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Proposition 12. The following are equivalent:\n\n(1) \( \lambda \) is an eigenvalue of \( T \)\n\n(2) \( {\lambda I} - T \) is a singular linear transformation of \( V \)\n\n(3) \( \det \left( {{\lambda I} - T}\right) = 0 \). | Proof: Since \( \lambda \) is an eigenvalue of \( T \) with corresponding eigenvector \( v \) if and only if \( v \) is a nonzero vector in the kernel of \( {\lambda I} - T \), it follows that (1) and (2) are equivalent.\n\n(2) and (3) are equivalent by our results on determinants. | Yes |
Theorem 15. Let \( S \) and \( T \) be linear transformations of \( V \) . Then the following are equivalent:\n\n(1) \( S \) and \( T \) are similar linear transformations\n\n(2) the \( F\left\lbrack x\right\rbrack \) -modules obtained from \( V \) via \( S \) and via \( T \) are isomorphic \( F\left\lbrack x\right\rbr... | Proof: [(1) implies (2)] Assume there is a nonsingular linear transformation \( U \) such that \( S = {UT}{U}^{-1} \) . The vector space isomorphism \( U : V \rightarrow V \) is also an \( F\left\lbrack x\right\rbrack \) -module homomorphism, where \( x \) acts on the first \( V \) via \( T \) and on the second via \( ... | Yes |
Corollary 18. Let \( A \) and \( B \) be two \( n \times n \) matrices over a field \( F \) and suppose \( F \) is a subfield of the field \( K \). (1) The rational canonical form of \( A \) is the same whether it is computed over \( K \) or over \( F \). The minimal and characteristic polynomials and the invariant fac... | Proof: (1) Let \( M \) be the rational canonical form of \( A \) when computed over the smaller field \( F \). Since \( M \) satisfies the conditions in the definition of the rational canonical form over \( K \), the uniqueness of the rational canonical form implies that \( M \) is also the rational canonical form of \... | Yes |
Lemma 19. Let \( a\left( x\right) \in F\left\lbrack x\right\rbrack \) be any monic polynomial.\n\n(1) The characteristic polynomial of the companion matrix of \( a\left( x\right) \) is \( a\left( x\right) \).\n\n(2) If \( M \) is the block diagonal matrix\n\n\[ M = \left( \begin{matrix} {A}_{1} & 0 & \ldots & 0 \\ 0 & ... | Proof: These are both straightforward exercises. | No |
Proposition 20. Let \( A \) be an \( n \times n \) matrix over the field \( F \) .\n\n(1) The characteristic polynomial of \( A \) is the product of all the invariant factors of A.\n\n(2) (The Cayley-Hamilton Theorem) The minimal polynomial of \( A \) divides the characteristic polynomial of \( A \) .\n\n(3) The charac... | Proof: Let \( B \) be the rational canonical form of \( A \) . By the previous lemma the block diagonal form of \( B \) shows that the characteristic polynomial of \( B \) is the product of the characteristic polynomials of the companion matrices of the invariant factors of \( A \) . By the first part of the lemma abov... | Yes |
Theorem 21. Let \( A \) be an \( n \times n \) matrix over the field \( F \) . Using the three elementary row and column operations above, the \( n \times n \) matrix \( {xI} - A \) with entries from \( F\left\lbrack x\right\rbrack \) can be put into the diagonal form (called the Smith Normal Form for \( A \) ) \[ \lef... | Proof: cf. the exercises. Sec. 12.2 The Rational Canonical Form | No |
Corollary 25. If \( A \) is an \( n \times n \) matrix with entries from \( F \) and \( F \) contains all the eigenvalues of \( A \), then \( A \) is similar to a diagonal matrix over \( F \) if and only if the minimal polynomial of \( A \) has no repeated roots. | Proof: Suppose \( A \) is similar to a diagonal matrix. The minimal polynomial of a diagonal matrix has no repeated roots (its roots are precisely the distinct elements along the diagonal). Since similar matrices have the same minimal polynomial it follows that the minimal polynomial for \( A \) has no repeated roots.\... | Yes |
Proposition 1. The characteristic of a field \( F,\operatorname{ch}\left( F\right) \), is either 0 or a prime \( p \) . If \( \operatorname{ch}\left( F\right) = p \) then for any \( \alpha \in F \) , | \[ p \cdot \alpha = \underset{p\text{ times }}{\underbrace{\alpha + \alpha + \cdots + \alpha }} = 0. \] Proof: Only the second statement has not been proved, and this follows immediately from the evident equality \( p \cdot \alpha = p \cdot \left( {{1}_{F}\alpha }\right) = \left( {p \cdot {1}_{F}}\right) \left( \alpha ... | No |
Theorem 3. Let \( F \) be a field and let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial. Then there exists a field \( K \) containing an isomorphic copy of \( F \) in which \( p\left( x\right) \) has a root. Identifying \( F \) with this isomorphic copy shows that there exists an ... | Proof: Consider the quotient\n\n\[ K = F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \]\n\nof the polynomial ring \( F\left\lbrack x\right\rbrack \) by the ideal generated by \( p\left( x\right) \). Since by assumption \( p\left( x\right) \) is an irreducible polynomial in the P.I.D. \( F\left\lbrack ... | Yes |
Theorem 4. Let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial of degree \( n \) over the field \( F \) and let \( K \) be the field \( F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \) . Let \( \theta = x{\;\operatorname{mod}\;\left( {p\left( x\right) }\right) } \i... | Proof: Let \( a\left( x\right) \in F\left\lbrack x\right\rbrack \) be any polynomial with coefficients in \( F \) . Since \( F\left\lbrack x\right\rbrack \) is a Euclidean Domain (this is Theorem 3 of Chapter 9), we may divide \( a\left( x\right) \) by \( p\left( x\right) \) :\n\n\[ a\left( x\right) = q\left( x\right) ... | Yes |
Corollary 5. Let \( K \) be as in Theorem 4, and let \( a\left( \theta \right), b\left( \theta \right) \in K \) be two polynomials of degree \( < n \) in \( \theta \) . Then addition in \( K \) is defined simply by usual polynomial addition and multiplication in \( K \) is defined by\n\n\[ a\left( \theta \right) b\left... | By the results proved above, this definition of addition and multiplication on the polynomials of degree \( < n \) in \( \theta \) make \( K \) into a field, so that one can also divide by nonzero elements as well, which is not so immediately obvious from the definitions of the operations. | Yes |
Theorem 6. Let \( F \) be a field and let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial. Suppose \( K \) is an extension field of \( F \) containing a root \( \alpha \) of \( p\left( x\right) : p\left( \alpha \right) = 0 \) . Let \( F\left( \alpha \right) \) denote the subfield of... | Proof: There is a natural homomorphism\n\n\[ \varphi : F\left\lbrack x\right\rbrack \rightarrow F\left( \alpha \right) \subseteq K \]\n\n\[ a\left( x\right) \mapsto a\left( \alpha \right) \]\n\nobtained by mapping \( F \) to \( F \) by the identity map and sending \( x \) to \( \alpha \) and then extending so that the ... | Yes |
