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Corollary 3.5. Let \( R \) be a commutative ring and \( A \in {\mathcal{M}}_{n}\left( R\right) \) . Then\n\n\[ A \cdot \left( \begin{matrix} {A}^{\left( {11}\right) } & \cdots & {A}^{\left( n1\right) } \\ \vdots & \ddots & \vdots \\ {A}^{\left( 1n\right) } & \cdots & {A}^{\left( nn\right) } \end{matrix}\right) = \left(... | Proof. Along the diagonal of the right-hand side, this is a restatement of Lemma 3.4, Off the diagonal, one is evaluating (for example)\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{{i}^{\prime }j}{A}^{\left( ij\right) } \]\n\nfor \( {i}^{\prime } \neq i \) . By Lemma 3.4 this is the same as the determinant of the matr... | Yes |
Proposition 3.6 (Cramer’s rule). Assume \( \det \left( A\right) \) is a unit, and let \( {A}^{\left( j\right) } \) be the matrix obtained by replacing the \( j \) -th column of \( A \) by the column vector \( b \) . Then\n\n\[ \n{x}_{j} = \det {\left( A\right) }^{-1}\det \left( {A}^{\left( j\right) }\right) .\n\] | Proof. Using Lemma 3.4, expand \( \det \left( {A}^{\left( j\right) }\right) \) with respect to the \( j \) -th column:\n\n\[ \n\det \left( {A}^{\left( j\right) }\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{A}^{\left( ij\right) }{b}_{i} \n\]\n\nTherefore\n\n\[ \n\left( \begin{matrix} {x}_{1} \\ \vdots \\ {x}_{n} \end{m... | Yes |
Proposition 3.7. The row rank of a matrix over a field \( k \) equals its column rank. | Proof. Equivalent matrices have the same ranks. Indeed, let \( P \in {\mathcal{M}}_{m, n}\left( k\right) \) ; the row space of \( P \) consists of all row vectors\n\n\[ \left( \begin{array}{lll} {a}_{1} & \cdots & {a}_{m} \end{array}\right) = \left( \begin{array}{lll} {v}_{1} & \cdots & {v}_{m} \end{array}\right) \cdot... | Yes |
Lemma 4.2. Submodules and direct sums of torsion-free modules are torsion-free. Free modules over an integral domain are torsion-free. | Proof. The first statement is immediate; the second follows from the first, since an integral domain is torsion-free as a module over itself. | No |
Example 4.3. Let \( R = \mathbb{Z}\left\lbrack x\right\rbrack \), and let \( I = \left( {2, x}\right) \) . Then \( I \) is not a free \( R \) -module. More generally, let \( I \) be any nonprincipal ideal of an integral domain \( R \) ; then \( I \) is a torsion-free module which is not free. | Indeed, if \( I \) were free, then its rank would have to be 1 at most, by Proposition 1.9 (a basis for \( I \) would be a linearly independent subset of \( R \), and \( R \) has rank 1 over itself); thus one element would suffice to generate \( I \), and \( I \) would be principal. | Yes |
Lemma 4.5. Let \( R \) be an integral domain. Assume that every cyclic \( R \) -module is torsion-free. Then \( R \) is a field. | Proof. Let \( c \in R, c \neq 0 \) ; then \( M = R/\left( c\right) \) is a cyclic module. Note that \( \operatorname{Tor}\left( M\right) = \) \( M \) : indeed, the class of 1 generates \( R/\left( c\right) \) and belongs to \( \operatorname{Tor}\left( M\right) \) since \( c \cdot 1 \) is 0 \( {\;\operatorname{mod}\;\le... | Yes |
Lemma 4.8. If \( R \) is a Noetherian ring, then every finitely generated \( R \) -module is finitely presented. | Proof. If \( M \) is a finitely generated module, there is an exact sequence\n\n\[ \n{R}^{m}\overset{\pi }{ \rightarrow }M \rightarrow 0 \n\]\n\nfor some \( m \) . Since \( R \) is Noetherian, \( {R}^{m} \) is Noetherian as an \( R \) -module (Corollary III 6.8). Thus \( \ker \pi \) is finitely generated; that is, ther... | Yes |
Proposition 4.10. Let \( R \) be an integral domain. Then \( R \) is a field if and only if every finitely generated \( R \) -module is free. | Proof. If \( R \) is a field, then every \( R \) -module is free, by Proposition 1.7. For the converse, assume that every finitely generated \( R \) -module is free; in particular, every cyclic module is free; in particular, every cyclic module is torsion-free. But then \( R \) is a field, by Lemma 4.5 | Yes |
Lemma 4.12. Let \( A, B \) be matrices with entries in an integral domain \( R \), and let \( M, N \) denote the corresponding \( R \) -modules. Then \( M \oplus N \) corresponds to the block matrix\n\n\[ \left( \begin{matrix} A & 0 \\ 0 & B \end{matrix}\right) \] | Proof. This follows immediately from Exercise 4.16 | No |
Proposition 4.13. Let \( A \) be a matrix with entries in an integral domain \( R \), and let \( B \) be obtained from \( A \) by any sequence of the following operations:\n\n- switch two rows or two columns;\n\n- add to one row (resp., column) a multiple of another row (resp., column);\n\n- multiply all entries in one... | Proof. The first three operations are the 'elementary operations' of [2.3] and they transform a matrix into an equivalent one (by Proposition 2.7); as observed above, this does not affect the corresponding module, up to isomorphism.\n\nAs for the fourth operation, if \( u \) is a unit and the only nonzero entry in (say... | Yes |
Example 4.14. The matrix with integer entries\n\n\\[ \n\\left( \\begin{array}{ll} 1 & 3 \\\\ 2 & 3 \\\\ 5 & 9 \\end{array}\\right) \n\\]\n\ndetermines an abelian group \( G \) . | Subtract three times the first column from the second column, obtaining\n\n\\[ \n\\left( \\begin{matrix} 1 & 0 \\\\ 2 & - 3 \\\\ 5 & - 6 \\end{matrix}\\right) \n\\]\n\nthe \( \\left( {1,1}\\right) \) entry is a unit and the only nonzero entry in the first row, so we can remove the first row and column:\n\n\\[ \n\\left(... | Yes |
Proposition 5.1. Let \( R \) be a PID, let \( F \) be a finitely generated free module over \( R \) , and let \( M \subseteq F \) be a submodule. Then \( M \) is free. | We will actually prove a more precise result, in view of the full statement of the classification theorem: we will show that there is a basis \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) of \( F \) and elements \( {a}_{1},\ldots ,{a}_{m} \) of \( R \) (with \( m \leq n \) ) such that\n\n\[ {y}_{1} = {a}_{1}{x}_{1},\ld... | Yes |
Lemma 5.2. Let \( R \) be a PID, let \( F \) be a finitely generated free module over \( R \), and let \( M \subseteq F \) be a nonzero submodule. Then there exist \( a \in R, x \in F, y \in M \), and submodules \( {F}^{\prime } \subseteq F \) and \( {M}^{\prime } \subseteq M \), such that \( y = {ax} \neq 0,{M}^{\prim... | Proof. For all \( \varphi \in {\operatorname{Hom}}_{R}\left( {F, R}\right) ,\varphi \left( M\right) \) is a submodule of \( R \), that is, an ideal. The family of all these ideals is nonempty, and PIDs are Noetherian; therefore (by Proposition VI1.1 there exists a maximal element in the family, say \( \alpha \left( M\r... | Yes |
