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Corollary 10.3 Let a field \( \mathbb{F} \) be finitely generated as an algebra over a subfield \( k \subset \mathbb{F} \). Then \( \mathbb{F} \) has finite dimension as a vector space over \( k \). | Proof If \( \mathbb{F} \) is generated as a \( \mathbb{k} \)-algebra by algebraic elements \( {b}_{1},{b}_{2},\ldots ,{b}_{m} \), then the monomials \( {b}_{1}^{{s}_{1}}{b}_{2}^{{s}_{2}}\cdots {b}_{m}^{{s}_{m}} \) with \( 0 \leq {s}_{i} < {\deg }_{\mathbb{k}}{b}_{i} \) span \( \mathbb{F} \) linearly over \( \mathbb{k} ... | Yes |
Lemma 10.2 (Exchange Lemma) Let elements \( {a}_{1},{a}_{2},\ldots ,{a}_{m} \) be transcendence generators of \( A \) over \( \mathbb{k} \), and let \( {b}_{1},{b}_{2},\ldots ,{b}_{n} \in A \) be algebraically independent over \( k \) . Then \( n \leq m \), and after appropriate renumbering of the \( {a}_{i} \) and rep... | Proof Since \( {b}_{1} \) is algebraic over \( \mathbb{k}\left( {{a}_{1},{a}_{2},\ldots ,{a}_{m}}\right) \), there is a polynomial relation\n\n\[ f\left( {{b}_{1},{a}_{1},{a}_{2},\ldots ,{a}_{m}}\right) = 0,\;f \in \mathbb{k}\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{m + 1}}\right\rbrack . \]\n\nSince \( {b}_{1} \) is ... | Yes |
Example 10.4 Let \( A \subset \mathbb{k}\left( t\right) \) be a \( \mathbb{k} \) -subalgebra different from \( \mathbb{k} \) . Then \( \operatorname{tr}{\deg }_{\mathbb{k}}A = 1 \) . | Indeed, for every \[ \psi = f\left( t\right) /g\left( t\right) \in A \smallsetminus \mathbb{k}, \] the element \( t \) satisfies the algebraic equation \( \psi \cdot g\left( x\right) - f\left( x\right) = 0 \) with coefficients in \( \mathbb{k}\left( \psi \right) \) . This forces the whole of \( \mathbb{k}\left( t\right... | Yes |
Corollary 11.1 Let \( A \) be a finitely generated algebra over an algebraically closed field \( \mathbb{k} \) . Then \( \mathfrak{n}\left( A\right) = \mathfrak{r}\left( A\right) \) . In other words, the nilradical of \( A \) coincides with the kernel of the homomorphism \( A \rightarrow {\mathbb{k}}^{{\operatorname{Sp... | Proof Since \( A/\mathfrak{m} \) is a field for all \( \mathfrak{m} \in {\operatorname{Spec}}_{\mathrm{m}}A \), all nilpotent elements of \( A \) are annihilated by every quotient map \( A \rightarrow A/\mathfrak{m} \) with \( \mathfrak{m} \in {\operatorname{Spec}}_{\mathfrak{m}}A \) . Therefore, \( \mathfrak{n}\left( ... | Yes |
The points of \( {\operatorname{Spec}}_{\mathrm{m}}\mathbb{k}\left\lbrack t\right\rbrack \) are in bijection with the points of the affine line \( {\mathbb{A}}^{1} = \mathbb{k} \) . | Indeed, every homomorphism ev : \( \mathbb{k}\left\lbrack t\right\rbrack \rightarrow \mathbb{k} \) is uniquely determined by its value at the generator \( t \), that is, uniquely determined by the point \( \operatorname{ev}\left( t\right) = p \in \mathbb{k} \) . In other words, every maximal ideal \( \mathfrak{m} \subs... | Yes |
Exercise 11.6 Verify that \( A \otimes B \) becomes a commutative \( \mathbb{k} \) -algebra with unit \( 1 \otimes 1 \) , and the \( \mathbb{k} \) -algebra homomorphisms \( A \hookrightarrow A \otimes B \hookleftarrow B, a \mapsto a \otimes 1, b \mapsto 1 \otimes b \), give the direct coproduct in the category of commu... | It follows from the universal property of the coproduct that there exists a bijection\n\n\[ \n{\operatorname{Spec}}_{\mathrm{m}}\left( A\right) \times {\operatorname{Spec}}_{\mathrm{m}}\left( B\right) \simeq {\operatorname{Spec}}_{\mathrm{m}}\left( {A \otimes B}\right) \n\]\n\nsending a pair of homomorphisms \( {\opera... | Yes |
Proposition 11.1 (Base for Open Sets and Compactness) Every Zariski open subset \( U \) of an affine algebraic variety \( X \) is a finite union of principal open sets\n\n\[ \mathcal{D}\left( f\right) \overset{\text{ def }}{ = }X \smallsetminus V\left( f\right) = \{ x \in X \mid f\left( x\right) \neq 0\} \]\n\nfor some... | Proof Let \( U = X \smallsetminus V\left( I\right) \) . Since \( \mathbb{k}\left\lbrack X\right\rbrack \) is Noetherian, \( I = \left( {{f}_{1},{f}_{2},\ldots ,{f}_{m}}\right) \) for some \( {f}_{i} \in \mathbb{k}\left\lbrack X\right\rbrack \) . Therefore \( V\left( I\right) = \bigcap V\left( {f}_{i}\right) \) and \( U... | Yes |
Proposition 11.2 (Continuity of Regular Maps) Every regular map of affine algebraic varieties \( \varphi : X \rightarrow Y \) is continuous in the Zariski topology. | Proof For every closed set \( V\left( I\right) \subset Y \), the preimage \( {\varphi }^{-1}\left( {V\left( I\right) }\right) \) consists of the points \( x \in X \) such that \( 0 = f\left( {\varphi \left( x\right) }\right) = {\varphi }^{ * }f\left( x\right) \) for all \( f \in I \) . Therefore, it coincides with \( V... | Yes |
Proposition 11.3 An affine algebraic variety \( X \) is irreducible if and only if its coordinate algebra \( \mathbb{k}\left\lbrack X\right\rbrack \) has no zero divisors. | Proof If \( X = {X}_{1} \cup {X}_{2} \) with proper closed \( {X}_{1},{X}_{2} \), then there exist nonzero regular functions \( {f}_{1},{f}_{2} \in \mathbb{k}\left\lbrack X\right\rbrack \) such that \( {f}_{1} \in I\left( {X}_{1}\right) ,{f}_{2} \in I\left( {X}_{2}\right) \) . Since \( {f}_{1}{f}_{2} \) vanishes at eve... | Yes |
Corollary 11.2 Given a polynomial \( g \in \mathbb{k}\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \), the affine hypersurface \( V\left( g\right) \subset {\mathbb{A}}^{n} \) is irreducible if and only if \( g = {q}^{n} \) for some irreducible \( q \in \mathbb{k}\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{... | Proof Since the polynomial ring \( \mathbb{k}\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) is a unique factorization domain, \( {}^{13} \) a polynomial \( f \in \mathbb{k}\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) is irreducible if and only if the quotient ring \( \mathbb{k}\left\lb... | No |
Example 11.4 (Big Open Sets) If \( X \) is irreducible, then every two nonempty open sets \( {U}_{1},{U}_{2} \subset X \) have nonempty intersection, because otherwise, \( X \) could be | decomposed as \( X = \left( {X \smallsetminus {U}_{1}}\right) \cup \left( {X \smallsetminus {U}_{2}}\right) \) . In other words, every nonempty open subset of an irreducible variety \( X \) is dense in \( X \) . Thus, the Zariski topology is quite far from being Hausdorff. | Yes |
Theorem 11.3 Every affine algebraic variety \( X \) admits a decomposition\n\n\[ X = {X}_{1} \cup {X}_{2} \cup \cdots \cup {X}_{k} \]\n\nthat is unique up to renumbering of components, where all \( {X}_{i} \subset X \) are closed and irreducible, and \( {X}_{i} ⊄ {X}_{j} \) for all \( i \neq j \) . | Proof The existence of the decomposition is proved similarly to the existence of irreducible factorization in a Noetherian ring. \( {}^{15} \) If \( X \) is reducible, write \( X \) as a union \( X = {Z}_{1} \cup {Z}_{2} \) of proper closed subsets \( {Z}_{1},{Z}_{2} \subset X \) and repeat the procedure recursively fo... | Yes |
