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Corollary 8. Let \( \varphi : V \rightarrow U \) be a linear transformation of vector spaces over \( F \) . Then \( \ker \varphi \) is a subspace of \( V,\varphi \left( V\right) \) is a subspace of \( U \) and \( \dim V = \dim \ker \varphi + \dim \varphi \left( V\right) \) . | Proof: This follows immediately from Theorem 7. Note that the proof of Theorem 7 is in fact the special case of Corollary 8 where \( U \) is the quotient \( V/W \) and \( \varphi \) is the natural projection homomorphism. | No |
Corollary 9. Let \( \varphi : V \rightarrow W \) be a linear transformation of vector spaces of the same finite dimension. Then the following are equivalent:\n\n(1) \( \varphi \) is an isomorphism\n\n(2) \( \varphi \) is injective, i.e., \( \ker \varphi = 0 \)\n\n(3) \( \varphi \) is surjective, i.e., \( \varphi \left(... | Proof: The equivalence of these conditions follows from Corollary 8 by counting dimensions. | No |
Theorem 10. Let \( V \) be a vector space over \( F \) of dimension \( n \) and let \( W \) be a vector space over \( F \) of dimension \( m \), with bases \( \mathcal{B},\mathcal{E} \) respectively. Then the map \( {\operatorname{Hom}}_{F}\left( {V, W}\right) \rightarrow \) \( {M}_{m \times n}\left( F\right) \) from t... | Proof: The columns of the matrix \( {M}_{\mathcal{B}}^{\mathcal{E}}\left( \varphi \right) \) are determined by the action of \( \varphi \) on the basis \( \mathcal{B} \) as in equation (3). This shows in particular that the map \( \varphi \mapsto {M}_{\mathcal{B}}^{\mathcal{E}}\left( \varphi \right) \) is an \( F \) -l... | Yes |
Corollary 11. The dimension of \( {\operatorname{Hom}}_{F}\left( {V, W}\right) \) is \( \left( {\dim V}\right) \left( {\dim W}\right) \) . | Proof: The dimension of \( {M}_{m \times n}\left( F\right) \) is \( {mn} \) . | No |
Corollary 13. Matrix multiplication is associative and distributive (whenever the dimensions are such as to make products defined). An \( n \times n \) matrix \( A \) is nonsingular if and only if it is invertible. | Proof: Let \( A, B \) and \( C \) be matrices such that the products \( \left( {AB}\right) C \) and \( A\left( {BC}\right) \) are defined, and let \( S, T \) and \( R \) denote the associated linear transformations. By Theorem 12, the linear transformation corresponding to \( {AB} \) is the composite \( S \circ T \) so... | Yes |
Proposition 17. Let \( \varphi : V \rightarrow X \) and \( \psi : W \rightarrow Y \) be linear transformations of finite dimensional vector spaces. Then the Kronecker product of matrices representing \( \varphi \) and \( \psi \) is a matrix representation of \( \varphi \otimes \psi \) . | ## Example\n\nLet \( V = X = {\mathbb{R}}^{3} \), both with basis \( {v}_{1},{v}_{2},{v}_{3} \), and \( W = Y = {\mathbb{R}}^{2} \), both with basis \( {w}_{1},{w}_{2} \) . Suppose \( \varphi : {\mathbb{R}}^{3} \rightarrow {\mathbb{R}}^{3} \) is the linear transformation given by \( \varphi \left( {a{v}_{1} + b{v}_{2} ... | Yes |
Proposition 18. With notations as above, \( \left\{ {{v}_{1}^{ * },{v}_{2}^{ * },\ldots ,{v}_{n}^{ * }}\right\} \) is a basis of \( {V}^{ * } \) . In particular, if \( V \) is finite dimensional then \( {V}^{ * } \) has the same dimension as \( V \) . | Proof: Observe that since \( V \) is finite dimensional, \( \dim {V}^{ * } = \dim {\operatorname{Hom}}_{F}\left( {V, F}\right) = \) \( \dim V = n \) (Corollary 11), so since there are \( n \) of the \( {v}_{i}^{ * } \) ’s it suffices to prove that they are linearly independent. If\n\n\[ \n{\alpha }_{1}{v}_{1}^{ * } + {... | Yes |