Corollary 5.3. Let \( R \) be a PID, let \( F \) be a finitely generated free module over \( R \) , and let \( M \subseteq F \) be a submodule. Then there exist a basis \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) of \( F \) and nonzero elements \( {a}_{1},\ldots ,{a}_{m} \) of \( R\left( {m \leq n}\right) \) such tha... | Proof. Now that we know that submodules of a free module are free, we see that the submodule \( {F}^{\prime } \subseteq F \) produced in Lemma 5.2 is free. The first part of the statement then follows from Lemma 5.2, by an inductive argument analogous to the proof of Proposition 5.1, and is left to the reader.\n\nThe m... | No |
Proposition 5.4. Let \( R \) be an integral domain. Then \( R \) is a PID if and only if for every finitely generated \( R \) -module \( M \) and every epimorphism \[ {R}^{{m}_{0}}\overset{{\pi }_{0}}{ \rightarrow }M \rightarrow 0 \] there exist a free \( R \) -module \( {R}^{{m}_{1}} \) and a homomorphism \( {\pi }_{1... | Proof. The fact that the stated condition implies that \( R \) is a PID was proved in Claim 4.11. For the converse, let \( {\pi }_{0} : {R}^{{m}_{0}} \rightarrow M \) be an epimorphism; then \( \ker {\pi }_{0} \) is free by Proposition 5.1 the result follows by choosing any isomorphism \( {\pi }_{1} : {R}^{{m}_{1}} \ri... | Yes |
Lemma 5.7. Let \( M \) be a torsion module, expressed as in Theorem 5.6 (with \( \operatorname{rk}M = \) \( 0) \) . Then \( \operatorname{Ann}\left( M\right) = \left( {a}_{m}\right) \) . Further, the prime ideals \( \left( {q}_{i}\right) \) are precisely the prime ideals of \( R \) containing \( \operatorname{Ann}\left... | Proof. By hypothesis\n\n\[ M \cong \frac{R}{\left( {a}_{1}\right) } \oplus \cdots \oplus \frac{R}{\left( {a}_{m}\right) } \]\n\nwith \( {a}_{1}\left| \cdots \right| {a}_{m} \) . If \( r \in \operatorname{Ann}\left( M\right) \), then\n\n\[ 0 = r\left( {1,\ldots ,1}\right) = \left( {r,\ldots, r}\right) . \]\n\nIn particu... | Yes |
Proposition 6.2. Two matrices \( A, B \in {\mathcal{M}}_{n}\left( R\right) \) are similar if and only if there exists an invertible matrix \( P \) such that\n\n\[ B = {PA}{P}^{-1}\text{.} \] | The reader who has really understood Proposition 2.5 will not need any detailed proof of this statement: it should be apparent from staring at the butterfly diagram\n\n\n\nwhich I am essentially copying from [2.2] Th... | No |
Proposition 6.5. Let \( \alpha \) be a linear transformation of a free \( R \) -module \( F \cong {R}^{n} \) . Then \( \det \left( \alpha \right) \neq 0 \) if and only if \( \alpha \) is injective. | Proof. Embed \( R \) in its field of fractions \( K \), and view \( \alpha \) as a linear transformation of \( {K}^{n} \) ; note that the determinant of \( \alpha \) is the same whether it is computed over \( R \) or over \( K \) . Then \( \alpha \) is injective as a linear transformation \( {R}^{n} \rightarrow {R}^{n}... | Yes |
Lemma 6.7. Let \( A, B \in {\mathcal{M}}_{n}\left( R\right) \) . Then \( \operatorname{tr}\left( {AB}\right) = \operatorname{tr}\left( {BA}\right) \) . | Proof. Let \( A = \left( {a}_{ij}\right), B = \left( {b}_{ij}\right) \) . Then \( {AB} = \left( {\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik}{b}_{kj}}\right) \) ; hence\n\n\[ \operatorname{tr}\left( {AB}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik}{b}_{ki} \]\n\nThis expression... | Yes |
Proposition 6.9. Let \( F \) be a free \( R \) -module of rank \( n \), and let \( \alpha \in {\operatorname{End}}_{R}\left( F\right) \) .\n\n- The characteristic polynomial \( {P}_{\alpha }\left( t\right) \) is a monic polynomial of degree \( n \) .\n\n- The coefficient of \( {t}^{n - 1} \) in \( {P}_{\alpha }\left( t... | Proof. The first point is immediate, and the third is checked by setting \( t = 0 \) . To verify the second assertion, let \( A = \left( {a}_{ij}\right) \) be a matrix representing \( \alpha \) with respect to any basis for \( F \), so that\n\n\[ \n{P}_{\alpha }\left( t\right) = \det \left( \begin{matrix} t - {a}_{11} ... | Yes |
Lemma 6.10. If \( \alpha \) and \( \beta \) are similar, then \( {\mathcal{I}}_{\alpha } = {\mathcal{I}}_{\beta } \) . | Proof. By hypothesis there exists an invertible \( \pi \) such that \( \beta = \pi \circ \alpha \circ {\pi }^{-1} \) . As\n\n\[ \n{\beta }^{k} = {\left( \pi \circ \alpha \circ {\pi }^{-1}\right) }^{k} = \left( {\pi \circ \alpha \circ {\pi }^{-1}}\right) \circ \left( {\pi \circ \alpha \circ {\pi }^{-1}}\right) \circ \cd... | Yes |
Theorem 6.11 (Cayley-Hamilton). Let \( {P}_{\alpha }\left( t\right) \) be the characteristic polynomial of the linear transformation \( \alpha \in {\operatorname{End}}_{R}\left( F\right) \) . Then\n\n\[ \n{P}_{\alpha }\left( \alpha \right) = 0 \n\] | This beautiful observation can be proved directly by judicious use of Cramer's rule \( {30} \), in the form of Corollary 3.5, cf. Exercise 6.9. In any case, the Cayley-Hamilton theorem will become essentially evident once we connect these linear algebra considerations with the classification theorem for finitely genera... | No |
Lemma 6.14. Let \( F \) be a finitely generated \( R \) -module, and let \( \alpha \in {\operatorname{End}}_{R}\left( F\right) \) . Then the set of eigenvalues of \( \alpha \) is precisely the set of roots in \( R \) of the characteristic polynomial \( {P}_{\alpha }\left( t\right) \) . | Proof. This is a straightforward consequence of Proposition 6.5.\n\n\[ \lambda \text{is an eigenvalue for}\alpha \Leftrightarrow \exists \mathbf{v} \neq 0\text{such that}\alpha \left( \mathbf{v}\right) = {\lambda I}\left( \mathbf{v}\right) \]\n\n\[ \Leftrightarrow \exists \mathbf{v} \neq 0\text{such that}\left( {{\lamb... | Yes |
The matrix\n\n\[ \left( \begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}\right) \]\n\nhas no eigenvalues over \( \mathbb{R} \), while it has eigenvalues over \( \mathbb{C} \) : indeed, the characteristic polynomial \( {t}^{2} + 1 \) has no real roots and two complex roots. The reader should observe that, as a linear transf... | The characteristic polynomial \( {t}^{2} + 1 \) has no real roots and two complex roots. | Yes |
Corollary 6.18. The number of eigenvalues of a linear transformation of \( {R}^{n} \) is at most \( n \) . If the base ring \( R \) is an algebraically closed field, then every linear transformation has exactly \( n \) eigenvalues (counted with algebraic multiplicity). | Proof. Immediate from Lemmas 6.14, V15.1, and V15.10. | No |
Lemma 7.2. Let \( \alpha ,\beta \) be linear transformations of a free \( R \) -module \( F \) . Then the corresponding \( R\left\lbrack t\right\rbrack \) -module structures on \( F \) are isomorphic if and only if \( \alpha \) and \( \beta \) are similar. | Proof. Denote by \( {F}_{\alpha },{F}_{\beta } \) the two \( R\left\lbrack t\right\rbrack \) -modules defined on \( F \) by \( \alpha ,\beta \) as per Claim 7.1\n\nAssume first that \( \alpha \) and \( \beta \) are similar. Then there exists an invertible \( R \) -linear transformation \( \pi : F \rightarrow F \) such ... | No |
