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Proposition 8.1.3. If \( A \) is an abelian group and \( \phi : G \rightarrow A \) is a homomorphism, then \( {G}^{\prime } \subseteq \ker \phi \) . Moreover, \( \phi \) factors as the composition of\n\n(8.1.3.1)\n\n\[ G\overset{\pi }{ \rightarrow }G/{G}^{\prime }\overset{\bar{\phi }}{ \rightarrow }A \]\n\n\[ g \mapsto... | Proof. Note that for any \( x, y \in G,\phi \left( \left\lbrack {x, y}\right\rbrack \right) = \phi \left( {{x}^{-1}{y}^{-1}{xy}}\right) = \phi {\left( x\right) }^{-1}\phi {\left( y\right) }^{-1}\phi \left( x\right) \phi \left( y\right) = 1 \) because \( A \) is abelian. So \( {G}^{\prime } \subseteq \ker \phi \) . It f... | Yes |
For \( G = {D}_{2n} = \left\langle {r, s \mid {r}^{n} = {s}^{2} = 1,{srs} = {r}^{-1}}\right\rangle \), compute all homomorphisms \( {\operatorname{Hom}}_{\mathrm{{gp}}}\left( {G,{\mathbb{C}}^{ \times }}\right) \). | First note that \( {G}^{\prime } \) contains \( {sr}{s}^{-1}{r}^{-1} = {r}^{-2} \). We separate two cases.\n\n- If \( n \) is odd, then \( \langle r\rangle = \left\langle {r}^{-2}\right\rangle \subseteq {G}^{\prime } \). We claim that \( {G}^{\prime } = \langle r\rangle \). Instead of checking every pair of elements, w... | Yes |
A group is solvable if and only if \( {G}^{\left( n\right) } = \{ 1\} \) for some finite \( n \in \mathbb{N} \) . | Proof. \ | No |
Lemma 8.2.5. All \( {G}^{\left( i\right) } \) are normal subgroups of \( G \) . In fact, they are characteristic subgroups of \( G \) . | Proof. Recall that \( {G}^{\left( 1\right) } = \left\lbrack {G, G}\right\rbrack = \left\langle {{x}^{-1}{y}^{-1}{xy} \mid x, y \in G}\right\rangle \) . If \( \phi : G \rightarrow G \) is an automorphism, we have\n\n\[ \phi \left( {G}^{\left( 1\right) }\right) = \left\langle {\phi \left( {{x}^{-1}{y}^{-1}{xy}}\right) \m... | Yes |
Corollary 8.3.3. If \( G \) is nilpotent, then \( G \) is solvable. | Proof. If \( {G}^{c} = \{ 1\} \) for some \( c \in \mathbb{N} \), then \( {G}^{\left( c\right) } \leq {G}^{c} = \{ 1\} \) . So \( {G}^{\left( c\right) } = 1 \) . | Yes |
Theorem 8.3.7. A group \( G \) is nilpotent if and only if \( {Z}_{c}\left( G\right) = G \) for some \( c \in \mathbb{N} \). | Proof. We prove this by induction on the minimal \( c \) such that either \( {G}^{c} = \{ 1\} \) or \( {Z}_{c}\left( G\right) = G \) . When \( c = 1 \), either conditions \( {G}^{1} = \{ 1\} \) and \( {Z}_{1}\left( G\right) = G \) is equivalent to the condition that \( G \) is abelian. The statement is clear.\n\nNow su... | Yes |
Proposition 8.4.1. All p-groups are nilpotent. | Proof. This is because for every \( p \) -group \( P, Z\left( P\right) \) is nontrivial by Proposition 6.3.4. | No |