Theorem 8. Let \( \varphi : F\overset{ \sim }{ \rightarrow }{F}^{\prime } \) be an isomorphism of fields. Let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial and let \( {p}^{\prime }\left( x\right) \in {F}^{\prime }\left\lbrack x\right\rbrack \) be the irreducible polynomial obtaine... | Proof: As noted above, the isomorphism \( \varphi \) induces a natural isomorphism from \( F\left\lbrack x\right\rbrack \) to \( {F}^{\prime }\left\lbrack x\right\rbrack \) which maps the maximal ideal \( \left( {p\left( x\right) }\right) \) to the maximal ideal \( \left( {{p}^{\prime }\left( x\right) }\right) \) . Tak... | Yes |
Proposition 9. Let \( \alpha \) be algebraic over \( F \) . Then there is a unique monic irreducible polynomial \( {m}_{\alpha, F}\left( x\right) \in F\left\lbrack x\right\rbrack \) which has \( \alpha \) as a root. A polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) has \( \alpha \) as a root if and o... | Proof: Let \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) be a polynomial of minimal degree having \( \alpha \) as a root. Multiplying \( g\left( x\right) \) by a constant, we may assume \( g\left( x\right) \) is monic. Suppose \( g\left( x\right) \) were reducible in \( F\left\lbrack x\right\rbrack \), say \(... | Yes |
Corollary 10. If \( L/F \) is an extension of fields and \( \alpha \) is algebraic over both \( F \) and \( L \) , then \( {m}_{\alpha, L}\left( x\right) \) divides \( {m}_{\alpha, F}\left( x\right) \) in \( L\left\lbrack x\right\rbrack \) . | Proof: This is immediate from the second statement in Proposition 9 applied to \( L \) , since \( {m}_{\alpha, F}\left( x\right) \) is a polynomial in \( L\left\lbrack x\right\rbrack \) having \( \alpha \) as a root. | Yes |
Proposition 11. Let \( \alpha \) be algebraic over the field \( F \) and let \( F\left( \alpha \right) \) be the field generated by \( \alpha \) over \( F \) . Then\n\n\[ F\left( \alpha \right) \cong F\left\lbrack x\right\rbrack /\left( {{m}_{\alpha }\left( x\right) }\right) \]\n\nso that in particular\n\n\[ \left\lbra... | Proof: This follows immediately from Theorem 6. * | No |
The element \( \alpha \) is algebraic over \( F \) if and only if the simple extension \( F\left( \alpha \right) /F \) is finite. More precisely, if \( \alpha \) is an element of an extension of degree \( n \) over \( F \) then \( \alpha \) satisfies a polynomial of degree at most \( n \) over \( F \) and if \( \alpha ... | Proof: If \( \alpha \) is algebraic over \( F \), then the degree of the extension \( F\left( \alpha \right) /F \) is the degree of the minimal polynomial for \( \alpha \) over \( F \) . Hence the extension is finite, of degree \( \leq n \) if \( \alpha \) satisfies a polynomial of degree \( n \) . Conversely, suppose ... | Yes |
Corollary 13. If the extension \( K/F \) is finite, then it is algebraic. | Proof: If \( \alpha \in K \), then the subfield \( F\left( \alpha \right) \) is in particular a subspace of the vector space \( K \) over \( F \) . Hence \( \left\lbrack {F\left( \alpha \right) : F}\right\rbrack \leq \left\lbrack {K : F}\right\rbrack \) and so \( \alpha \) is algebraic over \( F \) by the proposition. | Yes |
Corollary 15. Suppose \( L/F \) is a finite extension and let \( K \) be any subfield of \( L \) containing \( F, F \subseteq K \subseteq L \) . Then \( \left\lbrack {K : F}\right\rbrack \) divides \( \left\lbrack {L : F}\right\rbrack \) . | Proof: This is immediate. | No |
Lemma 16. \( F\left( {\alpha ,\beta }\right) = \left( {F\left( \alpha \right) }\right) \left( \beta \right) \), i.e., the field generated over \( F \) by \( \alpha \) and \( \beta \) is the field generated by \( \beta \) over the field \( F\left( \alpha \right) \) generated by \( \alpha \) . | Proof: This follows by the minimality of the fields in question. The field \( F\left( {\alpha ,\beta }\right) \) contains \( F \) and \( \alpha \), hence contains the field \( F\left( \alpha \right) \), and since it also contains \( \beta \), we have the inclusion \( \left( {F\left( \alpha \right) }\right) \left( \beta... | Yes |
Theorem 17. The extension \( K/F \) is finite if and only if \( K \) is generated by a finite number of algebraic elements over \( F \) . More precisely, a field generated over \( F \) by a finite number of algebraic elements of degrees \( {n}_{1},{n}_{2},\ldots ,{n}_{k} \) is algebraic of degree \( \leq {n}_{1}{n}_{2}... | Proof: If \( K/F \) is finite of degree \( n \), let \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n} \) be a basis for \( K \) as a vector space over \( F \) . By Corollary 15, \( \left\lbrack {F\left( {\alpha }_{i}\right) : F}\right\rbrack \) divides \( \left\lbrack {K : F}\right\rbrack = n \) for \( i = 1,2,\ldo... | Yes |
Corollary 18. Suppose \( \alpha \) and \( \beta \) are algebraic over \( F \) . Then \( \alpha \pm \beta ,{\alpha \beta },\alpha /\beta \) (for \( \beta \neq 0 \) ), (in particular \( {\alpha }^{-1} \) for \( \alpha \neq 0 \) ) are all algebraic. | Proof: All of these elements lie in the extension \( F\left( {\alpha ,\beta }\right) \), which is finite over \( F \) by the theorem, hence they are algebraic by Corollary 13. | Yes |
Corollary 19. Let \( L/F \) be an arbitrary extension. Then the collection of elements of \( L \) that are algebraic over \( F \) form a subfield \( K \) of \( L \) . | Proof: This is immediate from the previous corollary. | No |
Theorem 20. If \( K \) is algebraic over \( F \) and \( L \) is algebraic over \( K \), then \( L \) is algebraic over \( F \) . | Proof: Let \( \alpha \) be any element of \( L \) . Then \( \alpha \) is algebraic over \( K \), so \( \alpha \) satisfies some polynomial equation\n\n\[ \n{a}_{n}{\alpha }^{n} + {a}_{n - 1}{\alpha }^{n - 1} + \cdots + {a}_{1}\alpha + {a}_{0} = 0 \n\]\n\nwhere the coefficients \( {a}_{0},{a}_{1},\ldots ,{a}_{n} \) are ... | Yes |
Proposition 21. Let \( {K}_{1} \) and \( {K}_{2} \) be two finite extensions of a field \( F \) contained in \( K \) . Then\n\n\[ \left\lbrack {{K}_{1}{K}_{2} : F}\right\rbrack \leq \left\lbrack {{K}_{1} : F}\right\rbrack \left\lbrack {{K}_{2} : F}\right\rbrack \]\n\nwith equality if and only if an \( F \) -basis for o... | Proof: From \( {K}_{1}{K}_{2} = F\left( {{\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n},{\beta }_{1},{\beta }_{2},\ldots ,{\beta }_{m}}\right) = {K}_{1}\left( {{\beta }_{1},{\beta }_{2},\ldots ,{\beta }_{m}}\right) , \) we see as above that \( {\beta }_{1},{\beta }_{2},\ldots ,{\beta }_{m} \) span \( {K}_{1}{K}_{2} ... | Yes |
Corollary 22. Suppose that \( \left\lbrack {{K}_{1} : F}\right\rbrack = n,\left\lbrack {{K}_{2} : F}\right\rbrack = m \) in Proposition 21, where \( n \) and \( m \) are relatively prime: \( \left( {n, m}\right) = 1 \) . Then \( \left\lbrack {{K}_{1}{K}_{2} : F}\right\rbrack = \left\lbrack {{K}_{1} : F}\right\rbrack \l... | Proof: In general the extension degree \( \left\lbrack {{K}_{1}{K}_{2} : F}\right\rbrack \) is divisible by both \( n \) and \( m \) since \( {K}_{1} \) and \( {K}_{2} \) are subfields of \( {K}_{1}{K}_{2} \), hence is divisible by their least common multiple. In this case, since \( \left( {n, m}\right) = 1 \), this me... | Yes |