Theorem 7.5. Let \( k \) be a field, and let \( V \) be a finite-dimensional vector space. Let \( \alpha \) be a linear transformation on \( V \), and endow \( V \) with the corresponding \( k\left\lbrack t\right\rbrack \) -module structure, as in Claim 7.1. Then the following hold:\n\n- There exist distinct monic irre... | Proof. Since \( \dim V \) is finite, \( V \) is finitely generated as a \( k \) -module and a fortiori as a \( k\left\lbrack t\right\rbrack \) -module. The two isomorphisms are then obtained by applying Theorem 5.6 All the relevant polynomials may be chosen to be monic since every polynomial over a field is the associa... | No |
Proposition 7.9. Let \( {f}_{1}\left( t\right) \left| \cdots \right| {f}_{m}\left( t\right) \) be the invariant factors of a linear transformation \( \alpha \) on a vector space \( V \) . Then the minimal polynomial \( {m}_{\alpha }\left( t\right) \) equals \( {f}_{m}\left( t\right) \) , and the characteristic polynomi... | Proof. Tracing definitions, \( \left( {{m}_{\alpha }\left( t\right) }\right) \) is the annihilator ideal of \( V \) when this is viewed as a \( k\left\lbrack t\right\rbrack \) -module via \( \alpha \) (as in Claim 7.1). Therefore the equality of \( {m}_{\alpha }\left( t\right) \) and \( {f}_{m}\left( t\right) \) is a r... | No |
Corollary 7.10 (Cayley-Hamilton). The minimal polynomial of a linear transformation divides its characteristic polynomial. | Proof. This has now become evident, as promised in [6.2, | No |
Proposition 7.11. Let \( A \in {\mathcal{M}}_{n}\left( k\right) \) be a square matrix. Then \( A \) is similar to its transpose. | Proof. If \( B \) is similar to \( A \) and we can prove that \( B \) is similar to its transpose \( {B}^{t} \) , then \( A \) is similar to its transpose \( {A}^{t} \) : because \( B = {PA}{P}^{-1},{B}^{t} = {QB}{Q}^{-1} \) give\n\n\[ \n{A}^{t} = \left( {{P}^{t}{QP}}\right) A{\left( {P}^{t}QP\right) }^{-1}.\n\]\n\nThe... | Yes |
Lemma 7.12. Assume that the characteristic polynomial \( {P}_{\alpha }\left( t\right) \) factors completely; that is,\n\n\[ \n{P}_{\alpha }\left( t\right) = \mathop{\prod }\limits_{{i = 1}}^{s}{\left( t - {\lambda }_{i}\right) }^{{m}_{i}}\n\]\n\nwhere \( {\lambda }_{i}, i = 1,\ldots, s \), are the distinct eigenvalues ... | Proof. The first statement follows from uniqueness of factorizations. The statement about the minimal polynomial is immediate from Proposition 7.9 and the bookkeeping giving the equivalence of the two formulations in Theorem 7.5 | Yes |
One use of the Jordan canonical form is the enumeration of all possible similarity classes of transformations with given eigenvalues. For example, there are 5 similarity classes of linear transformations with a single eigenvalue \( \lambda \) with algebraic multiplicity 4, over a 4-dimensional vector space: indeed, the... | \[ \left( \begin{matrix} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{matrix}\right) ,\;\left( \begin{matrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{matrix}\right) ,\;\left( \begin{matrix} \lambda & 1 & 0 & 0 \\ 0 & ... | Yes |
Proposition 7.16. The geometric multiplicity of \( \lambda \) as an eigenvalue of \( \alpha \) equals the number of Jordan blocks corresponding to \( \lambda \) in the Jordan canonical form of \( \alpha \) . | Proof. As the geometric multiplicity is clearly additive in direct sums, it suffices to show that the geometric multiplicity of \( \lambda \) for the transformation corresponding to a single Jordan block\n\n\[ J = \left( \begin{matrix} \lambda & 1 & \ldots & 0 & 0 \\ 0 & \lambda & \ldots & 0 & 0 \\ \vdots & \vdots & \d... | Yes |
Proposition 7.18. Assume the characteristic polynomial of \( \alpha \in {\operatorname{End}}_{k}\left( V\right) \) factors completely over \( k \) . Then \( \alpha \) is diagonalizable if and only if the minimal polynomial of \( \alpha \) has no multiple roots. | Proof. Again, diagonalizability is equivalent to having all Jordan blocks of size 1 in the Jordan canonical form of \( \alpha \) . Therefore, if the characteristic polynomial of \( \alpha \) factors completely, then \( \alpha \) is diagonalizable if and only if all exponents \( {r}_{ij} \) appearing in Theorem 7.5 equa... | Yes |
Proposition 1.3. Let \( k \subseteq k\left( \alpha \right) \) be a simple extension. Consider the evaluation map \( \epsilon : k\left\lbrack t\right\rbrack \rightarrow k\left( \alpha \right) \), defined by \( f\left( t\right) \mapsto f\left( \alpha \right) \). Then we have the following:\n\n- \( \epsilon \) is injectiv... | Proof. Let \( F = k\left( \alpha \right) \). By the ’first isomorphism theorem’, the image of \( \epsilon : k\left\lbrack t\right\rbrack \rightarrow F \) is isomorphic to \( k\left\lbrack t\right\rbrack /\ker \left( \epsilon \right) \). Since \( F \) is an integral domain, so is \( k\left\lbrack t\right\rbrack /\ker \l... | Yes |
Consider the extension \( \mathbb{Q} \subseteq \mathbb{R} \). The polynomial \( {x}^{2} - 2 \in \mathbb{Q}\left\lbrack x\right\rbrack \) has roots in \( \mathbb{R} \): therefore, by Proposition V15.7 there exists a homomorphism (hence a field extension) | \[ \bar{\epsilon } : \;\frac{\mathbb{Q}\left\lbrack t\right\rbrack }{\left( {t}^{2} - 2\right) } \hookrightarrow \mathbb{R} \] such that the image of (the coset of) \( t \) is a root \( \alpha \) of \( {x}^{2} - 2 \). Proposition 1.3 simply identifies the image of this homomorphism with \( \mathbb{Q}\left( \alpha \righ... | No |
Proposition 1.5. Let \( {k}_{1} \subseteq {F}_{1} = {k}_{1}\left( {\alpha }_{1}\right) ,{k}_{2} \subseteq {F}_{2} = {k}_{2}\left( {\alpha }_{2}\right) \) be two finite simple extensions. Let \( {p}_{1}\left( t\right) \in {k}_{1}\left\lbrack t\right\rbrack \), resp., \( {p}_{2}\left( t\right) \in {k}_{2}\left\lbrack t\r... | Proof. Since every element of \( {k}_{1}\left( {\alpha }_{1}\right) \) is a linear combination of powers of \( {\alpha }_{1} \) with coefficients in \( {k}_{1}, j \) is determined by its action on \( {k}_{1} \) (which agrees with \( i \) ) and by \( j\left( {\alpha }_{1}\right) \), which is prescribed to be \( {\alpha ... | Yes |
Corollary 1.7. Let \( k \subseteq F = k\\left( \\alpha \\right) \) be a simple finite extension, and let \( p\\left( x\\right) \) be the minimal polynomial of \( \\alpha \) over \( k \) . Then \( \\left| {{\\operatorname{Aut}}_{k}\\left( F\\right) }\\right| \) equals the number of distinct roots of \( p\\left( x\\right... | Proof. Let \( j \\in {\\operatorname{Aut}}_{k}\\left( F\\right) \) . Since every element of \( F \) is a polynomial expression in \( \\alpha \) with coefficients in \( k \), and \( j \) extends the identity on \( k, j \) is determined by \( j\\left( \\alpha \\right) \) . Now\n\n\\[p\\left( {j\\left( \\alpha \\right) }\... | Yes |