Proposition 11.4 A nonzero element \( f \in \mathbb{k}\left\lbrack X\right\rbrack \) is a zero divisor if and only if it has the zero restriction on some irreducible component of \( X \) . | Proof Let \( {fg} = 0 \) for some \( g \neq 0 \) . Write \( {f}_{i},{g}_{i} \in \mathbb{k}\left\lbrack {X}_{i}\right\rbrack \) for the restrictions of \( f, g \) to the irreducible component \( {X}_{i} \subset X \) . Since \( \mathbb{k}\left\lbrack {X}_{i}\right\rbrack \) has no zero divisors, at least one of \( {f}_{i... | Yes |
Proposition 11.5 Let \( X \) be an affine algebraic variety over an infinite field, and \( f \in \mathbb{k}\left( X\right) \) a rational function. Then \( \left( {1/f}\right) \overset{\text{ def }}{ = }\{ g \in \mathbb{k}\left\lbrack X\right\rbrack \mid {gf} \in \mathbb{k}\left\lbrack X\right\rbrack \} \) is an ideal i... | Proof The intersection \( \left( {1/f}\right) \cap \mathbb{k}{\left\lbrack X\right\rbrack }^{ \circ } \) is exactly the set of all denominators \( q \) appearing in various fractional representations \( f = p/q \) . Thus, the closed set \( X \smallsetminus \operatorname{Dom}\left( f\right) \) is determined by the syste... | No |
Proposition 11.7 Let \( X = {X}_{1} \cup {X}_{2} \cup \cdots \cup {X}_{k} \) be the irreducible decomposition of an affine algebraic variety \( X \) . Then \( k\left( X\right) = k\left( {X}_{1}\right) \times k\left( {X}_{2}\right) \times \cdots \times k\left( {X}_{k}\right) \) . | Proof Write \( I = I\left( {\mathop{\bigcup }\limits_{{i \neq j}}\left( {{X}_{i} \cap {X}_{j}}\right) }\right) \subset \mathbb{k}\left\lbrack X\right\rbrack \) for the ideal of all regular functions on \( X \) vanishing on every intersection \( {X}_{i} \cap {X}_{j}, i \neq j \) .\n\nLet us choose some regular function ... | Yes |
Proposition 11.8 (Closeness of Finite Morphisms) Let \( \varphi : X \rightarrow Y \) be a finite morphism of affine algebraic varieties, and \( Z \subset X \) a closed subset. Then \( \varphi \left( Z\right) \subset Y \) is also closed, and the restriction \( {\left. \varphi \right| }_{Z} : Z \rightarrow \varphi \left(... | Proof Write \( I = I\left( Z\right) \subset \mathbb{k}\left\lbrack X\right\rbrack \) for the ideal of \( Z \) . The pullback homomorphism of the restricted map \( {\left. \varphi \right| }_{Z} : Z \rightarrow Y \) can be factorized as\n\n\[ \n{\left. \varphi \right| }_{Z}^{ * } : k\left\lbrack Y\right\rbrack \overset{{... | No |
Problem 11.17 (Quotient by a Finite Group Action) Let \( \\mathbb{k} \) be an algebraically closed field of characteristic zero, \( X \) an affine algebraic variety over \( \\mathbb{k} \), and \( G \) a finite group acting on \( X \) by regular automorphisms. Then \( G \) acts on \( \\mathbb{k}\\left\\lbrack X\\right\\... | Use them to prove that \( R \) is a finitely generated reduced \( \\mathbb{k} \) -algebra, and \( {\\operatorname{Spec}}_{\\mathrm{m}}R \) can be identified with the set of \( G \) -orbits \( X/G \) in such a way that the quotient map \( \\pi : X \\rightarrow X/G \) becomes a finite regular surjection of affine algebra... | No |
Example 12.1 (Projective Spaces) The projective space \( {}^{3}{\mathbb{P}}_{n} = \mathbb{P}\left( {\mathbb{k}}^{n + 1}\right) \) with homogeneous coordinates \( x = \left( {{x}_{0} : {x}_{1} : \cdots : {x}_{n}}\right) \) is covered by the \( \left( {n + 1}\right) \) standard affine charts \( {}^{4}{U}_{i} = \left\{ {\... | The preimage of the intersection \( {U}_{i} \cap {U}_{j} \) under this bijection is the principal open set \( D\left( {t}_{i, j}\right) \subset {X}_{i} \). | Yes |
Example 12.6 (Blowup of a Point on \( {\mathbb{P}}_{n} \) ) All the lines passing through a given point \( p \in {\mathbb{P}}_{n} \) form the projective space \( E \simeq {\mathbb{P}}_{n - 1} \) . The incidence graph | \[ {\mathcal{B}}_{p} = \left\{ {\left( {q,\ell }\right) \in {\mathbb{P}}_{n} \times E \mid q \in \ell }\right\} \] is called the blowup of the point \( p \in {\mathbb{P}}_{n} \) . The projection \( {\sigma }_{p} : {\mathcal{B}}_{p} \rightarrow {\mathbb{P}}_{n} \) is one-toone over \( {\mathbb{P}}_{n} \smallsetminus \{ ... | Yes |
Example 12.7 (Illustration to the Proof of Lemma 12.1) The zero set of the homogeneous polynomial \( {x}_{0}{x}_{1}{x}_{2} \) on \( {\mathbb{P}}_{2} \) is the union of three lines complementary to the standard affine charts. The affine equations of this set in the charts \( {U}_{0},{U}_{1} \) , \( {U}_{2} \) are, respe... | Applied to this \( X \), the previous proof transforms the left-hand sides of the local affine equations to the homogeneous polynomials \( {\bar{f}}_{0,1} = {x}_{1}{x}_{2},{\bar{f}}_{1,1} = {x}_{0}{x}_{2},{\bar{f}}_{2,1} = {x}_{0}{x}_{1} \), and then gives \( {x}_{0} \cdot {\bar{f}}_{0,1} = 0,{x}_{1} \cdot {\bar{f}}_{1... | Yes |
Lemma 12.2 The projection \( \pi : {\mathbb{P}}_{m} \times {\mathbb{A}}^{n} \rightarrow {\mathbb{A}}^{n} \) is closed, i.e., \( \pi \left( Z\right) \subset {\mathbb{A}}^{n} \) is closed for every closed \( Z \subset {\mathbb{P}}_{m} \times {\mathbb{A}}^{n} \) . | Proof Write \( x = \left( {{x}_{0} : {x}_{1} : \cdots : {x}_{m}}\right) \) and \( t = \left( {{t}_{1},{t}_{2},\ldots ,{t}_{n}}\right) \) for the homogeneous and affine coordinates on \( {\mathbb{P}}_{m} \) and \( {\mathbb{A}}^{n} \) respectively. Let a closed subset \( Z \subset {\mathbb{P}}_{m} \times {\mathbb{A}}^{n}... | Yes |
Corollary 12.1 Let \( X \) be a projective algebraic variety. Then the projection\n\n\[ X \times Y \rightarrow Y \]\n\nis closed for every algebraic manifold \( Y \) . | Proof It is enough to prove this statement separately for every affine chart of \( Y \) instead of the whole of \( Y \) . Thus, we may assume that \( Y \) is affine. In this case, \( X \times Y \) is a closed subset in \( {\mathbb{P}}_{m} \times {\mathbb{A}}^{n} \), and the projection in question is the restriction of ... | Yes |
Theorem 12.1 For every projective variety \( X \) and separated manifold \( Y \), every regular morphism \( \varphi : X \rightarrow Y \) is closed. | Proof Let \( {\Gamma }_{\varphi } \subset X \times Y \) be the graph \( {}^{15} \) of the morphism \( \varphi : X \rightarrow Y \) . The image \( \varphi \left( Z\right) \subset Y \) of every subset \( Z \subset X \) can be described as the image of the intersection \( {\Gamma }_{\varphi } \cap \left( {Z \times Y}\righ... | Yes |
Corollary 12.2 Let \( X \) be a connected \( {}^{16} \) projective variety. Then every regular map from \( X \) to an arbitrary affine algebraic variety \( Y \) contracts \( X \) to one point of \( Y \) . In particular, \( {\mathcal{O}}_{X}\left( X\right) = \mathbb{k} \) is exhausted by constants. | Proof Let \( \varphi : X \rightarrow Y \) be a regular map to an affine variety \( Y \subset {\mathbb{A}}^{n} \) . Composing it with the projections of \( Y \) onto the \( n \) coordinate axes of \( {\mathbb{A}}^{n} \) reduces the statement to the case \( Y = {\mathbb{A}}^{1} \) . Composing a regular map \( X \rightarr... | Yes |
