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Theorem 7.5. Let \( k \) be a field, and let \( V \) be a finite-dimensional vector space. Let \( \alpha \) be a linear transformation on \( V \) , and endow \( V \) with the corresponding \( k\left\lbrack t\right\rbrack \) -module structure, as in Claim 7.1. Then the following hold:\n\n- There exist distinct monic irr... | Proof. Since \( \dim V \) is finite, \( V \) is finitely generated as a \( k \) -module and a fortiori as a \( k\left\lbrack t\right\rbrack \) -module. The two isomorphisms are then obtained by applying Theorem 5.6. All the relevant polynomials may be chosen to be monic since every polynomial over a field is the associ... | Yes |
Proposition 7.9. Let \( {f}_{1}\left( t\right) \left| \cdots \right| {f}_{m}\left( t\right) \) be the invariant factors of a linear transformation \( \alpha \) on a vector space \( V \) . Then the minimal polynomial \( {m}_{\alpha }\left( t\right) \) equals \( {f}_{m}\left( t\right) \) , and the characteristic polynomi... | Proof. Tracing definitions, \( \left( {{m}_{\alpha }\left( t\right) }\right) \) is the annihilator ideal of \( V \) when this is viewed as a \( k\left\lbrack t\right\rbrack \) -module via \( \alpha \) (as in Claim 7.1). Therefore the equality of \( {m}_{\alpha }\left( t\right) \) and \( {f}_{m}\left( t\right) \) is a r... | Yes |
Corollary 7.10 (Cayley-Hamilton). The minimal polynomial of a linear transformation divides its characteristic polynomial. | Proof. This has now become evident, as promised in \( §{6.2} \) . | No |
Proposition 7.11. Let \( A \in {\mathcal{M}}_{n}\left( k\right) \) be a square matrix. Then \( A \) is similar to its transpose. | Proof. If \( B \) is similar to \( A \) and we can prove that \( B \) is similar to its transpose \( {B}^{t} \) , then \( A \) is similar to its transpose \( {A}^{t} \) : because \( B = {PA}{P}^{-1},{B}^{t} = {QB}{Q}^{-1} \) give\n\n\[ \n{A}^{t} = \left( {{P}^{t}{QP}}\right) A{\left( {P}^{t}QP\right) }^{-1}.\n\]\n\nThe... | Yes |
Lemma 7.12. Assume that the characteristic polynomial \( {P}_{\alpha }\left( t\right) \) factors completely; that is,\n\n\[ \n{P}_{\alpha }\left( t\right) = \mathop{\prod }\limits_{{i = 1}}^{s}{\left( t - {\lambda }_{i}\right) }^{{m}_{i}}\n\]\n\nwhere \( {\lambda }_{i}, i = 1,\ldots, s \), are the distinct eigenvalues ... | Proof. The first statement follows from uniqueness of factorizations. The statement about the minimal polynomial is immediate from Proposition 7.9 and the bookkeeping giving the equivalence of the two formulations in Theorem 7.5. | Yes |
One use of the Jordan canonical form is the enumeration of all possible similarity classes of transformations with given eigenvalues. For example, there are 5 similarity classes of linear transformations with a single eigenvalue \( \lambda \) with algebraic multiplicity 4, over a 4-dimensional vector space: indeed, the... | \[ \left( \begin{matrix} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{matrix}\right) ,\;\left( \begin{matrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{matrix}\right) ,\;\left( \begin{matrix} \lambda & 1 & 0 & 0 \\ 0 & ... | Yes |
Proposition 7.16. The geometric multiplicity of \( \lambda \) as an eigenvalue of \( \alpha \) equals the number of Jordan blocks corresponding to \( \lambda \) in the Jordan canonical form of \( \alpha \) . | Proof. As the geometric multiplicity is clearly additive in direct sums, it suffices to show that the geometric multiplicity of \( \lambda \) for the transformation corresponding to a single Jordan block\n\n\[ J = \left( \begin{matrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0... | Yes |
Proposition 7.18. Assume the characteristic polynomial of \( \alpha \in {\operatorname{End}}_{k}\left( V\right) \) factors completely over \( k \) . Then \( \alpha \) is diagonalizable if and only if the minimal polynomial of \( \alpha \) has no multiple roots. | Proof. Again, diagonalizability is equivalent to having all Jordan blocks of size 1 in the Jordan canonical form of \( \alpha \) . Therefore, if the characteristic polynomial of \( \alpha \) factors completely, then \( \alpha \) is diagonalizable if and only if all exponents \( {r}_{ij} \) appearing in Theorem 7.5 equa... | Yes |
Proposition 1.3. Let \( k \subseteq k\left( \alpha \right) \) be a simple extension. Consider the evaluation map \( \epsilon : k\left\lbrack t\right\rbrack \rightarrow k\left( \alpha \right) \), defined by \( f\left( t\right) \mapsto f\left( \alpha \right) \). Then we have the following:\n\n- \( \epsilon \) is injectiv... | Proof. Let \( F = k\left( \alpha \right) \). By the ’first isomorphism theorem’, the image of \( \epsilon : k\left\lbrack t\right\rbrack \rightarrow F \) is isomorphic to \( k\left\lbrack t\right\rbrack /\ker \left( \epsilon \right) \). Since \( F \) is an integral domain, so is \( k\left\lbrack t\right\rbrack /\ker \l... | Yes |
Consider the extension \( \mathbb{Q} \subseteq \mathbb{R} \). The polynomial \( {x}^{2} - 2 \in \mathbb{Q}\left\lbrack x\right\rbrack \) has roots in \( \mathbb{R} \): therefore, by Proposition V.5.7 there exists a homomorphism (hence a field extension) | \[ \bar{\epsilon } : \;\frac{\mathbb{Q}\left\lbrack t\right\rbrack }{\left( {t}^{2} - 2\right) } \hookrightarrow \mathbb{R} \] such that the image of (the coset of) \( t \) is a root \( \alpha \) of \( {x}^{2} - 2 \). Proposition 1.3 simply identifies the image of this homomorphism with \( \mathbb{Q}\left( \alpha \righ... | No |
Proposition 1.5. Let \( i : {k}_{1} \subseteq {F}_{1} = {k}_{1}\left( {\alpha }_{1}\right) ,{k}_{2} \subseteq {F}_{2} = {k}_{2}\left( {\alpha }_{2}\right) \) be two finite simple extensions. Let \( {p}_{1}\left( t\right) \in {k}_{1}\left\lbrack t\right\rbrack \), resp., \( {p}_{2}\left( t\right) \in {k}_{2}\left\lbrack... | Proof. Since every element of \( {k}_{1}\left( {\alpha }_{1}\right) \) is a linear combination of powers of \( {\alpha }_{1} \) with coefficients in \( {k}_{1}, j \) is determined by its action on \( {k}_{1} \) (which agrees with \( i \) ) and by \( j\left( {\alpha }_{1}\right) \), which is prescribed to be \( {\alpha ... | Yes |
Corollary 1.7. Let \( k \subseteq F = k\\left( \\alpha \\right) \) be a simple finite extension, and let \( p\\left( x\\right) \) be the minimal polynomial of \( \\alpha \) over \( k \) . Then \( \\left| {{\\operatorname{Aut}}_{k}\\left( F\\right) }\\right| \) equals the number of distinct roots of \( p\\left( x\\right... | Proof. Let \( j \\in {\\operatorname{Aut}}_{k}\\left( F\\right) \) . Since every element of \( F \) is a polynomial expression in \( \\alpha \) with coefficients in \( k \), and \( j \) extends the identity on \( k, j \) is determined by \( j\\left( \\alpha \\right) \) . Now\n\n\\[ \np\\left( {j\\left( \\alpha \\right)... | Yes |
