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Theorem 5.11. Let \( R \) and \( S \) be rings and let \( \varphi : R \rightarrow S \) be a homomorphism of rings. Let \( s \) be an element of \( S \) that commutes with \( \varphi \left( r\right) \) for every \( r \in R \) (an arbitrary element if \( S \) is commutative). There is a unique homomorphism of rings \( \p... | Proof. The mapping \( \psi \) is a homomorphism since it is the composition of the homomorphisms \( A \mapsto \mathcal{A} \) in 5.10 and \( B \mapsto B\left( s\right) ,\left( {\operatorname{Im}\varphi }\right) \left\lbrack X\right\rbrack \rightarrow S \) in 5.6. We see that \( \psi \) extends \( \varphi \) and sends \(... | Yes |
Proposition 5.12. For every field \( K : K\left\lbrack X\right\rbrack \) is a domain; every ideal of \( K\left\lbrack X\right\rbrack \) is principal; in fact, every nonzero ideal of \( K\left\lbrack X\right\rbrack \) is generated by a unique monic polynomial. | Proof. The trivial ideal \( 0 = \{ 0\} \) is generated by the zero polynomial 0 . Now, let \( \mathfrak{A} \neq 0 \) be a nonzero ideal of \( K\left\lbrack X\right\rbrack \) . There is a polynomial \( B \in \mathfrak{A} \) such that \( B \neq 0 \) and \( B \) has the least possible degree. Dividing \( B \) by its leadi... | Yes |
Proposition 6.1. For every ring \( R \) [with identity], \( R\left\lbrack {\left( {X}_{i}\right) }_{i \in I}\right\rbrack \), with the operations above, is a ring. | Proof. Let \( A, B \in R\left\lbrack {\left( {X}_{i}\right) }_{i \in I}\right\rbrack \) . Since \( {a}_{k} + {b}_{k} \neq 0 \) implies \( {a}_{k} \neq 0 \) or \( {b}_{k} \neq 0 \) , the set \( \left\{ {k \in M \mid {a}_{k} + {b}_{k} \neq 0}\right\} \) is finite and \( A + B \in R\left\lbrack {\left( {X}_{i}\right) }_{i... | No |
Proposition 6.2. \( A = \mathop{\sum }\limits_{k}{a}_{k}{X}^{k} \), for every \( A \in R\left\lbrack {\left( {X}_{i}\right) }_{i \in I}\right\rbrack \) . | Proof. The infinite sum \( \mathop{\sum }\limits_{k}{a}_{k}{X}^{k} \) exists since \( {a}_{k}{X}^{k} = 0 \) for almost all \( k \) . We find its coefficients. By induction on \( {m}_{i} \), the coefficient of \( {X}^{k} \) in \( {X}_{i}^{{m}_{i}} \) is 1 if \( {k}_{i} = {m}_{i} \) and \( {k}_{j} = 0 \) for all \( j \ne... | Yes |
Proposition 6.3. \( R\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \cong \left( {R\left\lbrack {{X}_{1},\ldots ,{X}_{n - 1}}\right\rbrack }\right) \left\lbrack {X}_{n}\right\rbrack \) when \( n \geqq 2 \) . | Proof. Every polynomial in \( R\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) can be rearranged by increasing powers of \( {X}_{n} \), and thereby written uniquely in the form \( {A}_{0} + {A}_{1}{X}_{n} + \cdots + {A}_{q}{X}_{n}^{q} \) , with \( {A}_{1},\ldots ,{A}_{q} \in R\left\lbrack {{X}_{1},\ldots ,{X}_{n... | Yes |
Proposition 6.6. If \( R \) is commutative, then evaluation at \( {\left( {r}_{i}\right) }_{i \in I} \in R \) is a homomorphism of \( R\left\lbrack X\right\rbrack \) into \( R \) . More generally, if \( R \) is a subring of \( S \) and \( {\left( {s}_{i}\right) }_{i \in I} \) are elements of \( S \) that commute with e... | This is proved like 5.6 ; we encourage our tireless readers to provide the details. | No |
Proposition 6.7. Every homomorphism of rings \( \varphi : R \rightarrow S \) extends uniquely to a homomorphism of rings \( A \mapsto \mathcal{A} \) of \( R\left\lbrack {\left( {X}_{i}\right) }_{i \in I}\right\rbrack \) into \( S\left\lbrack {\left( {X}_{i}\right) }_{i \in I}\right\rbrack \) that sends every \( {X}_{i}... | \[ {}^{\varphi }\left( {\mathop{\sum }\limits_{k}{a}_{k}{X}^{k}}\right) = \mathop{\sum }\limits_{k}\varphi \left( {a}_{k}\right) {X}^{k}. \] | "No" |
Theorem 6.8. Let \( R \) and \( S \) be rings and let \( \varphi : R \rightarrow S \) be a homomorphism of rings. Let \( {\left( {s}_{i}\right) }_{i \in I} \) be elements of \( S \) that commute with each other and with \( \varphi \left( r\right) \) for every \( r \in R \) (arbitrary elements of \( S \) if \( S \) is c... | This is proved like Theorem 5.11. | No |
Proposition 7.3. If \( R \) is commutative, then \( A = \mathop{\sum }\limits_{{n \geqq 0}}{a}_{n}{X}^{n} \) is a unit of \( R\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) if and only if \( {a}_{0} \) is a unit of \( R \) . | Proof. If \( A \) is a unit of \( R\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \), then \( {AB} = 1 \) for some \( B \in R\left\lbrack \left\lbrack X\right\rbrack \right\rbrack ,{a}_{0}{b}_{0} = 1 \) , and \( {a}_{0} \) is a unit of \( R \) .\n\nWe first prove the converse when \( {a}_{0} = 1 \) . Let \( A =... | Yes |
Proposition 7.4. For every ring \( R \) [with identity], the Laurent series over \( R \) with one indeterminate \( X \) constitute a ring \( R\left( \left( X\right) \right) \) . | The order ord \( A \) of a Laurent series \( A = \mathop{\sum }\limits_{n}{a}_{n}{X}^{n} \neq 0 \) is the smallest integer \( n \) such that \( {a}_{n} \neq 0 \) ; as before, we let ord \( 0 = \infty \) . Thus, a Laurent series \( A = \mathop{\sum }\limits_{{n \geqq 0}}{a}_{n}{X}^{n} \) has order at least \( n \in \mat... | No |
Proposition 7.5. For every field \( K, K\left( \left( X\right) \right) \) is a field; in fact, \( K\left( \left( X\right) \right) \) is isomorphic to the field of fractions of \( K\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) . | Proof. Let \( A = \mathop{\sum }\limits_{n}{a}_{n}{X}^{n} \in K\left( \left( X\right) \right), A \neq 0 \) . Then \( A \) has order \( m \in \mathbb{Z} \) and \( A = {X}^{m}B \) for some \( B \in K\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) whose constant term is \( {b}_{0} = {a}_{m} \neq 0 \) . By 7.3, \... | Yes |
Lemma 8.1. In a domain \( R,{Ra} = {Rb} \) if and only if \( a = {ub} \) for some unit \( u \) . | Proof. If \( u \) is a unit, then \( {Ru} = R \) and \( {Rub} = {Rb} \) . Conversely, if \( {Ra} = {Rb} \) , then \( a = {ub}, b = {va} \) for some \( u, v \in R \) ; if \( a = 0 \), then \( b = 0 \) and \( a = {1b} \) ; otherwise, \( {uva} = a \neq 0 \) implies \( {uv} = 1 \), so that \( u \) is a unit. | Yes |
Proposition 8.3. In a principal ideal domain \( R \), the following conditions on an element \( p \in R \) are equivalent: (i) \( p \) is irreducible; (ii) \( p \) is prime; (iii) \( {Rp} \) is a nonzero prime ideal; (iv) \( {Rp} \) is a nonzero maximal ideal. | Proof. (iv) implies (iii), by 4.7; (iii) implies (ii) trivially; and (ii) implies (i): if \( p \) is prime and \( p = {ab} \), then \( p \) divides, say, \( a \) ; since \( a \) already divides \( p, b \) is a unit, by 8.1 . We show that (i) implies (iv). Assume that \( {Rp} \) is contained in an ideal \( \mathfrak{a} ... | Yes |