Theorem 19. There is a natural injective linear transformation from \( V \) to \( {V}^{* * } \) . If \( V \) is finite dimensional then this linear transformation is an isomorphism. | Proof: Let \( v \in V \) . Define the map (evaluation at \( v \) ) \[ {E}_{v} : {V}^{ * } \rightarrow F\;\text{ by }\;{E}_{v}\left( f\right) = f\left( v\right) . \] Then \( {E}_{v}\left( {f + {\alpha g}}\right) = \left( {f + {\alpha g}}\right) \left( v\right) = f\left( v\right) + {\alpha g}\left( v\right) = {E}_{v}\lef... | Yes |
Theorem 20. With notations as above, \( {\varphi }^{ * } \) is a linear transformation from \( {W}^{ * } \) to \( {V}^{ * } \) and \( {M}_{{\mathcal{E}}^{ * }}^{{\mathcal{B}}^{ * }}\left( {\varphi }^{ * }\right) \) is the transpose of the matrix \( {M}_{\mathcal{B}}^{\mathcal{E}}\left( \varphi \right) \) (recall that t... | Proof: The map \( {\varphi }^{ * } \) is linear because \( \left( {f + {\alpha g}}\right) \circ \varphi = \left( {f \circ \varphi }\right) + \alpha \left( {g \circ \varphi }\right) \) . The equations which define \( \varphi \) are (from its matrix)\n\n\[ \varphi \left( {v}_{j}\right) = \mathop{\sum }\limits_{{i = 1}}^{... | Yes |
Proposition 22. Let \( \varphi \) be an \( n \) -multilinear alternating function on \( V \) . Then\n\n(1) \( \varphi \left( {{v}_{1},\ldots ,{v}_{i - 1},{v}_{i + 1},{v}_{i},{v}_{i + 2},\ldots ,{v}_{n}}\right) = - \varphi \left( {{v}_{1},{v}_{2},\ldots ,{v}_{n}}\right) \) for any \( i \in \) \( \{ 1,2,\ldots, n - 1\} \... | Proof: (1) Let \( \psi \left( {x, y}\right) \) be the function \( \varphi \) with variable entries \( x \) and \( y \) in positions \( i \) and \( i + 1 \) respectively and fixed entries \( {v}_{j} \) in position \( j \), for all other \( j \) . Thus (1) is the same as showing \( \psi \left( {y, x}\right) = - \psi \lef... | Yes |
Theorem 24. There is a unique \( n \times n \) determinant function on \( R \) and it can be computed for any \( n \times n \) matrix \( \left( {\alpha }_{ij}\right) \) by the formula:\n\n\[ \det \left( {\alpha }_{ij}\right) = \mathop{\sum }\limits_{{\sigma \in {S}_{n}}}\epsilon \left( \sigma \right) {\alpha }_{\sigma ... | Proof: Let \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) be the column vectors in a general \( n \times n \) matrix \( \left( {\alpha }_{ij}\right) \) . We leave it as an exercise to check that the formula given in the statement of the theorem does satisfy the axioms of a determinant function - this gives existence of a determ... | No |
The determinant is an \( n \) -multilinear function of the rows of \( {M}_{n \times n}\left( R\right) \) and for any \( n \times n \) matrix \( A \) , \( \det A = \det \left( {A}^{t}\right) \), where \( {A}^{t} \) is the transpose of \( A \) . | The first statement is an immediate consequence of the second, so it suffices to prove that a matrix and its transpose have the same determinant. For \( A = \left( {\alpha }_{ij}\right) \) one calculates that\n\n\[ \det {A}^{t} = \mathop{\sum }\limits_{{\sigma \in {S}_{n}}}\epsilon \left( \sigma \right) {\alpha }_{{1\s... | Yes |
Theorem 26. (Cramer’s Rule) If \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) are the columns of an \( n \times n \) matrix \( A \) and \( B = {\beta }_{1}{A}_{1} + {\beta }_{2}{A}_{2} + \cdots + {\beta }_{n}{A}_{n} \), for some \( {\beta }_{1},\ldots ,{\beta }_{n} \in R \), then\n\n\[{\beta }_{i}\det A = \det \left( {{A}_{1},\... | Proof: This follows immediately from Proposition 22(3) on replacing the given expression for \( B \) in the \( {i}^{\text{th }} \) position and expanding by multilinearity in that position. | No |