Lemma 1.9. Let \( k \subseteq F \) be a finite extension. Then every \( \alpha \in F \) is algebraic over \( k, \) of degree \( \leq \left\lbrack {F : k}\right\rbrack \) . | Proof. Since \( k \subseteq k\left( \alpha \right) \subseteq F \), the dimension of \( k\left( \alpha \right) \) as a \( k \) -vector space is bounded by \( {\dim }_{k}F = \left\lbrack {F : k}\right\rbrack \). Concretely, if \( k \subseteq F \) is finite and \( \alpha \in F \), then the powers \( 1,\alpha ,{\alpha }^{2... | Yes |
Proposition 1.10. Let \( k \subseteq E \subseteq F \) be field extensions. Then \( k \subseteq F \) is finite if and only if both \( k \subseteq E \) and \( E \subseteq F \) are finite. In this case, \[ \left\lbrack {F : k}\right\rbrack = \left\lbrack {F : E}\right\rbrack \left\lbrack {E : k}\right\rbrack \] | Proof. If \( F \) is finite-dimensional as a vector space over \( k \), then so is its subspace \( E \) ; and any linear dependence relation of elements of \( F \) over the field \( k \) gives one over the larger field \( E \) . It follows that if \( k \subseteq F \) is finite, then so are \( k \subseteq E \) and \( E ... | Yes |
Let \( k \subseteq F \) be a field extension, and let \( \alpha \in F \) be an algebraic element over \( k \), of \( {odd} \) order. Then we claim that \( \alpha \) may be written as a polynomial in \( {\alpha }^{2} \), with coefficients in \( k \) . | Indeed, \( k\left( {\alpha }^{2}\right) \) is intermediate between \( k \) and \( k\left( \alpha \right) \): \n\n\[ \n k \subseteq k\left( {\alpha }^{2}\right) \subseteq k\left( \alpha \right) \n\] \n\nwhat can we say about the degree \( d \) of \( k\left( \alpha \right) \) over \( k\left( {\alpha }^{2}\right) \) ? Sin... | Yes |
Proposition 1.15. Let \( k \subseteq F = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) be a finitely generated field extension. Then the following are equivalent:\n\n(i) \( k \subseteq F \) is a finite extension.\n\n(ii) \( k \subseteq F \) is an algebraic extension.\n\n(iii) Each \( {\alpha }_{i} \) is algeb... | Proof. Lemma 1.9 shows that (i) \( \Rightarrow \) (ii); (ii) \( \Rightarrow \) (iii) trivially. Thus, we only need to prove that (iii) \( \Rightarrow \) (i), and to bound the degree of \( F \) over \( k \) in the process.\n\nAssume that each \( {\alpha }_{i} \) is algebraic over \( k \), and let \( {d}_{i} \) be the de... | Yes |
Let \( \overline{\mathbb{Q}} \subseteq \mathbb{C} \) be the set of complex numbers that are algebraic over \( \mathbb{Q} \) ; then \( \overline{\mathbb{Q}} \) is a field, by Corollary 1.16, and the extension \( \mathbb{Q} \subseteq \overline{\mathbb{Q}} \) is (tautologically) algebraic. | Note that \( \mathbb{Q} \subseteq \overline{\mathbb{Q}} \) is not a finite extension, because in it there are elements of arbitrarily high degree over \( \mathbb{Q} \) : indeed, there exist irreducible polynomials in \( \mathbb{Q}\left\lbrack x\right\rbrack \) of arbitrarily high degree, as we observed in Corollary V.5... | No |
Corollary 1.18. Let \( k \subseteq E \subseteq F \) be field extensions. Then \( k \subseteq F \) is algebraic if and only if both \( k \subseteq E \) and \( E \subseteq F \) are algebraic. | Proof. If \( k \subseteq F \) is algebraic, then every element of \( F \) is algebraic over \( k \), hence over \( E \), and every element of \( E \) is algebraic over \( k \) ; thus \( E \subseteq F \) and \( k \subseteq E \) are algebraic.\n\nConversely, assume \( k \subseteq E \) and \( E \subseteq F \) are both alg... | Yes |
Consider the extension \( \mathbb{Q} \subseteq \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \). | -By Proposition 1.15 we know that this is a finite (hence algebraic) extension, of degree at most 4 .\n\n-Thus any five elements in \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \) must be linearly dependent over \( \mathbb{Q} \) . We consider powers of \( \sqrt{2} + \sqrt{3} \) :\n\n\[ 1,\;{\left( \sqrt{2} + \sqrt{3}... | Yes |
Lemma 2.1. For a field \( K \), the following are equivalent:\n\n- \( K \) is algebraically closed.\n\n- \( K \) has no nontrivial algebraic extensions.\n\n- If \( K \subseteq L \) is any extension and \( \alpha \in L \) is algebraic over \( K \), then \( \alpha \in K \) . | The proof is a straightforward application of the definitions and a good exercise (Exercise 2.1). | No |
Theorem 2.3. Every field \( k \) admits an algebraic closure \( k \subseteq \bar{k} \) ; this extension is unique up to isomorphism. | Concerning existence, the idea is to construct ’by hand’ a huge extension \( K \) of \( k \) where every polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) factors completely. The elements of \( K \) which are algebraic over \( k \) will form an algebraic closure of \( k \) .\n\nThe construction is done... | No |
Lemma 2.4. Let \( k \) be a field. Then there exists an extension \( k \subseteq K \) such that every nonconstant polynomial \( f\\left( x\\right) \\in k\\left\\lbrack x\\right\\rbrack \) has at least one root in \( K \) . | Proof. (This construction is apparently due to Emil Artin.) Consider a set \( \\mathcal{T} = \) \( \\left\\{ {t}_{f}\\right\\} \) in bijection with the set of nonconstant monic polynomials \( f\\left( x\\right) \\in k\\left\\lbrack x\\right\\rbrack \\), and let \( k\\left\\lbrack \\mathcal{T}\\right\\rbrack \) be the c... | Yes |
Lemma 2.6. Let \( k \subseteq L \) be a field extension, with \( L \) algebraically closed. Let\n\n\[ \bar{k} \mathrel{\text{:=}} \{ \alpha \in L \mid \alpha \text{ is algebraic over }k\} .\n\]\n\nThen \( \bar{k} \) is an algebraic closure of \( k \) . | By Corollary 1.16, \( \bar{k} \) is a field, and the extension \( k \subseteq \bar{k} \) is tautologically algebraic. To verify that \( \bar{k} \) is algebraically closed, let \( \alpha \) be algebraic over \( \bar{k} \) ; then\n\n\[ k \subseteq \bar{k} \subseteq \bar{k}\left( \alpha \right)\n\]\n\nis a composition of ... | Yes |
Lemma 2.8. Let \( k \subseteq L \) be a field extension, with \( L \) algebraically closed. Let \( k \subseteq F \) be any algebraic extension. Then there exists a morphism of extensions \( i : F \rightarrow L \) . | Proof. This argument also relies on Zorn’s lemma. Consider the set \( Z \) of homomorphisms\n\n\[ \n{i}_{K} : K \rightarrow L \n\]\n\nwhere \( K \) is an intermediate field, \( k \subseteq K \subseteq F \), and \( {i}_{K} \) restricts to the identity on \( k;Z \) is nonempty, since the extension \( {i}_{k} : k \subsete... | Yes |