Theorem 24. None of the classical Greek problems: (I) Doubling the Cube, (II) Trisecting an Angle, and (III) Squaring the Circle, is possible. | Proof: (I) Doubling the cube amounts to constructing \( \sqrt[3]{2} \) in the reals starting with the unit 1. Since \( \left\lbrack {\mathbb{Q}\left( \sqrt[3]{2}\right) : \mathbb{Q}}\right\rbrack = 3 \) is not a power of 2, this is impossible. | Yes |
Theorem 25. For any field \( F \), if \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) then there exists an extension \( K \) of \( F \) which is a splitting field for \( f\left( x\right) \) . | Proof: We first show that there is an extension \( E \) of \( F \) over which \( f\left( x\right) \) splits completely into linear factors by induction on the degree \( n \) of \( f\left( x\right) \) . If \( n = 1 \), then take \( E = F \) . Suppose now that \( n > 1 \) . If the irreducible factors of \( f\left( x\righ... | Yes |
Corollary 28. (Uniqueness of Splitting Fields) Any two splitting fields for a polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) over a field \( F \) are isomorphic. | Proof: Take \( \varphi \) to be the identity mapping from \( F \) to itself and \( E \) and \( {E}^{\prime } \) to be two splitting fields for \( f\left( x\right) \left( { = {f}^{\prime }\left( x\right) }\right) \). | Yes |
Proposition 29. Let \( \overline{F} \) be an algebraic closure of \( F. \) Then \( \overline{F} \) is algebraically closed. | Proof: Let \( f\left( x\right) \) be a polynomial in \( \bar{F}\left\lbrack x\right\rbrack \) and let \( \alpha \) be a root of \( f\left( x\right) \) . Then \( \alpha \) generates an algebraic extension \( \bar{F}\left( \alpha \right) \) of \( \bar{F} \), and \( \bar{F} \) is algebraic over \( F \) . By Theorem \( {20... | Yes |
Proposition 31. Let \( K \) be an algebraically closed field and let \( F \) be a subfield of \( K \) . Then the collection of elements \( \bar{F} \) of \( K \) that are algebraic over \( F \) is an algebraic closure of \( F \) . An algebraic closure of \( F \) is unique up to isomorphism. | Proof: By definition, \( \bar{F} \) is an algebraic extension of \( F \) . Every polynomial \( f\left( x\right) \in \) \( F\left\lbrack x\right\rbrack \) splits completely over \( K \) into linear factors \( x - \alpha \) (the same is true for every polynomial even in \( K\left\lbrack x\right\rbrack ) \) . But each \( ... | Yes |
A polynomial \( f\left( x\right) \) has a multiple root \( \alpha \) if and only if \( \alpha \) is also a root of \( {D}_{x}f\left( x\right) \), i.e., \( f\left( x\right) \) and \( {D}_{x}f\left( x\right) \) are both divisible by the minimal polynomial for \( \alpha \) . In particular, \( f\left( x\right) \) is separa... | Proof: Suppose first that \( \alpha \) is a multiple root of \( f\left( x\right) \) . Then over a splitting field,\n\n\[ f\left( x\right) = {\left( x - \alpha \right) }^{n}g\left( x\right) \]\n\nfor some integer \( n \geq 2 \) and some polynomial \( g\left( x\right) \) . Taking derivatives we obtain\n\n\[ {D}_{x}f\left... | Yes |
Every irreducible polynomial over a field of characteristic 0 (for example, \( \mathbb{Q}) \) is separable. A polynomial over such a field is separable if and only if it is the product of distinct irreducible polynomials. | Proof: Suppose \( F \) is a field of characteristic 0 and \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) is irreducible of degree \( n \) . Then the derivative \( {D}_{x}p\left( x\right) \) is a polynomial of degree \( n - 1 \) . Up to constant factors the only factors of \( p\left( x\right) \) in \( F\left\lb... | Yes |
Proposition 35. Let \( F \) be a field of characteristic \( p \) . Then for any \( a, b \in F \) , \[ {\left( a + b\right) }^{p} = {a}^{p} + {b}^{p},\;\text{ and }\;{\left( ab\right) }^{p} = {a}^{p}{b}^{p}. \] Put another way, the \( {p}^{\text{th }} \) -power map defined by \( \varphi \left( a\right) = {a}^{p} \) is a... | Proof: The Binomial Theorem for expanding \( {\left( a + b\right) }^{n} \) for any positive integer \( n \) holds (by the standard induction proof) over any commutative ring: \[ {\left( a + b\right) }^{n} = {a}^{n} + \left( \begin{array}{l} n \\ 1 \end{array}\right) {a}^{n - 1}b + \cdots + \left( \begin{array}{l} n \\ ... | Yes |
Corollary 36. Suppose that \( \mathbb{F} \) is a finite field of characteristic \( p \) . Then every element of \( \mathbb{F} \) is a \( {p}^{\text{th }} \) power in \( \mathbb{F} \) (notationally, \( \mathbb{F} = {\mathbb{F}}^{p} \) ). | Proof: The injectivity of the Frobenius endomorphism of \( \mathbb{F} \) implies that it is also surjective when \( \mathbb{F} \) is finite, which is the statement of the corollary. | Yes |
Every irreducible polynomial over a finite field \( \mathbb{F} \) is separable. | The important part of the proof of this result is the fact that every element in the characteristic \( p \) field \( \mathbb{F} \) was a \( {p}^{\text{th }} \) power in \( \mathbb{F} \) . This suggests the following definition:\n\nDefinition. A field \( K \) of characteristic \( p \) is called perfect if every element ... | No |
Lemma 40. The cyclotomic polynomial \( {\Phi }_{n}\left( x\right) \) is a monic polynomial in \( \mathbb{Z}\left\lbrack x\right\rbrack \) of degree \( \varphi \left( n\right) \) . | Proof: It is clear that \( {\Phi }_{n}\left( x\right) \) is monic and has degree \( \varphi \left( n\right) \) . We must show the coefficients lie in \( \mathbb{Z} \) . We use induction on \( n \) . The result is true for \( n = 1 \) (and \( n \leq {12} \) ). Assume by induction that \( {\Phi }_{d}\left( x\right) \in \... | Yes |
Corollary 42. The degree over \( \mathbb{Q} \) of the cyclotomic field of \( {n}^{\text{th }} \) roots of unity is \( \varphi \left( n\right) \) : | \[ \left\lbrack {\mathbb{Q}\left( {\zeta }_{n}\right) : \mathbb{Q}}\right\rbrack = \varphi \left( n\right) \] Proof: By the theorem, \( {\Phi }_{n}\left( x\right) \) is the minimal polynomial for any primitive \( {n}^{\text{th }} \) root of unity \( {\zeta }_{n} \) . | Yes |
Proposition 1. Aut \( \left( K\right) \) is a group under composition and \( \operatorname{Aut}\left( {K/F}\right) \) is a subgroup. | Proof: It is clear that \( \operatorname{Aut}\left( K\right) \) is a group. If \( \sigma \) and \( \tau \) are automorphisms of \( K \) which fix \( F \) then also \( {\sigma \tau } \) and \( {\sigma }^{-1} \) are the identity on \( F \), which shows that \( \operatorname{Aut}\left( {K/F}\right) \) is a subgroup. | Yes |