Corollary 27. If \( R \) is an integral domain, then \( \det A = 0 \) for \( A \in {M}_{n}\left( R\right) \) if and only if the columns of \( A \) are \( R \) -linearly dependent as elements of the free \( R \) -module of rank \( n \) . Also, \( \det A = 0 \) if and only if the rows of \( A \) are \( R \) -linearly dep... | Proof: Since \( \det A = \det {A}^{t} \) the first sentence implies the second.\n\nAssume first that the columns of \( A \) are linearly dependent and\n\n\[ 0 = {\beta }_{1}{A}_{1} + {\beta }_{2}{A}_{2} + \cdots + {\beta }_{n}{A}_{n} \]\n\nis a dependence relation on the columns of \( A \) with, say, \( {\beta }_{i} \n... | Yes |
Theorem 28. For matrices \( A, B \in {M}_{n \times n}\left( R\right) \) , \( \det {AB} = \left( {\det A}\right) \left( {\det B}\right) \) . | Proof: Let \( B = \left( {\beta }_{ij}\right) \) and let \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) be the columns of \( A \) . Then \( C = {AB} \) is the \( n \times n \) matrix whose \( {j}^{\text{th }} \) column is \( {C}_{j} = {\beta }_{1j}{A}_{1} + {\beta }_{2j}{A}_{2} + \cdots + {\beta }_{nj}{A}_{n} \) . By Propositio... | Yes |
Theorem 29. (The Cofactor Expansion Formula along the \( {i}^{\text{th }} \) row) If \( A = \left( {\alpha }_{ij}\right) \) is an \( n \times n \) matrix, then for each fixed \( i \in \{ 1,2,\ldots, n\} \) the determinant of \( A \) can be computed from the formula\n\n\[ \det A = {\left( -1\right) }^{i + 1}{\alpha }_{i... | Proof: For each \( A \) let \( D\left( A\right) \) be the element of \( R \) obtained from the cofactor expansion formula described above. We prove that \( D \) satisfies the axioms of a determinant function, hence is the determinant function. Proceed by induction on \( n \) . If \( n = 1 \) , \( D\left( \left( \alpha ... | Yes |
Theorem 30. (Cofactor Formula for the Inverse of a Matrix) Let \( A = \left( {\alpha }_{ij}\right) \) be an \( n \times n \) matrix and let \( B \) be the transpose of its matrix of cofactors, i.e., \( B = \left( {\beta }_{ij}\right) \), where \( {\beta }_{ij} = {\left( -1\right) }^{i + j}\det {A}_{ji},1 \leq i, j \leq... | Proof: The \( i, j \) entry of \( {AB} \) is \( {\alpha }_{i1}{\beta }_{1j} + {\alpha }_{i2}{\beta }_{2j} + \cdots + {\alpha }_{in}{\beta }_{nj} \) . By definition of the entries of \( B \) this equals\n\n\[{\alpha }_{i1}{\left( -1\right) }^{j + 1}D\left( {A}_{j1}\right) + {\alpha }_{i2}{\left( -1\right) }^{j + 2}D\lef... | Yes |
Theorem 31. If \( M \) is any \( R \) -module over the commutative ring \( R \) then\n\n(1) \( \mathcal{T}\left( M\right) \) is an \( R \) -algebra containing \( M \) with multiplication defined by mapping\n\n\[ \left( {{m}_{1} \otimes \cdots \otimes {m}_{i}}\right) \left( {{m}_{1}^{\prime } \otimes \cdots \otimes {m}_... | Proof: The map\n\n\[ \underset{i\text{ factors }}{\underbrace{M \times M \times \cdots \times M}} \times \underset{j\text{ factors }}{\underbrace{M \times M \times \cdots \times M}} \rightarrow {\mathcal{T}}^{i + j}\left( M\right) \]\n\ndefined by\n\n\[ \left( {{m}_{1},\ldots ,{m}_{i},{m}_{1}^{\prime },\ldots ,{m}_{j}^... | Yes |
Proposition 32. Let \( V \) be a finite dimensional vector space over the field \( F \) with basis \( \mathcal{B} = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) . Then the \( k \) -tensors\n\n\[{v}_{{i}_{1}} \otimes {v}_{{i}_{2}} \otimes \cdots \otimes {v}_{{i}_{k}}\;\text{with}{v}_{{i}_{j}} \in \mathcal{B}\]\n\nare a ... | Proof: This follows immediately from Proposition 16 of Section 2. | No |