Theorem 2.9 (Nullstellensatz). Let \( k \subseteq F \) be a field extension, and assume that \( F \) is a finite-type \( k \) -algebra. Then \( k \subseteq F \) is a finite (hence algebraic) extension. | ## Proof for uncountable fields. Assume that \( k \) is uncountable.\n\nLet \( k \subseteq F \) be a field extension, and assume that \( F \) is finitely generated as an algebra over \( k \) ; in particular, it is finitely generated as a field extension. We have to prove that \( k \subseteq F \) is a finite extension, ... | No |
Corollary 2.10. Let \( K \) be an algebraically closed field, and let \( I \) be an ideal of \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then \( I \) is maximal if and only if\n\n\[ I = \left( {{x}_{1} - {c}_{1},\ldots ,{x}_{n} - {c}_{n}}\right) \]\n\nfor \( {c}_{1},\ldots ,{c}_{n} \in K \) . | Proof. For \( {c}_{1},\ldots ,{c}_{n} \in K \)\n\n\[ \frac{K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack }{\left( {x}_{1} - {c}_{1},\ldots ,{x}_{n} - {c}_{n}\right) } \cong K \]\n\n(Exercise III.4.12) is a field; therefore \( \left( {{x}_{1} - {c}_{1},\ldots ,{x}_{n} - {c}_{n}}\right) \) is maximal. Conversely, ... | No |
Lemma 2.15. Let \( K \) be a field, and let \( S \) be a subset of \( {\mathbb{A}}_{K}^{n} \) . Then the ideal \( \mathcal{I}\left( S\right) \) is a radical ideal of \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . | Proof. The inclusion \( \mathcal{I}\left( S\right) \subseteq \sqrt{\mathcal{I}\left( S\right) } \) holds for every ideal, so it is trivially satisfied. To verify the inclusion \( \sqrt{\mathcal{I}\left( S\right) } \subseteq \mathcal{I}\left( S\right) \), let \( f \in \sqrt{\mathcal{I}\left( S\right) } \) . Then there i... | Yes |
Proposition 2.16 (Weak Nullstellensatz). Let \( K \) be an algebraically closed field, and let \( I \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an ideal. Then \( \mathcal{V}\left( I\right) = \varnothing \) if and only if \( I = \left( 1\right) \) . | Proof. If \( I = \left( 1\right) \), then \( \mathcal{V}\left( I\right) = \varnothing \) by definition.\n\nConversely, assume that \( I \neq \left( 1\right) \) . By Proposition V.3.5, \( I \) is then contained in a maximal ideal \( \mathfrak{m} \) . Since \( K \) is algebraically closed, by Corollary 2.10 we have\n\n\[... | Yes |
Corollary 2.18. Let \( K \) be an algebraically closed field. Then for any \( n \geq 0 \) the functions \[ \left\{ {\text{ algebraic subsets of }{\mathbb{A}}_{K}^{n}}\right\} \overset{\not{} }{\overset{\mathcal{I}}{ \leftrightarrow }}\left\{ {\text{ radical ideals in }K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrac... | Proof. Proposition 2.17 shows that \( \mathcal{I} \circ \mathcal{V} \) is the identity on radical ideals, so \( \mathcal{V} \) is injective, and \( \mathcal{V} \) is surjective by definition of affine algebraic set. It follows that \( \mathcal{V} \) is a bijection and \( \mathcal{I} \) is its inverse. | Yes |
Lemma 3.3. The subset \( {\mathcal{C}}_{\mathbb{R}} \subseteq \mathbb{R} \) of constructible numbers is a subfield of \( \mathbb{R} \). Likewise, \( {\mathcal{C}}_{\mathbb{C}} \) is a subfield of \( \mathbb{C} \), and in fact \( {\mathcal{C}}_{\mathbb{C}} = {\mathcal{C}}_{\mathbb{R}}\left( i\right) \). | Proof. The set \( {\mathcal{C}}_{\mathbb{R}} \subseteq \mathbb{R} \) is nonempty, so in order to show it is a field, we only need to show that it is closed with respect to subtraction and division by a nonzero constructible number (cf. Exercise II.6.2).\n\nThe reader will check that \( {\mathcal{C}}_{\mathbb{R}} \) is ... | No |
Theorem 3.4. Let \( \gamma \in \mathbb{R} \) . Then \( \gamma \in {\mathcal{C}}_{\mathbb{R}} \) if and only if there exist real numbers \( {\delta }_{1},\ldots ,{\delta }_{k} \) such that \( \forall j = 1,\ldots, k \)\n\n\[ \left\lbrack {\mathbb{Q}\left( {{\delta }_{1},\ldots ,{\delta }_{j}}\right) : \mathbb{Q}\left( {... | Proof. Let's first argue in the 'geometry to algebra' direction. A configuration of points, lines, and circles obtained by a straightedge-and-compass construction may be described by the coordinates of the points and the equations of the lines and circles. Suppose that at one stage in a given construction all coordinat... | No |
Regular pentagons are constructible. | Indeed, it suffices to construct the point \( A = \left( {\cos \left( {{2\pi }/5}\right) ,0}\right) \), and it so happens that \( \gamma = \cos \left( {{2\pi }/5}\right) \) satisfies\n\n(*) \n\n\[ 4{\gamma }^{2} + {2\gamma } - 1 = 0 \]\n\n(in fact, \( \gamma \) is half of the inverse of the golden ratio: \( \gamma = \f... | No |
Corollary 3.6. Let \( \gamma \in {\mathcal{C}}_{\mathbb{C}} \) be a constructible number. Then \( \left\lbrack {\mathbb{Q}\left( \gamma \right) : \mathbb{Q}}\right\rbrack \) is a power of 2. | Proof. By Lemma 3.3 and Theorem 3.4, there exist \( {\delta }_{1}\ldots ,{\delta }_{k} \in \mathbb{R} \) such that\n\n\[ \gamma \in \mathbb{Q}\left( {{\delta }_{1},\ldots ,{\delta }_{k}, i}\right) \]\n\nand each \( {\delta }_{j} \) has degree \( \leq 2 \) over \( \mathbb{Q}\left( {{\delta }_{1},\ldots ,{\delta }_{j - 1... | Yes |
The splitting field \( F \) of \( {x}^{8} - 1 \) over \( \mathbb{Q} \) is generated by \( \zeta \mathrel{\text{:=}} {e}^{{2\pi i}/8} \) | indeed, the roots of \( {x}^{8} - 1 \) are all the 8-th roots of 1, and all of them are powers of \( \zeta \) : In fact, \( \zeta \) is a root of the polynomial \( {x}^{4} + 1 \), which is irreducible over \( \mathbb{Q} \) ; therefore \( F = \mathbb{Q}\left( \zeta \right) \) is ’already’ the splitting field of \( {x}^{... | Yes |
Variation on the theme: \( {x}^{4} + 2 \) . The overoptimistic reader may now hope that the difference between the splitting fields of \( {x}^{4} + 1 \) vs. \( {x}^{4} - 1 \) over \( \mathbb{Q} \) is just due to the fact that the first polynomial is irreducible over \( \mathbb{Q} \) and the second is not. This example ... | With notation as in Example 4.4, the roots of \( {x}^{4} + 2 \) are\n\n\[ \sqrt[4]{2}\zeta ,\sqrt[4]{2}{\zeta }^{3},\sqrt[4]{2}{\zeta }^{5},\sqrt[4]{2}{\zeta }^{7} \]\n\nTherefore, with \( K = \mathbb{Q}\left( {\sqrt[4]{2}\zeta ,\sqrt[4]{2}{\zeta }^{3},\sqrt[4]{2}{\zeta }^{5},\sqrt[4]{2}{\zeta }^{7}}\right) \) the spli... | Yes |
If a complex root of an irreducible polynomial \( p\\left( x\\right) \\in \\mathbb{Q}\\left\\lbrack x\\right\\rbrack \) may be expressed as a polynomial in \( i \) and \( \\sqrt[4]{2} \) with rational coefficients, then all roots of \( p\\left( x\\right) \) may be expressed likewise in terms of \( i \) and \( \\sqrt[4]... | Indeed, we have checked (Example 4.6) that \( \\mathbb{Q}\\left( {i,\\sqrt[4]{2}}\\right) \) is a splitting field over \( \\mathbb{Q} \) ; hence it is a normal extension of \( \\mathbb{Q} \) . | Yes |