Corollary 12.3 Every projective variety admits a regular finite surjection onto projective space. | Proof Let \( X \subset {\mathbb{P}}_{n} \) be a projective variety. Make a finite projection \( {\pi }_{1} : X \rightarrow {H}_{1} \) from some point \( {p}_{1} \in {\mathbb{P}}_{n} \smallsetminus X \) to some hyperplane \( {H}_{1} \subset {\mathbb{P}}_{n} \) . If \( {\pi }_{1}\left( X\right) \neq {H}_{1} \), make a se... | Yes |
Corollary 12.4 Every affine algebraic variety \( X \) admits a regular finite surjection onto affine space. | Proof Let \( X \subsetneq {\mathbb{A}}^{n} \), where \( {\mathbb{A}}^{n} \) is placed in \( {\mathbb{P}}_{n} \) as the standard affine chart \( {U}_{0} \) . Put \( {H}_{\infty }\overset{\text{ def }}{ = }{\mathbb{P}}_{n} \smallsetminus {U}_{0} \) and write \( \bar{X} \subset {\mathbb{P}}_{n} \) for the projective closu... | Yes |
Example 12.8 (Noether's Normalization) Given an arbitrary polynomial\n\n\\[ \nf \in \mathbb{k}\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \n\\]\n\nwrite it as \\( f = {f}_{0} + {f}_{1} + \\cdots + {f}_{d} \\), where each \\( {f}_{k} \\) is homogeneous of degree \\( k \\) . Let \\( {\\mathbb{A}}^{n} \\su... | A point \\( p = \\left( {0 : {p}_{1} : {p}_{2} : \\ldots : {p}_{n}}\\right) \\) at infinity with respect to the chart \\( {U}_{0} \\) does not belong to \\( \\bar{X} \\) if and only if \\( {f}_{d}\\left( {{p}_{1},{p}_{2},\\ldots ,{p}_{n}}\\right) \\neq 0 \\) . Since \\( {f}_{d} \\neq 0 \\), such a point \\( p \\notin \... | Yes |
Proposition 12.2 For every regular surjection of irreducible manifolds \( \varphi : Y \rightarrow X \) , the inequality \( {\dim }_{y}Y \geq {\dim }_{\varphi \left( y\right) }X \) holds at every point \( y \in Y \) . | Proof Given a chain (12.11) with \( x = \varphi \left( y\right) \), for every \( i \) the closed submanifold \( {\varphi }^{-1}\left( {X}_{i}\right) \subset Y \) has an irreducible component \( {Y}_{i} \) such that the restricted map\n\n\[ \n{\left. \varphi \right| }_{{Y}_{i}} : {Y}_{i} \rightarrow {X}_{i}\n\]\n\n is d... | Yes |
Proposition 12.3 Given a finite morphism of irreducible algebraic varieties\n\n\[ \varphi : X \rightarrow Y \]\n\nthen \( {\dim }_{x}X \leq {\dim }_{\varphi \left( x\right) }Y \) for all \( x \in X \), and equality holds for some \( x \in X \) if and only if \( \varphi \left( X\right) = Y \) . | Proof Replacing \( Y \) by an affine neighborhood of \( \varphi \left( x\right) \) and \( X \) by the preimage of this neighborhood allows us to assume, by Exercise 12.16, that both \( X \) and \( Y \) are affine. It follows from Proposition 11.8 on p. 258 that every chain (12.11) in \( X \) is mapped to a strictly inc... | No |
Proposition 12.4 \( {\dim }_{x}{\mathbb{A}}^{n} = n \) for all \( x \in {\mathbb{A}}^{n} \) . | Proof Since for every \( x \in {\mathbb{A}}^{n} \) there is a chain (12.11) of strictly increasing affine subspaces \( {X}_{i} = {\mathbb{A}}^{i} \) passing through \( x \), the inequality \( {\dim }_{x}{\mathbb{A}}^{n} \geq n \) holds. The opposite inequality is established by induction on \( n \) . It is obvious for ... | Yes |
For every irreducible affine algebraic variety \( X \), the equality \[ {\dim }_{x}X = \operatorname{tr}{\deg }_{k}\mathbb{k}\left\lbrack X\right\rbrack \] holds for all \( x \in X \), where \( {\operatorname{trdeg}}_{k}k\left\lbrack X\right\rbrack \) means the transcendence degree of the coordinate algebra of \( X \) ... | Proof A finite surjection \( \pi : X \rightarrow {\mathbb{A}}^{m} \) forces \( \mathbb{k}\left\lbrack X\right\rbrack \) to be an integral extension of the subalgebra \( {\pi }^{ * }\left( {\mathbb{k}\left\lbrack {\mathbb{A}}^{m}\right\rbrack }\right) \simeq \mathbb{k}\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{m}}\right... | Yes |
Proposition 12.5 Let \( X \) be an irreducible algebraic manifold and \( f \in {\mathcal{O}}_{X}\left( X\right) \) a nonconstant global regular function on \( X \) . Then \( V\left( f\right) \neq \varnothing \) and \[ {\dim }_{p}V\left( f\right) = {\dim }_{p}\left( X\right) - 1\text{ for all }p \in V\left( f\right) . | Proof Exercise 12.16 allows us to assume that \( X \) is affine. For \( X = {\mathbb{A}}^{n} \), the statement follows from Example 12.8. The general case is reduced to affine spaces by the same geometric construction as in the proof of Proposition 11.9 on p. 260. Namely, fix a finite surjection \( \pi : X \rightarrow ... | Yes |
Corollary 12.8 For affine algebraic varieties \( {X}_{1},{X}_{2} \subset {\mathbb{A}}^{n} \) and every point \( x \in {X}_{1} \cap {X}_{2} \), the inequality \( {\dim }_{x}\left( {{X}_{1} \cap {X}_{2}}\right) \geq {\dim }_{x}\left( {X}_{1}\right) + {\dim }_{x}\left( {X}_{2}\right) - n \) holds. | Proof Let \( {\varphi }_{i} : {X}_{i} \hookrightarrow {\mathbb{A}}^{n}, i = 1,2 \), be the closed immersions corresponding to the quotient maps \( \mathbb{k}\left\lbrack {X}_{1}\right\rbrack \leftarrow \mathbb{k}\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \rightarrow \mathbb{k}\left\lbrack {X}_{2}\right... | Yes |
Proposition 12.6 For irreducible projective varieties \( {X}_{1},{X}_{2} \subset {\mathbb{P}}_{n} \), the inequality \( \dim \left( {X}_{1}\right) + \dim \left( {X}_{2}\right) \geq n \) forces \( {X}_{1} \cap {X}_{2} \neq \varnothing . \) | Proof Let \( {\mathbb{P}}_{n} = \mathbb{P}\left( V\right) \) and \( {\mathbb{A}}^{n + 1} = \mathbb{A}\left( V\right) \) . Given a nonempty irreducible projective variety \( Z \subset {\mathbb{P}}_{n} \), write \( {Z}^{\prime } \subset {\mathbb{A}}^{n + 1} \) for the affine cone over \( Z \) provided by the same homogen... | Yes |
Corollary 12.9 (Semicontinuity Theorem) For every regular map of algebraic manifolds \( \varphi : X \rightarrow Y \), the sets \[ {X}_{k}\overset{\text{ def }}{ = }\left\{ {x \in X \mid {\dim }_{x}{\varphi }^{-1}\left( {\varphi \left( x\right) }\right) \geq k}\right\} \] are closed in \( X \) for all \( k \in \mathbb{Z... | Proof If \( \dim Y = 0 \), then this is trivially true for all \( X \) and \( k \) . For \( \dim Y = m > 0 \) , we may assume by induction that the statement holds for all \( X, k \), and all \( Y \) with \( \dim Y < m \) . Replacing \( Y \) and \( X \) by irreducible components of maximal dimension passing through \( ... | Yes |
Theorem 12.3 (Dimension Criterion of Irreducibility) Assume that a closed regular surjection of algebraic manifolds \( \varphi : X \rightarrow Y \) has irreducible fibers of the same constant dimension. Then \( X \) is irreducible if \( Y \) is. | Proof Let \( X = {X}_{1} \cup {X}_{2} \) be reducible. Since every fiber of \( \varphi \) is irreducible, it is entirely contained in \( {X}_{1} \) or in \( {X}_{2} \) . Put \( {Y}_{i}\overset{\text{ def }}{ = }\left\{ {y \in Y \mid {\varphi }^{-1}\left( y\right) \subset {X}_{i}}\right\} \) for \( i = 1,2 \) . Then \( ... | Yes |