Lemma 1.9. Let \( k \subseteq F \) be a finite extension. Then every \( \alpha \in F \) is algebraic over \( k \), of degree \( \leq \left\lbrack {F : k}\right\rbrack \) . | Proof. Since \( k \subseteq k\left( \alpha \right) \subseteq F \), the dimension of \( k\left( \alpha \right) \) as a \( k \) -vector space is bounded by \( {\dim }_{k}F = \left\lbrack {F : k}\right\rbrack \). Concretely, if \( k \subseteq F \) is finite and \( \alpha \in F \), then the powers \( 1,\alpha ,{\alpha }^{2... | Yes |
Proposition 1.10. Let \( k \subseteq E \subseteq F \) be field extensions. Then \( k \subseteq F \) is finite if and only if both \( k \subseteq E \) and \( E \subseteq F \) are finite. In this case,\n\n\[ \left\lbrack {F : k}\right\rbrack = \left\lbrack {F : E}\right\rbrack \left\lbrack {E : k}\right\rbrack . \] | Proof. If \( F \) is finite-dimensional as a vector space over \( k \), then so is its subspace \( E \) ; and any linear dependence relation of elements of \( F \) over the field \( k \) gives one over the larger field \( E \) . It follows that if \( k \subseteq F \) is finite, then so are \( k \subseteq E \) and \( E ... | Yes |
Let \( k \subseteq F \) be a field extension, and let \( \alpha \in F \) be an algebraic element over \( k \), of \( {odd} \) degree. Then I claim that \( \alpha \) may be written as a polynomial in \( {\alpha }^{2} \), with coefficients in \( k \) . | Indeed, \( k\left( {\alpha }^{2}\right) \) is intermediate between \( k \) and \( k\left( \alpha \right) \) :\n\n\[ k \subseteq k\left( {\alpha }^{2}\right) \subseteq k\left( \alpha \right) \]\n\nwhat can we say about the degree \( d \) of \( k\left( \alpha \right) \) over \( k\left( {\alpha }^{2}\right) \) ? Since \( ... | Yes |
Proposition 1.15. Let \( k \subseteq F = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) be a finitely generated field extension. Then the following are equivalent:\n\n(i) \( k \subseteq F \) is a finite extension.\n\n(ii) \( k \subseteq F \) is an algebraic extension.\n\n(iii) Each \( {\alpha }_{i} \) is algeb... | Proof. Lemma 1.9 shows that (i) \( \Rightarrow \) (ii); (ii) \( \Rightarrow \) (iii) trivially. Thus, we only need to prove that (iii) \( \Rightarrow \) (i), and to bound the degree of \( F \) over \( k \) in the process.\n\nAssume that each \( {\alpha }_{i} \) is algebraic over \( k \), and let \( {d}_{i} \) be the de... | No |
Let \( \overline{\mathbb{Q}} \subseteq \mathbb{C} \) be the set of complex numbers that are algebraic over \( \mathbb{Q} \); then \( \overline{\mathbb{Q}} \) is a field, by Corollary 1.16, and the extension \( \mathbb{Q} \subseteq \overline{\mathbb{Q}} \) is (tautologically) algebraic. | Note that \( \mathbb{Q} \subseteq \overline{\mathbb{Q}} \) is not a finite extension, because in it there are elements of arbitrarily high degree over \( \mathbb{Q} \): indeed, there exist irreducible polynomials in \( \mathbb{Q}\left\lbrack x\right\rbrack \) of arbitrarily high degree, as we observed in Corollary V15.... | Yes |
Corollary 1.18. Let \( k \subseteq E \subseteq F \) be field extensions. Then \( k \subseteq F \) is algebraic if and only if both \( k \subseteq E \) and \( E \subseteq F \) are algebraic. | Proof. If \( k \subseteq F \) is algebraic, then every element of \( F \) is algebraic over \( k \), hence over \( E \), and every element of \( E \) is algebraic over \( k \) ; thus \( E \subseteq F \) and \( k \subseteq E \) are algebraic.\n\nConversely, assume \( k \subseteq E \) and \( E \subseteq F \) are both alg... | Yes |
Consider the extension \( \mathbb{Q} \subseteq \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \). | -By Proposition 1.15 we know that this is a finite (hence algebraic) extension, of degree at most 4 .\n\n- Thus any five elements in \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \) must be linearly dependent over \( \mathbb{Q} \) . We consider powers of \( \sqrt{2} + \sqrt{3} \) :\n\n\[ 1,\;{\left( \sqrt{2} + \sqrt{3... | Yes |
Lemma 2.1. For a field \( K \), the following are equivalent:\n\n- \( K \) is algebraically closed.\n\n- \( K \) has no nontrivial algebraic extensions.\n\n- If \( K \subseteq L \) is any extension and \( \alpha \in L \) is algebraic over \( K \), then \( \alpha \in K \) . | The proof is a straightforward application of the definitions and a good exercise (Exercise 2.1). | No |
Theorem 2.3. Every field \( k \) admits an algebraic closure \( k \subseteq \bar{k} \) ; this extension is unique up to isomorphism. | Concerning existence, the idea is to construct ’by hand’ a huge extension \( K \) of \( k \) where every polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) factors completely. The elements of \( K \) which are algebraic over \( k \) will form an algebraic closure of \( k \) .\n\nThe construction is done... | No |
Lemma 2.4. Let \( k \) be a field. Then there exists an extension \( k \subseteq K \) such that every nonconstant polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) has at least one root in \( K \) . | Proof. (This construction is apparently due to Emil Artin.) Consider a set \( \mathcal{T} = \) \( \left\{ {t}_{f}\right\} \) in bijection with the set of nonconstant monic polynomials \( f\left( x\right) \in k\left\lbrack x\right\rbrack \), and let \( k\left\lbrack \mathcal{T}\right\rbrack \) be the corresponding polyn... | Yes |
Lemma 2.6. Let \( k \subseteq L \) be a field extension, with \( L \) algebraically closed. Let\n\n\[ \bar{k} \mathrel{\text{:=}} \{ \alpha \in L \mid \alpha \text{ is algebraic over }k\} .\n\]\n\nThen \( \bar{k} \) is an algebraic closure of \( k \) . | The construction reviewed above provides us with an algebraically closed field \( L \) containing any given field \( k \), so the lemma is all we need to prove.\n\nBy Corollary 1.16, \( \bar{k} \) is a field, and the extension \( k \subseteq \bar{k} \) is tautologically algebraic. To verify that \( \bar{k} \) is algebr... | Yes |
Lemma 2.8. Let \( k \subseteq L \) be a field extension, with \( L \) algebraically closed. Let \( k \subseteq F \) be any algebraic extension. Then there exists a morphism of extensions \( i : F \rightarrow L \) . | Proof. This argument also relies on Zorn’s lemma. Consider the set \( Z \) of homomorphisms\n\n\[ \n{i}_{K} : K \rightarrow L \n\] \n\nwhere \( K \) is an intermediate field, \( k \subseteq K \subseteq F \), and \( {i}_{K} \) restricts to the identity on \( k \) ; \( Z \) is nonempty, since the extension \( {i}_{k} : k... | Yes |