Let \( p \) be a prime, and consider the field \( {\mathbb{F}}_{p}\left( t\right) \) of rational functions over \( {\mathbb{F}}_{p} \). Then the polynomial \[ {x}^{p} - t \in {\mathbb{F}}_{p}\left( t\right) \left\lbrack x\right\rbrack \] is irreducible. | by Eisenstein’s criterion it is irreducible in \( {\mathbb{F}}_{p}\left\lbrack t\right\rbrack \left\lbrack x\right\rbrack \) (since \( \left( t\right) \) is prime in \( {\mathbb{F}}_{p}\left\lbrack t\right\rbrack \) ), hence in \( {\mathbb{F}}_{p}\left( t\right) \left\lbrack x\right\rbrack \) by Proposition V.4.16. Let... | No |
Lemma 4.13. Let \( k \) be a field, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) . Then \( f\left( x\right) \) is separable if and only if \( f\left( x\right) \) and \( {f}^{\prime }\left( x\right) \) are relatively prime. | Proof. First assume that \( f\left( x\right) \) is not separable. Then \( f\left( x\right) \) has a multiple root in a splitting field \( F \) ; that is,\n\n\[ f\left( x\right) = {\left( x - \alpha \right) }^{m}g\left( x\right) \]\n\nfor some \( \alpha \in F, g\left( x\right) \in F\left\lbrack x\right\rbrack \), and \(... | Yes |
Lemma 4.14. Let \( k \) be a field, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be an inseparable irreducible polynomial. Then \( {f}^{\prime }\left( x\right) = 0 \) . | Proof. Since \( f\left( x\right) \) is inseparable, \( f\left( x\right) \) and \( {f}^{\prime }\left( x\right) \) have a common irreducible factor \( q\left( x\right) \) by Lemma 4.13; but as \( f\left( x\right) \) is itself irreducible, \( q\left( x\right) \) must be an associate of \( f\left( x\right) \), and in part... | Yes |
Proposition 4.17. Let \( k \) be a field. Then \( k \) is perfect if and only if all irreducible polynomials in \( k\left\lbrack x\right\rbrack \) are separable. | Proof. We will prove that irreducible polynomials over a perfect field are separable, leaving the other implication to the reader (Exercise 4.12).\n\nWe have already noted that irreducible polynomials are separable over fields of characteristic zero. In positive characteristic \( p \), we have observed that an insepara... | No |
Corollary 4.18. Finite fields are perfect. Therefore, over finite fields, irreducible polynomials are separable. | Proof. The Frobenius map is injective (because it is a homomorphism of fields), so it is surjective over finite fields, by the pigeon-hole principle. Therefore finite fields are perfect, and the second part of the statement follows from Proposition 4.17. | Yes |
Lemma 4.22. Let \( k \subseteq k\left( \alpha \right) \) be a simple algebraic extension. Then \( {\left\lbrack k\left( \alpha \right) : k\right\rbrack }_{s} \) equals the number of distinct roots in \( \bar{k} \) of the minimal polynomial of \( \alpha \) . In particular, \( {\left\lbrack k\left( \alpha \right) : k\rig... | Proof. The proof is essentially (and not by coincidence) a rehash of the proof of Corollary 1.7. Associate with each \( \iota : k\left( \alpha \right) \rightarrow \bar{k} \) extending \( {\operatorname{id}}_{k} \) the image \( \iota \left( \alpha \right) \), which must be a root of the minimal polynomial of \( \alpha \... | Yes |
Lemma 4.23. Let \( k \subseteq E \subseteq F \) be algebraic extensions. Then \( {\left\lbrack F : k\right\rbrack }_{s} \) is finite if and only if both \( {\left\lbrack F : E\right\rbrack }_{s},{\left\lbrack E : k\right\rbrack }_{s} \) are finite, and in this case \[ {\left\lbrack F : k\right\rbrack }_{s} = {\left\lbr... | Proof. Different embeddings of \( E \) into \( \bar{k} \) extend to different embeddings of \( F \) into \( \bar{k} \) , by Lemma 2.8; and embeddings of \( F \) into \( \bar{E} = \bar{k} \) extending the identity on \( E \) extend a fortiori the identity on \( k \) . Therefore, if any of \( {\left\lbrack F : E\right\rb... | Yes |
Proposition 4.24. Let \( k \subseteq F \) be a finite extension. Then \( {\left\lbrack F : k\right\rbrack }_{s} \leq \left\lbrack {F : k}\right\rbrack \), and the following are equivalent:\n\n(i) \( F = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \), where each \( {\alpha }_{i} \) is separable over \( k \) ;\n... | Proof. Since \( F \) is finite over \( k \), it is finitely generated. Let \( F = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) . Then using Lemma 4.23, Lemma 4.22, and Proposition 1.10,\n\n\[{\left\lbrack F : k\right\rbrack }_{s} = {\left\lbrack k\left( {\alpha }_{1},\ldots ,{\alpha }_{r - 1}\right) \left( {... | No |
Theorem 5.1. Let \( q = {p}^{d} \) be a power of a prime integer \( p \) . Then the splitting field of the polynomial \( {x}^{q} - x \) over \( {\mathbb{F}}_{p} \) is a field with precisely \( q \) elements.\n\nConversely, let \( F \) be a field with exactly \( q \) elements; then \( F \) is a splitting field for \( {x... | Proof. Let \( F \) be the splitting field of \( {x}^{q} - x \) over \( {\mathbb{F}}_{p} \) . Let \( E \) be the set of roots of \( f\left( x\right) = {x}^{q} - x \) in \( F \) . Since \( {f}^{\prime }\left( x\right) = q{x}^{q - 1} - 1 = - 1 \) (as \( q = 0 \) in characteristic \( p \) ), we have \( \left( {f\left( x\ri... | Yes |
Corollary 5.2. For every prime power \( q \) there exists one and only one finite field of order \( q \), up to isomorphism. | Proof. This follows immediately from Theorem 5.1 and the uniqueness of splitting fields (Lemma 4.2). | Yes |
Let \( p \) be a prime integer. Then we claim that the polynomial \( {x}^{4} + 1 \) is reducible over \( {\mathbb{F}}_{p} \) (and therefore over every finite field). | Since \( {x}^{4} + 1 = {\left( x + 1\right) }^{4} \) in \( {\mathbb{F}}_{2}\left\lbrack x\right\rbrack \), the statement holds for \( p = 2 \) . Thus, we may assume that \( p \) is an odd prime. Then we claim that \( {x}^{4} + 1 \) divides \( {x}^{{p}^{2}} - x \) . Indeed, the square of every odd number is congruent to... | No |
Corollary 5.4. Let \( p \) be a prime, and let \( d \leq e \) be positive integers. Then there is an extension \( {\mathbb{F}}_{{p}^{d}} \subseteq {\mathbb{F}}_{{p}^{e}} \) if and only if \( d \mid e \) . Further, if \( d \mid e \), then there is exactly one such extension, in the sense that \( {\mathbb{F}}_{{p}^{e}} \... | Proof. If there is an extension as stated, then \( {\mathbb{F}}_{p} \subseteq {\mathbb{F}}_{{p}^{d}} \subseteq {\mathbb{F}}_{{p}^{e}} \) ; hence \( \left\lbrack {{\mathbb{F}}_{{p}^{d}} : {\mathbb{F}}_{p}}\right\rbrack \) divides \( \left\lbrack {{\mathbb{F}}_{{p}^{e}} : {\mathbb{F}}_{p}}\right\rbrack \) by Corollary 1.... | Yes |