Given a collection of positive integers \( {d}_{0},{d}_{1},\ldots ,{d}_{n} \) , write \( {\mathbb{P}}_{{N}_{i}} = \mathbb{P}\left( {{S}^{{d}_{i}}{V}^{ * }}\right) \) for the space of degree- \( {d}_{i} \) hypersurfaces in \( {\mathbb{P}}_{n} = \mathbb{P}\left( V\right) \) . We are going to show that the resultant varie... | Consider the incidence variety\n\n\[ \Gamma \overset{\text{ def }}{ = }\left\{ {\left( {{S}_{1},{S}_{2},\ldots ,{S}_{n}, p}\right) \in {\mathbb{P}}_{{N}_{0}} \times \cdots \times {\mathbb{P}}_{{N}_{n}} \times {\mathbb{P}}_{n} \mid p \in \cap {S}_{i}}\right\} . \]\n\nSince the equation \( f\left( p\right) = 0 \) is line... | No |
Exercise 12.23 Convince yourself that \( \Gamma \subset {\mathbb{P}}_{N} \times \operatorname{Gr}\left( {2,4}\right) \) is a projective algebraic variety. | The projection \( {\pi }_{2} : \Gamma \rightarrow {Q}_{P} \) is surjective, and all its fibers are projective spaces of the same constant dimension. Indeed, the line \( \ell \) given by the equations \( {x}_{0} = {x}_{1} = 0 \)\n\n\( {}^{23} \) Compare with Problem 17.20 of Algebra I.\n\nlies on a surface \( Z\left( f\... | No |
Theorem 13.1 For every finite field extension \( \mathbb{F} \supset \mathbb{k} \), there exists a tower of simple field extensions\n\n\[ k = {\mathbb{L}}_{0} \subset {\mathbb{L}}_{1} \subset {\mathbb{L}}_{2} \subset \cdots \subset {\mathbb{L}}_{k - 1} \subset {\mathbb{L}}_{k} = \mathbb{F} \]\n\n(13.4)\n\nsuch that \( {... | Proof Assume by induction that the field \( {\mathbb{L}}_{i} \subset \mathbb{F} \) of level \( i \) has been constructed. If \( {\mathbb{L}}_{i} \neq \mathbb{F} \), let \( \vartheta \in \mathbb{F} \smallsetminus {\mathbb{L}}_{i} \), and let \( {f}_{i + 1} \in {\mathbb{L}}_{i}\left\lbrack x\right\rbrack \) be the minima... | Yes |
Corollary 13.1 If \( \mathbb{K} \supset \mathbb{k} \) is a separable algebraic extension and the degrees of elements \( {}^{6} \) of \( \mathbb{K} \) over \( \mathbb{k} \) are bounded above, then \( \mathbb{K} \) is finite over \( \mathbb{k} \) and\n\n\[ \deg \mathbb{K}/k = \mathop{\max }\limits_{{\vartheta \in \mathbb... | Proof Let \( \alpha \in \mathbb{K} \) be an element of maximal degree over \( \mathbb{k} \) . If there exists some \( \beta \in \mathbb{K} \smallsetminus \mathbb{k}\left\lbrack \alpha \right\rbrack \), then \( \deg \mathbb{k}\left\lbrack {a,\beta }\right\rbrack /\mathbb{k} > \deg \mathbb{k}\left\lbrack a\right\rbrack /... | Yes |
Lemma 13.1 Let \( \mathbb{K} = \mathbb{k}\left\lbrack x\right\rbrack /\left( f\right) \) be a simple extension of a field \( \mathbb{k} \), and \( \varphi : \mathbb{k} \hookrightarrow \mathbb{F} \) an embedding of \( \mathbb{k} \) into an arbitrary field \( \mathbb{F} \) . The embeddings \( \widetilde{\varphi } : \math... | Proof Associated with every element \( \alpha \in \mathbb{F} \) is the map\n\n\[{\varphi }_{\alpha } : \mathbb{k}\left\lbrack x\right\rbrack \rightarrow \mathbb{F},\;g\left( x\right) \mapsto {g}^{\varphi }\left( \alpha \right) .\]\n\nIf \( \alpha \) is a root of the polynomial \( {f}^{\varphi } \in \mathbb{F}\left\lbra... | Yes |
Lemma 13.2 Let \( \mathbb{K} \supset \mathbb{k} \) be an algebraic field extension, not necessarily finite, and \( \varphi : \mathbb{k} \hookrightarrow \mathbb{F} \) an embedding of fields such that for every \( \vartheta \in \mathbb{K} \) with minimal polynomial \( {\mu }_{\vartheta } \) over \( k \), the polynomial \... | Proof By Lemma 13.1, the embedding \( \varphi : \mathbb{k} \hookrightarrow \mathbb{F} \) can be extended to an embedding \( {\varphi }_{\xi } : \mathbb{k}\left\lbrack \vartheta \right\rbrack \hookrightarrow \mathbb{F},\;\vartheta \mapsto \xi \) . Consider the set \( \mathcal{S} \) of all embeddings \( \psi : \mathbb{L}... | Yes |
Proposition 13.1 Let \( \varphi : \mathbb{k} \hookrightarrow \mathbb{F} \) be an arbitrary embedding of fields, and \( \mathbb{K} \supset \mathbb{k} \) a finite extension. Then there exist at most \( \deg \mathbb{K}/\mathbb{k} \) distinct embeddings \( \psi : \mathbb{K} \hookrightarrow \mathbb{F} \) extending \( \varph... | Proof Consider a finite tower of simple field extensions (13.4),\n\n\[ \n\mathbb{k} = {\mathbb{L}}_{0} \subset {\mathbb{L}}_{1} \subset {\mathbb{L}}_{2} \subset \cdots \subset {\mathbb{L}}_{k - 1} \subset {\mathbb{L}}_{k} = \mathbb{K},\n\]\n\n(13.6)\n\nwhere \( {\mathbb{L}}_{i} = {\mathbb{L}}_{i - 1}\left\lbrack {\vart... | Yes |
Proposition 13.2 Let \( \mathbb{K} \supset k \) be an algebraic field extension, not necessarily finite. Then every embedding \( \varphi : \mathbb{K} \hookrightarrow \mathbb{K} \) that acts identically on the subfield \( \mathbb{k} \) is an automorphism of \( \mathbb{K} \) . | Proof It is enough to check that \( \varphi \left( \mathbb{K}\right) = \mathbb{K} \) . Let \( \vartheta \in \mathbb{K} \) have minimal polynomial \( f \in \mathbb{k}\left\lbrack x\right\rbrack \) over \( \mathbb{k} \) . Since \( \varphi \) maps every root of \( f \) to a root of \( f \), the equality \( {\varphi }^{m}\... | Yes |
Theorem 13.3 Every polynomial \( f \in \mathbb{k}\left\lbrack x\right\rbrack \) has a splitting field \( {\mathbb{L}}_{f} \), and every two splitting fields are (noncanonically) isomorphic over \( {}^{10}\mathbb{k} \) . | Proof (of Theorem 13.3) Let \( \mathbb{F} \supset \mathbb{k} \) be a finite extension such that \( f \) splits into a product of deg \( f \) linear factors \( {}^{11} \) in \( \mathbb{F}\left\lbrack x\right\rbrack \) . Write \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{m} \in \mathbb{F} \) for the roots of \( f \)... | Yes |
Example 13.4 (The Classification of Finite Fields Revisited) Every finite field \( \mathbb{F} \) of characteristic \( p \) is a finite extension of the prime subfield \( {\mathbb{F}}_{p} = \mathbb{Z}/\left( p\right) \subset \mathbb{F} \) and consists of \( q = {p}^{n} \) elements for \( n = \deg \mathbb{F}/{\mathbb{F}}... | Since the nonzero elements of \( \mathbb{F} \) form a multiplicative group of order \( q - 1 \), they are exactly the roots of the polynomial \( {x}^{q - 1} - 1 \in {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) in the field \( \mathbb{F} \) . Therefore, \( \mathbb{F} \) is a splitting field of the polynomial \( f\left... | Yes |
Theorem 13.4 Every field \( \mathbb{k} \) has an algebraic closure, unique up to a (noncanonical) isomorphism that acts identically on \( \mathbb{k} \) . | Proof Given two algebraic closures \( {\mathbb{L}}^{\prime },{\mathbb{L}}^{\prime \prime } \) of the field \( \mathbb{k} \), then by Lemma 13.2 on p. 300, the embedding \( \mathbb{k} \subset {\mathbb{L}}^{\prime } \) can be extended to an embedding \( {\varphi }^{\prime } : {\mathbb{L}}^{\prime } \hookrightarrow {\math... | Yes |
Corollary 13.2 For every tower of finite field extensions \( {\mathbb{L}}_{1} \subset {\mathbb{L}}_{2} \subset {\mathbb{L}}_{3} \), the extension \( {\mathbb{L}}_{1} \subset {\mathbb{L}}_{3} \) is separable if and only if the extensions \( {\mathbb{L}}_{1} \subset {\mathbb{L}}_{2},{\mathbb{L}}_{2} \subset {\mathbb{L}}_... | Proof If \( {\mathbb{L}}_{3} \) is separable over \( {\mathbb{L}}_{1} \), then in particular, the subfield \( {\mathbb{L}}_{2} \) is separable, and \( {\mathbb{L}}_{3} \) is separable over \( {\mathbb{L}}_{2} \), because the minimal polynomial of every element \( \vartheta \in {\mathbb{L}}_{3} \) over \( {\mathbb{L}}_{... | Yes |