Theorem 2.9 (Nullstellensatz). Let \( k \subseteq F \) be a field extension, and assume that \( F \) is a finite-type \( k \) -algebra. Then \( k \subseteq F \) is a finite (hence algebraic) extension. | ## Proof for uncountable fields. Assume that \( k \) is uncountable.\n\nLet \( k \subseteq F \) be a field extension, and assume that \( F \) is finitely generated as an algebra over \( k \) ; in particular, it is finitely generated as a field extension. We have to prove that \( k \subseteq F \) is a finite extension, ... | No |
Corollary 2.10. Let \( K \) be an algebraically closed field, and let \( I \) be an ideal of \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then \( I \) is maximal if and only if\n\n\[ I = \left( {{x}_{1} - {c}_{1},\ldots ,{x}_{n} - {c}_{n}}\right) \]\n\nfor \( {c}_{1},\ldots ,{c}_{n} \in K \) . | Proof. For \( {c}_{1},\ldots ,{c}_{n} \in K \)\n\n\[ \frac{K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack }{\left( {x}_{1} - {c}_{1},\ldots ,{x}_{n} - {c}_{n}\right) } \cong K \]\n\n(Exercise III 4.12) is a field; therefore \( \left( {{x}_{1} - {c}_{1},\ldots ,{x}_{n} - {c}_{n}}\right) \) is maximal. Conversely, ... | No |
What feature of the ideal \( \left( {{y}^{2} - {x}^{3}}\right) \) is responsible for the ’cusp’ at \( \left( {0,0}\right) \) in the second picture? | The reader will find out in any course in elementary algebraic geometry. | No |
Lemma 2.15. Let \( K \) be a field, and let \( S \) be a subset of \( {\mathbb{A}}_{K}^{n} \) . Then the ideal \( \mathcal{I}\left( S\right) \) is a radical ideal of \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . | Proof. The inclusion \( \mathcal{I}\left( S\right) \subseteq \sqrt{\mathcal{I}\left( S\right) } \) holds for every ideal, so it is trivially satisfied. To verify the inclusion \( \sqrt{\mathcal{I}\left( S\right) } \subseteq \mathcal{I}\left( S\right) \), let \( f \in \sqrt{\mathcal{I}\left( S\right) } \) . Then there i... | Yes |
Proposition 2.16 (Weak Nullstellensatz). Let \( K \) be an algebraically closed field, and let \( I \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an ideal. Then \( \mathcal{V}\left( I\right) = \varnothing \) if and only if \( I = \left( 1\right) \) . | Proof. If \( I = \left( 1\right) \), then \( \mathcal{V}\left( I\right) = \varnothing \) by definition.\n\nConversely, assume that \( I \neq \left( 1\right) \) . By Proposition V13.5, \( I \) is then contained in a maximal ideal \( \mathfrak{m} \) . Since \( K \) is algebraically closed, by Corollary 2.10 we have\n\n\[... | Yes |
Corollary 2.18. Let \( K \) be an algebraically closed field. Then for any \( n \geq 0 \) the functions\n\n\[ \left\{ {\text{ algebraic subsets of }{\mathbb{A}}_{K}^{n}}\right\} \overset{\mathcal{I}}{ \leftrightarrow }\left\{ {\text{ radical ideals in }K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack }\right\} \]\n... | Proof. Proposition 2.17 shows that \( \mathcal{I} \circ \mathcal{V} \) is the identity on radical ideals, so \( \mathcal{V} \) is injective, and \( \mathcal{V} \) is surjective by definition of affine algebraic set. It follows that \( \mathcal{V} \) is a bijection and \( \mathcal{I} \) is its inverse. | Yes |
Lemma 3.3. The subset \( {\mathcal{C}}_{\mathbb{R}} \subseteq \mathbb{R} \) of constructible numbers is a subfield of \( \mathbb{R} \). Likewise, \( {\mathcal{C}}_{\mathbb{C}} \) is a subfield of \( \mathbb{C} \), and in fact \( {\mathcal{C}}_{\mathbb{C}} = {\mathcal{C}}_{\mathbb{R}}\left( i\right) \). | Proof. The set \( {\mathcal{C}}_{\mathbb{R}} \subseteq \mathbb{R} \) is nonempty, so in order to show it is a field, we only need to show that it is closed with respect to subtraction and division by a nonzero constructible number (cf. Proposition III6.2).\n\nThe reader will check that \( {\mathcal{C}}_{\mathbb{R}} \) ... | No |
Theorem 3.4. Let \( \gamma \in \mathbb{R} \) . Then \( \gamma \in {\mathcal{C}}_{\mathbb{R}} \) if and only if there exist real numbers \( {\delta }_{1},\ldots ,{\delta }_{k} \) such that \( \forall j = 1,\ldots, k \)\n\n\[ \left\lbrack {\mathbb{Q}\left( {{\delta }_{1},\ldots ,{\delta }_{j}}\right) : \mathbb{Q}\left( {... | Proof. Let's first argue in the 'geometry to algebra' direction. A configuration of points, lines, and circles obtained by a straightedge-and-compass construction may be described by the coordinates of the points and the equations of the lines and circles. Suppose that at one stage in a given construction all coordinat... | Yes |
Example 3.5. Regular pentagons are constructible. | Indeed, it suffices to construct the point \( A = \left( {\cos \left( {{2\pi }/5}\right) ,0}\right) \), and it so happens that \( \gamma = \cos \left( {{2\pi }/5}\right) \) satisfies\n\n(*) \n\n\[ 4{\gamma }^{2} + {2\gamma } - 1 = 0 \]\n\n(in fact, \( \gamma \) is half of the inverse of the golden ratio: \( \gamma = \f... | No |
Corollary 3.6. Let \( \gamma \in {\mathcal{C}}_{\mathbb{C}} \) be a constructible number. Then \( \left\lbrack {\mathbb{Q}\left( \gamma \right) : \mathbb{Q}}\right\rbrack \) is a power of 2 . | Proof. By Lemma 3.3 and Theorem 3.4, there exist \( {\delta }_{1}\ldots ,{\delta }_{k} \in \mathbb{R} \) such that\n\n\[ \gamma \in \mathbb{Q}\left( {{\delta }_{1},\ldots ,{\delta }_{k}, i}\right) \]\n\nand each \( {\delta }_{j} \) has degree \( \leq 2 \) over \( \mathbb{Q}\left( {{\delta }_{1},\ldots ,{\delta }_{j - 1... | Yes |
The splitting field \( F \) of \( {x}^{8} - 1 \) over \( \mathbb{Q} \) is generated by \( \zeta \mathrel{\text{:=}} {e}^{{2\pi i}/8} \) : indeed, the roots of \( {x}^{8} - 1 \) are all the 8-th roots of 1, and all of them are powers of \( \zeta \) : | In fact, \( \zeta \) is a root of the polynomial \( {x}^{4} + 1 \), which is irreducible over \( \mathbb{Q} \) ; therefore \( F = \mathbb{Q}\left( \zeta \right) \) is ’already’ the splitting field of \( {x}^{4} + 1 \) . The degree of \( F \) over \( \mathbb{Q} \) is \[ \left\lbrack {\mathbb{Q}\left( \zeta \right) : \ma... | Yes |
Example 4.5. Variation on the theme: \( {x}^{4} - 1 \) . The situation changes if instead of \( {x}^{4} + 1 \) we consider \( {x}^{4} - 1 \) : this polynomial factors over \( \mathbb{Q} \) , | \[ {x}^{4} - 1 = \left( {x - 1}\right) \left( {x + 1}\right) \left( {{x}^{2} + 1}\right) \] and it follows that the splitting field is the same as for \( {x}^{2} + 1 \), that is, just \( \mathbb{Q}\left( i\right) \) . | Yes |