Proposition 8.5. In a principal ideal domain \( R \), every \( a, b \in R \) have a least common multiple and a greatest common divisor. Moreover, \( m = \operatorname{lcm}\left( {a, b}\right) \) if and only if \( {Rm} = {Ra} \cap {Rb} \), and \( d = \gcd \left( {a, b}\right) \) if and only if \( {Rd} = {Ra} + {Rb} \) ... | Proof. By definition, \( m = \operatorname{lcm}\left( {a, b}\right) \) ( \( m \) is an l.c.m. of \( a \) and \( b \) ) if and only if \( m \in {Ra} \cap {Rb} \), and \( c \in {Ra} \cap {Rb} \) implies \( c \in {Rm} \) ; if and only if \( {Rm} = {Ra} \cap {Rb} \) . An l.c.m. exists since the ideal \( {Ra} \cap {Rb} \) m... | Yes |
Proposition 8.9. Let \( K \) be a field. In \( K\left\lbrack X\right\rbrack \) :\n\n(1) every polynomial of degree 1 is irreducible;\n\n(2) an irreducible polynomial of degree at least 2 has no root in \( K \) ;\n\n(3) a polynomial of degree 2 or 3 with no root in \( K \) is irreducible. | On the other hand, \( {\left( {X}^{2} + 1\right) }^{2} \in \mathbb{R}\left\lbrack X\right\rbrack \) has no root in \( \mathbb{R} \) but is not irreducible. | No |
Theorem 8.11 (Fundamental Theorem of Algebra). Every nonconstant polynomial over \( \mathbb{C} \) has a root in \( \mathbb{C} \) . | This result is due to Gauss [1799]. In 1799, algebra was primarily concerned with polynomial equations, and Theorem 8.11 was indeed of fundamental importance. Complex analysis provides the best proof of Theorem 8.11 (a much more algebraic proof is given in Section VI.2). Assume that \( f \in \mathbb{C}\left\lbrack X\ri... | Yes |
Proposition 8.12. A polynomial over \( \mathbb{R} \) is irreducible if and only if it has either degree 1, or degree 2 and no root in \( \mathbb{R} \) . | Proof. Polynomials with these properties are irreducible, by 8.9. Conversely, let \( f \in \mathbb{R}\left\lbrack X\right\rbrack, f \neq 0 \) . As a polynomial over \( \mathbb{C}, f \) is, by 8.8 and 8.10, the product of a constant and monic polynomials of degree 1 :\n\n\[ f\left( X\right) = {a}_{n}\left( {X - {r}_{1}}... | Yes |
Proposition 8.13. For every field \( K, K\left\lbrack X\right\rbrack \) contains infinitely many monic irreducible polynomials. | Proof. This proof is due to Euclid, who used a similar argument to show that \( \mathbb{Z} \) contains infinitely many primes. We show that no finite sequence \( {q}_{1},{q}_{2},\ldots ,{q}_{n} \) can contain every monic irreducible polynomial of \( K\left\lbrack X\right\rbrack \) . Indeed, \( f = 1 + \) \( {q}_{1}{q}_... | Yes |
Theorem 9.1. Every rational fraction over a field can be written uniquely as the sum of a polynomial and partial fractions with distinct denominators. | The proof of Theorem 9.1 has three parts. The first part reduces rational fractions and ejects the polynomial part. A rational fraction \( f/g \in K\left( X\right) \) is in reduced form when \( g \) is monic and \( \gcd \left( {f, g}\right) = 1 \) . | No |
Lemma 9.2. Every rational fraction can be written uniquely in reduced form. | Proof. Given \( f/g \), divide \( f \) and \( g \) by the leading coefficient of \( g \) and then by a monic g.c.d. of \( f \) and \( g \) ; the result is in reduced form.\n\nLet \( f/g = p/q,{fq} = {gp} \), with \( g, q \) monic and \( \gcd \left( {f, g}\right) = \gcd \left( {p, q}\right) = 1 \) . Then \( q \) divides... | Yes |
Lemma 9.3. Every rational fraction can be written uniquely as the sum of a polynomial and a polynomial-free fraction in reduced form. | Proof. By 9.2 we may start with a rational fraction \( f/g \) in reduced form. Polynomial division yields \( f = {gq} + r \) with \( q, r \in K\left\lbrack X\right\rbrack \) and \( \deg r < \deg g \) . Then \( f/g = q + r/g;r/g \) is polynomial-free and is in reduced form, since \( g \) is monic and \( \gcd \left( {r, ... | Yes |
Lemma 9.4. If \( \deg f < \deg {gh} \) and \( \gcd \left( {g, h}\right) = 1 \), then there exist unique polynomials \( a, b \) such that \( \deg a < \deg g,\deg b < \deg h \), and \( f/\left( {gh}\right) = \) \( \left( {a/g}\right) + \left( {b/h}\right) \) . If \( \gcd \left( {f,{gh}}\right) = 1 \), then \( \gcd \left(... | Proof. Since \( \gcd \left( {g, h}\right) = 1 \), there exist polynomials \( s, t \) such that \( {gs} + {ht} = f \) . Polynomial division yields \( t = {gp} + a, s = {hq} + b \), where \( \deg a < \deg g \) and \( \deg b < \deg h \) . Then \( f = {gh}\left( {p + q}\right) + {ah} + {bg} \), with \( \deg \left( {{ah} + ... | Yes |
Lemma 9.5. If \( \\deg f < \\deg g \) and \( \\gcd \\left( {f, g}\\right) = 1 \), then there exist unique integers \( n \\geqq 0,{k}_{1},\\ldots ,{k}_{n} > 0 \) and unique polynomials \( {a}_{1},\\ldots ,{a}_{n},{q}_{1},\\ldots ,{q}_{n} \) such that \( {q}_{1},\\ldots ,{q}_{n} \) are distinct monic irreducible polynomi... | If \( g \) is monic in Lemma 9.5, readers will see that \( g = {q}_{1}^{{k}_{1}}{q}_{2}^{{k}_{2}}\\cdots {q}_{n}^{{k}_{n}} \) is the unique factorization of \( g \) into a product of positive powers of distinct monic irreducible polynomials; then 9.5 follows from 9.4 by induction on \( n \). | No |
Lemma 9.6. If \( \deg q > 0, k > 0 \), and \( \deg a < \deg {q}^{k} \), then there exist unique polynomials \( {a}_{1},\ldots ,{a}_{k} \) such that \( \deg {a}_{i} < \deg q \) for all \( i \) and\n\n\[ \frac{a}{{q}^{k}} = \frac{{a}_{1}}{q} + \frac{{a}_{2}}{{q}^{2}} + \cdots + \frac{{a}_{k}}{{q}^{k}}. \] | Readers will easily prove Lemma 9.6 by induction on \( k \), using polynomial division. | No |
Proposition 10.1. In a unique factorization domain, let \( a = u{p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\cdots {p}_{n}^{{a}_{n}} \) and \( b = v{p}_{1}^{{b}_{1}}{p}_{2}^{{b}_{2}}\cdots {p}_{n}^{{b}_{n}} \) be products of a unit and nonnegative powers of the same distinct representative irreducible elements. Then:\n\n(1) a di... | On the other hand, in a UFD, the g.c.d. of \( a \) and \( b \) is not necessarily in the form \( {xa} + {yb} \) . Proposition 10.1 is proved like its particular case Proposition 8.6. More generally, every family of elements has a g.c.d., and every finite family of elements has an l.c.m.; the proofs of these statements ... | No |
Proposition 10.3. In a UFD, if \( \gcd \left( {a, b}\right) = \gcd \left( {a, c}\right) = 1 \), then \( \gcd \left( {a,{bc}}\right) = \) \( 1 \) ; if \( a \) divides \( {bc} \) and \( \gcd \left( {a, b}\right) = 1 \), then \( a \) divides \( c \) . | This result is proved like its particular case Proposition 8.7. | No |