Lemma 13.3 Let \( \bar{k} \supset k \) be an algebraic closure of a field \( k \) . An algebraic field extension \( \mathbb{k} \subset \mathbb{K} \) is normal if and only if all embeddings \( \mathbb{K} \hookrightarrow \overline{\mathbb{k}} \) extending the inclusion \( k \subset \bar{k} \) have the same image. | Proof Fix one such an embedding \( \varphi : \mathbb{K} \hookrightarrow \overline{\mathbb{k}} \), which exists by Lemma 13.2 on p. 300, and identify \( \mathbb{K} \) with the subfield \( \varphi \left( \mathbb{K}\right) \subset \overline{\mathbb{k}} \) by means of this embedding. Thus, we have a tower \( \mathbb{k} \su... | Yes |
Lemma 13.4 Let \( k \subset \mathbb{L} \subset \mathbb{K} \) be a tower of algebraic extensions of fields such that \( \mathbb{K} \) is normal over \( \mathbb{k} \) . Then \( \mathbb{K} \) is normal over \( \mathbb{L} \) as well, whereas \( \mathbb{L} \) is normal over \( k \) if and only if the image of every embeddin... | Proof For every element \( \vartheta \in \mathbb{K} \), the minimal polynomial of \( \vartheta \) over \( \mathbb{L} \) divides in \( \mathbb{L}\left\lbrack x\right\rbrack \) the minimal polynomial of \( \vartheta \) over \( \mathbb{k} \) . Thus, if the minimal polynomial of \( \vartheta \) over \( \mathbb{k} \) is com... | Yes |
Proposition 13.3 A finite field extension \( \mathbb{K} \supset \mathbb{k} \) is normal if and only if \( \mathbb{K} \) is a splitting field of some not necessarily irreducible polynomial \( f \in \mathbb{k}\left\lbrack x\right\rbrack \) . | Proof Let \( \mathbb{K} \) be normal over \( \mathbb{k} \), and suppose \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{k} \in \mathbb{K} \) generate \( \mathbb{K} \) as a \( \mathbb{k} \) -algebra. Write \( {f}_{i} \in \mathbb{k}\left\lbrack x\right\rbrack \) for the minimal polynomial of \( {\alpha }_{i} \) over \(... | Yes |
Proposition 13.4 Let \( \mathbb{F},\mathbb{K} \subset \overline{\mathbb{k}} \) be two fields containing \( \mathbb{k} \) . If \( \mathbb{K} \) is normal (respectively separable) over \( \mathbb{k} \), then the compositum \( \mathbb{K}\mathbb{F} \) is normal (respectively separable) over \( \mathbb{F} \) . | Proof The embeddings \( \mathbb{K}\mathbb{F} \hookrightarrow \overline{\mathbb{F}} = \overline{\mathbb{k}} \) that act identically on \( \mathbb{F} \subset \mathbb{K}\mathbb{F} \) are in bijection with the embeddings \( \mathbb{K} \hookrightarrow \overline{\mathbb{k}} \) that act identically on \( \mathbb{k} \subset \m... | Yes |
Theorem 13.5 (Normal Closure) Let \( \\mathbb{F} \\supset \\mathbb{k} \) be a finite separable field extension. There is a field \( \\mathbb{K} \\supset \\mathbb{F} \) normal and separable over \( \\mathbb{k} \) such that for every field \( \\mathbb{L} \\supset \\mathbb{F} \) normal and separable over \( \\mathbb{k} \)... | Proof Let \( \\overline{\\mathbb{k}} \\supset \\mathbb{k} \) be an algebraic closure of \( \\mathbb{k} \), and \( n = \\deg \\mathbb{F}/\\mathbb{k} \). Put \( \\mathbb{K} \) as the compositum of all \( n \) distinct embeddings \( \\mathbb{F} \\hookrightarrow \\overline{\\mathbb{k}} \) that act identically on \( \\mathb... | Yes |
Theorem 13.6 Let \( \mathbb{K} \) be an arbitrary field, and \( G \) a finite group of automorphisms \( \mathbb{K} \simeq \mathbb{K} \) . Then the extension \( \mathbb{K} \supset {\mathbb{K}}^{G} \) is a Galois extension of degree \( \left| G\right| \), and \( \operatorname{Gal}\mathbb{K}/k = G \) . | Proof Given an element \( \vartheta \in \mathbb{K} \), write \( {\vartheta }_{1},{\vartheta }_{2},\ldots ,{\vartheta }_{m} \in \mathbb{K} \) for all the distinct elements of the \( G \) -orbit of \( \vartheta = {\vartheta }_{1} \) . Then the polynomial\n\n\[ \n{f}_{\vartheta }\left( x\right) = \left( {x - {\vartheta }_... | Yes |
For every finite field extension \( \mathbb{k} \subset \mathbb{K} \) and subgroup \( G \subset {\operatorname{Aut}}_{\mathbb{k}}\mathbb{K} \) , the equalities \( {\mathbb{K}}^{G} = \mathbb{k} \) and \( \left| G\right| = \deg \mathbb{K}/\mathbb{k} \) are equivalent. If they hold, then \( G = {\operatorname{Aut}}_{k}\mat... | Proof Applying Theorem 13.6 to the tower \( \mathbb{k} \subset {\mathbb{K}}^{G} \subset \mathbb{K} \) leads to the equality \( \deg \mathbb{K}/{\mathbb{K}}^{G} = \left| G\right| \), which immediately implies all the statements.\n\nHowever, it is quite instructive to give a direct proof of Corollary 13.3 without using t... | Yes |
Example 13.6 (Automorphisms and Embeddings of Finite Fields) Let \( q = {p}^{n} \) for a prime \( p \in \mathbb{N} \) . Since the extension \( {\mathbb{F}}_{p} \subset {\mathbb{F}}_{q} \) is finite, normal, and separable, we have \( \left| {{\operatorname{Aut}}_{{\mathbb{F}}_{p}}{\mathbb{F}}_{q}}\right| = \deg {\mathbb... | Write \( {F}_{p}^{0} = \mathrm{{Id}},{F}_{p},{F}_{p}^{2},\ldots ,{F}_{p}^{n - 1} \) for iterations of the Frobenius automorphism \( {F}_{p} : \vartheta \mapsto {\dot{\vartheta }}^{p} \) . They all are distinct, because an equality \( {F}_{p}^{k} = {F}_{p}^{m} \) would force all the \( {p}^{n} \) elements of \( {\mathbb... | Yes |
Theorem 13.7 (Galois Correspondence) Let \( k \subset \mathbb{K} \) be a finite Galois extension with Galois group \( G = {\operatorname{Aut}}_{k}\mathbb{K} \). Then there is a canonical bijection between the subgroups \( H \subset G \) and the subfields \( \mathbb{L} \subset \mathbb{K} \) such that \( \mathbb{k} \subs... | Proof Given a tower of fields \( \mathbb{k} \subset \mathbb{L} \subset \mathbb{K} \), the extension \( \mathbb{L} \subset \mathbb{K} \) is normal by Lemma 13.4 and separable by Corollary 13.2. Thus, \( \mathbb{L} \subset \mathbb{K} \) is a Galois extension with Galois group \( H = {\operatorname{Aut}}_{\mathbb{L}}\math... | Yes |
Proposition 14.1 A finite Galois extension \( \mathbb{K} \supset \mathbb{k} \) is contained in a field \( \mathbb{L} \supset \mathbb{k} \) obtained by a tower of quadratic extensions\n\n\[ k = {\mathbb{L}}_{0} \subset {\mathbb{L}}_{1} \subset {\mathbb{L}}_{2} \subset \cdots \subset {\mathbb{L}}_{m - 1} \subset {\mathbb... | Proof If \( \mathbb{K} \subset \mathbb{L} \) for \( \mathbb{L} \) from (14.2), then \( \deg \mathbb{K}/\mathbb{k} \) divides \( \deg \mathbb{L}/\mathbb{k} = {2}^{m} \), and therefore, \( \deg \mathbb{K}/\mathbb{k} = {2}^{n} \) for some \( n \leq m \) . Conversely, let \( \deg \mathbb{K}/\mathbb{k} = \left| {\operatorna... | Yes |
Theorem 14.1 A complex root of an irreducible polynomial \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) can be constructed by straightedge and compass starting from the points \( 0,1 \in \mathbb{C} \) if and only if the degree of the splitting field off over \( \mathbb{Q} \) is a power of two. In this... | Proof Write \( \mathbb{K} \subset \mathbb{C} \) for the splitting field of \( f \) . Then \( \mathbb{K} \supset \mathbb{Q} \) is a finite Galois extension by Proposition 13.3. For \( \deg \mathbb{K}/\mathbb{Q} = {2}^{m} \), we have seen in the proof of Proposition 14.1 that \( \mathbb{K} \) can be achieved by quadratic... | Yes |