Proposition 33. Let \( S \) be a graded ring, let \( I \) be a graded ideal in \( S \) and let \( {I}_{k} = I \cap {S}_{k} \) for all \( k \geq 0 \) . Then \( S/I \) is naturally a graded ring whose homogeneous component of degree \( k \) is isomorphic to \( {S}_{k}/{I}_{k} \) . | Proof: The map\n\n\[ S = { \oplus }_{k = 0}^{\infty }{S}_{k} \rightarrow { \oplus }_{k = 0}^{\infty }\left( {{S}_{k}/{I}_{k}}\right) \]\n\n\[ \left( {\ldots ,{s}_{k},\ldots }\right) \mapsto \left( {\ldots ,{s}_{k}{\;\operatorname{mod}\;{I}_{k}},\ldots }\right) \]\n\n is surjective with kernel \( I = { \oplus }_{k = 0}^... | No |
Theorem 34. Let \( M \) be an \( R \) -module over the commutative ring \( R \) and let \( \mathcal{S}\left( M\right) \) be its symmetric algebra.\n\n(1) The \( {k}^{\text{th }} \) symmetric power, \( {\mathcal{S}}^{k}\left( M\right) \), of \( M \) is equal to \( M \otimes \cdots \otimes M \) ( \( k \) factors) modulo ... | Proof: The \( k \) -tensors \( {\mathcal{C}}^{k}\left( M\right) \) in the ideal \( \mathcal{C}\left( M\right) \) are finite sums of elements of the form\n\n\[ {m}_{1} \otimes \ldots \otimes {m}_{i - 1} \otimes \left( {{m}_{i} \otimes {m}_{i + 1} - {m}_{i + 1} \otimes {m}_{i}}\right) \otimes {m}_{i + 2} \otimes \ldots \... | Yes |
Let \( V \) be an \( n \) -dimensional vector space over the field \( F \) . Then \( \mathcal{S}\left( V\right) \) is isomorphic as a graded \( F \) -algebra to the ring of polynomials in \( n \) variables over \( F \) (i.e., the isomorphism is also a vector space isomorphism from \( {\mathcal{S}}^{k}\left( V\right) \)... | Proof: Let \( \mathcal{B} = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a basis of \( V \) . By Proposition 32 there is a bijection between a basis of \( {\mathcal{T}}^{k}\left( V\right) \) and the set \( {\mathcal{B}}^{k} \) of ordered \( k \) -tuples of elements from \( \mathcal{B} \) . Define two \( k \) -tuples... | No |
Theorem 36. Let \( M \) be an \( R \) -module over the commutative ring \( R \) and let \( \bigwedge \left( M\right) \) be its exterior algebra.\n\n(1) The \( {k}^{\text{th }} \) exterior power, \( \mathop{\bigwedge }\limits^{k}\left( M\right) \), of \( M \) is equal to \( M \otimes \cdots \otimes M \) ( \( k \) factor... | Proof: The \( k \) -tensors \( {\mathcal{A}}^{k}\left( M\right) \) in the ideal \( \mathcal{A}\left( M\right) \) are finite sums of elements of the form\n\n\[ \n{m}_{1} \otimes \ldots \otimes {m}_{i - 1} \otimes \left( {m \otimes m}\right) \otimes {m}_{i + 2} \otimes \ldots \otimes {m}_{k}\n\]\n\nwith \( {m}_{1},\ldots... | Yes |
Corollary 37. Let \( V \) be a finite dimensional vector space over the field \( F \) with basis \( \mathcal{B} = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) . Then the vectors\n\n\[ \n{v}_{{i}_{1}} \land {v}_{{i}_{2}} \land \cdots \land {v}_{{i}_{k}}\;\text{ for }1 \leq {i}_{1} < {i}_{2} < \cdots < {i}_{k} \leq n \n\... | Proof: As the proof of Theorem 36 shows, modulo \( {\mathcal{A}}^{k}\left( M\right) \), the order of the terms in any simple \( k \) -tensor can be rearranged up to introducing a sign change. It follows that the \( k \) -tensors in the corollary (which have been arranged with increasing subscripts on the \( {v}_{i} \) ... | No |
Proposition 39. Let \( \sigma \) be an element in the symmetric group \( {S}_{k} \) and let \( \epsilon \left( \sigma \right) \) be the sign of the permutation \( \sigma \) . Then\n\n(1) for every \( w \in {\mathcal{S}}^{k}\left( M\right) \) we have \( {\sigma w} = w \), and\n\n(2) for every \( w \in \mathop{\bigwedge ... | Proof: The first statement is immediate from (1) in Theorem 34. We showed in the course of the proof of Theorem 36 that\n\n\[ \n{m}_{1} \land \cdots \land {m}_{i} \land {m}_{i + 1} \land \cdots \land {m}_{k} = - {m}_{1} \land \cdots \land {m}_{i + 1} \land {m}_{i} \land \cdots \land {m}_{k}, \n\] \n\nwhich shows that t... | Yes |