Proposition 2. Let \( K/F \) be a field extension and let \( \alpha \in K \) be algebraic over \( F \) . Then for any \( \sigma \in \operatorname{Aut}\left( {K/F}\right) ,{\sigma \alpha } \) is a root of the minimal polynomial for \( \alpha \) over \( F \) i.e., Aut \( \left( {K/F}\right) \) permutes the roots of irred... | Proof: Suppose \( \alpha \) satisfies the equation\n\n\[ \n{\alpha }^{n} + {a}_{n - 1}{\alpha }^{n - 1} + \cdots + {a}_{1}\alpha + {a}_{0} = 0 \n\]\n\nwhere \( {a}_{0},{a}_{1},\ldots ,{a}_{n - 1} \) are elements of \( F \) . Applying the automorphism \( \sigma \) we obtain (using the fact that \( \sigma \) is an additi... | Yes |
Proposition 3. Let \( H \leq \operatorname{Aut}\left( K\right) \) be a subgroup of the group of automorphisms of \( K \) . Then the collection \( F \) of elements of \( K \) fixed by all the elements of \( H \) is a subfield of \( K \) . | Proof: Let \( h \in H \) and let \( a, b \in F \) . Then by definition \( h\left( a\right) = a, h\left( b\right) = b \) so that \( h\left( {a \pm b}\right) = h\left( a\right) \pm h\left( b\right) = a \pm b, h\left( {ab}\right) = h\left( a\right) h\left( b\right) = {ab} \) and \( h\left( {a}^{-1}\right) = h{\left( a\rig... | Yes |
Proposition 4. The association of groups to fields and fields to groups defined above is inclusion reversing, namely\n\n(1) if \( {F}_{1} \subseteq {F}_{2} \subseteq K \) are two subfields of \( K \) then \( \operatorname{Aut}\left( {K/{F}_{2}}\right) \leq \operatorname{Aut}\left( {K/{F}_{1}}\right) \), and\n\n(2) if \... | Proof: Any automorphism of \( K \) that fixes \( {F}_{2} \) also fixes its subfield \( {F}_{1} \), which gives (1). The second assertion is proved similarly. | No |
Theorem 7. (Linear Independence of Characters) If \( {\chi }_{1},{\chi }_{2},\ldots ,{\chi }_{n} \) are distinct characters of \( G \) with values in \( L \) then they are linearly independent over \( L \) . | Proof: Suppose the characters were linearly dependent. Among all the linear dependence relations (2) above, choose one with the minimal number \( m \) of nonzero coefficients \( {a}_{i} \) . We may suppose (by renumbering, if necessary) that the \( m \) nonzero coefficients are \( {a}_{1},{a}_{2},\ldots ,{a}_{m} \) :\n... | Yes |
Corollary 10. Let \( K/F \) be any finite extension. Then\n\n\[ \left| {\operatorname{Aut}\left( {K/F}\right) }\right| \leq \left\lbrack {K : F}\right\rbrack \]\n\nwith equality if and only if \( F \) is the fixed field of \( \operatorname{Aut}\left( {K/F}\right) \) . Put another way, \( K/F \) is Galois if and only if... | Proof: Let \( {F}_{1} \) be the fixed field of \( \operatorname{Aut}\left( {K/F}\right) \), so that\n\n\[ F \subseteq {F}_{1} \subseteq K \]\n\nBy Theorem 9, \( \left\lbrack {K : {F}_{1}}\right\rbrack = \left| {\operatorname{Aut}\left( {K/F}\right) }\right| \) . Hence \( \left\lbrack {K : F}\right\rbrack = \left| {\ope... | Yes |
Let \( G \) be a finite subgroup of automorphisms of a field \( K \) and let \( F \) be the fixed field. Then every automorphism of \( K \) fixing \( F \) is contained in \( G \), i.e., \( \operatorname{Aut}\left( {K/F}\right) = G \), so that \( K/F \) is Galois, with Galois group \( G \). | Proof: By definition \( F \) is fixed by all the elements of \( G \) so we have \( G \leq \operatorname{Aut}\left( {K/F}\right) \) (and the question is whether there are any automorphisms of \( K \) fixing \( F \) not in \( G \) i.e., whether this containment is proper). Hence \( \left| G\right| \leq \left| {\operatorn... | Yes |
Corollary 12. If \( {G}_{1} \neq {G}_{2} \) are distinct finite subgroups of automorphisms of a field \( K \) then their fixed fields are also distinct. | Proof: Suppose \( {F}_{1} \) is the fixed field of \( {G}_{1} \) and \( {F}_{2} \) is the fixed field of \( {G}_{2} \) . If \( {F}_{1} = {F}_{2} \) then by definition \( {F}_{1} \) is fixed by \( {G}_{2} \) . By the previous corollary any automorphism fixing \( {F}_{1} \) is contained in \( {G}_{1} \), hence \( {G}_{2}... | Yes |
Proposition 15. Any finite field is isomorphic to \( {\mathbb{F}}_{{p}^{n}} \) for some prime \( p \) and some integer \( n \geq 1 \). | The field \( {\mathbb{F}}_{{p}^{n}} \) is the splitting field over \( {\mathbb{F}}_{p} \) of the polynomial \( {x}^{{p}^{n}} - x \), with cyclic Galois group of order \( n \) generated by the Frobenius automorphism \( {\sigma }_{p} \). The subfields of \( {\mathbb{F}}_{{p}^{n}} \) are all Galois over \( {\mathbb{F}}_{p... | No |
Corollary 16. The irreducible polynomial \( {x}^{4} + 1 \in \mathbb{Z}\left\lbrack x\right\rbrack \) is reducible modulo every prime \( p \) . | Proof: Consider the polynomial \( {x}^{4} + 1 \) over \( {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) for the prime \( p \) . If \( p = 2 \) we have \( {x}^{4} + 1 = {\left( x + 1\right) }^{4} \) and the polynomial is reducible. Assume now that \( p \) is odd. Then \( {p}^{2} - 1 \) is divisible by 8 since \( p \) is... | Yes |
Proposition 17. The finite field \( {\mathbb{F}}_{{p}^{n}} \) is simple. In particular, there exists an irreducible polynomial of degree \( n \) over \( {\mathbb{F}}_{p} \) for every \( n \geq 1 \) . | We have described the finite fields \( {\mathbb{F}}_{{p}^{n}} \) above as the splitting fields of the polynomials \( {x}^{{p}^{n}} - x \) . By the previous proposition, this field can also be described as a quotient of \( {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \), namely by the minimal polynomial for \( \theta \) ... | Yes |
Proposition 19. Suppose \( K/F \) is a Galois extension and \( {F}^{\prime }/F \) is any extension. Then \( K{F}^{\prime }/{F}^{\prime } \) is a Galois extension, with Galois group \[ \operatorname{Gal}\left( {K{F}^{\prime }/{F}^{\prime }}\right) \cong \operatorname{Gal}\left( {K/K \cap {F}^{\prime }}\right) \] isomorp... | Proof: If \( K/F \) is Galois, then \( K \) is the splitting field of some separable polynomial \( f\left( x\right) \) in \( F\left\lbrack x\right\rbrack \) . Then \( K{F}^{\prime }/{F}^{\prime } \) is the splitting field of \( f\left( x\right) \) viewed as a polynomial in \( {F}^{\prime }\left\lbrack x\right\rbrack \)... | Yes |
Corollary 20. Suppose \( K/F \) is a Galois extension and \( {F}^{\prime }/F \) is any finite extension.\n\nThen\n\[ \left\lbrack {K{F}^{\prime } : F}\right\rbrack = \frac{\left\lbrack {K : F}\right\rbrack \left\lbrack {{F}^{\prime } : F}\right\rbrack }{\left\lbrack K \cap {F}^{\prime } : F\right\rbrack }.\] | Proof: This follows by the proposition from the equality \( \left\lbrack {K{F}^{\prime } : {F}^{\prime }}\right\rbrack = \left\lbrack {K : K \cap {F}^{\prime }}\right\rbrack \) given by the orders of the Galois groups in the proposition. | No |