Corollary 14.1 A number \( \zeta \in \mathbb{C} \) can be constructed by straightedge and compass starting from the points \( 0,1 \in \mathbb{C} \) only if \( \zeta \) is algebraic over \( \mathbb{Q} \) and \( {\deg }_{\mathbb{Q}}\zeta = {2}^{n} \) for some \( n \in \mathbb{N} \) . | Proof Since the simple extension \( \mathbb{Q}\left\lbrack \zeta \right\rbrack \) is contained in the splitting field of the minimal polynomial for \( \zeta \), the degree of \( \zeta \) over \( \mathbb{Q} \) divides the degree of that splitting field. | Yes |
The angle \( \pi /3 \) cannot be subdivided into three equal angles \( \pi /9 \) by straightedge and compass. | Indeed, such a possibility would allow the construction of the number \( \zeta = \cos \left( {\pi /9}\right) \), which is a root of the polynomial \( {}^{5}4{x}^{3} - {3x} - 1/2 \). Since this polynomial has no rational roots, it is irreducible over \( \mathbb{Q} \), and therefore proportional to the minimal polynomial... | No |
Proposition 14.2 (Accessory Irrationalities Theorem) Let \( \mathbb{F},\mathbb{K} \supset \mathbb{k} \) be fields contained in a common algebraically closed field \( \mathbb{L} \) . If the extension \( \mathbb{K} \supset \mathbb{k} \) is a finite Galois extension, then the extension \( \mathbb{F}\mathbb{K} \supset \mat... | Proof By Proposition 13.3, \( \mathbb{K} \subset \mathbb{L} \) is the splitting field of some separable polynomial \( f \in \mathbb{k}\left\lbrack x\right\rbrack \) . As a \( \mathbb{k} \) -algebra, \( \mathbb{K} \) is generated by the roots \( {\vartheta }_{1},{\vartheta }_{2},\ldots ,{\vartheta }_{n} \in \mathbb{L} \... | Yes |
Proposition 14.3 The affine algebraic variety \( V\left( {{I}_{k}\left( \vartheta \right) }\right) \subset {\mathbb{A}}^{n}\left( \bar{k}\right) \) consists of\n\n\[ m = \deg {\mathbb{L}}_{f}/k = \left| {\operatorname{Gal}f/k}\right| \]\n\ndistinct points \( \left( {{\vartheta }_{g\left( 1\right) },{\vartheta }_{g\left... | Proof Let \( f = {x}^{n} + {a}_{1}{x}^{n - 1} + \cdots + {a}_{n - 1}x + {a}_{n} \). Write \( {e}_{i}\left( {{t}_{1},{t}_{2},\ldots ,{t}_{n}}\right) \) for the elementary symmetric polynomials. Note that \( {e}_{i}\left( {{t}_{1},{t}_{2},\ldots ,{t}_{n}}\right) - {\left( -1\right) }^{i}{a}_{i} \in {I}_{\mathrm{k}}\left(... | Yes |
Theorem 14.2 If the polynomial \( \bar{f} \in {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) is separable, then there exists an injective group homomorphism\n\n\[ \operatorname{Gal}\bar{f}/{\mathbb{F}}_{p} \hookrightarrow \operatorname{Gal}f/\mathbb{Q}. \] | Proof Since the roots \( {\vartheta }_{1},{\vartheta }_{2},\ldots ,{\vartheta }_{n} \) of \( f \) are integral over \( \mathbb{Z} \), all the coefficients of every polynomial \( {F}_{h} \) in the irreducible decomposition (14.9) belong to the ring of integers \( O \subset {\mathbb{L}}_{f} \) . This forces the coefficie... | Yes |
Corollary 14.2 Let \( f \in \mathbb{Z}\left\lbrack x\right\rbrack \) be an irreducible monic polynomial, and \( \bar{f} \in {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) its reduction modulo \( p \) . If\n\n\[ \bar{f} = {q}_{1}{q}_{2}\cdots {q}_{m} \]\n\nfor irreducible polynomials \( {q}_{1},{q}_{2},\ldots ,{q}_{m} \... | Proof The splitting field of \( \bar{f} \) over \( {\mathbb{F}}_{p} \) is a finite field with a cyclic \( {}^{7} \) Galois group \( G \) over \( {\mathbb{F}}_{p} \) . Since \( G \) acts transitively on the roots of each irreducible polynomial \( {q}_{i} \), the generator of \( G \) acts on the roots of \( \bar{f} \) by... | Yes |
Example 14.4 (Quintic Polynomial with Galois Group \( {S}_{5} \) ) Let us compute the Galois group of the polynomial \( f\left( x\right) = {x}^{5} - x - 1 \) over \( \mathbb{Q} \) . | Consider the irreducible factorizations of \( \bar{f} \) in \( {\mathbb{F}}_{2}\left\lbrack x\right\rbrack \) and in \( {\mathbb{F}}_{3}\left\lbrack x\right\rbrack \) . Every nontrivial factorization of \( f \) contains a factor of degree at most 2 . By Exercise 13.13, the product of all monic irreducible polynomials o... | Yes |
Proposition 14.5 The embedding (14.13) is a group isomorphism, i.e., \[ \text{Gal}{\Phi }_{n} \simeq {\left( \mathbb{Z}/\left( n\right) \right) }^{ * }\text{.} \] | Proof Since the Galois group \( \operatorname{Gal}{\Phi }_{n} \) acts transitively on the roots of \( {\Phi }_{n} \), the inequality \( \left| {\operatorname{Gal}{\Phi }_{n}}\right| \geq \deg {\Phi }_{n} = \varphi \left( n\right) = \left| {\left( \mathbb{Z}/\left( n\right) \right) }^{ * }\right| \) holds. | Yes |
Example 14.5 (Gaussian Sum) Let \( p > 2 \) be a rational prime. If a subgroup \( H \subset {\mathbb{F}}_{p}^{ * } \) has index two, then \( H \) contains all nonzero squares in \( {\mathbb{F}}_{p} \) | \[ {\xi }^{2}H = {\xi H} \cdot {\xi H} = H \] in the quotient group \( {\mathbb{F}}_{p}^{ * }/H \simeq \mathbb{Z}/\left( 2\right) \) . This forces \( H \) to coincide with the multiplicative group of quadratic residues \( {}^{9} \) modulo \( p \) . Therefore, the Galois group of the cyclotomic field contains a unique s... | Yes |
Theorem 14.3 Let \( \mathbb{k} \) be an arbitrary field containing a primitive mth root of unity, and \( a \in {\mathbb{k}}^{ * } \) . Then the binomial \( f\left( x\right) = {x}^{m} - a \) has cyclic Galois group over \( \mathbb{k} \) | Proof Let \( \overline{\mathbb{k}} \supset \mathbb{k} \) be an algebraic closure, and \( \alpha \in \overline{\mathbb{k}} \) a root of \( f \) . Then the roots of \( f \) in \( \overline{\mathbb{k}} \) are in bijection with the group \( {\mathbf{\mu }}_{m} \subset \mathbb{k} \) and are equal to \( {\xi \alpha },\xi \in... | No |
Theorem 14.4 Let \( \mathbb{k} \) be an arbitrary field containing a primitive mth root of unity. Then the cyclic extensions of \( \mathbb{k} \) of degree \( m \) are exhausted by the splitting fields of irreducible binomials \( {x}^{m} - a \in \mathbb{k}\left\lbrack x\right\rbrack \), i.e., by the simple extensions \(... | Proof Let \( \mathbb{K} \subset \mathbb{k} \) be a cyclic extension of degree \( m \) with Galois group \( G = \operatorname{Gal}\mathbb{K}/\mathbb{k} \) generated by an automorphism \( \sigma \in {\operatorname{Aut}}_{\mathbb{k}}\mathbb{K} \) of order \( m \), and let \( \zeta \in \mathbb{k} \) be a primitive \( m \) ... | No |
Lemma 14.1 A finite group \( G \) is solvable if and only if there exists a decreasing series of subgroups\n\n\[ G = {G}_{0} \supset {G}_{1} \supset {G}_{2} \supset \cdots \supset {G}_{m - 1} \supset {G}_{m} = \{ e\} \]\n\n(14.17)\n\nsuch that \( {G}_{i + 1} \vartriangleleft {G}_{i} \) and the quotient \( {G}_{i}/{G}_{... | Proof By definition, a composition series of a solvable group satisfies the condition of the lemma. Conversely, given a series (14.17), a composition series for \( G \) can be constructed by taking a composition series of every quotient \( {G}_{i}/{G}_{i + 1} \), \n\n\[ {G}_{i}/{G}_{i + 1} = {H}_{i,0} \supset {H}_{i,1}... | Yes |