Theorem 10.4. If \( R \) is a unique factorization domain, then \( R\left\lbrack X\right\rbrack \) is a unique factorization domain. | The proof of Theorem 10.4 uses the quotient field \( Q \) of \( R \), and studies irreducible polynomials to show how \( R\left\lbrack X\right\rbrack \) inherits unique factorization from \( Q\left\lbrack X\right\rbrack \) . | Yes |
Every nonzero polynomial \( f\left( X\right) \in Q\left( X\right) \) can be written in the form \( f\left( X\right) = t{f}^{ * }\left( X\right) \), where \( t \in Q, t \neq 0 \), and \( {f}^{ * }\left( X\right) \in R\left\lbrack X\right\rbrack \) is primitive; moreover, \( t \) and \( {f}^{ * } \) are unique up to mult... | Proof. We have \( f\left( X\right) = \left( {{a}_{0}/{b}_{0}}\right) + \left( {{a}_{1}/{b}_{1}}\right) X + \cdots + \left( {{a}_{n}/{b}_{n}}\right) {X}^{n} \), where \( {a}_{i},{b}_{i} \in R \) and \( {b}_{i} \neq 0 \) . Let \( b \) be a common denominator (for instance, \( b = \) \( \left. {{b}_{0}{b}_{1}\cdots {b}_{n... | Yes |
Lemma 10.6 (Gauss). If \( f \) and \( g \in R\left\lbrack X\right\rbrack \) are primitive, then \( {fg} \) is primitive. | Proof. Let \( f\left( X\right) = {a}_{0} + {a}_{1}X + \cdots + {a}_{m}{X}^{m} \) and \( g\left( X\right) = {b}_{0} + {b}_{1}X + \cdots + \) \( {b}_{n}{X}^{n} \), so that \( \left( {fg}\right) \left( X\right) = {c}_{0} + {c}_{1}X + \cdots + {c}_{m + n}{X}^{m + n} \), where \( {c}_{k} = \mathop{\sum }\limits_{{i + j = k}... | Yes |
Corollary 10.7. In Lemma 10.5, \( f \) is irreducible in \( Q\left\lbrack X\right\rbrack \) if and only if \( {f}^{ * } \) is irreducible in \( R\left\lbrack X\right\rbrack \) . | Proof. We may assume that \( \deg f \geqq 1 \) . If \( f \) is not irreducible, then \( f \) has a factorization \( f = {gh} \) in \( Q\left\lbrack X\right\rbrack \) where \( \deg g \) , \( \deg h \geqq 1 \) . Let \( g\left( X\right) = v{g}^{ * }\left( X\right), h\left( X\right) = w{h}^{ * }\left( X\right) \), with \( ... | Yes |
In \( R\left\lbrack X\right\rbrack \), every polynomial, other than 0 and units of \( R \), is a nonempty product of irreducible elements of \( R \) and irreducible primitive polynomials. Hence the irreducible elements of \( R\left\lbrack X\right\rbrack \) are the irreducible elements of \( R \) and the irreducible pri... | Assume that \( f \in R\left\lbrack X\right\rbrack \) is not zero and not a unit of \( R \) . Let \( d \) be a g.c.d. of the coefficients of \( f \) . Then \( f\left( X\right) = d{f}^{ * }\left( X\right) \), where \( {f}^{ * } \in R\left\lbrack X\right\rbrack \) is primitive; moreover, \( d \) and \( {f}^{ * } \) are no... | Yes |
Proposition 10.9 (Eisenstein’s Criterion). Let \( R \) be a UFD and let \( f\left( X\right) = \) \( {a}_{0} + {a}_{1}X + \cdots + {a}_{n}{X}^{n} \in R\left\lbrack X\right\rbrack \) . If \( f \) is primitive and there exists an irreducible element \( p \) of \( R \) such that \( p \) divides \( {a}_{i} \) for all \( i <... | Proof. Suppose that \( f = {gh} \) ; let \( g\left( X\right) = {b}_{0} + {b}_{1}X + \cdots + {b}_{r}{X}^{r} \) and \( h\left( X\right) = \) \( {c}_{0} + {c}_{1}X + \cdots + {c}_{s}{X}^{s} \in R\left\lbrack X\right\rbrack \), where \( r = \deg g \) and \( s = \deg h \) . Then \( {a}_{k} = \mathop{\sum }\limits_{{i + j =... | Yes |
Proposition 11.1. A commutative ring \( R \) is Noetherian if and only if every ideal of \( R \) is finitely generated (as an ideal). | Proof. Let \( \mathfrak{a} \) be an ideal of \( R \) . Let \( \mathcal{S} \) be the set of all finitely generated ideals of \( R \) contained in \( \mathfrak{a} \) . Then \( \mathcal{S} \) contains principal ideals and is not empty. If \( R \) is Noetherian, then \( \mathcal{S} \) has a maximal element \( \mathfrak{s} ... | Yes |
Corollary 11.4. Let \( R \subseteq S \) be commutative rings. If \( R \) is Noetherian, and \( S \) is generated by \( R \) and finitely many elements of \( S \), then \( S \) is Noetherian. | We leave this to our readers, who deserve some fun. | No |
Proposition 12.1. The lexicographic order, degree lexicographic order, and degree reverse lexicographic order are monomial orders on \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . | In any monomial order, \( {X}^{a}{X}^{b} > {X}^{a} \) whenever \( {X}^{b} \neq 1 \) ; hence \( {X}^{c} \geqq {X}^{a} \) whenever \( {X}^{c} \) is a multiple of \( {X}^{a} \) . | No |
Proposition 12.2. In any monomial order, there is no infinite strictly decreasing sequence \( {X}^{{m}_{1}} > {X}^{{m}_{2}} > \cdots > {X}^{{m}_{k}} > {X}^{{m}_{k + 1}} > \cdots \) . | Proof. Suppose that \( {X}^{{m}_{1}} > {X}^{{m}_{2}} > \cdots > {X}^{{m}_{k}} > {X}^{{m}_{k + 1}} > \cdots \) By 12.3 below, the ideal of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) generated by all \( {X}^{{m}_{i}} \) is generated by finitely many \( {X}^{{m}_{i}} \) ’s. Let \( {X}^{t} \) be the least o... | Yes |
Lemma 12.3. An ideal of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) that is generated by a set \( \mathcal{S} \) of monomials is generated by a finite subset of \( \mathcal{S} \) . | Proof. By the Hilbert basis theorem, the ideal (S) generated by S is generated by finitely many polynomials \( {f}_{1},\ldots ,{f}_{r} \) . Every nonzero term of \( {f}_{j} \) is a multiple of some \( {X}^{s} \in \mathcal{S} \) . Let \( \mathcal{T} \) be the set of all \( {X}^{s} \in \mathcal{S} \) that divide a nonzer... | Yes |
Proposition 12.4. Let \( K \) be a field. Let \( f,{g}_{1},\ldots ,{g}_{k} \in K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) , \( {g}_{1},\ldots ,{g}_{k} \neq 0 \) . Relative to any monomial order on \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \), there exist \( {q}_{1},\ldots ,{q}_{k}, r \in K\le... | Proof. Let \( {f}_{0} = f \) . If none of \( \operatorname{ldm}{g}_{1},\ldots ,\operatorname{ldm}{g}_{k} \) divides a nonzero term of \( f \), then \( {q}_{1} = \cdots = {q}_{k} = 0 \) and \( r = f \) serve. Otherwise, there is a greatest monomial \( {X}^{m} \) that appears in a term \( {a}_{m}{X}^{m} \neq 0 \) of \( f... | Yes |
Proposition 12.5. Let \( K \) be a field, let \( \mathfrak{A} \) be an ideal of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \), let \( {g}_{1},\ldots ,{g}_{k} \) be a Gröbner basis of \( \mathfrak{A} \) relative to a monomial order \( < \), and let \( f \in K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbra... | Proof. Let \( f \in \mathfrak{A} \) . Let \( r \) be the remainder in a division of \( f \) by \( {g}_{1},\ldots ,{g}_{k} \) . Then \( r \in \mathfrak{A} \) . If \( r \neq 0 \), then \( \operatorname{ldt}r \in \operatorname{ldm}\mathfrak{A} \) is a linear combination of \( \operatorname{ldm}{g}_{1} \) , \( \ldots ,\ope... | Yes |