Proposition 40. Suppose \( k \) ! is a unit in the ring \( R \) and \( M \) is an \( R \) -module. Then\n\n(1) The map \( \left( {1/k!}\right) {Sym} \) induces an \( R \) -module isomorphism between the \( {k}^{\text{th }} \) symmetric power of \( M \) and the \( R \) -submodule of symmetric \( k \) -tensors:\n\n\[ \n\... | Proof: We have seen that the respective maps are surjective \( R \) -homomorphisms from \( {\mathcal{T}}^{k}\left( M\right) \) so to prove the proposition it suffices to check that their kernels are \( {\mathcal{C}}^{k}\left( M\right) \) and \( {\mathcal{A}}^{k}\left( M\right) \), respectively. We show the first and le... | No |
Theorem 1. Let \( R \) be a ring and let \( M \) be a left \( R \) -module. Then the following are equivalent:\n\n(1) \( M \) is a Noetherian \( R \) -module.\n\n(2) Every nonempty set of submodules of \( M \) contains a maximal element under inclusion.\n\n(3) Every submodule of \( M \) is finitely generated. | Proof: [(1) implies (2)] Assume \( M \) is Noetherian and let \( \sum \) be any nonempty collection of submodules of \( M \) . Choose any \( {M}_{1} \in \sum \) . If \( {M}_{1} \) is a maximal element of \( \sum \) ,(2) holds, so assume \( {M}_{1} \) is not maximal. Then there is some \( {M}_{2} \in \sum \) such that \... | Yes |
Corollary 2. If \( R \) is a P.I.D. then every nonempty set of ideals of \( R \) has a maximal element and \( R \) is a Noetherian ring. | Proof: The P.I.D. \( R \) satisfies condition (3) in the theorem with \( M = R \) . | No |
Proposition 3. Let \( R \) be an integral domain and let \( M \) be a free \( R \) -module of rank \( n < \infty \) . Then any \( n + 1 \) elements of \( M \) are \( R \) -linearly dependent, i.e., for any \( {y}_{1},{y}_{2},\ldots ,{y}_{n + 1} \in M \) there are elements \( {r}_{1},{r}_{2},\ldots ,{r}_{n + 1} \in R \)... | Proof: The quickest way of proving this is to embed \( R \) in its quotient field \( F \) (since \( R \) is an integral domain) and observe that since \( M \cong R \oplus R \oplus \cdots \oplus R \) ( \( n \) times) we obtain \( M \subseteq F \oplus F \oplus \cdots \oplus F \) . The latter is an \( n \) -dimensional ve... | Yes |
Theorem 5. (Fundamental Theorem, Existence: Invariant Factor Form) Let \( R \) be a P.I.D. and let \( M \) be a finitely generated \( R \)-module.\n\n(1) Then \( M \) is isomorphic to the direct sum of finitely many cyclic modules. More precisely,\n\n\[ M \cong {R}^{r} \oplus R/\left( {a}_{1}\right) \oplus R/\left( {a}... | Proof: The module \( M \) can be generated by a finite set of elements by assumption so let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be a set of generators of \( M \) of minimal cardinality. Let \( {R}^{n} \) be the free \( R \)-module of rank \( n \) with basis \( {b}_{1},{b}_{2},\ldots ,{b}_{n} \) and define the homomor... | Yes |
Theorem 6. (Fundamental Theorem, Existence: Elementary Divisor Form) Let \( R \) be a P.I.D. and let \( M \) be a finitely generated \( R \) -module. Then \( M \) is the direct sum of a finite number of cyclic modules whose annihilators are either (0) or generated by powers of primes in \( R \), i.e., \[ M \cong {R}^{r... | We proved Theorem 6 by using the prime power factors of the invariant factors for \( M \) . In fact we shall see that the decomposition of \( M \) into a direct sum of cyclic modules whose annihilators are (0) or prime powers as in Theorem 6 is unique, i.e., the integer \( r \) and the ideals \( \left( {p}_{1}^{{\alpha... | No |