Lemma 14.2 All the subgroups and quotient groups of a solvable group are solvable. Conversely, if a group \( G \) has a solvable normal subgroup \( N \vartriangleleft G \) with solvable quotient \( G/N \), then \( G \) is solvable. | Proof Let \( G \) possess a decreasing series of subgroups\n\n\[ G = {G}_{0} \supset {G}_{1} \supset {G}_{2} \supset \cdots \supset {G}_{m - 1} \supset {G}_{m} = \{ e\} \]\n\n(14.19)\n\nsatisfying the conditions of Lemma 14.1. Intersecting this series with an arbitrary subgroup \( H \subset G \) leads to a chain\n\n\[ ... | Yes |
Theorem 14.6 Let \( \mathbb{k} \) be an arbitrary field of characteristic zero, and \( f \in \mathbb{k}\left\lbrack x\right\rbrack \) a monic irreducible polynomial. If the Galois group \( \mathrm{{Gal}}f/\mathbb{k} \) is solvable, then every root off can be expressed in radicals in terms of the elements of \( k \) . | Proof Fix an algebraic closure \( \overline{\mathbb{k}} \supset \mathbb{k} \), and write \( \mathbb{K} \subset \overline{\mathbb{k}} \) for the splitting field of \( f \), and \( \mathbb{L} \subset \overline{\mathbb{k}} \) for the extension of \( \mathbb{k} \) by a primitive root of unity of degree \( n = \left| {\oper... | Yes |
Corollary 1. If \( \mathrm{F} \) is the fixed field for the finite group \( \mathrm{G} \), then each automorphism \( \sigma \) that leaves \( \mathrm{F} \) fixed must belong to \( \mathrm{G} \) . | \( \left( {E/F}\right) = \) order of \( G = n \) . Assume there is a \( \sigma \) not in \( G \) . Then \( F \) would remain fixed under the \( n + 1 \) elements consisting of \( \sigma \) and the elements of \( G \), thus contradicting the corollary to Theorem 13. | Yes |
Corollary 2. There are no two finite groups \( {G}_{1} \) and \( {G}_{2} \) with the same fixed field. | This follows immediately from Corollary 1. | No |
Lemma 1. If in an abelian group \( A \) and \( B \) are two elements of orders \( a \) and \( b \), and if \( c \) is the least common multiple of \( a \) and \( b \), then there is an element \( \mathrm{C} \) of order \( \mathrm{c} \) in the group. | Proof: (a) If a and b are relatively prime, \( C = {AB} \) has the required order ab. The order of \( {C}^{a} = {B}^{a} \) is \( b \) and therefore \( c \) is divisible by b. Similarly it is divisible by a. Since \( {C}^{ab} = 1 \) it follows \( c = {ab} \) .\n\n(b) If \( d \) is a divisor of \( a \), we can find in th... | Yes |
Lemma 2. If there is an element \( \mathrm{C} \) in an abelian group whose order \( c \) is maximal (as is always the case if the group is finite) then \( c \) is divisible by the order a of every element \( A \) in the group; hence \( {x}^{c} = 1 \) is satisfied by each element in the group. | Proof: If a does not divide \( c \), the greatest common multiple of a and \( c \) would be larger than \( c \) and we could find an element of that order, thus contradicting the choice of \( c \) . | Yes |
Corollary 1. If \( \mathrm{H} \) is a subgroup and \( \mathrm{N} \) a normal subgroup of the group \( G \), then \( H/H \cap N \) is isomorphic to \( {HN}/N \), a subgroup of \( G/N \) . | Proof: Set \( G = U, N = u, H = V \) and the identity \( 1 = v \) in Theorem 1. | No |
Corollary 1. If \( \mathrm{G} \) is a solvable transitive substitution group on \( \mathrm{q} \) letters ( \( q \) prime), then the only substitution of \( G \) which leaves two or more letters fixed is the identity. | This follows from the fact that each substitution is linear modulo \( q \) and bi \( + c \equiv i\left( {\;\operatorname{mod}\;q}\right) \) has either no solution \( \left( {b \equiv 1, c ≢ 0}\right) \) or exactly one solution \( \left( {b \neq 1}\right) \) unless \( b \equiv 1, c \equiv 0 \) in which case the substitu... | Yes |
Corollary 2. A solvable, irreducible equation of prime degree in a field which is a subset of the real numbers has either one real root or all its roots are real. | The group of the equation is a solvable transitive substitution group on \( q \) (prime) letters. In the splitting field (contained in the field of complex numbers) the automorphism which maps a number into its complex conjugate would leave fixed all the real numbers. By Corollary 1 , if two roots are left fixed, then ... | Yes |
The addition and multiplication defined by (1.16) are well-defined. | Suppose that \( a/b = {a}^{\prime }/{b}^{\prime } \) and \( c/d = {c}^{\prime }/{d}^{\prime } \) . Then \( a{b}^{\prime } = {a}^{\prime }b \) and \( c{d}^{\prime } = {c}^{\prime }d \), and so\n\n\[ \left( {{ad} + {bc}}\right) {b}^{\prime }{d}^{\prime } = a{b}^{\prime }d{d}^{\prime } + b{b}^{\prime }c{d}^{\prime } = {a}... | Yes |
Describe the group \( \operatorname{Gal}\left\lbrack {\mathbb{Q}\left( {\sqrt{2}, i\sqrt{3}}\right) : \mathbb{Q}}\right\rbrack \) . For each of its subgroups \( H \), determine \( \Phi \left( H\right) \) . | The elements of \( \mathbb{Q}\left( {\sqrt{2}, i\sqrt{3}}\right) \) are of the form \( a + b\sqrt{2} + {ci}\sqrt{3} + {di}\sqrt{6} \) . By Theorem 7.9, if \( \alpha \in \operatorname{Gal}\left( {\mathbb{Q}\left( {\sqrt{2}, i\sqrt{3}}\right) ,\mathbb{Q}}\right) \), then \( \alpha \left( \sqrt{2}\right) = \pm \sqrt{2},\a... | No |
Find the three roots of\n\n\\[ \n{X}^{3} + {6X} + 2 = 0.\n\\] | ## Solution\n\nHere \( a = b = 2 \), and so \( \Delta = {b}^{2} + 4{a}^{3} = {36} \) . It follows from (8.5) that \( {q}_{1} = {2}^{1/3} \) and \( {r}_{1} = - {4}^{1/3} = - {2}^{2/3} \) . (Note that \( {q}_{1}{r}_{1} = - 2 \) .) The three solutions are\n\n\\[ \n{q}_{1} + {r}_{1},\;{q}_{1}\omega + {r}_{1}{\omega }^{2},\... | No |
The minimum polynomial of \( \omega \) is\n\n\[ f = 1 + {X}^{{p}^{m - 1}} + {X}^{2{p}^{m - 1}} + \cdots + {X}^{\left( {p - 1}\right) {p}^{m - 1}}. \] | Proof\n\nWriting \( {X}^{{p}^{m - 1}} \) as \( Z \), we easily see that\n\n\[ f = 1 + Z + \cdots + {Z}^{p - 1} = \frac{{Z}^{p} - 1}{Z - 1} = \frac{{X}^{{p}^{m}} - 1}{{X}^{{p}^{m - 1}} - 1}, \]\n\nand from (11.1) we see that \( f\left( \omega \right) = 0 \) . It remains to show that \( f \) is irreducible over \( \mathb... | Yes |
Theorem 1. Let \( R \) be a ring.\n\n(i) \( {0r} = 0 \) for every \( r \in R \) ;\n\n(ii) \( - r = \left( {-1}\right) r \) for every \( r \in R \) (where \( - r \) is the additive inverse of \( r \) ; that is, \( - r + r = 0 \) );\n\n(iii) \( \left( {-1}\right) \left( {-r}\right) = r \) for every \( r \in R \) (in part... | Proof. (i) The distributive law gives\n\n\[ \n{0r} = \left( {0 + 0}\right) r = {0r} + {0r} \n\]\n\nand subtracting \( {0r} \) from both sides gives \( {0r} = 0 \) .\n\n(ii) \( 0 = {0r} = \left( {-1 + 1}\right) r = \left( {-1}\right) r + r \) ; now add \( - r \) to both sides of the equation.\n\n(iii)\n\n\[ \n0 = 0\left... | Yes |
Theorem 2. A nonzero ring \( R \) is a domain if and only if it satisfies the cancellation law: if \( {ra} = {rb} \) and \( r \neq 0 \), then \( a = b \) . | Proof. Assume \( R \) is a domain, \( r \neq 0 \), and \( {ra} = {rb} \) . Then \( r\left( {a - b}\right) = 0 \) . Since \( R \) is a domain, \( a - b \neq 0 \) is untenable; hence \( a - b = 0 \) and \( a = b \) .\n\nConversely, assume the cancellation law holds. If \( r \neq 0, a \neq 0 \), and \( {ra} = 0 \), then \... | Yes |
