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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 \( {f}^{\prime }\left( a\right) = 0 \) . | 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 }... | Yes |
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... | 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 c... | Yes |
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\[ \operatorname{cont}\left( {fg}\right) = \operatorname{cont}\left( f\right) \operatorname{cont}\left( g\right) \] | 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 \... | Yes |
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... | 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. | Yes |
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 . | 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},\l... | Yes |
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\l... | 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. | No |
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\;... | 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\l... | Yes |
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} ... | 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... | Yes |
Theorem 4.1. Let \( A \) be a commutative Noetherian ring. Then the polynomial ring \( A\left\lbrack X\right\rbrack \) is also Noetherian. | 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... | Yes |
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. | Proof. Use Theorem 4.1 and the preceding remark, representing \( B \) as a factor ring of a polynomial ring. | No |
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 \( j\left( p\right) \geqq 0 \)... | 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 ... | Yes |
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 | 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... | Yes |
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 th... | 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 < \... | Yes |
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 \( ... | 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 - {... | Yes |
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\[ \max \deg \{ a, b, c\} \leqq {n}_{0}\left( {abc}\right) - 1. \] | 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}... | No |
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 \( R\left( {a, b}\right) = 0 \) . | 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 \( R\left( {a, b}\right) = 0 \) . | Yes |
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 ,{... | Proof. Exercise for the reader. | No |
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\\] | 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}_... | Yes |
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 \(... | 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 ho... | Yes |
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}\ri... | Proof. One substitutes the expression obtained for \( {f}_{v}^{\prime }\left( {t}_{i}\right) \) into the product (4). The result follows at once. | No |
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 ... | 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 \... | Yes |
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 \le... | 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 \( {c}_{0} \) ... | Yes |
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. | 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\[ \n{x}_{i} = \sum {c}_{ij}{Y}_{j}\;\text{ with }\;{c}_{ij} \in k, \n\]... | Yes |
Theorem 9.4. If \( A \) is Noetherian, then \( A\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) is also Noetherian. | 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 \( {\mathf... | Yes |
Proposition 1.1. Let \( E \) be a finite extension of \( F \) . Then \( E \) is algebraic over \( F \) . | 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 s... | Yes |
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}... | 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... | Yes |
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( {\... | 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 \( g\left( X\right) \)... | Yes |
Proposition 1.5. Let \( E \) be a finite extension of \( k \). Then \( E \) is finitely generated. | Proof. Let \( \\left\{ {{\\alpha }_{1},\\ldots ,{\\alpha }_{n}}\\right\} \) be a basis of \( E \) as vector space over \( k \). Then certainly\n\n\[ E = k\\left( {{\\alpha }_{1},\\ldots ,{\\alpha }_{n}}\\right) \] | Yes |
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 \( k \) . | 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. | Yes |
Proposition 1.7. The class of algebraic extensions is distinguished, and so is the class of finite extensions. | 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 ,{... | Yes |
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. | 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 \( p\left( X\right) \) ... | Yes |
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 \[ \sigma \left( {{E}_{1}{E}_{2}}\right) = \sigma \left( {E}_{1}\right) \sigma \left( {E}_{2}\right) \] | 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 }... | Yes |
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. | 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 complemen... | Yes |
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, \( i = 1,\ldots, n \) . | 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. | Yes |
Corollary 2.6. Let \( k \) be a field. There exists an extension \( {k}^{\mathrm{a}} \) which is algebraic over \( k \) and algebraically closed. | 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}}... | Yes |
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... | 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 p... | Yes |
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 \( k \) . | Proof. Extend the identity mapping on \( k \) to an embedding of \( E \) into \( {E}^{\prime } \) and apply the theorem. | No |
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 \( k \subset K \subset {k}^{\mathrm{a}} \), wh... | 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... | Yes |
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 \( K \) . | 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.... | Yes |
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 spl... | 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 ... | Yes |
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}... | 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 \( k \), a... | Yes |
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\lbrac... | 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 ext... | Yes |
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 \( F/k \) . | Proof. Clear. | No |
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 \( k \) . | 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 \( \left\lbrack {k\left( \alpha \right) : k}\right\rbrack = {\left\lbrack k\left( \alpha \right) : k\right\rbrack }_{s} \) when... | Yes |
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 \( k \) . | 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. | No |
Theorem 4.5. Separable extensions form a distinguished class of extensions. | 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 \... | Yes |
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 \( {\mathbf{F}}_{{q}^{n}} \) . | 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}}... | Yes |
Theorem 5.3. The multiplicative group of a finite field is cyclic. | Proof. This has already been proved in Chapter IV, Theorem 1.9. | No |