Variation on the theme: \( {x}^{4} + 2 \) . The overoptimistic reader may now hope that the difference between the splitting fields of \( {x}^{4} + 1 \) vs. \( {x}^{4} - 1 \) over \( \mathbb{Q} \) is just due to the fact that the first polynomial is irreducible over \( \mathbb{Q} \) and the second is not. This example ... | With notation as in Example 4.4, the roots of \( {x}^{4} + 2 \) are\n\n\[ \sqrt[4]{2}\zeta ,\sqrt[4]{2}{\zeta }^{3},\sqrt[4]{2}{\zeta }^{5},\sqrt[4]{2}{\zeta }^{7} \]\n\nTherefore, with \( K = \mathbb{Q}\left( {\sqrt[4]{2}\zeta ,\sqrt[4]{2}{\zeta }^{3},\sqrt[4]{2}{\zeta }^{5},\sqrt[4]{2}{\zeta }^{7}}\right) \) the spli... | Yes |
If a complex root of an irreducible polynomial \( p\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) may be expressed as a polynomial in \( i \) and \( \sqrt[4]{2} \) with rational coefficients, then all roots of \( p\left( x\right) \) may be expressed likewise in terms of \( i \) and \( \sqrt[4]{2} \). | Indeed, we have checked (Example 4.6) that \( \mathbb{Q}\left( {i,\sqrt[4]{2}}\right) \) is a splitting field over \( \mathbb{Q} \) ; hence it is a normal extension of \( \mathbb{Q} \). | Yes |
Let \( p \) be a prime, and consider the field \( {\mathbb{F}}_{p}\left( t\right) \) of rational functions over \( {\mathbb{F}}_{p} \). Then the polynomial \[ {x}^{p} - t \in {\mathbb{F}}_{p}\left( t\right) \left\lbrack x\right\rbrack \] is irreducible. | by Eisenstein’s criterion it is irreducible in \( {\mathbb{F}}_{p}\left\lbrack t\right\rbrack \left\lbrack x\right\rbrack \) (since \( \left( t\right) \) is prime in \( {\mathbb{F}}_{p}\left\lbrack t\right\rbrack \) ), hence in \( {\mathbb{F}}_{p}\left( t\right) \left\lbrack x\right\rbrack \) by Proposition V14.16 Let ... | Yes |
Lemma 4.13. Let \( k \) be a field, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) . Then \( f\left( x\right) \) is separable if and only if \( f\left( x\right) \) and \( {f}^{\prime }\left( x\right) \) are relatively prime. | Proof. First assume that \( f\left( x\right) \) is not separable. Then \( f\left( x\right) \) has a multiple root in a splitting field \( F \) ; that is,\n\n\[ f\left( x\right) = {\left( x - \alpha \right) }^{m}g\left( x\right) \]\n\nfor some \( \alpha \in F, g\left( x\right) \in F\left\lbrack x\right\rbrack \), and \(... | Yes |
Lemma 4.14. Let \( k \) be a field, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be an inseparable irreducible polynomial. Then \( {f}^{\prime }\left( x\right) = 0 \) . | Proof. Since \( f\left( x\right) \) is inseparable, \( f\left( x\right) \) and \( {f}^{\prime }\left( x\right) \) have a common irreducible factor \( q\left( x\right) \) by Lemma 4.13 but as \( f\left( x\right) \) is itself irreducible, \( q\left( x\right) \) must be an associate of \( f\left( x\right) \), and in parti... | Yes |
Proposition 4.17. Let \( k \) be a field. Then \( k \) is perfect if and only if all irreducible polynomials in \( k\left\lbrack x\right\rbrack \) are separable. | Proof. I will prove that irreducible polynomials over a perfect field are separable, leaving the other implication to the reader (Exercise 4.12).\n\nWe have already noted that irreducible polynomials are separable over fields of characteristic zero. In positive characteristic \( p \), we have observed that an inseparab... | No |
Corollary 4.18. Finite fields are perfect. Therefore, over finite fields, irreducible polynomials are separable. | Proof. The Frobenius map is injective (because it is a homomorphism of fields), so it is surjective over finite fields, by the pigeon-hole principle. Therefore finite fields are perfect, and the second part of the statement follows from Proposition 4.17. | Yes |
Lemma 4.22. Let \( k \subseteq k\left( \alpha \right) \) be a simple algebraic extension. Then \( {\left\lbrack k\left( \alpha \right) : k\right\rbrack }_{s} \) equals the number of distinct roots in \( \bar{k} \) of the minimal polynomial of \( \alpha \) . In particular, \( {\left\lbrack k\left( \alpha \right) : k\rig... | Proof. The proof is essentially (and not by coincidence) a rehash of the proof of Corollary 1.7. Associate with each \( \iota : k\left( \alpha \right) \rightarrow \bar{k} \) extending \( {\operatorname{id}}_{k} \) the image \( \iota \left( \alpha \right) \), which must be a root of the minimal polynomial of \( \alpha \... | Yes |
Lemma 4.23. Let \( k \subseteq E \subseteq F \) be algebraic extensions. Then \( {\left\lbrack F : k\right\rbrack }_{s} \) is finite if and only if both \( {\left\lbrack F : E\right\rbrack }_{s},{\left\lbrack E : k\right\rbrack }_{s} \) are finite, and in this case \[ {\left\lbrack F : k\right\rbrack }_{s} = {\left\lbr... | Proof. Different embeddings of \( E \) into \( \bar{k} \) extend to different embeddings of \( F \) into \( \bar{k} \) , by Lemma 2.8, and embeddings of \( F \) into \( \bar{E} = \bar{k} \) extending the identity on \( E \) extend a fortiori the identity on \( k \) . Therefore, if any of \( {\left\lbrack F : E\right\rb... | Yes |
Proposition 4.24. Let \( k \subseteq F \) be a finite extension. Then \( {\left\lbrack F : k\right\rbrack }_{s} \leq \left\lbrack {F : k}\right\rbrack \), and the following are equivalent:\n\n(i) \( F = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \), where each \( {\alpha }_{i} \) is separable over \( k \) ;\n... | Proof. Since \( F \) is finite over \( k \), it is finitely generated. Let \( F = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) . Then using Lemma 4.23, Lemma 4.22, and Proposition 1.10,\n\n\[{\left\lbrack F : k\right\rbrack }_{s} = {\left\lbrack k\left( {\alpha }_{1},\ldots ,{\alpha }_{r - 1}\right) \left( {... | Yes |
Theorem 5.1. Let \( q = {p}^{d} \) be a power of a prime integer \( p \) . Then the polynomial \( {x}^{q} - x \) is separable over \( {\mathbb{F}}_{p} \), and the splitting field of the polynomial \( {x}^{q} - x \) over \( {\mathbb{F}}_{p} \) is a field with precisely \( q \) elements. Conversely, let \( F \) be a fiel... | Proof. Let \( F \) be the splitting field of \( {x}^{q} - x \) over \( {\mathbb{F}}_{p} \) . Let \( E \) be the set of roots of \( f\left( x\right) = {x}^{q} - x \) in \( F \) . Since \( {f}^{\prime }\left( x\right) = q{x}^{q - 1} - 1 = - 1 \) (as \( q = 0 \) in characteristic \( p \) ), we have \( \left( {f\left( x\ri... | Yes |
Corollary 5.2. For every prime power \( q \) there exists one and only one finite field of order \( q \), up to isomorphism. | Proof. This follows immediately from Theorem 5.1 and the uniqueness of splitting fields (Lemma 4.2). | Yes |
Let \( p \) be a prime integer. Then I claim that the polynomial \( {x}^{4} + 1 \) is reducible over \( {\mathbb{F}}_{p} \) (and therefore over every finite field). | Since \( {x}^{4} + 1 = {\left( x + 1\right) }^{4} \) in \( {\mathbb{F}}_{2}\left\lbrack x\right\rbrack \), the statement holds for \( p = 2 \) . Thus, we may assume that \( p \) is an odd prime. Then I claim that \( {x}^{4} + 1 \) divides \( {x}^{{p}^{2}} - x \) . Indeed, the square of every odd number is congruent to ... | No |