Theorem 3. Let \( I \) be an ideal in a ring \( R \) . Then the abelian group \( R/I \) can be equipped with a multiplication which makes it a ring and which makes the natural map \( \pi : R \rightarrow R/I \) a ring homomorphism. | Proof. Define multiplication on \( R/I \) by\n\n\[ \left( {r + I}\right) \left( {{r}^{\prime } + I}\right) = r{r}^{\prime } + I. \]\n\nTo see that this is well defined, suppose that \( r + I = s + I \) and that \( {r}^{\prime } + I = {s}^{\prime } + I \) ; we must show that \( r{r}^{\prime } + I = s{s}^{\prime } + I \)... | Yes |
Theorem 4. If \( F \) is a field, then every ideal in \( F\left\lbrack x\right\rbrack \) is a principal ideal. | Proof. Let \( I \) be an ideal in \( F\left\lbrack x\right\rbrack \) . If \( I = \{ 0\} \), then \( I = \left( 0\right) \) is principal with generator 0 . If \( I \neq \{ 0\} \), choose a polynomial \( m\left( x\right) \) in \( I \) having smallest degree; we claim that \( I = \left( {m\left( x\right) }\right) \) . Cle... | No |
Theorem 5. Let \( F \) be a field and let \( f\left( x\right), g\left( x\right) \in F\left\lbrack x\right\rbrack \) with \( g\left( x\right) \neq 0 \) . Then \( \left( {f\left( x\right), g\left( x\right) }\right) = d\left( x\right) \) exists and it is a linear combination of \( f\left( x\right) \) and \( g\left( x\righ... | Proof. By Exercise 32,\n\n\[ I = \{ a\left( x\right) f\left( x\right) + b\left( x\right) g\left( x\right) : a\left( x\right), b\left( x\right) \in F\left\lbrack x\right\rbrack \} \]\n\nis an ideal in \( F\left\lbrack x\right\rbrack \) containing both \( f\left( x\right) \) and \( g\left( x\right) \) . Since \( F \) is ... | No |
Corollary 6 (Euclid’s Lemma). Let \( F \) be a field. If \( \left( {f\left( x\right), g\left( x\right) }\right) = 1 \) and \( f\left( x\right) \) divides \( g\left( x\right) h\left( x\right) \), then \( f\left( x\right) \) divides \( h\left( x\right) \) in \( F\left\lbrack x\right\rbrack \) . | Proof. There are polynomials \( a\left( x\right) \) and \( b\left( x\right) \) with \( 1 = {af} + {bg} \) . Hence \( h = {afh} + {bgh} \) . But \( {gh} = {fk} \) for some polynomial \( k \), so that \( h = f\left( {{ah} + {bk}}\right) \) and \( f \) divides \( h \) . | Yes |
Theorem 7 (Euclidean Algorithm). There are algorithms to compute the gcd and to express it as a linear combination. | Proof. The idea is just to iterate the division algorithm. Consider the list of equations (we abbreviate \( f\left( x\right) \) to \( f \), for example):\n\n\[ f = {q}_{1}g + {r}_{1}\;\partial {r}_{1} < \partial g \]\n\n\[ g = {q}_{2}{r}_{1} + {r}_{2} \]\n\n\[ \partial {r}_{2}\; < \partial {r}_{1} \]\n\n\[ {r}_{1} = {q... | Yes |
Corollary 8. Let \( F \subset E \) be fields, and let \( f\left( x\right), g\left( x\right) \in F\left\lbrack x\right\rbrack \subset E\left\lbrack x\right\rbrack \) . Then the \( \gcd \) of \( f \) and \( g \) computed in \( F\left\lbrack x\right\rbrack \) is the same as the \( \gcd \) of \( f \) and \( g \) computed i... | Proof. Regard \( f\left( x\right), g\left( x\right) \) as lying in \( E\left\lbrack x\right\rbrack \) . The Euclidean algorithm computes their gcd in \( E\left\lbrack x\right\rbrack \) . But the list of equations (obtained by iterating the division algorithm) has all its terms involving polynomials over \( F \), and he... | Yes |
Theorem 9. Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) and let \( a \in F \) . Then there is \( q\left( x\right) \in F\left\lbrack x\right\rbrack \) with\n\n\[ f\left( x\right) = q\left( x\right) \left( {x - a}\right) + f\left( a\right) . | Proof. Use the division algorithm. Dividing \( f\left( x\right) \) by \( x - a \) gives a quotient and a constant remainder (because \( x - a \) has degree 1):\n\n\[ f\left( x\right) = q\left( x\right) \left( {x - a}\right) + r.\]\n\nEvaluating at \( a \) gives \( f\left( a\right) = q\left( a\right) \left( {a - a}\righ... | Yes |
Corollary 10. Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) . Then \( a \in F \) is a root of \( f\left( x\right) \) if and only if \( x - a \) divides \( f\left( x\right) \) . | Proof. If \( a \) is a root of \( f\left( x\right) \), then \( f\left( a\right) = 0 \), and the theorem gives \( f\left( x\right) = \) \( q\left( x\right) \left( {x - a}\right) \) . Conversely, if \( f\left( x\right) = q\left( x\right) \left( {x - a}\right) \), then evaluating at \( a \) gives \( f\left( a\right) = 0 \... | Yes |
Theorem 11. If \( F \) is a field and \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) has degree \( n \), then \( F \) contains at most \( n \) roots of \( f\left( x\right) \) . | Proof. Suppose that \( F \) contains \( n + 1 \) distinct roots of \( f\left( x\right) \), say, \( {a}_{1},\ldots ,{a}_{n + 1} \) . By the corollary, \( f\left( x\right) = \left( {x - {a}_{1}}\right) {g}_{1}\left( x\right) \) (for some \( {g}_{1}\left( x\right) \in F\left\lbrack x\right\rbrack \) ). Now \( x - {a}_{2} ... | Yes |
Theorem 12. An ideal \( I \) in \( R \) with \( I \neq R \) is a prime ideal if and only if \( R/I \) is a domain. | Proof. Let \( I \) be a prime ideal. Suppose that \( a + I \neq 0 \) and \( b + I \neq 0 \) ; that is, neither \( a \) nor \( b \) lies in \( I \) . If \( \left( {a + I}\right) \left( {b + I}\right) = {ab} + I = 0 \), then \( {ab} \in I \) , contradicting \( I \) being prime. The converse is just as easy. | Yes |
Theorem 13. An ideal \( I \) with \( I \neq R \) in a ring \( R \) is a maximal ideal if and only if \( R/I \) is a field. | Proof. The Correspondence Theorem (Exercise 37) shows that \( I \) is a maximal ideal if and only if \( R/I \) has no ideals other than \( \{ 0\} \) and \( R/I \) itself; Exercise 30 shows that this property holds if and only if \( R/I \) is a field. 口 | No |
Corollary 14. Every maximal ideal is a prime ideal. | Proof. Every field is a domain. | No |
Theorem 15. If \( R \) is a principal ideal domain, then every nonzero prime ideal \( I \) is a maximal ideal. | Proof. Assume there is an ideal \( J \neq I \) with \( I \subset J \subset R \) . Since \( R \) is a PID, \( I = \left( a\right) \) and \( J = \left( b\right) \) for some \( a, b \in R \) . Now \( a \in J \) implies that \( a = {rb} \) for some \( r \in R \), and so \( {rb} \in I \) . Since \( I \) is prime, either \( ... | No |
Corollary 16. If \( F \) is a field and \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) is irreducible, then \( F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \) is a field containing (an isomorphic copy of) \( F \) and a root of \( p\left( x\right) \) . | Proof. Since \( p\left( x\right) \) is irreducible, the principal ideal \( I = \left( {p\left( x\right) }\right) \) is a nonzero prime ideal; since \( F\left\lbrack x\right\rbrack \) is a PID, \( I \) is a maximal ideal, and so \( E = F\left\lbrack x\right\rbrack /I \) is a field. It is easy to see that \( a \mapsto a ... | Yes |
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