Proposition 12.6. Every ideal of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) has a Gröbner basis. | Proof. Let \( \mathfrak{A} \) be an ideal of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . By \( {12.3},\operatorname{ldm}\mathfrak{A} \) is generated by \( \operatorname{ldm}{g}_{1},\ldots ,\operatorname{ldm}{g}_{k} \) for some nonzero \( {g}_{1},\ldots ,{g}_{k} \in \mathfrak{A} \) . Let \( f \in \mathf... | Yes |
Proposition 12.8 (Buchberger’s Algorithm). Let \( K \) be a field and let \( {g}_{1},\ldots ,{g}_{k} \) \( \in K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) be nonzero polynomials. Compute a sequence \( B \) of polynomials as follows. Start with \( B = {g}_{1},\ldots ,{g}_{k} \) . Compute all polynomials \( {... | Proof. Let \( \mathfrak{A} = \left( {{g}_{1},\ldots ,{g}_{k}}\right) \) . Since \( {r}_{i, j} \) is the remainder of some \( {d}_{i, j} \in \mathfrak{A} \) in a division by \( {g}_{1},\ldots ,{g}_{k} \), we have \( {r}_{i, j} \in \mathfrak{A} \), but, if \( {r}_{i, j} \neq 0 \), no \( \operatorname{ldm}{g}_{t} \) divid... | Yes |
Let \( {g}_{1} = {XY} - X,{g}_{2} = Y - {X}^{2} \in \mathbb{C}\left\lbrack {X, Y}\right\rbrack \) . Use the lexicographic order with \( Y > X \) . Start with \( B = {g}_{1},{g}_{2} \). | We have \( \operatorname{ldm}{g}_{1} = {XY},\operatorname{ldm}{g}_{2} = Y,{\ell }_{12} = {XY} \), and\n\n\[ \n{d}_{1,2} = \left( {{\ell }_{12}/\operatorname{ldt}{g}_{1}}\right) {g}_{1} - \left( {{\ell }_{12}/\operatorname{ldt}{g}_{2}}\right) {g}_{2} = {g}_{1} - X{g}_{2} = {X}^{3} - X = {r}_{1,2}, \n\] \n\nsince \( {XY}... | Yes |
Proposition 1.1. Every homomorphism of fields is injective. | This is Proposition III.4.9. Consequently, a homomorphism of a field \( K \) into a field \( L \) induces a homomorphism of multiplicative groups of \( K \smallsetminus \{ 0\} \) into \( L \smallsetminus \{ 0\} \) , and preserves powers and inverses. | No |
Proposition 1.5. Every field \( K \) has a smallest subfield, which is isomorphic to \( \mathbb{Q} \) if \( K \) has characteristic \( 0 \), to \( {\mathbb{Z}}_{p} \) if \( K \) has characteristic \( p \neq 0 \) . | Proofs. If \( K \) has characteristic \( p \neq 0 \), then \( p \) is prime, by III.4.3; hence the smallest subring of \( K \) is a field, by III.4.1, and is the smallest subfield of \( K \) . If \( K \) has characteristic 0, then, by III.4.11, the injection \( m \mapsto {m1} \) of \( \mathbb{Z} \) into \( K \) extends... | Yes |
Proposition 1.6. Every finite multiplicative subgroup of a field is cyclic. | Proof. Let \( K \) be a field and let \( G \) be a finite subgroup of the multiplicative group \( K \smallsetminus \{ 0\} \) . Write \( \left| G\right| \) as a product \( \left| G\right| = {p}_{1}^{{k}_{1}}{p}_{2}^{{k}_{2}}\cdots {p}_{r}^{{k}_{r}} \) of positive powers of distinct primes. By II.1.3, \( G \) is a direct... | Yes |
Proposition 1.7. Every intersection of subfields of a field \( F \) is a subfield of \( F \) . | The proof is an exercise. | No |
Proposition 1.9. Let \( K \) be a subfield of a field \( F \) and let \( S \) be a subset of \( F \). The subring \( K\left\lbrack S\right\rbrack \) of \( F \) generated by \( K \cup S \) is the set of all finite linear combinations with coefficients in \( K \) of finite products of powers of elements of \( S \). The s... | Proof. Let \( {\left( {X}_{s}\right) }_{s \in S} \) be a family of indeterminates, one for each \( s \in S \) . By III.6.6 there is an evaluation homomorphism \( \varphi : K\left\lbrack {\left( {X}_{s}\right) }_{s \in S}\right\rbrack \rightarrow F \): \[ \varphi \left( {\mathop{\sum }\limits_{k}\left( {{a}_{k}\mathop{\... | Yes |
Proposition 1.11. Let \( {\left( {K}_{i}\right) }_{i \in I} \) be a nonempty family of subfields of a field \( F \) . Then \( x \in \mathop{\prod }\limits_{{i \in I}}{K}_{i} \) if and only if \( x = a{b}^{-1} \in F \) for some \( a, b \in R, b \neq 0 \), where \( R \) is the set of all finite sums of finite products of... | Proof. We have \( \mathop{\prod }\limits_{{i \in I}}{K}_{i} = {K}_{0}\left( {\mathop{\bigcup }\limits_{{i \in I}}{K}_{i}}\right) \), where \( {K}_{0} \) is the smallest subfield of \( F \) . Multiplying an element of \( {K}_{0} \) by a finite product of powers of elements of \( \mathop{\bigcup }\limits_{{i \in I}}{K}_{... | Yes |
Proposition 2.1. If \( K \subseteq E \subseteq F \), then \( \left\lbrack {F : K}\right\rbrack = \left\lbrack {F : E}\right\rbrack \left\lbrack {E : K}\right\rbrack \) . | Proof. Let \( {\left( {\alpha }_{i}\right) }_{i \in I} \) be a basis of \( E \) over \( K \) and let \( {\left( {\beta }_{j}\right) }_{j \in J} \) be a basis of \( F \) over \( E \) . Every element of \( F \) is a linear combination of \( {\beta }_{j} \) ’s with coefficients in \( E \), which are themselves linear comb... | Yes |
Proposition 2.2. Let \( K \subseteq E \) be a field extension and let \( \alpha \in E \). Either \( f\left( \alpha \right) \neq 0 \) for every nonzero polynomial \( f\left( X\right) \in K\left\lbrack X\right\rbrack \), in which case there is a \( K \) -isomorphism \( K\left( \alpha \right) \cong K\left( X\right) \); or... | Proof. Let \( \psi : K\left\lbrack X\right\rbrack \rightarrow E, f\left( X\right) \mapsto f\left( \alpha \right) \) be the evaluation homomorphism. By 1.10, \( \operatorname{Im}\psi = K\left\lbrack \alpha \right\rbrack \subseteq E \). If \( \operatorname{Ker}\psi = 0 \), then \( K\left\lbrack \alpha \right\rbrack \cong... | Yes |
Proposition 2.3. Let \( K \) be a field and let \( q \in K\left\lbrack X\right\rbrack \) be irreducible. Up to isomorphism, \( E = K\left\lbrack X\right\rbrack /\left( q\right) \) is a simple field extension of \( K : E = K\left( \alpha \right) \), where \( \alpha = X + \left( q\right) \) . Moreover, \( \left\lbrack {E... | Proof. By III.8.3, \( \left( q\right) \) is a maximal ideal of \( K\left\lbrack X\right\rbrack \) ; hence \( E = K\left\lbrack X\right\rbrack /\left( q\right) \) is a field. Then \( x \mapsto x + \left( q\right) \) is a homomorphism of \( K \) into \( E \) ; we may identify \( x \in K \) and \( x + \left( q\right) \in ... | Yes |
