \), and is therefore Galois over \( K \) . The Galois group of \( M \) over \( K \) is a subgroup of the product\n\n\[ \mathop{\prod }\limits_{\sigma }G\left( {{\sigma L}/K}\right) \]\n\nby Theorem 1.14. Hence it is solvable. We have a surjective homomorphism \( G\left( {M/k}\right) \rightarrow G\left( {K/k}\right) \) by Theorem 1.10. Hence the Galois group of \( M/k \) has a solvable normal subgroup whose factor group is solvable. It is therefore solvable. Since \( E \subset M \), our proof is complete.
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Yes
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