Proposition 21. Let \( {K}_{1} \) and \( {K}_{2} \) be Galois extensions of a field \( F \) . Then\n\n(1) The intersection \( {K}_{1} \cap {K}_{2} \) is Galois over \( F \) .\n\n(2) The composite \( {K}_{1}{K}_{2} \) is Galois over \( F \) . The Galois group is isomorphic to the subgroup\n\n\[ H = \left\{ {\left( {\sig... | Proof: (1) Suppose \( p\left( x\right) \) is an irreducible polynomial in \( F\left\lbrack x\right\rbrack \) with a root \( \alpha \) in \( {K}_{1} \cap {K}_{2} \) . Since \( \alpha \in {K}_{1} \) and \( {K}_{1}/F \) is Galois, all the roots of \( p\left( x\right) \) lie in \( {K}_{1} \) . Similarly all the roots lie i... | Yes |
Corollary 22. Let \( {K}_{1} \) and \( {K}_{2} \) be Galois extensions of a field \( F \) with \( {K}_{1} \cap {K}_{2} = F \) . Then\n\n\[ \operatorname{Gal}\left( {{K}_{1}{K}_{2}/F}\right) \cong \operatorname{Gal}\left( {{K}_{1}/F}\right) \times \operatorname{Gal}\left( {{K}_{2}/F}\right) . \] | Proof: The first part follows immediately from the proposition. For the second, let \( {K}_{1} \) be the fixed field of \( {G}_{1} \subset G \) and let \( {K}_{2} \) be the fixed field of \( {G}_{2} \subset G \) . Then \( {K}_{1} \cap {K}_{2} \) is the field corresponding to the subgroup \( {G}_{1}{G}_{2} \), which is ... | No |
Corollary 23. Let \( E/F \) be any finite separable extension. Then \( E \) is contained in an extension \( K \) which is Galois over \( F \) and is minimal in the sense that in a fixed algebraic closure of \( K \) any other Galois extension of \( F \) containing \( E \) contains \( K \) . | Proof: There exists a Galois extension of \( F \) containing \( E \), for example the composite of the splitting fields of the minimal polynomials for a basis for \( E \) over \( F \) (which are all separable since \( E \) is separable over \( F \) ). Then the intersection of all the Galois extensions of \( F \) contai... | Yes |
Proposition 24. Let \( K/F \) be a finite extension. Then \( K = F\left( \theta \right) \) if and only if there exist only finitely many subfields of \( K \) containing \( F \). | Proof: Suppose first that \( K = F\left( \theta \right) \) is simple. Let \( E \) be a subfield of \( K \) containing \( F : F \subseteq E \subseteq K \) . Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be the minimal polynomial for \( \theta \) over \( F \) and let \( g\left( x\right) \in E\left\lbrack x\... | Yes |
Theorem 25. (The Primitive Element Theorem) If \( K/F \) is finite and separable, then \( K/F \) is simple. In particular, any finite extension of fields of characteristic 0 is simple. | Proof: Let \( L \) be the Galois closure of \( K \) over \( F \) . Then any subfield of \( K \) containing \( F \) corresponds to a subgroup of the Galois group \( \operatorname{Gal}\left( {L/F}\right) \) by the Fundamental Theorem. Since there are only finitely many such subgroups, the previous proposition shows that ... | Yes |
The Galois group of the cyclotomic field \( \mathbb{Q}\left( {\zeta }_{n}\right) \) of \( {n}^{\text{th }} \) roots of unity is isomorphic to the multiplicative group \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \) . The isomorphism is given explicitly by the map\n\n\[ \n{\left( \mathbb{Z}/n\mathbb{Z}\right) }... | Proof: The discussion above shows that \( {\sigma }_{a} \) is an automorphism for any \( a\left( {\;\operatorname{mod}\;n}\right) \) , so the map above is well defined. It is a homomorphism since \n\n\[ \n\left( {{\sigma }_{a}{\sigma }_{b}}\right) \left( {\zeta }_{n}\right) = {\sigma }_{a}\left( {\zeta }_{n}^{b}\right)... | Yes |
Corollary 27. Let \( n = {p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\cdots {p}_{k}^{{a}_{k}} \) be the decomposition of the positive integer \( n \) into distinct prime powers. Then the cyclotomic fields \( \mathbb{Q}\left( {\zeta }_{{p}_{i}^{{a}_{i}}}\right), i = 1,2,\ldots, k \) intersect only in the field \( \mathbb{Q} \) an... | Proof: The only statement which has not been proved is the identification of the isomorphism of Galois groups with the statement of the Chinese Remainder Theorem on the group \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \), which is quite simple and is left for the exercises. | No |
Proposition 29. The regular \( n \) -gon can be constructed by straightedge and compass if and only if \( n = {2}^{k}{p}_{1}\cdots {p}_{r} \) is the product of a power of 2 and distinct Fermat primes. | The proof above actually indicates a procedure for constructing the regular \( n \) -gon as a succession of square roots. For example, the construction of the regular 17-gon (solved by Gauss in 1796 at age 19) requires the construction of the subfields of degrees \( 2,4,8 \) and 16 in \( \mathbb{Q}\left( {\zeta }_{17}\... | No |
Corollary 31. (Fundamental Theorem on Symmetric Functions) Any symmetric function in the variables \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) is a rational function in the elementary symmetric functions \( {s}_{1},{s}_{2},\ldots ,{s}_{n} \) . | Proof: A symmetric function lies in the fixed field of \( {S}_{n} \) above, hence is a rational function in \( {s}_{1},\ldots ,{s}_{n} \) . | No |
Proposition 34. The Galois group of \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) is a subgroup of \( {A}_{n} \) if and only if the discriminant \( D \in F \) is the square of an element of \( F \) . | Proof: This is a restatement of Proposition 33 in this case. The Galois group is contained in \( {A}_{n} \) if and only if every element of the Galois group fixes\n\n\[ \sqrt{D} = \mathop{\prod }\limits_{{i < j}}\left( {{\alpha }_{i} - {\alpha }_{j}}\right) \]\n\ni.e., if and only if \( \sqrt{D} \in F \) . | No |
Proposition 36. Let \( F \) be a field of characteristic not dividing \( n \) which contains the \( {n}^{\text{th }} \) roots of unity. Then the extension \( F\left( \sqrt[n]{a}\right) \) for \( a \in F \) is cyclic over \( F \) of degree dividing \( n \) . | Proof: The extension \( K = F\left( \sqrt[n]{a}\right) \) is Galois over \( F \) if \( F \) contains the \( {n}^{\text{th }} \) roots of unity since it is the splitting field for \( {x}^{n} - a \) . For any \( \sigma \in \operatorname{Gal}\left( {K/F}\right) ,\sigma \left( \sqrt[n]{a}\right) \) is another root of this ... | Yes |
Lemma 38. If \( \alpha \) is contained in a root extension \( K \) as in (21) above, then \( \alpha \) is contained in a root extension which is Galois over \( F \) and where each extension \( {K}_{i + 1}/{K}_{i} \) is cyclic. | Proof: Let \( L \) be the Galois closure of \( K \) over \( F \) . For any \( \sigma \in \operatorname{Gal}\left( {L/F}\right) \) we have the chain of subfields\n\n\[ F = \sigma {K}_{0} \subset \sigma {K}_{1} \subset \cdots \subset \sigma {K}_{i} \subset \sigma {K}_{i + 1} \subset \cdots \subset \sigma {K}_{s} = {\sigm... | Yes |
Theorem 39. The polynomial \( f\left( x\right) \) can be solved by radicals if and only if its Galois group is a solvable group. | Proof: Suppose first that \( f\left( x\right) \) can be solved by radicals. Then each root of \( f\left( x\right) \) is contained in an extension as in the lemma. The composite \( L \) of such extensions is\n\nagain of the same type by Proposition 21. Let \( {G}_{i} \) be the subgroups corresponding to the subfields \(... | Yes |