Proposition 8.4.2. Let \( P \) be a p-group.\n\n(2) If \( H \trianglelefteq P \) is a nontrivial normal subgroup, then \( H \cap Z\left( P\right) \neq \{ 1\} \) . | Proof. (1) is proved in Proposition 6.3.4.\n\n(2) Consider the conjugation action of \( P \) on \( H \) :\n\n\[ P\overset{\mathrm{{Ad}}}{ \hookrightarrow }H,\;{\operatorname{Ad}}_{p}\left( h\right) = {ph}{p}^{-1}. \]\n\nFor this action, we write \( H \) as the disjoint union of orbits:\n\n\[ H = \mathop{\coprod }\limit... | Yes |
Theorem 8.4.4 (Classification theorem for nilpotent groups). Let \( G \) be a finite group of order \( n = {p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}\cdots {p}_{r}^{{\alpha }_{r}} \) and \( {P}_{i} \in {\operatorname{Syl}}_{{p}_{i}}\left( G\right) \) . The following are equivalent.\n\n(1) \( G \) is nilpotent.\n\n(... | Proof. \( \left( 3\right) \Rightarrow \left( 4\right) \) By criterion of direct products:\n\n\[ \n{P}_{1}{P}_{2} = {P}_{1} \times {P}_{2},\;{P}_{1}{P}_{2}{P}_{3} = {P}_{1}{P}_{2} \times {P}_{3} = {P}_{1} \times {P}_{2} \times {P}_{3},\ldots \n\] \n\n\( \left( 4\right) \Rightarrow \left( 1\right) \) as each \( {P}_{i} \... | Yes |
1-dimensional representations are in one-to-one correspondence with homomorphisms \( \rho : G \rightarrow \mathrm{{GL}}\left( \mathbb{C}\right) = {\mathbb{C}}^{ \times } \) . | We have seen in Proposition 8.1.3 that\n\n\( {\operatorname{Hom}}_{\mathrm{{gp}}}\left( {G,{\mathbb{C}}^{ \times }}\right) \; = \;{\operatorname{Hom}}_{\mathrm{{gp}}}\left( {G/{G}^{\prime },{\mathbb{C}}^{ \times }}\right) \)\n\n\( \{ 1 \) -dimensional representations of \( G\} \leftrightarrow \{ 1 \) -dimensional repre... | No |
Assume that \( G \) is finite. Consider the regular representation in Example 9.1.3(2). Then the subspace \( W = \left\{ {\left. {a \cdot \mathop{\sum }\limits_{{h \in G}}\left\lbrack h\right\rbrack }\right| \;a \in \mathbb{C}}\right\} \subseteq \mathbb{C}\left\lbrack G\right\rbrack \) is one-dimensional, and can be se... | The corresponding representation \( {\rho }_{W} : G \rightarrow \mathrm{{GL}}\left( W\right) = {\mathbb{C}}^{ \times } \) is the trivial representation \( {\rho }_{W}\left( g\right) = 1 \) for all \( g \in G \) . | Yes |
Theorem 9.2.5. Let \( G \) be a finite group. If \( W \subseteq V \) is a subrepresentation of \( G \), then there exists a subrepresentation \( {W}^{ \circ } \subseteq V \) such that \[ V = W \oplus {W}^{ \circ } \] We call \( {W}^{ \circ } \) a complementary representation of \( W \) in \( V \). | 9.2.7. Proof of Theorem 9.2.5. Pick an arbitrary subspace \( {W}^{\prime } \subseteq V \) such that \( V = W \oplus {W}^{\prime } \) as \( \mathbb{C} \) -vector spaces. This is equivalent to give a map \( \operatorname{pr} : V \rightarrow W \) (projection along \( {W}^{\prime } \) ) such that \( {\left. \operatorname{p... | No |