Corollary 5.5. Let \( F \) be a finite field. Then for all integers \( n \geq 1 \) there exist irreducible polynomials of degree \( n \) in \( F\left\lbrack x\right\rbrack \) . | Proof. We know \( F = {\mathbb{F}}_{{p}^{d}} \) for some prime \( p \) and some \( d \geq 1 \) . By Corollary 5.4 there is an extension \( {\mathbb{F}}_{{p}^{d}} \subseteq {\dot{\mathbb{F}}}_{{p}^{dn}} \), generated by an element \( \alpha \) . Then \( \left\lbrack {{\mathbb{F}}_{{p}^{dn}} : {\mathbb{F}}_{{p}^{d}}}\rig... | Yes |
Let \( F = {\mathbb{F}}_{q} \) be a finite field, and let \( n \) be a positive integer. Then the factorization of \( {x}^{{q}^{n}} - x \) in \( F\left\lbrack x\right\rbrack \) consists of all irreducible monic polynomials of degree \( d \), as \( d \) ranges over the positive divisors of \( n \) . In particular, all t... | Proof. By Theorem 5.1, \( {\mathbb{F}}_{{q}^{n}} \) is the splitting field of \( {x}^{{q}^{n}} - x \) over \( {\mathbb{F}}_{p} \), and hence over \( {\mathbb{F}}_{q} = F \) .\n\nIf \( f\left( x\right) \) is a monic irreducible polynomial of degree \( d \), then \( F\left\lbrack x\right\rbrack /\left( {f\left( x\right) ... | Yes |
Let’s contemplate the case \( q = 2 : {\mathbb{F}}_{2} = \mathbb{Z}/2\mathbb{Z} \). | - \( n = 1 \) : the polynomial \( {x}^{2} - x \) factors as the product of \( x \) and \( \left( {x - 1}\right) \) (which we could write as \( \left( {x + 1}\right) \) just as well, since we are working over \( {\mathbb{F}}_{2} \) ). These are all the irreducible polynomials of degree 1 over \( {\mathbb{F}}_{2} \).\n\n... | No |
Proposition 5.8. \( {\operatorname{Aut}}_{{\mathbb{F}}_{p}}\left( {\mathbb{F}}_{{p}^{d}}\right) \) is cyclic, generated by the Frobenius isomorphism. | Proof. Let \( \varphi \) be the Frobenius homomorphism \( {\mathbb{F}}_{{p}^{d}} \rightarrow {\mathbb{F}}_{{p}^{d}} : \varphi \left( x\right) = {x}^{p} \) . The Frobenius homomorphism is an isomorphism on a finite field (Corollary 4.18) and restricts to the identity on \( {\mathbb{F}}_{p} \) (Exercise 4.11), so \( \var... | Yes |
If \( n = p \) is prime, then every nonidentity element of \( {\mu }_{p} \cong {C}_{p} \) is a generator: every \( p \) -th root of 1 is primitive except 1 itself. | Therefore\n\n\[ \n{\Phi }_{p}\left( x\right) = \frac{{x}^{p} - 1}{x - 1} = {x}^{p - 1} + \cdots + 1 \n\] \n\nis the particular case encountered in Example V.5.19, where we proved that \( {\Phi }_{p}\left( x\right) \) is indeed irreducible. | No |
Lemma 5.11. For all positive integers \( n \) , \n\n\[ \n{x}^{n} - 1 = \mathop{\prod }\limits_{{1 \leq d \mid n}}{\Phi }_{d}\left( x\right) \n\] | Proof. If \( n = {de} \), then every \( d \) -th root \( \zeta \) of 1 is an \( n \) -th root of 1, because \( {\zeta }^{n} = \) \( {\zeta }^{de} = {\left( {\zeta }^{d}\right) }^{e} = 1 \) . In particular, every primitive \( d \) -th root \( \zeta \) of 1 is an \( n \) -th root of 1.\n\nOn the other hand, every \( \zet... | Yes |
Corollary 5.12. The cyclotomic polynomials \( {\Phi }_{n}\left( x\right) \) have integer coefficients. | Proof. Use induction on \( n \) . Note that \( {\Phi }_{1}\left( x\right) = x - 1 \), and assume we have shown that all \( {\Phi }_{m}\left( x\right) \) have integer coefficients for \( m < n \) . In particular, \( f\left( x\right) \mathrel{\text{:=}} \) \( \mathop{\prod }\limits_{{1 \leq d|n, d < n}}{\Phi }_{d}\left( ... | Yes |
The reader can spend some quality time computing explicitly the cyclotomic polynomials \( {\Phi }_{n}\left( x\right) \) for several nonprime numbers \( n \), working inductively and capitalizing on the fact that we know explicitly \( {\Phi }_{p}\left( x\right) \) for prime \( p \) . | For example, \( {x}^{4} - 1 = {\Phi }_{1}\left( x\right) {\Phi }_{2}\left( x\right) {\Phi }_{4}\left( x\right) \) ; therefore\n\n\[ \n{\Phi }_{4}\left( x\right) = \frac{{x}^{4} - 1}{{x}^{2} - 1} = {x}^{2} + 1 \n\]\n\nSince \( {x}^{6} - 1 = {\Phi }_{1}\left( x\right) {\Phi }_{2}\left( x\right) {\Phi }_{3}\left( x\right)... | Yes |
Proposition 5.14. For all positive \( n,{\Phi }_{n}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) is irreducible over \( \mathbb{Q} \) . | Proof. Arguing by contradiction, assume \( {\Phi }_{n}\left( x\right) \) is reducible. Then its roots \( {\zeta }_{n}^{m} \) , with \( \left( {m, n}\right) = 1 \), are divided among the factors; we can choose a root \( {\zeta }_{n}^{m} \) of one irreducible monic factor \( f\left( x\right) \), such that another root \(... | Yes |
Proposition 5.16. \( {\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) }\right) \) is isomorphic to the group of units in \( \mathbb{Z}/n\mathbb{Z} \) . | Proof. We know that \( {\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) }\right) \) has cardinality \( \phi \left( n\right) \) (Corollary 1.7; the roots are distinct since \( \Phi \left( n\right) \) is separable), so all we need to do is exhibit an injective homomorphism\n\n\[ j : {\left( \... | Yes |
Proposition 5.19. Every finite separable extension is simple. | Proof. Arguing inductively as in the proof of Proposition 5.18, we may assume \( F = k\left( {\alpha ,\beta }\right) \), with \( \alpha \) and \( \beta \) separable (and in particular algebraic) over \( k \), and we may assume \( k \) is an infinite field.\n\nConsider the set \( I \) of embeddings \( \iota : F \hookrig... | Yes |
Corollary 5.20. Let \( k \subseteq F \) be a finite, separable extension. Then\n\n\[ \left| {{\operatorname{Aut}}_{k}\left( F\right) }\right| \leq \left\lbrack {F : k}\right\rbrack \]\n\nwith equality if and only if \( k \subseteq F \) is a normal extension. | Proof. Since \( k \subseteq F \) is finite and separable, Proposition 5.19 implies it is simple: \( F = k\left( \alpha \right) \) for some \( \alpha \in F \) . The inequality follows immediately from Corollary 1.7, and equality holds if and only if the minimal polynomial \( f\left( x\right) \) of \( \alpha \) factors i... | Yes |
Example 5.21. By Proposition 5.19, if we want to construct a nonsimple finite extension, we have to use inseparable elements; to produce an example, we jazz up Example 4.11 a little. Consider the field of rational functions \( F = {\mathbb{F}}_{p}\left( {u, v}\right) \) in two variables over \( {\mathbb{F}}_{p} \) and ... | \[ \left\lbrack {F : k}\right\rbrack = {p}^{2}. \] As \( c \) ranges in \( k \), we obtain intermediate fields \[ k \subseteq k\left( {{cu} + v}\right) \subseteq F\text{.} \] If any two choices \( c,{c}^{\prime } \) led to the same intermediate field, then we would deduce that \( k\left( {u, v}\right) = k\left( {{cu} +... | Yes |