Theorem 7. (The Primary Decomposition Theorem) Let \( R \) be a P.I.D. and let \( M \) be a nonzero torsion \( R \) -module (not necessarily finitely generated) with nonzero annihilator \( a \) . Suppose the factorization of \( a \) into distinct prime powers in \( R \) is\n\n\[ a = u{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\a... | Proof: We have already proved these results in the case where \( M \) is finitely generated over \( R \) . In the general case it is clear that \( {N}_{i} \) is a submodule of \( M \) with annihilator dividing \( {p}_{i}^{{\alpha }_{i}} \) . Since \( R \) is a P.I.D. the ideals \( \left( {p}_{i}^{{\alpha }_{i}}\right) ... | Yes |
Lemma 8. Let \( R \) be a P.I.D. and let \( p \) be a prime in \( R \) . Let \( F \) denote the field \( R/\left( p\right) \) .\n\n(1) Let \( M = {R}^{r} \) . Then \( M/{pM} \cong {F}^{r} \) . | Proof: (1) There is a natural map from \( {R}^{r} \) to \( {\left( R/\left( p\right) \right) }^{r} \) defined by mapping \( \left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) to \( \left( {{\alpha }_{1}{\;\operatorname{mod}\;\left( p\right) },\ldots ,{\alpha }_{r}{\;\operatorname{mod}\;\left( p\right) }}\right) \) ... | Yes |
Corollary 10. Let \( R \) be a P.I.D. and let \( M \) be a finitely generated \( R \)-module.\n\n(1) The elementary divisors of \( M \) are the prime power factors of the invariant factors of \( M \).\n\n(2) The largest invariant factor of \( M \) is the product of the largest of the distinct prime powers among the ele... | Proof: The procedure in (1) gives \( a \) set of elementary divisors and since the elementary divisors for \( M \) are unique by the theorem, it follows that the procedure in (1) gives the set of elementary divisors. Similarly for (2). | No |
Corollary 11. (The Fundamental Theorem of Finitely Generated Abelian Groups) \( \; \) See Theorem 5.3 and Theorem 5.5. | Proof: Take \( R = \mathbb{Z} \) in Theorems 5,6 and 9 (note however that the invariant factors are listed in reverse order in Chapter 5 for computational convenience). | No |
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 \( A \) ) \[ \left... | 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.\n\n(II) If an angle \( \theta \) can be constructed, then determining the po... | 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 \( f\left( x\right) \left( { = {f}^{\prime }\left( x\right) }\right) \) . | No |
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 |
Proposition 37. Every irreducible polynomial over a finite field \( \mathbb{F} \) is separable. A polynomial in \( \mathbb{F}\left\lbrack x\right\rbrack \) is separable if and only if it is the product of distinct irreducible polynomials in \( \mathbb{F}\left\lbrack x\right\rbrack \) . | 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 \( {\chi }_{1},{\chi }_{2},\ldots ,{\chi }_{m} \) 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\n\n\[ \operatorname{Gal}\left( {K{F}^{\prime }/{F}^{\prime }}\right) \cong \operatorname{Gal}\left( {K/K \cap {F}^{\prime }}\right) \]\n\ni... | 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 \[ \operatorname{Gal}\left( {{K}_{1}{K}_{2}/F}\right) \cong \operatorname{Gal}\left( {{K}_{1}/F}\right) \times \operatorname{Gal}\left( {{K}_{2}/F}\right) . \] | 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 ... | No |
Theorem 26. 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... | 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 |
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