Theorem 5.4. The group of automorphisms of \( {\mathbf{F}}_{q} \) is cyclic of degree \( n \) , generated by \( \varphi \) . | 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... | Yes |
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 \( {\... | Proof. All the statements are trivial consequences of what has already been proved and will be left to the reader. | No |
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\[ \text{char}k = p > 0\text{,}\n\]\nthe... | 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 ... | Yes |
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 \( p > 0 \) . | 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 |
Corollary 6.3. A finite extension is separable if and only if \( {\left\lbrack E : k\right\rbrack }_{i} = 1 \) . | Proof. By definition. | No |
Corollary 6.4 If \( E \supset F \supset k \) are two finite extensions, then\n\n\[{\left\lbrack E : k\right\rbrack }_{i} = {\left\lbrack E : F\right\rbrack }_{i}{\left\lbrack F : k\right\rbrack }_{i}.\] | Proof. Immediate by Theorem 4.1. | No |
Proposition 6.5. Purely inseparable extensions form a distinguished class of extensions. | Proof. The tower theorem is clear from Theorem 4.1, and the lifting property is clear from condition P. Ins. 4. | No |
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 \( {E}_{0} \) . | 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 \( ... | Yes |
Corollary 6.7. If an algebraic extension \( E \) of \( k \) is both separable and purely inseparable, then \( E = k \) . | Proof. Obvious. | No |
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 \( k \) . | 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 ... | Yes |
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\[ \left\lbrack {{EF} : F}\right\rbrack = \left\lbrack {E : k}\right\rbrack = {\left\lbrack EF : k\right\rbrack }_{s}, \]... | 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. | 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 \( n \geqq 1 \) . | 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,\ldot... | 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 separa... | 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 def... | No |
Corollary 6.12. If \( k \) is perfect, then every algebraic extension of \( k \) is separable, and every algebraic extension of \( k \) is perfect. | Proof. Every finite algebraic extension is contained in a normal extension, and we apply Proposition 6.11 to get what we want. | No |
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 su... | 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 automorphis... | 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 \( F{F}^{\prime } \) . | 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 \( {H}^{\prime } \subset H \) . | 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 \(... | 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 \( k \subset F \subset E \) . | 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 \( K \) containin... | 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 \( \left\lbrack {E : k}\right\rbrack \leqq n \) . | 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( \a... | 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 \( \left\lbrack {K : k}\right\rbrack = n \) . | 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 \... | 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 \( k \subset F \subset K \) . | 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 \... | 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}^{\prim... | 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... | 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... | 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 \( \left\lbrack {K : k}\right\rbrack \) . | 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 {... | 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 ind... | Yes |
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... | Proof. Induction. | No |
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\nwh... | 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 ... | Yes |
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 \( {KL} \) . | Proof. Immediate from Theorems 1.12 and 1.14. | No |
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 \( ... | 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 \( \... | No |
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 d... | 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 .... | Yes |
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. | 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) \... | Yes |
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... | 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\lef... | Yes |
We shall prove that the complex numbers are algebraically closed. | 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 ... | Yes |
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 \( {S}_{p} \) . | 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}\rig... | Yes |
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... | 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 kern... | Yes |
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... | 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 i... | Yes |
Corollary 3.2. If \( n, m \) are relative prime integers \( \geqq 1 \), then\n\n\[ \mathbf{Q}\left( {\zeta }_{n}\right) \cap \mathbf{Q}\left( {\zeta }_{m}\right) = \mathbf{Q} \] | 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( {\... | Yes |
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\[ {S}^{2} = \left( \frac{-1}{p}\right) p \] | 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 \( \sqrt{-1} \) . Furthe... | Yes |
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 \( K \). | 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 \( {... | Yes |
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} =... | Proof. We apply the theorem to the distinct homomorphisms\n\n\[ \nv \mapsto {\alpha }_{i}^{v} \n\]\n\nof \( {\mathbf{Z}}_{ \geqq 0} \) into \( {K}^{ * } \) . | Yes |
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. | 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 ... | Yes |
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 \( \operatorname{Tr}\left( {{\omega }_{i}{\omega }_{j}^{\prime }}\right) = {\delta }_{ij} \) . | 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. | Yes |
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}... | 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 }_... | Yes |
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 + ... | 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... | Yes |
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) . | 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 \( \left\{ {{\alpha }^{i}{w}_{j}}\right\} \) \... | Yes |
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 ... | 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 con... | Yes |
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( \al... | 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 th... | Yes |
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 ... | 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... | Yes |
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 \... | 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 \a... | Yes |
Proposition 7.1. Solvable extensions form a distinguished class of extensions. | 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( {... | Yes |
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