Proposition 2.4. Let \( \alpha \) be algebraic over \( K \) and let \( q = \operatorname{Irr}\left( {\alpha : K}\right) \) . If \( \psi : K\left( \alpha \right) \rightarrow L \) is a field homomorphism and \( \varphi \) is the restriction of \( \psi \) to \( K \), then \( \psi \left( \alpha \right) \) is a root of \( {... | Proof. Let \( \psi : K\left( \alpha \right) \rightarrow L \) be a field homomorphism. Its restriction \( \varphi \) to \( K \) is a field homomorphism. For each \( f\left( X\right) = {a}_{0} + {a}_{1}X + \cdots + {a}_{m}{X}^{m} \in K\left\lbrack X\right\rbrack \) , we have\n\n\[ \psi \left( {f\left( \alpha \right) }\ri... | Yes |
Proposition 3.2. If \( E = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) and every \( {\alpha }_{i} \) is algebraic over \( K \), then \( E \) is finite (hence algebraic) over \( K \) . | Proof. We give this proof as an example. Let \( E = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \), where all \( {\alpha }_{i} \) are algebraic over \( K \) . We prove by induction on \( n \) that \( E \) is finite over \( K \) . If \( n = 0 \), then \( E = K \) is finite over \( K \) . If \( n > 0 \), then \(... | Yes |
Proposition 3.8. If \( E \) is finite over \( K \) and the composite \( {EF} \) exists, then \( {EF} \) is finite over \( {KF} \) . Hence the composite of finitely many finite extensions of \( K \) is a finite extension of \( K \) . | Proof. We prove the first statement and leave the second as an exercise. Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be a basis of \( E \) over \( K \) . Then \( E = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) and every \( {\alpha }_{i} \) is algebraic over \( K \) . Hence \( {EF} = {KF}\left( {{\alpha }_... | No |
Proposition 3.9. In any field extension \( K \subseteq E \), the elements that are algebraic over \( K \) constitute a field. | Proof. First,0 and \( 1 \in K \) are algebraic over \( K \) . Now let \( \alpha ,\beta \in E \) be algebraic over \( K \) . By 3.3, \( K\left( {\alpha ,\beta }\right) \subseteq E \) is algebraic over \( K \) . Hence \( \alpha - \beta \in \) \( K\left( {\alpha ,\beta }\right) \) and \( \alpha {\beta }^{-1} \in K\left( {... | Yes |
Proposition 4.1. For a field \( K \) the following properties are equivalent:\n\n(1) the only algebraic extension of \( K \) is \( K \) itself;\n\n(2) in \( K\left\lbrack X\right\rbrack \), every irreducible polynomial has degree 1 ;\n\n(3) every nonconstant polynomial in \( K\left\lbrack X\right\rbrack \) has a root i... | Proof. (1) implies (2): when \( q \in K\left\lbrack X\right\rbrack \) is irreducible, then \( E = K\left\lbrack X\right\rbrack /\left( q\right) \) has degree \( \left\lbrack {E : K}\right\rbrack = \deg q \), by 2.3; hence (1) implies \( \deg q = 1 \) .\n\n(2) implies (3) since every nonconstant polynomial \( f \in K\le... | Yes |
Theorem 4.2. Every homomorphism of a field \( K \) into an algebraically closed field can be extended to every algebraic extension of \( K \) . | Proof. Let \( E \) be an algebraic extension of \( K \) and let \( \varphi \) be a homomorphism of \( K \) into an algebraically closed field \( L \) . If \( E = K\left( \alpha \right) \) is a simple extension of \( K \), and \( q = \operatorname{Irr}\left( {\alpha : K}\right) \), then \( {}^{\varphi }q \in L\left\lbra... | Yes |
Theorem 4.4. Every field \( K \) has an algebraic extension \( \bar{K} \) that is algebraically closed. Moreover, \( \bar{K} \) is unique up to \( K \) -isomorphism. | Proof. There is a very tall tower of fields\n\n\[ K = {E}_{0} \subseteq {E}_{1} \subseteq \cdots \subseteq {E}_{n} \subseteq {E}_{n + 1} \subseteq \cdots \]\n\nin which \( {E}_{n + 1} \) is the algebraic extension of \( {E}_{n} \) in Lemma 4.3, which contains a root of every nonconstant polynomial with coefficients in ... | Yes |
Proposition 4.6. Every \( K \) -endomorphism of \( \bar{K} \) is a \( K \) -automorphism. | Proof. Let \( \varphi : \bar{K} \rightarrow \bar{K} \) is a \( K \) -homomorphism. As in the proof of 4.4, \( \operatorname{Im}\varphi \cong \bar{K} \) is algebraically closed, \( \bar{K} \) is algebraic over \( \operatorname{Im}\varphi \), by 3.4, hence \( \bar{K} = \operatorname{Im}\varphi \) and \( \varphi \) is a \... | Yes |
Proposition 4.7. If \( K \subseteq E \subseteq \bar{K} \) is an algebraic extension of \( K \), then every \( K \) -homomorphism of \( E \) into \( \bar{K} \) extends to a \( K \) -automorphism of \( \bar{K} \) . | Proof. By 4.2, every \( K \) -homomorphism of \( E \) into \( \bar{K} \) extends to a \( K \) -endomorphism of \( \bar{K} \), which is a \( K \) -automorphism of \( \bar{K} \) by 4.6. \( ▱ \) | Yes |
Proposition 5.1. Let \( q \in K\left\lbrack X\right\rbrack \) be irreducible.\n\n(1) If \( K \) has characteristic 0, then \( q \) is separable.\n\n(2) If \( K \) has characteristic \( p \neq 0 \), then all roots of \( q \) in \( \bar{K} \) have the same multiplicity, which is a power \( {p}^{m} \) of \( p \), and ther... | Proof. We may assume that \( q \) is monic. If \( q \) has a multiple root \( \alpha \) in \( \bar{K} \), then \( {q}^{\prime }\left( \alpha \right) = 0 \) by III.5.9. Now, \( \alpha \) is algebraic over \( K \), with \( q = \operatorname{Irr}\left( {\alpha : K}\right) \) since \( q\left( \alpha \right) = 0 \) ; hence ... | Yes |
Proposition 5.3 (Tower Property). If \( F \) is algebraic over \( K \) and \( K \subseteq E \subseteq F \) , then \( {\left\lbrack F : K\right\rbrack }_{s} = {\left\lbrack F : E\right\rbrack }_{s}{\left\lbrack E : K\right\rbrack }_{s} \) . | Proof. By 3.13 we may assume that \( E \subseteq \bar{K} \) . Let \( \varphi : E \rightarrow \bar{K} \) be a \( K \) - homomorphism. By 3.14, there is a \( K \) -automorphism \( \sigma \) of \( \bar{K} \) that extends \( \varphi \) . If now \( \psi : F \rightarrow \bar{K} \) is an \( E \) -homomorphism, then \( \sigma ... | Yes |
Proposition 5.4. For every finite extension \( E \) of \( K,{\left\lbrack E : K\right\rbrack }_{s} \leqq \left\lbrack {E : K}\right\rbrack \) ; if \( K \) has characteristic 0, then \( {\left\lbrack E : K\right\rbrack }_{s} = \left\lbrack {E : K}\right\rbrack \) ; if \( K \) has characteristic \( p \neq 0 \) , then \( ... | Proof. \( E = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) for some \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in E \), which yields a tower \( K = {E}_{0} \subseteq {E}_{1} \subseteq \cdots \subseteq {E}_{n} = E \) of simple extensions \( {E}_{i} = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{i}}\right) = \) \( {... | No |