Proposition 10.1.6. Let \( \left( {{\rho }_{1},{V}_{1}}\right) \) and \( \left( {{\rho }_{2},{V}_{2}}\right) \) be two representations of \( G \) . Then for every \( g \in G \) ,\n\n(1) \( {\chi }_{{\rho }_{1} \oplus {\rho }_{2}}\left( g\right) = {\chi }_{{\rho }_{1}}\left( g\right) + {\chi }_{{\rho }_{2}}\left( g\righ... | Proof. For (1), it follows from that\n\n\[ \left( {{\rho }_{1} \oplus {\rho }_{2}}\right) \left( g\right) = \left( \begin{matrix} {\rho }_{1}\left( g\right) & 0 \\ 0 & {\rho }_{2}\left( g\right) \end{matrix}\right) \]\n\nFor (2), we write matrix form of \( {\rho }_{1} \) and \( {\rho }_{2} \) by \( {\rho }_{1}\left( g\... | Yes |
Proposition 10.2.4 (Schur’s lemma). Let \( \left( {{\rho }_{1},{V}_{1}}\right) \) and \( \left( {{\rho }_{2},{V}_{2}}\right) \) be irreducible \( \mathbb{C} \) -representations of \( G \) and let \( \phi : {V}_{1} \rightarrow {V}_{2} \) be a homomorphism. Then\n\n(1) If \( \left( {{\rho }_{1},{V}_{1}}\right) \) and \( ... | Proof. The case \( \phi = 0 \) being trivial, we now assume that \( \phi \neq 0 \) .\n\n(a) The kernel of \( \phi \) is a subrepresentation of \( {V}_{1} \) (and \( \phi \) is not zero); so \( \ker \left( \phi \right) = 0 \) . Thus \( \phi \) is injective.\n\n(b) The image of \( \phi \) is a subrepresentation of \( {V}... | Yes |
Corollary 10.2.5. Let \( \left( {{\rho }_{1},{V}_{1}}\right) \) and \( \left( {{\rho }_{2},{V}_{2}}\right) \) be two irreducible representations of \( G \) . Let \( \phi : {V}_{1} \rightarrow {V}_{2} \) be a \( \mathbb{C} \) -linear map (not necessarily \( G \) -linear). Then\n\n\[ \widetilde{\phi } \mathrel{\text{:=}}... | Proof. Construction 9.2.6 implies that \( \widetilde{\phi } \) is a homomorphism. Then Schur’s lemma implies (1) immediate. For (2), Schur’s lemma implies that \( \widetilde{\phi } = \lambda \cdot {\operatorname{id}}_{{V}_{1}} \) . The first equality in (10.2.5.1) holds because \( \widetilde{\phi } \) is scalar multipl... | Yes |
Proposition 10.2.6. Assume that \( \left( {{\rho }_{1},{V}_{1}}\right) ≄ \left( {{\rho }_{2},{V}_{2}}\right) \) . Identify \( {V}_{1} \) with \( {\mathbb{C}}^{m} \) and \( {V}_{2} \) with \( {\mathbb{C}}^{n} \) , and thus, the two representations can be viewed as\n\n\[ g \mapsto {\left( {\rho }_{1}{\left( g\right) }_{i... | Proof. Consider \( \phi : {V}_{1} \rightarrow {V}_{2} \) given by the matrix \( {E}_{\ell i} \), which is zero in all entries except having 1 at the \( \left( {\ell, i}\right) \) -entry. Then Corollary 10.2.5(1) implies that\n\n\[ \mathop{\sum }\limits_{{g \in G}}{\rho }_{2}\left( g\right) {E}_{\ell i}{\rho }_{1}\left(... | Yes |
Proposition 10.3.1. For a general representation \( \left( {\rho, V}\right) \) of \( G \) with character \( {\chi }_{V} \), it is the direct sum of irreducible representations \( V = {W}_{1}^{\oplus {m}_{1}} \oplus {W}_{2}^{\oplus {m}_{2}} \oplus \cdots \oplus {W}_{r}^{\oplus {m}_{r}} \) (where \( {W}_{1},\ldots ,{W}_{... | \[ \left\langle {{\chi }_{V},{\chi }_{{W}_{i}}}\right\rangle = \left\langle {\mathop{\sum }\limits_{j}{m}_{j}{\chi }_{{W}_{j}},{\chi }_{{W}_{i}}}\right\rangle = \mathop{\sum }\limits_{j}{m}_{j} \cdot \left\{ \begin{array}{ll} 1 & \text{ if }{W}_{j} \simeq {W}_{i} \\ 0 & \text{ if }{W}_{j} \simeq {W}_{i} \end{array}\rig... | Yes |