Corollary 41. For any prime \( p \) not dividing the discriminant of \( f\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \), the Galois group of \( f\left( x\right) \) over \( \mathbb{Q} \) contains an element with cycle decomposition \( \left( {{n}_{1},{n}_{2},\ldots ,{n}_{k}}\right) \) where \( {n}_{1},{n}_... | ## Example\n\nConsider the polynomial \( {x}^{5} - x - 1 \) . The discriminant of this polynomial is \( {2869} = {19} \cdot {151} \) so we reduce at primes \( \neq {19},{151} \) . Reducing mod 2 the polynomial \( {x}^{5} - x - 1 \) factors as \( \left( {{x}^{2} + x + 1}\right) \left( {{x}^{3} + {x}^{2} + 1}\right) \lef... | Yes |
Proposition 1. If \( I \) is an ideal of the Noetherian ring \( R \), then the quotient \( R/I \) is a Noetherian ring. Any homomorphic image of a Noetherian ring is Noetherian. | Proof: If \( R \) is a ring and \( I \) is an ideal in \( R \), then any infinite ascending chain of ideals in the quotient \( R/I \) would correspond by the Lattice Isomorphism Theorem to an infinite ascending chain of ideals in \( R \) . This gives the first statement, and the second follows by the first Isomorphism ... | Yes |
Theorem 2. The following are equivalent:\n\n(1) \( R \) is a Noetherian ring.\n\n(2) Every nonempty set of ideals of \( R \) contains a maximal element under inclusion.\n\n(3) Every ideal of \( R \) is finitely generated. | Proof: The proof is identical to that of Theorem 1 in Section 12.1 in the special case where the \( R \) -module \( M \) is \( R \) itself (and submodules are ideals). | No |
Corollary 5. The ring \( R \) is a finitely generated \( k \) -algebra if and only if there is some surjective \( k \) -algebra homomorphism\n\n\[ \varphi : k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \rightarrow R \]\n\nfrom the polynomial ring in a finite number of variables onto \( R \) that is the ... | Proof: If \( R \) is generated as a \( k \) -algebra by \( {r}_{1},\ldots ,{r}_{n} \), then we may define the map \( \varphi : k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow R \) by \( \varphi \left( {x}_{i}\right) = {r}_{i} \) for all \( i \) and \( \varphi \left( a\right) = a \) for all \( a \in k \... | Yes |
Theorem 6. Let \( V \subseteq {\mathbb{A}}^{n} \) and \( W \subseteq {\mathbb{A}}^{m} \) be affine algebraic sets. Then there is a bijective correspondence\n\n\[ \n\left\{ \begin{matrix} \text{ morphisms from }V\text{ to }W \\ \text{ as algebraic sets } \end{matrix}\right\} \leftrightarrow \left\{ \begin{matrix} k\text... | Proof: The proof of (3) is left as an exercise and (4) is then immediate. | No |
Proposition 8. With notation as above, let \( R = k\left\lbrack {{y}_{1},\ldots ,{y}_{m},{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) and let \( \mathcal{A} \) be the ideal generated by \( {y}_{1} - {\varphi }_{1},\ldots ,{y}_{m} - {\varphi }_{m} \) together with generators for \( I \) . Let \( G \) be the reduced Gröbner ... | Proof: If we show \( \ker \Phi = \mathcal{A} \cap k\left\lbrack {{y}_{1},\ldots ,{y}_{m}}\right\rbrack \) modulo \( J \) then (a) follows by Proposition 30 in Section 9.6. Suppose first that \( f \in \mathcal{A} \cap k\left\lbrack {{y}_{1},\ldots ,{y}_{m}}\right\rbrack \) . If \( {f}_{1},\ldots ,{f}_{s} \) are generato... | Yes |
Proposition 10. Suppose \( \alpha \) is a root of the irreducible polynomial \( p\left( x\right) \in k\left\lbrack x\right\rbrack \) and \( \beta \in k\left( \alpha \right) \), say \( \beta = f\left( \alpha \right) \) for the polynomial \( f \in k\left\lbrack x\right\rbrack \) . Let \( G \) be the reduced Gröbner basis... | Proof: The kernel of the \( k \) -algebra homomorphism \( k\left\lbrack y\right\rbrack \rightarrow k\left\lbrack x\right\rbrack /\left( p\right) \cong k\left( \alpha \right) \) defined by mapping \( y \) first to \( f \) and then to \( \beta \) is the principal ideal generated by the minimal polynomial of \( \beta \) i... | Yes |
Proposition 11. Let \( I \) be an ideal in the commutative ring \( R \) . Then rad \( I \) is an ideal containing \( I \), and \( \left( {\operatorname{rad}I}\right) /I \) is the nilradical of \( R/I \) . In particular, \( R/I \) has no nilpotent elements if and only if \( I = \operatorname{rad}I \) is a radical ideal. | Proof: It is clear that \( I \subseteq \operatorname{rad}I \) . By definition, the nilradical of \( R/I \) consists of the elements in the quotient some power of which is 0 . Under the Lattice Isomorphism Theorem for rings this collection of elements corresponds to the elements of \( R \) some power of which lie in \( ... | Yes |
The radical of a proper ideal \( I \) is the intersection of all prime ideals containing \( I \) . In particular, the nilradical is the intersection of all the prime ideals in \( R \) . | Proof: Passing to \( R/I \), Proposition 11 shows that it suffices to prove this result for \( I = 0 \), and in this case the statement is that the nilradical \( N \) of \( R \) is the intersection of all the prime ideals in \( R \) . Let \( {N}^{\prime } \) denote the intersection of all the prime ideals in \( R \) .\... | Yes |
Corollary 13. Prime (and hence also maximal) ideals are radical. | Proof: If \( P \) is a prime ideal, then \( P \) is clearly the intersection of all the prime ideals containing \( P \), so \( P = \operatorname{rad}P \) by the proposition. | Yes |
Proposition 14. If \( R \) is a Noetherian ring then for any ideal \( I \) some positive power of \( \operatorname{rad}I \) is contained in \( I \) . In particular, the nilradical, \( N \), of a Noetherian ring is a nilpotent ideal: \( {N}^{k} = 0 \) for some \( k \geq 1 \) . | Proof: For any ideal \( I \), the ideal rad \( I \) is finitely generated since \( R \) is Noetherian. If \( {a}_{1},\ldots ,{a}_{m} \) are generators of rad \( I \), then by definition of the radical, for each \( i \) we have \( {a}_{i}^{{k}_{i}} \in I \) for some positive integer \( {k}_{i} \) . Let \( k \) be the ma... | Yes |
Proposition 16. Suppose \( \varphi : V \rightarrow W \) is a morphism of algebraic sets and \( \widetilde{\varphi } : k\left\lbrack W\right\rbrack \rightarrow k\left\lbrack V\right\rbrack \) is the associated \( k \) -algebra homomorphism of coordinate rings. Then\n\n(1) The kernel of \( \widetilde{\varphi } \) is \( \... | Proof: Since \( \widetilde{\varphi } = f \circ \varphi \), we have \( \widetilde{\varphi }\left( f\right) = 0 \) if and only if \( \left( {f \circ \varphi }\right) \left( P\right) = 0 \) for all \( P \in V \), i.e., \( f\left( Q\right) = 0 \) for all \( Q = \varphi \left( P\right) \in \varphi \left( V\right) \), which ... | Yes |