Proposition 5.6. For a finite extension \( K \subseteq E \) the following conditions are equivalent:\n\n(1) \( E \) is separable over \( K \) (every element of \( E \) is separable over \( K \) );\n\n(2) \( E \) is generated by finitely many separable elements;\n\n(3) \( {\left\lbrack E : K\right\rbrack }_{s} = \left\l... | Proof. (1) implies (2), since \( E = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) for some \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in E \) .\n\n(2) implies (3). Let \( E = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \), where \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are separable over \( K \) . The... | Yes |
Proposition 5.12 (Primitive Element Theorem). Every finite separable extension is simple. | Proof. Let \( E \) be a finite separable extension of a field \( K \) . If \( K \) is finite, then \( E \) is finite, the multiplicative group \( E \smallsetminus \{ 0\} \) is cyclic by 1.6, and \( E \) is singly generated as an extension.\n\nNow let \( K \) be infinite. We show that every finite separable extension \(... | Yes |
Proposition 5.13. If \( E \) is separable over \( K \) and \( \operatorname{Irr}\left( {\alpha : K}\right) \) has degree at most \( n \) for every \( \alpha \in E \), then \( E \) is finite over \( K \) and \( \left\lbrack {E : K}\right\rbrack \leqq n \) . | Proof. Choose \( \alpha \in E \) so that \( m = \deg \operatorname{Irr}\left( {\alpha : K}\right) \) is maximal. For every \( \beta \in \) \( E \) we have \( K\left( {\alpha ,\beta }\right) = K\left( \gamma \right) \) for some \( \gamma \in E \), by 5.12. Then \( \deg \operatorname{Irr}\left( {\gamma : K}\right) \leqq ... | Yes |
For every algebraic extension \( E \) of \( K, S = \{ \alpha \in E \mid \alpha \) is separable over \( K\} \) is a subfield of \( E, S \) is separable over \( K \), and \( E \) is purely inseparable over \( S \) . | Proof. First,0 and \( 1 \in K \) are separable over \( K \) . If \( \alpha ,\beta \in E \) are separable over \( K \), then \( K\left( {\alpha ,\beta }\right) \) is separable over \( K \) by 5.7 and \( \alpha - \beta ,\alpha {\beta }^{-1} \in K\left( {\alpha ,\beta }\right) \) are separable over \( K \) . Thus \( S \) ... | Yes |
Proposition 6.2. If \( K \) has characteristic \( p \neq 0 \), then \( {K}^{1/{p}^{\infty }} = \{ \alpha \in \) \( \left. {\bar{K} \mid {\alpha }^{{p}^{{p}^{m}}} \in K\text{for some}m \geqq 0}\right\} \) is a purely inseparable field extension of \( K \) . | Proof. We have \( K \subseteq {K}^{1/{p}^{\infty }} \), in particular \( 0,1 \in {K}^{1/{p}^{\infty }} \) . If \( \alpha ,\beta \in {K}^{1/{p}^{\infty }} \) , then \( {\alpha }^{{p}^{m}},{\beta }^{{p}^{m}} \in K \) when \( m \) is large enough, and then \( {\left( \alpha - \beta \right) }^{{p}^{m}} = {\alpha }^{{p}^{m}... | Yes |
Lemma 6.3. If \( K \) has characteristic \( p \neq 0 \), and \( \alpha \) is algebraic over \( K \), then \( {\alpha }^{{p}^{n}} \in K \) for some \( n \geqq 0 \) if and only if \( \operatorname{Irr}\left( {\alpha : K}\right) = {X}^{{p}^{m}} - a \) for some \( m \geqq 0 \) , and \( a \in K \) . | Proof. Let \( q = \operatorname{Irr}\left( {\alpha : K}\right) \) . By 5.1, \( q\left( X\right) = s\left( {X}^{{p}^{m}}\right) \) for some \( m \geqq 0 \) and separable monic irreducible polynomial \( s \in K\left\lbrack X\right\rbrack \) . If \( {\alpha }^{{p}^{n}} = b \in K \) for some \( n \geqq 0 \), then \( q \) d... | Yes |
Proposition 6.4. Let \( K \) have characteristic \( p \neq 0 \) and let \( E \) be an algebraic extension of \( K \) . The following conditions are equivalent:\n\n(1) \( E \) is purely inseparable over \( K \) (no \( \alpha \in E \smallsetminus K \) is separable over \( K \) );\n\n(2) every element of \( E \) is purely... | Proof. (1) implies (2). Assume that \( E \) is purely inseparable over \( K \) . Let \( \alpha \in E \) and \( q = \operatorname{Irr}\left( {\alpha : K}\right) \) . By 5.1, \( q\left( X\right) = s\left( {X}^{{p}^{m}}\right) \) for some \( m \geqq 0 \) and separable monic irreducible polynomial \( s \in K\left\lbrack X\... | Yes |
Proposition 7.2. Let \( K \) be a field and \( f\left( X\right) = {a}_{m}{X}^{m} + \cdots + {a}_{0}, g\left( X\right) = {b}_{n}{X}^{n} + \cdots + {b}_{0} \in K\left\lbrack X\right\rbrack \) . If \( {a}_{m},{b}_{n} \neq 0 \), then \[ \operatorname{Res}\left( {f, g}\right) = \left| \begin{matrix} {a}_{m} & \ldots & \ldot... | Proof. If a polynomial \( p \in \mathbb{Z}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) becomes 0 when \( {X}_{j} \neq {X}_{i} \) is substituted for \( {X}_{i} \), then \( p \) is divisible by \( {X}_{i} - {X}_{j} \) : if, say, \( i = 1 \), then polynomial division in \( \mathbb{Z}\left\lbrack {{X}_{2},\ldots ... | Yes |
Proposition 7.4. Let \( K \) be a field. If \( f \in K\left\lbrack X\right\rbrack \) has degree \( n \geqq 2 \) and leading coefficient \( {a}_{n} \), then \( \operatorname{Res}\left( {f,{f}^{\prime }}\right) = {\left( -1\right) }^{n\left( {n - 1}\right) /2}{a}_{n}\operatorname{Dis}\left( f\right) \) . | Proof. In \( \bar{K}\left\lbrack X\right\rbrack, f\left( X\right) = {a}_{n}\mathop{\prod }\limits_{i}\left( {X - {\alpha }_{i}}\right) \) . By III.5.11,\n\n\[ \n{f}^{\prime }\left( X\right) = {a}_{n}\mathop{\sum }\limits_{i}\left( {\mathop{\prod }\limits_{{j \neq i}}\left( {X - {\alpha }_{j}}\right) }\right) .\n\]\n\nH... | Yes |
Proposition 7.5. Let \( K \) be a field. If \( f = {a}_{n}{X}^{n} + \cdots + {a}_{0} \in K\left\lbrack X\right\rbrack \) has degree \( n \geqq 2 \), then\n\n\[ \n\operatorname{Dis}\left( f\right) = {\left( -1\right) }^{n\left( {n - 1}\right) /2}\frac{1}{{a}_{n}}\left| \begin{matrix} {a}_{n} & \cdots & \cdots & {a}_{0} ... | For example, if \( f = a{X}^{2} + {bX} + c \), then\n\n\[ \n\operatorname{Res}\left( {f,{f}^{\prime }}\right) = \left| \begin{matrix} a & b & c \\ {2a} & b & 0 \\ 0 & {2a} & b \end{matrix}\right| = 4{a}^{2}c - a{b}^{2};\n\]\n\nhence \( \operatorname{Dis}\left( f\right) = {b}^{2} - {4ac} \) . If \( K \) does not have ch... | No |
Proposition 8.1. For every field \( K, K\left( {\left( {X}_{i}\right) }_{i \in I}\right) \) is totally transcendental over \( K \) . | Proof. First we show that \( K\left( X\right) \) is totally transcendental over \( K \) . For clarity’s sake we prove the equivalent result that \( K\left( \chi \right) \cong K\left( X\right) \) is totally transcendental over \( K \) when \( \chi \) is transcendental over \( K \) . Let \( \alpha \in K\left( \chi \right... | Yes |
Proposition 8.2. Every field extension is a totally transcendental extension of an algebraic extension. | Proof. In any field extension \( K \subseteq E \), the set \( A = \{ \alpha \in E \mid \alpha \) is algebraic over \( K\} \) is a subfield of \( E \) by 3.6, and contains \( K \) . Hence \( A \) is an algebraic extension of \( K \) . If now \( \alpha \in E \) is algebraic over \( A \), then \( A\left( \alpha \right) \)... | Yes |