Corollary 5.4. Let \( p \) be a prime, and let \( d \leq e \) be positive integers. Then there is an extension \( {\mathbb{F}}_{{p}^{d}} \subseteq {\mathbb{F}}_{{p}^{e}} \) if and only if \( d \mid e \) . Further, if \( d \mid e \), then there is exactly one such extension, in the sense that \( {\mathbb{F}}_{{p}^{e}} \... | Proof. If there is an extension as stated, then \( {\mathbb{F}}_{p} \subseteq {\mathbb{F}}_{{p}^{d}} \subseteq {\mathbb{F}}_{{p}^{e}} \) ; hence \( \left\lbrack {{\mathbb{F}}_{{p}^{d}} : {\mathbb{F}}_{p}}\right\rbrack \) divides \( \left\lbrack {{\mathbb{F}}_{{p}^{e}} : {\mathbb{F}}_{p}}\right\rbrack \) by Corollary 1.... | Yes |
Corollary 5.5. Let \( F \) be a finite field. Then for all integers \( n \geq 1 \) there exist irreducible polynomials of degree \( n \) in \( F\left\lbrack x\right\rbrack \) . | Proof. We know \( F = {\mathbb{F}}_{{p}^{d}} \) for some prime \( p \) and some \( d \geq 1 \) . By Corollary 5.4 there is an extension \( {\mathbb{F}}_{{p}^{d}} \subseteq {\mathbb{F}}_{{p}^{dn}} \), generated by an element \( \alpha \) . Then \( \left\lbrack {{\mathbb{F}}_{{p}^{dn}} : {\mathbb{F}}_{{p}^{d}}}\right\rbr... | Yes |
Corollary 5.6. Let \( F = {\mathbb{F}}_{q} \) be a finite field, and let \( n \) be a positive integer. Then the factorization of \( {x}^{{q}^{n}} - x \) in \( F\left\lbrack x\right\rbrack \) consists of all irreducible monic polynomials of degree \( d \), as \( d \) ranges over the positive divisors of \( n \) . In pa... | Proof. By Theorem 5.1, \( {\mathbb{F}}_{{q}^{n}} \) is the splitting field of \( {x}^{{q}^{n}} - x \) over \( {\mathbb{F}}_{p} \), and hence over \( {\mathbb{F}}_{q} = F \) . If \( f\left( x\right) \) is a monic irreducible polynomial of degree \( d \), then \( F\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\r... | Yes |
Let’s contemplate the case \( q = 2 : {\mathbb{F}}_{2} = \mathbb{Z}/2\mathbb{Z} \). | - \( n = 1 \) : the polynomial \( {x}^{2} - x \) factors as the product of \( x \) and \( \left( {x - 1}\right) \) (which we could write as \( \left( {x + 1}\right) \) just as well, since we are working over \( {\mathbb{F}}_{2} \) ). These are all the irreducible polynomials of degree 1 over \( {\mathbb{F}}_{2} \).\n\n... | No |
Proposition 5.8. \( {\operatorname{Aut}}_{{\mathbb{F}}_{p}}\left( {\mathbb{F}}_{{p}^{d}}\right) \) is cyclic, generated by the Frobenius isomorphism. | Proof. Let \( \varphi \) be the Frobenius homomorphism \( {\mathbb{F}}_{{p}^{d}} \rightarrow {\mathbb{F}}_{{p}^{d}} : \varphi \left( x\right) = {x}^{p} \) . The Frobenius homomorphism is an isomorphism on a finite field (Corollary 4.18) and restricts to the identity on \( {\mathbb{F}}_{p} \) (Exercise 4.11), so \( \var... | Yes |
If \( n = p \) is prime, then every nonidentity element of \( {\mu }_{p} \cong {C}_{p} \) is a generator: every \( p \) -th root of 1 is primitive except 1 itself. | Therefore\n\n\[ \n{\Phi }_{p}\left( x\right) = \frac{{x}^{p} - 1}{x - 1} = {x}^{p - 1} + \cdots + 1 \n\] \n\nis the particular case encountered in Example V15.19, where we proved that \( {\Phi }_{p}\left( x\right) \) is indeed irreducible. | No |
Lemma 5.11. For all positive integers \( n \) , \n\n\[ \n{x}^{n} - 1 = \mathop{\prod }\limits_{{1 \leq d \mid n}}{\Phi }_{d}\left( x\right) \n\] | Proof. If \( n = {de} \), then every \( d \) -th root \( \zeta \) of 1 is an \( n \) -th root of 1, because \( {\zeta }^{n} = \) \( {\zeta }^{de} = {\left( {\zeta }^{d}\right) }^{e} = 1 \) . In particular, every primitive \( d \) -th root \( \zeta \) of 1 is an \( n \) -th root of 1 . \n\nOn the other hand, every \( \z... | Yes |
Corollary 5.12. The cyclotomic polynomials \( {\Phi }_{n}\left( x\right) \) have integer coefficients. | Proof. Use induction on \( n \) . Note that \( {\Phi }_{1}\left( x\right) = x - 1 \), and assume we have shown that all \( {\Phi }_{m}\left( x\right) \) have integer coefficients for \( m < n \) . In particular, \( f\left( x\right) \mathrel{\text{:=}} \) \( \mathop{\prod }\limits_{{1 \leq d \mid n, d < n}}{\Phi }_{d}\l... | Yes |
The reader can spend some quality time computing explicitly the cyclotomic polynomials \( {\Phi }_{n}\left( x\right) \) for several nonprime numbers \( n \), working inductively and capitalizing on the fact that we know explicitly \( {\Phi }_{p}\left( x\right) \) for prime \( p \) . | For example, \( {x}^{4} - 1 = {\Phi }_{1}\left( x\right) {\Phi }_{2}\left( x\right) {\Phi }_{4}\left( x\right) \) ; therefore\n\n\[ \n{\Phi }_{4}\left( x\right) = \frac{{x}^{4} - 1}{{x}^{2} - 1} = {x}^{2} + 1.\n\]\n\nSince \( {x}^{6} - 1 = {\Phi }_{1}\left( x\right) {\Phi }_{2}\left( x\right) {\Phi }_{3}\left( x\right)... | Yes |
Proposition 5.14. For all positive \( n,{\Phi }_{n}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) is irreducible over \( \mathbb{Q} \) . | (Since \( {\Phi }_{n}\left( x\right) \) is monic, irreducibility in \( \mathbb{Q}\left\lbrack x\right\rbrack \) is equivalent to irreducibility in \( \mathbb{Z}\left\lbrack x\right\rbrack \) ; cf. Corollary V 4.17\n\nProof. Arguing by contradiction, assume \( {\Phi }_{n}\left( x\right) \) is reducible. Then its roots \... | Yes |
Proposition 5.16. \( {\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) }\right) \) is isomorphic to the group of units in \( \mathbb{Z}/n\mathbb{Z} \) . | Proof. We know that \( {\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) }\right) \) has cardinality \( \phi \left( n\right) \) (Corollary 1.7) the roots are distinct since \( {\Phi }_{n}\left( x\right) \) is separable), so all we need to do is exhibit an injective homomorphism\n\n\[ j : {\l... | Yes |
Proposition 5.19. Every finite separable extension is simple. | Proof. Arguing inductively as in the proof of Proposition 5.18 we may assume \( F = k\left( {\alpha ,\beta }\right) \), with \( \alpha \) and \( \beta \) separable (and in particular algebraic) over \( k \), and we may assume \( k \) is an infinite field.\n\nConsider the set \( I \) of embeddings \( \iota : F \hookrigh... | Yes |
Corollary 5.20. Let \( k \subseteq F \) be a finite, separable extension. Then\n\n\[ \left| {{\operatorname{Aut}}_{k}\left( F\right) }\right| \leq \left\lbrack {F : k}\right\rbrack \]\n\nwith equality if and only if \( k \subseteq F \) is a normal extension. | Proof. Since \( k \subseteq F \) is finite and separable, Proposition 5.19 implies it is simple: \( F = k\left( \alpha \right) \) for some \( \alpha \in F \) . The inequality follows immediately from Corollary 1.7, and equality holds if and only if the minimal polynomial \( f\left( x\right) \) of \( \alpha \) factors i... | Yes |