Lemma 6.3. The Galois correspondence is inclusion-reversing. Further, for all subgroups \( G \) of \( {\operatorname{Aut}}_{k}\left( F\right) \) and all intermediate fields \( k \subseteq E \subseteq F \) :\n\n- \( E \subseteq {F}^{{\operatorname{Aut}}_{E}\left( F\right) } \) ;\n\n- \( G \subseteq {\operatorname{Aut}}_... | Proof. Exercise 6.1. | No |
Consider the extension \( \mathbb{Q} \subseteq \mathbb{Q}\left( \sqrt[3]{2}\right) \). Since \[ \left\lbrack {\mathbb{Q}\left( \sqrt[3]{2}\right) : \mathbb{Q}}\right\rbrack = 3 \] is prime, the only intermediate fields are \( \mathbb{Q} \) and \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) (by Corollary 1.11). Concerning \(... | Thus, in this example the Galois correspondence acts between a set with two elements and a singleton: \[ \{ \mathbb{Q},\mathbb{Q}\left( \sqrt[3]{2}\right) \} \Leftrightarrow \left\{ {{\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\left( \sqrt[3]{2}\right) }\right) }\right\} = \{ e\} . \] In particular, the function... | Yes |
Proposition 6.5. Let \( k \subseteq F \) be a finite extension, and let \( G \) be a subgroup of \( {\operatorname{Aut}}_{k}\left( F\right) \) . Then \( \left| G\right| = \left\lbrack {F : {F}^{G}}\right\rbrack \), and | \[ G = {\operatorname{Aut}}_{{F}^{G}}\left( F\right) \] | No |
Lemma 6.6. Let \( k \subseteq F \) be a finite extension, and let \( G \) be a subgroup of \( {\operatorname{Aut}}_{k}\left( F\right) \) . Then \( {F}^{G} \subseteq F \) is a finite, simple, normal, separable extension. | Proof of Lemma 6.6. The extension \( {F}^{G} \subseteq F \) is finite because \( k \subseteq F \) is finite.\n\nLet \( \alpha \in F \) ; by the remark following the statement of the lemma, \( \alpha \) is a root of a separable polynomial \( {q}_{\alpha }\left( t\right) \) with coefficients in \( {F}^{G} \) . It follows... | Yes |
Theorem 6.9. Let \( k \subseteq F \) be a finite field extension. Then the following are equivalent:\n\n(1) \( F \) is the splitting field of a separable polynomial \( f\left( t\right) \in k\left\lbrack t\right\rbrack \) over \( k \) ;\n\n(2) \( k \subseteq F \) is normal and separable;\n\n(3) \( \left| {{\operatorname... | Proof. Most of the needed implications have been proven along the way.\n\n\( \left( 1\right) \Leftrightarrow \left( 2\right) \) by Theorem \( {4.8};\left( 2\right) \Leftrightarrow \left( 3\right) \) by Corollary \( {5.20}. \) (3) \( \Leftrightarrow \left( 4\right) \) follows from Proposition 6.5, applied to the extensi... | Yes |
Theorem 6.12. Let \( k \subseteq F \) be a Galois extension. The Galois correspondence is an inclusion-reversing isomorphism of the lattice of intermediate subfields of \( k \subseteq F \) with the lattice of subgroups of \( {\operatorname{Aut}}_{k}\left( F\right) \) . That is (with notation as in Lemma 6.3), if \( {E}... | Proof. This follows immediately from Theorem 6.9 and Lemma 6.3, which gives \[ {\operatorname{Aut}}_{{E}_{1}{E}_{2}}\left( F\right) = {G}_{1} \cap {G}_{2},\;{F}^{\left\langle {G}_{1},{G}_{2}\right\rangle } = {E}_{1} \cap {E}_{2} \] as needed. | Yes |
For finite fields, this coincidence of lattices was essentially proven ’by hand’ in Corollary 5.4 and Proposition 5.8. For example, the extension \( {\mathbb{F}}_{2} \subseteq {\mathbb{F}}_{64} \) is Galois, with cyclic Galois group \( {C}_{6} \), generated by the Frobenius automorphism \( \varphi \). The lattices are |   | Yes |
The extension \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) = \mathbb{Q}\left( {\sqrt{2} + \sqrt{3}}\right) \) studied in Example 1.19 is the splitting field of the polynomial \( {t}^{4} - {10}{t}^{2} + 1 \), so it is Galois. | We found that its Galois group is \( \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \) ; the lattice of this group has no mysteries for us: and therefore the lattice of intermediate fields is just as transparent: (The intermediate fields are determined by recalling the generators of the corresponding subgroups, a... | Yes |
Proposition 6.17. Suppose \( k \subseteq F \) is a Galois extension and \( k \subseteq K \) is any finite extension. Then \( K \subseteq {KF} \) is a Galois extension, and \( {\operatorname{Aut}}_{K}\left( {KF}\right) \cong {\operatorname{Aut}}_{F \cap K}\left( F\right) \) . | Proof. As \( k \subseteq F \) is Galois, it is the splitting field of a separable polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \subseteq K\left\lbrack x\right\rbrack \) . The roots of \( f\left( x\right) \) generate \( F \) over \( k \), so they generate \( {KF} \) over \( K \) ; in other words, \( {... | Yes |
We have studied cyclotomic fields \( \mathbb{Q}\left( {\zeta }_{n}\right) \) as extensions of \( \mathbb{Q};\mathbb{Q}\left( {\zeta }_{n}\right) \) is the splitting field of \( {x}^{n} - 1 \), so these extensions are Galois; we have proved (Proposition 5.16) that \( {\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\l... | Now let \( k \) be any field of characteristic zero. The splitting field of \( {x}^{n} - 1 \) over \( k \) is the composite \( k\left( \zeta \right) \) of \( k \) and \( \mathbb{Q}\left( \zeta \right) \) . By Proposition 6.17 the extension \( k \subseteq k\left( \zeta \right) \) is Galois, and \( {\operatorname{Aut}}_{... | Yes |
Proposition 6.19. Let \( k \subseteq F \) be an extension of degree \( m \) . Assume that \( k \) contains a primitive \( m \) -th root of 1 and \( \operatorname{char}k \) does not divide \( {}^{24}m \) . Then \( k \subseteq F \) is Galois and cyclic if and only if \( F = k\left( \delta \right) \), with \( {\delta }^{m... | Proof. Let \( \zeta \in k \) be a primitive \( m \) -th root of 1 . First assume that \( F = k\left( \delta \right) \), with \( {\delta }^{m} = c \in k \) . Then all \( m \) roots of the polynomial \( {x}^{m} - c \) , \[ \delta ,{\zeta \delta },{\zeta }^{2}\delta ,\cdots ,{\zeta }^{m - 1}\delta \] are in \( F \), and \... | Yes |
Theorem 7.1. \( \mathbb{C} \) is algebraically closed. | Proof. Let \( f\left( x\right) \in \mathbb{C}\left\lbrack x\right\rbrack \) be a nonconstant polynomial; we have to prove that \( f\left( x\right) \) has roots in \( \mathbb{C} \). Note that if \( f\left( x\right) \) has no roots in \( \mathbb{C} \), then neither does \( f\left( x\right) \overline{f\left( x\right) } \i... | No |