Lemma 8.5. For a subset \( S \) of a field extension \( K \subseteq E \) the following conditions are equivalent:\n\n(1) \( S \) is a maximal algebraically independent subset;\n\n(2) \( S \) is algebraically independent over \( K \) and \( E \) is algebraic over \( K\left( S\right) \) ;\n\n(3) \( S \) is minimal such t... | Proof. (1) and (2) are equivalent: by 8.3, if no \( S \cup \{ \beta \} \) with \( \beta \notin S \) is algebraically independent, then every \( \beta \in E \smallsetminus S \) is algebraic over \( K\left( S\right) \) ; conversely, if every \( \beta \in E \) is algebraic over \( K\left( S\right) \), then no \( S \cup \{... | Yes |
Theorem 8.6. Every field extension \( K \subseteq E \) has a transcendence base; in fact, when \( S \subseteq T \subseteq E, S \) is algebraically independent over \( K \), and \( E \) is algebraic over \( K\left( T\right) \), then \( E \) has a transcendence base \( S \subseteq B \subseteq T \) over \( K \) . | Proof. Readers will verify that the union of a chain of algebraically independent subsets is algebraically independent. The existence of a maximal algebraically independent subset then follows from Zorn's lemma. More generally, let \( S \subseteq T \subseteq E \), where \( S \) is algebraically independent over \( K \)... | Yes |
Theorem 8.7. In a field extension, all transcendence bases have the same number of elements. | Theorem 8.7 is similar to the statement that all bases of a vector space have the same number of elements, and is proved in much the same way. First we establish an exchange property. | No |
Lemma 8.8. Let \( B \) and \( C \) be transcendence bases of a field extension \( E \) of \( K \). For every \( \beta \in B \) there exists \( \gamma \in C \) such that \( \left( {B\smallsetminus \{ \beta \} }\right) \cup \{ \gamma \} \) is a transcendence base of \( E \) over \( K \), and either \( \gamma = \beta \) o... | Proof. If \( \beta \in C \), then \( \gamma = \beta \) serves. Now let \( \beta \notin C \). If every \( \gamma \in C \) is algebraic over \( K\left( {B\smallsetminus \{ \beta \} }\right) \), then, by \( {3.3},{3.5}, K\left( C\right) \) is algebraic over \( K\left( {B\smallsetminus \{ \beta \} }\right) \), and \( E \),... | Yes |
Proposition 9.2. Let \( K \subseteq E \subseteq L \) and \( K \subseteq F \subseteq L \) be fields. Let \( E \) be the quotient field of a ring \( K \subseteq R \subseteq E \) (for instance, let \( R = E \) ). If\n\n(1) \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in R \) linearly independent over \( K \) implies \( {\alpha... | Proof. Assume (1) and let \( {\left( {\alpha }_{i}\right) }_{i \in I} \in E \) be linearly independent over \( K \) . Then \( {\left( {\alpha }_{j}\right) }_{j \in J} \) is linearly independent over \( K \) for every finite subset \( J \) of \( I \) . If \( J \) is finite, then there exists \( r \in R, r \neq 0 \), suc... | Yes |
Corollary 9.3. If \( K \subseteq E \subseteq L \) and \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in L \) are algebraically independent over \( E \), then \( E \) and \( K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) are linearly disjoint over \( K \) . | Proof. \( K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \cong K\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) is the quotient field of \( K\left\lbrack {{\alpha }_{1},\ldots }\right. \) , \( \left. {\alpha }_{n}\right\rbrack \cong K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \), and the monomials \( {\alpha ... | Yes |
Proposition 9.4. Let \( K \subseteq E \subseteq L \) and \( K \subseteq F \subseteq {F}^{\prime } \subseteq L \) be fields. If \( E \) and \( F \) are linearly disjoint over \( K \), and \( {EF} \) and \( {F}^{\prime } \) are linearly disjoint over \( F \) , then \( E \) and \( {F}^{\prime } \) are linearly disjoint ov... | Proof. Take bases \( {\left( {\alpha }_{i}\right) }_{i \in I} \) of \( E \) over \( K,{\left( {\beta }_{j}\right) }_{j \in J} \) of \( F \) over \( K \), and \( {\left( {\gamma }_{h}\right) }_{h \in H} \) of \( {F}^{\prime } \) over \( F \) . Then \( {\left( {\beta }_{j}{\gamma }_{h}\right) }_{j \in J, h \in H} \) is a... | Yes |
Proposition 9.5. If \( K \) has characteristic \( p \neq 0 \) and \( E \) is purely transcendental over \( K \), then \( E \) and \( {K}^{1/{p}^{\infty }} \) are linearly disjoint over \( K \) . | Proof. Let \( E = K\left( {\left( {\chi }_{i}\right) }_{i \in I}\right) \cong K\left( {\left( {X}_{i}\right) }_{i \in I}\right) \), where \( {\left( {\chi }_{i}\right) }_{i \in I} \) are algebraically independent over \( K \) . Both \( E \) and \( {K}^{1/{p}^{\infty }} \) are contained in \( \bar{K}\left( {\left( {\chi... | Yes |
Proposition 9.8 (Tower Property). If \( F \) is separable over \( K \), and \( E \) is separable over \( F \), then \( E \) is separable over \( K \) . | The proof is an easy exercise, using 9.4. | No |
Lemma 1.1. If \( E \) and \( F \) are splitting fields of \( \mathcal{S} \subseteq K\left\lbrack X\right\rbrack \) over \( K \), and \( F \subseteq \bar{K} \) , then \( {\varphi E} = F \) for every \( K \) -homomorphism \( \varphi : E \rightarrow \overrightarrow{K} \) . | Proof. Every \( f \in \mathcal{S} \) has unique factorizations \( f\left( X\right) = a\left( {X - {\alpha }_{1}}\right) (X - \) \( \left. {\alpha }_{2}\right) \cdots \left( {X - {\alpha }_{n}}\right) \) in \( E\left\lbrack X\right\rbrack \) and \( f\left( X\right) = a\left( {X - {\beta }_{1}}\right) \left( {X - {\beta ... | Yes |
Theorem 1.3. For every prime \( p \) and every \( n > 0 \) there is, up to isomorphism, exactly one field \( F \) of order \( {p}^{n};F \) is a splitting field of \( {X}^{{p}^{n}} - X \) over \( {\mathbb{Z}}_{p} \), and all its elements are roots of \( {X}^{{p}^{n}} - X \) . | Proof. Let \( F \) be a field of order \( {p}^{n} \) . By IV.1.6, the multiplicative group \( {F}^{ * } = F \smallsetminus \{ 0\} \) is cyclic; since \( \left| {F}^{ * }\right| = {p}^{n} - 1 \) we have \( {x}^{{p}^{n} - 1} = 1 \) for all \( x \in {F}^{ * } \) and \( {x}^{{p}^{n}} = x \) for all \( x \in F \) . Thus the... | Yes |
Proposition 2.1. For an algebraic extension \( K \subseteq E \subseteq \bar{K} \) the following conditions are equivalent:\n\n(1) \( E \) is the splitting field over \( K \) of a set of polynomials;\n\n(2) \( {\varphi E} = E \) for every \( K \) -homomorphism \( \varphi : E \rightarrow \bar{K} \) ;\n\n(3) \( {\varphi E... | Proof. (1) implies (2) by 1.1; (2) implies (3) and (4) implies (5); (2) implies (4), and (3) implies (5), since every \( K \) -automorphism of \( \bar{K} \) induces a \( K \) -homomorphism of \( E \) into \( \bar{K} \) .\n\n(5) implies (6). Let \( q \in K\left\lbrack X\right\rbrack \) be irreducible, with a root \( \al... | Yes |