(1) The affine algebraic set \( V \) is irreducible if and only if \( \mathcal{I}\left( V\right) \) is a prime ideal. | Proof: Let \( I = \mathcal{I}\left( V\right) \) and suppose first that \( V = {V}_{1} \cup {V}_{2} \) is reducible, where \( {V}_{1} \) and \( {V}_{2} \) are proper closed subsets. Since \( {V}_{1} \neq V \), there is some function \( {f}_{1} \) that vanishes on \( {V}_{1} \) but not on \( V \), i.e., \( {f}_{1} \in \m... | Yes |
Corollary 18. An affine algebraic set \( V \) is a variety if and only if its coordinate ring \( k\left\lbrack V\right\rbrack \) is an integral domain. | Proof: This follows immediately since \( \mathcal{I}\left( V\right) \) is a prime ideal if and only if the quotient \( k\left\lbrack V\right\rbrack = k\left\lbrack {\mathbb{A}}^{n}\right\rbrack /\mathcal{I}\left( V\right) \) is an integral domain (Proposition 13 of Chapter 7). | Yes |
Prime ideals are primary. | The first two statements are immediate from the definition of a primary ideal. | No |
Proposition 20. Let \( R \) be a Noetherian ring. Then\n\n(1) every irreducible ideal is primary, and\n\n(2) every proper ideal in \( R \) is a finite intersection of irreducible ideals. | Proof: To prove (1) let \( Q \) be an irreducible ideal and suppose that \( {ab} \in Q \) and \( b \notin Q \) . It is easy to check that for any fixed \( n \) the set of elements \( x \in R \) with \( {a}^{n}x \in Q \) is an ideal, \( {A}_{n} \), in \( R \) . Clearly \( {A}_{1} \subseteq {A}_{2} \subseteq \ldots \) an... | Yes |
Theorem 21. (Primary Decomposition Theorem) Let \( R \) be a Noetherian ring. Then every proper ideal \( I \) in \( R \) has a minimal primary decomposition. If\n\n\[ I = \mathop{\bigcap }\limits_{{i = 1}}^{m}{Q}_{i} = \mathop{\bigcap }\limits_{{i = 1}}^{n}{Q}_{i}^{\prime } \]\n\nare two minimal primary decompositions ... | Proof: The proof of the uniqueness of the set of associated primes is outlined in the exercises, and the proof of the uniqueness of the primary components associated to the minimal primes will be given in Section 4. | No |
Let \( I \) be a proper ideal in the Noetherian ring \( R \). (1) A prime ideal \( P \) contains the ideal \( I \) if and only if \( P \) contains one of the associated primes of \( I \), hence if and only if \( P \) contains one of the isolated primes of \( I \), i.e., the isolated primes of \( I \) are precisely the ... | The first statement in (1) is an exercise (cf. Exercise 37), and the remainder of (1) follows. | No |
Proposition 23. Let \( R \) be a subring of the commutative ring \( S \) with \( 1 \in R \) and let \( s \in S \) . Then the following are equivalent: (1) \( s \) is integral over \( R \) , (2) \( R\left\lbrack s\right\rbrack \) is a finitely generated \( R \) -module (where \( R\left\lbrack s\right\rbrack \) is the ri... | Proof: Suppose first that (1) holds and let \( s \) be a root of the monic polynomial \( {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} \in R\left\lbrack x\right\rbrack \) . Then \[ {s}^{n} = - \left( {{a}_{n - 1}{s}^{n - 1} + {a}_{n - 2}{s}^{n - 2} + \cdots + {a}_{0}}\right) \] and so \( {s}^{n} \), and then all ... | Yes |
Corollary 24. Let \( R \subseteq S \) be as in Proposition 23 and let \( s, t \in S \) .\n\n(1) If \( s \) and \( t \) are integral over \( R \) then so are \( s \pm t \) and \( {st} \) . | Proof: Let \( s \) and \( t \) be integral over \( R \) . By Proposition 23 both \( R\left\lbrack s\right\rbrack \) and \( R\left\lbrack t\right\rbrack \) are finitely generated \( R \) -modules, say\n\n\[ R\left\lbrack s\right\rbrack = R{s}_{1} + R{s}_{2} + \cdots + R{s}_{n} \]\n\n\[ R\left\lbrack t\right\rbrack = R{t... | Yes |
Theorem 26. Let \( R \) be a subring of the commutative ring \( S \) with \( 1 \in R \) and suppose that \( S \) integral over \( R \) . (1) Assume that \( S \) is an integral domain. Then \( R \) is a field if and only if \( S \) is a field. | Proof: To prove (1) assume first that \( R \) is a field and let \( s \) be a nonzero element of \( S \) . Then \( s \) is integral over \( R \), so \[ {s}^{n} + {a}_{n - 1}{s}^{n - 1} + \cdots + {a}_{1}s + {a}_{0} = 0 \] for some \( {a}_{0},{a}_{1},\ldots ,{a}_{n - 1} \) in \( R \) . Since \( S \) is an integral domai... | Yes |
Corollary 27. Suppose \( R \) is a subring of the ring \( S \) with \( 1 \in R \) and assume \( S \) is integral and finitely generated (as a ring) over \( R \) . If \( P \) is a maximal ideal in \( R \) then there is a nonzero and finite number of maximal ideals \( Q \) of \( S \) with \( Q \cap R = P \) . | Proof: There exists at least one maximal ideal \( Q \) lying over \( P \) by (2) of the theorem, so we must see why there are only finitely many such maximal ideals in \( S \) . If \( Q \) is a maximal ideal of \( S \) with \( Q \cap R = P \) then \( S/Q \) is a field containing the field \( R/P \) . To prove that ther... | Yes |
Proposition 28. An element \( \alpha \) in some field extension of \( \mathbb{Q} \) is an algebraic integer if and only if \( \alpha \) is algebraic over \( \mathbb{Q} \) and its minimal polynomial \( {m}_{\alpha ,\mathbb{Q}}\left( x\right) \) has integer coefficients. In particular, the algebraic integers in \( \mathb... | Proof: If \( \alpha \) is algebraic over \( \mathbb{Q} \) with \( {m}_{\alpha .\mathbb{Q}}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \), then by definition \( \alpha \) is integral over \( \mathbb{Z} \) . Conversely, assume \( \alpha \) is integral over \( \mathbb{Z} \), and let \( f\left( x\right) \) be... | Yes |
Theorem 29. Let \( K \) be a number field of degree \( n \) over \( \mathbb{Q} \) . (1) The ring \( {\mathcal{O}}_{K} \) of integers in \( K \) is a Noetherian ring and is a free \( \mathbb{Z} \) -module of rank \( n \) . (2) For every \( \beta \in K \) there is some nonzero \( d \in \mathbb{Z} \) such that \( {d\beta ... | Proof: Note first that any \( \mathbb{Z} \) -linear dependence relation among elements in \( {\mathcal{O}}_{K} \) is a \( \mathbb{Q} \) -linear dependence relation in \( K \), and multiplying a \( \mathbb{Q} \) -linear dependence relation of elements of \( {\mathcal{O}}_{K} \) in \( K \) by a common denominator for the... | Yes |
Theorem 30. (Noether’s Normalization Lemma) Let \( k \) be a field and suppose that \( A = k\left\lbrack {{r}_{1},{r}_{2},\ldots ,{r}_{m}}\right\rbrack \) is a finitely generated \( k \) -algebra. Then for some \( q,0 \leq q \leq m \) , there are algebraically independent elements \( {y}_{1},{y}_{2},\ldots ,{y}_{q} \in... | Proof: Proceed by induction on \( m \) . If \( {r}_{1},\ldots ,{r}_{m} \) are algebraically independent over \( k \) then take \( {y}_{i} = {r}_{i}, i = 1,\ldots, m \) . Otherwise, there exists \( f\left( {{x}_{1},\ldots ,{x}_{m}}\right) \in \) \( k\left\lbrack {{x}_{1},\ldots ,{x}_{m}}\right\rbrack \) such that \( f\l... | No |