Proposition 7.2. Let \( k \subseteq F \) be a Galois extension, and assume \( \left\lbrack {F : k}\right\rbrack = {p}^{r} \) for some prime \( p \) and \( r \geq 0 \) . Then there exist intermediate fields\n\n\[ k = {E}_{0} \subseteq {E}_{1} \subseteq {E}_{2} \subseteq \cdots \subseteq {E}_{r} = F \]\n\nsuch that \( \l... | Proof. As the Galois correspondence is bijective for Galois extensions (Theorem 6.9, part (5)), this statement follows immediately from the fact that a group of order \( {p}^{r} \), with \( p \) prime, has a complete series of \( p \) -subgroups; cf. for example the discussion following the statement of Theorem IV.2.8. | No |
Theorem 7.3. The regular n-gon is constructible by straightedge and compass if and only if \( \phi \left( n\right) \) is a power of 2 . | Proof. As recalled above, we have already established the \( \Rightarrow \) direction.\n\nFor the converse, assume \( \phi \left( n\right) = {2}^{r} \) for some \( r \) . The extension \( \mathbb{Q} \subseteq \mathbb{Q}\left( {\zeta }_{n}\right) \) is Galois (it is the splitting field of \( {\Phi }_{n}\left( x\right) \... | No |
Theorem 7.4 (Fundamental theorem on symmetric functions). Let \( K \) be a field, and let \( \varphi \in K\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) . Then \( \varphi \) is symmetric if and only if it is a rational function (with coefficients in \( K \) ) of the elementary symmetric functions \( {s}_{1},\ldots ,{s}_{n}... | Proof. Let \( F = K\left( {{t}_{1},\ldots ,{t}_{n}}\right) \), and let \( k = K\left( {{s}_{1},\ldots ,{s}_{n}}\right) \) be the subfield generated by the elementary symmetric functions over \( K \) . Then \( F \) is a splitting field of the separable polynomial \( {P}_{n}\left( x\right) \) over \( k \) . In particular... | Yes |
Corollary 7.6. Let \( G \) be a finite group. Then there exists a Galois extension \( k \subseteq F \) such that \( {\operatorname{Aut}}_{k}\left( F\right) \cong G \) . | ## Proof. Exercise 7.4. | No |
Lemma 7.10. Every separable radical extension is contained in a Galois radical extension. | Proof. Let \( k \subseteq F \) be a separable radical extension. In particular \( k \subseteq F \) is finite and separable, so \( F = k\left( \alpha \right) \) for some \( \alpha \in F \) (Proposition 5.19). Let \( p\left( x\right) \) be the minimal polynomial of \( \alpha \) over \( k \) . The splitting field \( L \) ... | Yes |
Lemma 7.11. Let \( k \) be a field of characteristic 0, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be an irreducible polynomial. Then there exists a formula solving \( f\left( x\right) \) by radicals if and only if the splitting field of \( f\left( x\right) \) is contained in a Galois radical exten... | Proof. If the splitting field of \( f\left( x\right) \) is contained in a radical extension (Galois or not), then the roots may be written as combinations of field operations and radicals, as needed.\n\nFor the converse, assume \( f\left( x\right) \) is solvable by radical. A formula for a root \( {t}_{1} \) can be tur... | Yes |
Lemma 7.13. Let \( k \subseteq F \) be a Galois extension, with \( \operatorname{char}k = 0 \) . Provided it has enough roots of \( 1, k \subseteq F \) is radical if and only if it is solvable. | Proof. Modulo the fundamental theorem of Galois theory, this is a straightforward generalization of Proposition 6.19.\n\nIndeed, assume that \( k \subseteq F \) is radical:\n\n\[ k \subseteq k\left( {\delta }_{1}\right) \subseteq \cdots \subseteq k\left( {{\delta }_{1},\ldots ,{\delta }_{r}}\right) = F \]\n\nwith \( {\... | Yes |
Corollary 7.16. Let \( k \) be a field of characteristic 0, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be an irreducible polynomial. Then \( f\left( x\right) \) is solvable by radicals if and only if its Galois group is solvable. | Proof. This is an immediate consequence of Lemma 7.11 and Proposition 7.14. | No |
Example 7.19. Lemma 7.18 and a discriminant computation are all that is needed to compute the Galois group of an irreducible cubic polynomial \[ f\left( x\right) = {x}^{3} + a{x}^{2} + {bx} + c. \] | It would be futile to try to remember the discriminant \[ D = {a}^{2}{b}^{2} - 4{a}^{3}c - 4{b}^{3} + {18abc} - {27}{c}^{2} \] but one may remember the trick of shifting \( x \) by \( a/3 \) (in characteristic \( \neq 3 \) ), with the effect of killing the coefficient of \( {x}^{2} \) : \[ f\left( {x - \frac{a}{3}}\rig... | Yes |
Let \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) be an irreducible polynomial of degree \( p \), where \( p \) is prime. Assume that \( f\left( x\right) \) has \( p - 2 \) real roots and 2 nonreal, complex roots. Then the Galois group of \( f\left( x\right) \) is \( {S}_{p} \). | Indeed, complex conjugation induces an automorphism of the splitting field and acts by interchanging the two nonreal roots, so the Galois group \( G \), as a subgroup of \( {S}_{p} \), contains a transposition. On the other hand, the degree of the splitting field (and hence \( \left| G\right| \) ) is divisible by \( p ... | Yes |
The operation \( {\operatorname{Aut}}_{k}\left( \_ \right) \) from Galois field extensions to groups is contravariantly functorial. | Indeed, if \( k \subseteq E \subseteq F \) is viewed as a morphism of two Galois extensions \( \left( {k \subseteq E\text{to}k \subseteq F}\right) \), we have a corresponding group homomorphism\n\n\[ \n{\operatorname{Aut}}_{k}\left( F\right) \rightarrow {\operatorname{Aut}}_{k}\left( E\right) \n\]\n\ndefined by restric... | Yes |
In §III.4.3 we have defined the spectrum of a commutative ring \( R \) , Spec \( R \), as the set of prime ideals of \( R \) . If \( R, S \) are commutative rings and \( \varphi : R \rightarrow S \) is a ring homomorphism, then the inverse image \( {\varphi }^{-1}\left( \mathfrak{p}\right) \) of a prime ideal \( \mathf... | \[ {\varphi }^{ * } : \operatorname{Spec}\left( S\right) \rightarrow \operatorname{Spec}\left( R\right) \] This assignment is clearly functorial, so we can view Spec as a contravariant functor from the category of commutative rings to Set. | Yes |
Construct a category by taking the objects to be nonnegative integers and \( \operatorname{Hom}\left( {m, n}\right) \) to be the set of \( n \times m \) matrices with entries in a field \( k \), with composition defined by product of matrices (and suitable care concerning matrices with no rows or columns). | The resulting category is equivalent to the category of finite-dimensional \( k \) -vector spaces. Indeed, we obtain a functor from the former to the latter by sending \( n \) to the vector space \( {k}^{n} \) (endowed with the standard basis) and each matrix to the corresponding linear map. This functor is clearly ful... | Yes |