Proposition 2.2. Over a field \( K \), the conjugates of \( \alpha \in \bar{K} \) are the roots of \( \operatorname{Irr}\left( {\alpha : K}\right) \) in \( \bar{K} \) . | Proof. If \( \sigma \) is a \( K \) -automorphism of \( \bar{K} \), then \( {\sigma \alpha } \) is a root of \( q = \operatorname{Irr}\left( {\alpha : K}\right) \) , since \( q\left( {\sigma \alpha }\right) = {}^{\sigma }q\left( {\sigma \alpha }\right) = {\sigma q}\left( \alpha \right) = 0 \) . Conversely, if \( \beta ... | Yes |
Proposition 2.3. For an algebraic extension \( K \subseteq E \subseteq \bar{K} \) the following conditions are equivalent:\n\n(1) \( E \) is a normal extension of \( K \) ;\n\n(2) \( E \) contains all conjugates over \( K \) of all elements of \( E \) ;\n\n(3) \( E \) has only one conjugate. | Proof. (3) is part (4) of Proposition 2.1, and (2) is, by 2.2, equivalent to part (6) of 2.1. \( ▱ \) | No |
Proposition 2.9. Every finite (respectively separable, finite separable) extension \( E \subseteq \bar{K} \) of a field \( K \) is contained in a finite (separable, finite separable) normal extension of \( K \) . | Proof. If \( E \subseteq \bar{K} \) is finite, then, by IV.5.4, there are only finitely many \( K \) - homomorphisms of \( E \) into \( \bar{K} \) . Since the restriction to \( E \) of a \( K \) -automorphism of \( \bar{K} \) is a \( K \) -homomorphism, \( E \) has only finitely many conjugates; their composite \( F \)... | Yes |
Proposition 2.10. If \( E \subseteq \bar{K} \) is a normal extension of \( K \), then\n\n\[ F = \left\{ {\alpha \in E \mid {\sigma \alpha } = \alpha \text{ for every }K\text{-automorphism }\sigma \text{ of }\bar{K}}\right\} \]\n\nis a purely inseparable extension of \( K \), and \( E \) is a separable extension of \( F... | Proof. First, \( F \) is a subfield of \( E \) and \( K \subseteq F \) . If \( \alpha \in F \), then every \( K \) - homomorphism \( \varphi \) of \( K\left( \alpha \right) \) into \( \bar{K} \) extends to a \( K \) -automorphism of \( \bar{K} \), by IV.4.7; hence \( \varphi \left( \alpha \right) = \alpha \) and \( \va... | Yes |
Proposition 2.11. Finite fields and algebraically closed fields are perfect. | Proof. Algebraically closed fields are supremely perfect. If \( K \) is a finite field, then the characteristic of \( K \) is some prime \( p \neq 0,\pi : x \mapsto {x}^{p} \) is injective by III.4.4; therefore \( \pi \) is surjective and \( K \) is perfect. \( ▱ \) | Yes |
Lemma 2.12. A perfect field has no proper purely inseparable extension. | Proof. By IV.5.5 we may assume that \( K \) has characteristic \( p \neq 0 \) . If \( K \) is perfect, then \( K \) contains the \( p \) th root of every \( a \in K \) in \( \bar{K} \) ; by induction, \( K \) contains the \( {p}^{m} \) th root of every \( a \in K \) in \( \bar{K} \) . Therefore, only the elements of \(... | Yes |
Proposition 2.13. Every algebraic extension of a perfect field is separable. | Proof. Let \( K \) be perfect and let \( E \subseteq \bar{K} \) be an algebraic extension of \( K \) . By \( {2.8}, E \) is contained in a normal extension \( N \) of \( K \), which by 2.10 is a separable extension of a purely inseparable extension \( F \) of \( K \) . By 2.12, \( F = K \) ; hence \( E \subseteq N \) i... | Yes |
Proposition 2.14. Every algebraic extension of a perfect field is perfect. | The proof is an exercise. (Readers may not groan.) | No |
Proposition 3.6. If \( E \) is Galois over \( K \), then \( \left| {\operatorname{Gal}\left( {E : K}\right) }\right| = \left\lbrack {E : K}\right\rbrack \) . | Proof. If \( E \subseteq \bar{K} \) is normal over \( K \), then every \( K \) -homomorphism of \( E \) into \( \bar{K} \) sends \( E \) onto \( E \) and is (as a set of ordered pairs) a \( K \) -automorphism of \( E \) . Hence \( \left| {\operatorname{Gal}\left( {E : K}\right) }\right| = {\left\lbrack E : K\right\rbra... | Yes |
Proposition 3.7 (Artin). If \( G \) is a finite group of automorphisms of a field \( E \) , then \( E \) is a finite Galois extension of \( F = {\operatorname{Fix}}_{E}\left( G\right) \) and \( \operatorname{Gal}\left( {E : F}\right) = G \) . | Proof. Let \( \alpha \in E \) . Since \( G \) is finite, \( {G\alpha } \) is a finite set, \( {G\alpha } = \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \) , where \( n \leqq \left| G\right| ,{\alpha }_{1},\ldots ,{\alpha }_{n} \in E \) are distinct, and, say, \( {\alpha }_{1} = \alpha \) . Let \( {f}_{\alpha }... | Yes |
Proposition 3.8. If \( E \) is a Galois extension of \( K \), then the fixed field of \( \operatorname{Gal}\left( {E : K}\right) \) is \( K \) . | Proof. Let \( G = \operatorname{Gal}\left( {E : K}\right) \) . Then \( K \subseteq {\operatorname{Fix}}_{E}\left( G\right) \) . Conversely, let \( \alpha \in \) \( {\operatorname{Fix}}_{E}\left( G\right) \) . By IV.4.5, there is an algebraic closure \( \bar{K} \supseteq E \) . By IV.4.7, every \( K \) -homomorphism \( ... | Yes |
Proposition 3.10. Let \( {F}_{1},{F}_{2},{F}_{3} \) be intermediate fields of a finite Galois extension \( E \) of \( K \), with Galois groups \( {H}_{1},{H}_{2},{H}_{3} \). (4) when \( E \subseteq \bar{K} \), then \( {F}_{1} \) and \( {F}_{2} \) are conjugate if and only if \( {H}_{1} \) and \( {H}_{2} \) are conjugat... | Proof. We prove (4) and leave (1),(2),(3) as exercises. First, \( {F}_{1} \) and \( {F}_{2} \) are conjugate if and only if \( \tau {F}_{2} = {F}_{1} \) for some \( \tau \in \operatorname{Gal}\left( {E : K}\right) \) : indeed, \( \tau \) can be extended to a \( K \) -automorphism \( \sigma \) of \( \bar{K} \) ; convers... | No |
Proposition 3.11. If \( E \) is a finite Galois extension of \( K \), then an intermediate field \( K \subseteq F \subseteq E \) is normal over \( K \) if and only if \( \operatorname{Gal}\left( {E : F}\right) \) is normal in \( \operatorname{Gal}\left( {E : K}\right) \), and then \( \operatorname{Gal}\left( {F : K}\ri... | Proof. By part (4) of 3.10, \( F \) is normal over \( K \) ( \( F \) has only one conjugate) if and only if \( \operatorname{Gal}\left( {E : F}\right) \) is normal in \( \operatorname{Gal}\left( {E : K}\right) \) . Now let \( F \) be normal over \( K \) . By 3.2, \( F \) is Galois over \( K \) . Hence every \( \sigma \... | Yes |
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