Lemma 6.3. The Galois correspondence is inclusion-reversing. Further, for all subgroups \( G \) of \( {\operatorname{Aut}}_{k}\left( F\right) \) and all intermediate fields \( k \subseteq E \subseteq F \) :\n\n\[ \text{-}E \subseteq {F}^{{\operatorname{Aut}}_{E}\left( F\right) }\text{;}\]\n\n\[ \text{-}G \subseteq {\op... | ## Proof. Exercise 6.1 | No |
Consider the extension \( \mathbb{Q} \subseteq \mathbb{Q}\left( \sqrt[3]{2}\right) \). Since \[ \left\lbrack {\mathbb{Q}\left( \sqrt[3]{2}\right) : \mathbb{Q}}\right\rbrack = 3 \] is prime, the only intermediate fields are \( \mathbb{Q} \) and \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) (by Corollary 1.11). Concerning \(... | Thus, in this example the Galois correspondence acts between a set with two elements and a singleton: \[ \{ \mathbb{Q},\mathbb{Q}\left( \sqrt[3]{2}\right) \} \leftrightarrow \left\{ {{\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\left( \sqrt[3]{2}\right) }\right) }\right\} = \{ e\} . \] In particular, the function... | Yes |
Lemma 6.6. Let \( k \subseteq F \) be a finite extension, and let \( G \) be a subgroup of \( {\operatorname{Aut}}_{k}\left( F\right) \) . Then \( {F}^{G} \subseteq F \) is a finite, simple, normal, separable extension. | Proof of Lemma 6.6. The extension \( {F}^{G} \subseteq F \) is finite because \( k \subseteq F \) is finite.\n\nLet \( \alpha \in F \) ; by the remark following the statement of the lemma, \( \alpha \) is a root of a separable polynomial \( {q}_{\alpha }\left( t\right) \) with coefficients in \( {F}^{G} \) . It follows... | Yes |
Theorem 6.9. Let \( k \subseteq F \) be a finite field extension. Then the following are equivalent:\n\n(1) \( F \) is the splitting field of a separable polynomial \( f\left( t\right) \in k\left\lbrack t\right\rbrack \) over \( k \) ;\n\n(2) \( k \subseteq F \) is normal and separable;\n\n(3) \( \left| {{\operatorname... | Proof. Most of the needed implications have been proven along the way.\n\n\( \left( 1\right) \Leftrightarrow \left( 2\right) \) by Theorem 4.8. \( \left( 2\right) \Rightarrow \left( 3\right) \) by Corollary 5.20. \( \left( 3\right) \Leftrightarrow \left( 4\right) \) follows from Proposition 6.5, applied to the extensio... | Yes |
Theorem 6.12. Let \( k \subseteq F \) be a Galois extension. The Galois correspondence is an inclusion-reversing isomorphism of the lattice of intermediate subfields of \( k \subseteq F \) with the lattice of subgroups of \( {\operatorname{Aut}}_{k}\left( F\right) \) . That is (with notation as in Lemma 6.3), if \( {E}... | Proof. This follows immediately from Theorem 6.9 and Lemma 6.3, which gives \[ {\operatorname{Aut}}_{{E}_{1}{E}_{2}}\left( F\right) = {G}_{1} \cap {G}_{2},\;{F}^{\left\langle {G}_{1},{G}_{2}\right\rangle } = {E}_{1} \cap {E}_{2} \] as needed. | Yes |
The extension \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) = \mathbb{Q}\left( {\sqrt{2} + \sqrt{3}}\right) \) studied in Example 1.19 is the splitting field of the polynomial \( {t}^{4} - {10}{t}^{2} + 1 \), so it is Galois. | We found that its Galois group is \( \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \) ; the lattice of this group has no mysteries for us: and therefore the lattice of intermediate fields is just as transparent: (The intermediate fields are determined by recalling the generators of the corresponding subgroups, a... | Yes |
Proposition 6.17. Suppose \( k \subseteq F \) is a Galois extension and \( k \subseteq K \) is any finite extension. Then \( K \subseteq {KF} \) is a Galois extension, and \( {\operatorname{Aut}}_{K}\left( {KF}\right) \cong {\operatorname{Aut}}_{F \cap K}\left( F\right) \) . | Proof. As \( k \subseteq F \) is Galois, it is the splitting field of a separable polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \subseteq K\left\lbrack x\right\rbrack \) . The roots of \( f\left( x\right) \) generate \( F \) over \( k \), so they generate \( {KF} \) over \( K \) ; in other words, \( {... | Yes |
We have studied cyclotomic fields \( \mathbb{Q}\left( {\zeta }_{n}\right) \) as extensions of \( \mathbb{Q};\mathbb{Q}\left( {\zeta }_{n}\right) \) is the splitting field of \( {x}^{n} - 1 \), so these extensions are Galois; we have proved (Proposition 5.16) that \( {\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\l... | Now let \( k \) be any field of characteristic zero. The splitting field of \( {x}^{n} - 1 \) over \( k \) is the composite \( k\left( \zeta \right) \) of \( k \) and \( \mathbb{Q}\left( \zeta \right) \) . By Proposition 6.17 the extension \( k \subseteq k\left( \zeta \right) \) is Galois, and \( {\operatorname{Aut}}_{... | Yes |
Proposition 6.19. Let \( k \subseteq F \) be an extension of degree \( m \) . Assume that \( k \) contains a primitive \( m \) -th root of 1 and char \( k \) does not divide 24 \( m \) . Then \( k \subseteq F \) is Galois and cyclic if and only if \( F = k\left( \delta \right) \), with \( {\delta }^{m} \in k \) . | Proof. Let \( \zeta \in k \) be a primitive \( m \) -th root of 1 . First assume that \( F = k\left( \delta \right) \), with \( {\delta }^{m} = c \in k \) . Then all \( m \) roots of the polynomial \( {x}^{m} - c \) , \[ \delta ,{\zeta \delta },{\zeta }^{2}\delta ,\cdots ,{\zeta }^{m - 1}\delta \] are in \( F \), and \... | Yes |
Theorem 7.1. \( \mathbb{C} \) is algebraically closed. | Proof. Let \( f\left( x\right) \in \mathbb{C}\left\lbrack x\right\rbrack \) be a nonconstant polynomial; we have to prove that \( f\left( x\right) \) has roots in \( \mathbb{C} \). Note that if \( f\left( x\right) \) has no roots in \( \mathbb{C} \), then neither does \( f\left( x\right) \overline{f\left( x\right) } \i... | No |
Proposition 7.2. Let \( k \subseteq F \) be a Galois extension, and assume \( \left\lbrack {F : k}\right\rbrack = {p}^{r} \) for some prime \( p \) and \( r \geq 0 \) . Then there exist intermediate fields\n\n\[ k = {E}_{0} \subseteq {E}_{1} \subseteq {E}_{2} \subseteq \cdots \subseteq {E}_{r} = F \]\n\nsuch that \( \l... | Proof. As the Galois correspondence is bijective for Galois extensions (Theorem 6.9, part (5)), this statement follows immediately from the fact that a group of order \( {p}^{r} \), with \( p \) prime, has a complete series of \( p \) -subgroups; cf. for example the discussion following the statement of Theorem IV 2.8 | No |
Theorem 7.3. The regular n-gon is constructible by straightedge and compass if and only if \( \phi \left( n\right) \) is a power of 2 . | Proof. As recalled above, we have already established the \( \Rightarrow \) direction.\n\nFor the converse, assume \( \phi \left( n\right) = {2}^{r} \) for some \( r \) . The extension \( \mathbb{Q} \subseteq \mathbb{Q}\left( {\zeta }_{n}\right) \) is Galois (it is the splitting field of \( {\Phi }_{n}\left( x\right) \... | Yes |
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