Recall (Definition VII.2.19) that the coordinate ring of an affine algebraic set over a field \( K \) is a reduced, commutative, finite-type \( K \) -algebra. We can define the category \( K \) -Aff of affine \( K \) -algebraic sets by prescribing that the objects be algebraic subsets of some affine \( K \) -space and ... | Thus, \( K \) -Aff is defined in such a way that the functor \( K \) -Aff \( {}^{ \circ } \rightarrow K \) -Alg that maps an affine algebraic set \( S \) to its coordinate ring \( K\left\lbrack S\right\rbrack \) is an equivalence of the opposite category of \( K \) -Aff with the subcategory of reduced, commutative, fin... | Yes |
Example 1.10 (Products). Let I be the 'discrete category' consisting of two objects \( \mathbf{1},\mathbf{2} \), with only identity morphisms \( {}^{6} \), and let \( \mathcal{A} \) be a functor from 1 to any category \( \mathrm{C} \) ; let \( {A}_{1} = \mathcal{A}\left( \mathbf{1}\right) ,{A}_{2} = \mathcal{A}\left( \... | We can similarly define the product of any (possibly infinite) family of objects in a category as the limit over the corresponding discrete indexing category, provided of course that this limit exists. | No |
Claim 1.13. The limit \( \underset{i}{\mathop{\lim }\limits^{⏜}}{A}_{i} \) exists in \( R \) -Mod. | Proof. The product \( \mathop{\prod }\limits_{i}{A}_{i} \) consists of arbitrary sequences \( {\left( {a}_{i}\right) }_{i > 0} \) of elements \( {a}_{i} \in \) \( {A}_{i} \) . Say that a sequence \( {\left( {a}_{i}\right) }_{i > 0} \) is coherent if for all \( i > 0 \) we have \( {a}_{i} = {\varphi }_{i, i + 1}\left( {... | No |
If \( \mathrm{C} = \) Set and all the \( {\psi }_{ij} \) are injective, we are talking about a 'nested sequence of sets':\n\n\[ \n{A}_{1} \subseteq {A}_{2} \subseteq {A}_{3} \subseteq {A}_{4} \subseteq \cdots \n\] \n\nthe direct limit of this sequence would be the ’infinite union’ \( \mathop{\bigcup }\limits_{i}{A}_{i}... | More formally, \( \mathop{\bigcup }\limits_{i}{A}_{i} \) consists of equivalence classes of pairs \( \left( {i,{a}_{i}}\right) \), where \( {a}_{i} \in {A}_{i} \) and \( \left( {i,{a}_{i}}\right) \) is equivalent to \( \left( {j,{a}_{j}}\right) \) for \( i \leq j \) if \( {a}_{j} = {\psi }_{ij}\left( {a}_{i}\right) . \... | Yes |
For all \( R \) -modules \( N, R{ \otimes }_{R}N \cong N \) . | Indeed, every \( R \) -bilinear \( R \times N \rightarrow P \) factors through \( N \) (as is immediately verified):\n\n\n\nwhere \( \otimes \left( {r, n}\right) = {rn} \) . By the uniqueness property of universal ob... | No |
Lemma 2.4. For all \( R \) -modules \( M, N, P \), there is an isomorphism of \( R \) -modules\n\n\[{\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{R}\left( {N, P}\right) }\right) \cong {\operatorname{Hom}}_{R}\left( {M{ \otimes }_{R}N, P}\right) . | Proof. As noted before the statement, every \( \alpha \in {\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{R}\left( {N, P}\right) }\right) \) determines an \( R \) -bilinear map \( \varphi : M \times N \rightarrow P \), by\n\n\[ \left( {m, n}\right) \mapsto \alpha \left( m\right) \left( n\right) .\n\nBy the univ... | No |
For every \( R \) -module \( N \), the functor \( \_ { \otimes }_{R}N \) is left-adjoint to the functor \( {\operatorname{Hom}}_{R}\left( {N,\_ }\right) \) . | Proof. The claim is that the isomorphism found in Lemma 2.4 is natural in the sense hinted at, but not fully explained, in \( §{1.5} \) ; the interested reader should have no problems checking this naturality. | No |
Corollary 2.7. For any two sets \( A, B \) :\n\n\[ {R}^{\oplus A}{ \otimes }_{R}{R}^{\oplus B} \cong {R}^{\oplus A \times B}. \] | Indeed, 'distributing' the direct sum identifies the left-hand side with the direct sum \( {\left( {R}^{\oplus A}\right) }^{\oplus B} \), which is isomorphic to the right-hand side (Exercise III.6.5). For finitely generated free modules, this simply says that \( {R}^{\oplus m} \otimes {R}^{\oplus n} \cong {R}^{\oplus {... | No |
Corollary 2.8. For all \( R \) -modules \( N \) and all ideals \( I \) of \( R \) ,\n\n\[ \frac{R}{I}{ \otimes }_{R}N \cong \frac{N}{IN} \] | Indeed, \( {}_{ - }{ \otimes }_{R}N \) is right-exact; thus, the exact sequence\n\n\[ 0 \rightarrow I \rightarrow R \rightarrow \frac{R}{I} \rightarrow 0 \]\n\ninduces an exact sequence\n\n\[ I{ \otimes }_{R}N \rightarrow R{ \otimes }_{R}N \rightarrow \frac{R}{I}{ \otimes }_{R}N \rightarrow 0. \]\n\nThe image of \( I{ ... | Yes |
Corollary 2.9. For all ideals \( I, J \) of \( R \) ,\n\n\[ \frac{R}{I}{ \otimes }_{R}\frac{R}{J} \cong \frac{R}{I + J} \] | This follows immediately from Corollary 2.8 and the 'third isomorphism theorem', Proposition III.5.17. Indeed, \( {IR}/J = \left( {I + J}\right) /J \) . | Yes |
Example 2.10. \( \mathbb{Z}/m\mathbb{Z}{ \otimes }_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/\gcd \left( {m, n}\right) \mathbb{Z} \) . | Indeed, \( \left( m\right) + \left( n\right) = \left( {\gcd \left( {m, n}\right) }\right) \) in \( \mathbb{Z} \) . For instance, \[ \frac{\mathbb{Z}}{2\mathbb{Z}}{ \otimes }_{\mathbb{Z}}\frac{\mathbb{Z}}{3\mathbb{Z}} \cong 0 \] | Yes |
Multiplication by 2 gives an inclusion\n\n\[ \n{\mathbb{Z}}^{- \cdot 2} \rightarrow \mathbb{Z} \n\]\n\nidentifying the first copy of \( \mathbb{Z} \) with the ideal (2) in the second copy. Tensoring by \( \mathbb{Z}/2\mathbb{Z} \) over \( \mathbb{Z} \) (and keeping in mind that \( R{ \otimes }_{R}N \cong N \) ), we get... | which sends both [0] and [1] to zero. This is the zero-morphism, and in particular it is not injective. | Yes |
Consider the affine algebraic set \( \mathcal{V}\left( {xy}\right) \) in the plane \( {\mathbb{A}}^{2} \) (over a fixed field \( k \) ) and the ’projection on the first coordinate’ \( \mathcal{V}\left( {xy}\right) \rightarrow {\mathbb{A}}^{1},\left( {x, y}\right) \mapsto x \) : | In terms of coordinate rings (cf. §VII.2.3), this map corresponds to the homomorphism of \( k \) -algebras:\n\n\[ k\left\lbrack x\right\rbrack \rightarrow \frac{k\left\lbrack {x, y}\right\rbrack }{\left( xy\right) } \]\n\ndefined by mapping \( x \) to the coset \( x + \left( {xy}\right) \) (this will be completely clea... | No |
Lemma 3.2. Let \( R \) be a commutative ring; let \( M, N \) be \( R \) -modules, and let \( G \) be an abelian group. Then every \( \mathbb{Z} \) -bilinear, \( R \) -balanced map \( \varphi : M \times N \rightarrow G \) factors through \( M{ \otimes }_{R}N \) ; that is, there exists a unique group homomorphism \( \ove... | The universal property explored in \( §{2.1} \) is recovered as the statement that if \( G \) is an \( R \) -module and \( \varphi \) is \( R \) -bilinear, then the induced group homomorphism \( M{ \otimes }_{R}N \rightarrow G \) is in fact an \( R \) -linear map. | No |
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