Theorem 31. (Hilbert’s Nullstellensatz — Weak Form) Let \( k \) be an algebraically closed field. Then \( M \) is a maximal ideal in the polynomial ring \( k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) if and only if \( M = \left( {{x}_{1} - {a}_{1},\ldots ,{x}_{n} - {a}_{n}}\right) \) for some \( {a}... | Proof: Certainly \( \left( {{x}_{1} - {a}_{1},\ldots ,{x}_{n} - {a}_{n}}\right) \) is a maximal ideal in \( k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) . Conversely, for any maximal ideal \( M \) in \( k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) , let \( E = k\left\lbrack {{x}_{1... | Yes |
Corollary 33. (Variant of Hilbert’s Nullstellensatz) If \( k \) is any field with algebraic closure \( \bar{k} \) and \( I \) is an ideal in \( k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \), then \( {\mathcal{I}}_{k}\left( {{\mathcal{Z}}_{\bar{k}}\left( I\right) }\right) = \operatorname{rad}I \), wher... | Proof: Since \( \bar{k}\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) is an integral extension of \( k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) (generated by the integral elements \( \bar{k} \) ), the corollary follows immediately from Theorem 32 and the remarks on radicals above. | Yes |
Proposition 34. Suppose \( k \) is any field. If \( I = \left( {{f}_{1},\ldots ,{f}_{s}}\right) \) is a proper ideal in \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), then \( f \in \operatorname{rad}I \) if and only if \( \left( {{f}_{1},\ldots ,{f}_{s},1 - {yf}}\right) = k\left\lbrack {{x}_{1},\ldots ,{x}... | Proof: By Corollary 33, \( \left( {{f}_{1},\ldots ,{f}_{s},1 - {yf}}\right) = k\left\lbrack {{x}_{1},\ldots ,{x}_{n}, y}\right\rbrack \) if and only if the equations\n\n\[ 1 - {yf}\left( {{x}_{1},\ldots ,{x}_{n}}\right) = 0,\;{f}_{1}\left( {{x}_{1},\ldots ,{x}_{n}}\right) = 0,\;\ldots ,\;{f}_{s}\left( {{x}_{1},\ldots ,... | Yes |
Corollary 35. Suppose \( I = \left( {{f}_{1},\ldots ,{f}_{s}}\right) \) in \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then \( f \in \operatorname{rad}I \) if and only if \( \left\{ 1\right\} \) is the reduced Gröbner basis for the ideal \( \left( {{f}_{1},\ldots ,{f}_{s},1 - {yf}}\right) \) in \( k\le... | ## Example\n\nConsider \( I = \left( {{x}^{2} - {y}^{2},{xy}}\right) \) in \( k\left\lbrack {x, y}\right\rbrack \) . The reduced Gröbner basis for \( \left( {{x}^{2} - {y}^{2},{xy},1 - {tx}}\right) \) in \( k\left\lbrack {x, y, t}\right\rbrack \) with respect to the order \( x > y > t \) is \( \{ 1\} \), showing \( x \... | Yes |
Corollary 37. In the notation of Theorem 36,\n\n(1) \( \ker \pi = \{ r \in R \mid {xr} = 0 \) for some \( x \in D\} \) ; in particular, \( \pi : R \rightarrow {D}^{-1}R \) is an injection if and only if \( D \) contains no zero divisors of \( R \), and\n\n(2) \( {D}^{-1}R = 0 \) if and only if \( 0 \in D \), hence if a... | Proof: By definition, we have \( \pi \left( r\right) = 0 \) if and only if \( \left( {r,1}\right) \sim \left( {0,1}\right) \), i.e., if and only if \( {xr} = 0 \) for some \( x \in D \), which is (1). For (2), note that \( {D}^{-1}R = 0 \) if and only\n\nif the 1 of this ring is zero, i.e., \( \left( {1,1}\right) \sim ... | Yes |
Proposition 38. In the preceding notation we have\n\n(1) For any ideal \( J \) of \( {D}^{-1}R \) we have \( J = {}^{e}\left( {{}^{c}J}\right) \) . In particular, every ideal of \( {D}^{-1}R \) is the extension of some ideal of \( R \), and distinct ideals of \( {D}^{-1}R \) have distinct contractions in \( R \) . | Proof: We always have \( {}^{e}\left( {{}^{c}J}\right) \subseteq J \) . For the reverse inclusion let \( a/d \in J \) . Then \( a/1 = d\left( {a/d}\right) \in J \), and so \( a \in {\pi }^{-1}\left( J\right) = {}^{c}J \) . Thus \( a/1 \in {}^{e}\left( {{}^{c}J}\right) \), so we also have \( \left( {a/1}\right) \left( {... | Yes |
Proposition 39. Suppose \( R \) is a commutative ring with 1 and \( I \) is an ideal in \( R\left\lbrack x\right\rbrack \) . Then \( I \) is a prime ideal in \( R\left\lbrack x\right\rbrack \) if and only if\ni. \( J = I \cap R \) is a prime ideal in \( R \), i.e., \( S = R/J \) is an integral domain, and\nii. if \( \b... | Proof: Suppose \( I \) is a prime ideal in \( R\left\lbrack x\right\rbrack \), so that \( J = I \cap R \) is a prime ideal in \( R \) and \( S = R/J \) is an integral domain. By Proposition 2 in Chapter 9, the kernel of the reduction homomorphism \( R\left\lbrack x\right\rbrack \mapsto S\left\lbrack x\right\rbrack = \l... | Yes |
Proposition 40. Let \( S \) be an integral domain with fraction field \( F \) and let \( A \) be a nonzero ideal in \( S\left\lbrack x\right\rbrack \) . Suppose \( {AF}\left\lbrack x\right\rbrack = \left( {h\left( x\right) }\right) \) where \( h\left( x\right) \) is a polynomial in \( S\left\lbrack x\right\rbrack \) wi... | Proof: We first show \( {AF}\left\lbrack x\right\rbrack \cap {S}_{a}\left\lbrack x\right\rbrack = A{S}_{a}\left\lbrack x\right\rbrack \) . Since \( {S}_{a} \subseteq F \), the containment \( A{S}_{a}\left\lbrack x\right\rbrack \subseteq {AF}\left\lbrack x\right\rbrack \cap {S}_{a}\left\lbrack x\right\rbrack \) is immed... | Yes |
Proposition 41. Let \( D \) be a multiplicatively closed subset of \( R \) containing 1 and let \( M \) be an \( R \) -module. Then \( {D}^{-1}M \cong {D}^{-1}R{ \otimes }_{R}M \) as \( {D}^{-1}R \) -modules, i.e., \( {D}^{-1}M \) is the \( {D}^{-1}R \) -module obtained by extension of scalars from the \( R \) -module ... | Proof: The map from \( {D}^{-1}R \times M \) to \( {D}^{-1}M \) defined by mapping \( \left( {r/d, m}\right) \) to \( {rm}/d \) is well defined and \( R \) -balanced, so induces a homomorphism from \( {D}^{-1}R{ \otimes }_{R}M \) to \( {D}^{-1}M \) . The map sending \( m/d \) to \( \left( {1/d}\right) \otimes m \) give... | Yes |
Proposition 42. Let \( R \) be a commutative ring with 1 and let \( {D}^{-1}R \) be its localization with respect to the multiplicatively closed subset \( D \) of \( R \) containing 1 . (1) Localization commutes with finite sums and intersections of ideals: If \( I \) and \( J \) are ideals of \( R \), then \[ {D}^{-1}... | Proof: We first prove (6). Suppose that \( 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0 \) is a short exact sequence of \( R \) -modules. Every element of \( {D}^{-1}N \) is of the form \( n/d \) for some \( n \in N \) and \( d \in D \) . Since \( \varphi \) is surjecti... | No |
Proposition 43. Let \( R \) be a Noetherian ring and let\n\n\[ I = {Q}_{1} \cap \cdots \cap {Q}_{m} \]\n\nbe a minimal primary decomposition of the proper ideal \( I \), where \( {Q}_{i} \) is a \( {P}_{i} \) -primary ideal. Suppose \( D \) is a multiplicatively closed set of \( R \) containing 1 and the primary ideals... | Proof: By (3) of Proposition \( {42},{D}^{-1}{Q}_{i} = {D}^{-1}R \) for \( t + 1 \leq i \leq m \), and \( {D}^{-1}{Q}_{i} \) is a \( {D}^{-1}{P}_{i} \) -primary ideal with pullback \( {Q}_{i} \) for \( 1 \leq i \leq t \) . By (1) of the same proposition, \( {D}^{-1}I = {D}^{-1}{Q}_{1} \cap \cdots \cap {D}^{-1}{Q}_{t} \... | Yes |
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