For \( G = {\mathbf{Z}}_{n} \), the cyclic group. All irreducible representations are, for \( b = 0,1,\ldots, n - 1 \) : | \[ {\chi }_{b} : {\mathbf{Z}}_{n} \rightarrow {\mathbb{C}}^{ \times } \] \[ 1 \mapsto {e}^{{2\pi i}/n} \] \[ a{\;\operatorname{mod}\;n} \mapsto {e}^{{2\pi iab}/n}. \] We have \( \left| {\operatorname{Irr}\left( W\right) }\right| = n = \left| {\mathbf{Z}}_{n}\right| \) . (Since \( {\mathbf{Z}}_{n} \) is abelian, all con... | Yes |
Lemma 11.1.9. A finite integral domain \( R \) is a field. | Proof. For any nonzero element \( a \in R \), we need to find its inverse. Consider the following homomorphism of additive groups\n\n\[ \begin{matrix} {\phi }_{a} : \left( {R, + }\right) \rightarrow \left( {R, + }\right) \\ x \mapsto {ax} \end{matrix} \]\n\nThen \( \ker {\phi }_{a} = \{ x \in R \mid {ax} = 0\} = \{ 0\}... | Yes |
Theorem 11.2.4 (Isomorphism Theorems). (1) If \( \phi : R \rightarrow S \) is a ring homomorphism, then \( \ker \phi \) is a two-sided ideal and \( \phi \left( R\right) \) is a subring of \( S \) . | Moreover, \( \phi \) induces an isomorphism\n\n\[ R/\ker \phi \xrightarrow[]{ \cong }\phi \left( R\right) \]\n\n\[ x + \ker \phi \mapsto \phi \left( x\right) \] | Yes |
In \( R = \mathbb{Z} \), \(\left( {4,6}\right) = \{ {4x} + {6y} \mid x, y \in \mathbb{Z}\} = 2\mathbb{Z} = \left( 2\right) . | In general, \( \left( {{a}_{1},\ldots ,{a}_{s}}\right) = \left( {\gcd \left( {{a}_{1},\ldots ,{a}_{s}}\right) }\right) \). | Yes |
If \( R \) is a commutative ring and \( I = \left( {{a}_{1},\ldots ,{a}_{s}}\right) \) and \( J = \left( {{b}_{1},\ldots ,{b}_{t}}\right) \), then | \[ I + J = \left( {{a}_{1},\ldots ,{a}_{s},{b}_{1},\ldots ,{b}_{t}}\right) ,\;{IJ} = \left( {{a}_{1}{b}_{1},\ldots ,{a}_{1}{b}_{t},\ldots ,{a}_{i}{b}_{j},\ldots ,{a}_{s}{b}_{t}}\right) . \] | Yes |
Theorem 12.1.2. Let \( {I}_{1},\ldots ,{I}_{k} \) be ideals of a commutative ring \( R \) . Then the natural map\n\n\[ \n\phi : R \rightarrow R/{I}_{1} \times \cdots \times R/{I}_{k} \n\]\n\n\[ \nx \mapsto \left( {x{\;\operatorname{mod}\;{I}_{1}},\ldots, x{\;\operatorname{mod}\;{I}_{k}}}\right) \n\]\n\nis a ring homomo... | Proof. The first claim on \( \phi \) being a homomorphism with kernel \( {I}_{1} \cap \cdots \cap {I}_{k} \) is clear. We now prove (1) and (2).\n\nWe first assume that \( k = 2 \) . As \( {I}_{1}{I}_{2} \subseteq {I}_{1} \) and \( {I}_{1}{I}_{2} \subseteq {I}_{2} \), we have \( {I}_{1}{I}_{2} \subseteq {I}_{1} \cap {I... | Yes |
Lemma 12.2.4. For an element \( u \in \mathcal{O}, u \in {\mathcal{O}}^{ \times } \) if and only if \( \operatorname{Nm}\left( u\right) = \pm 1 \) . | Proof. \ | No |
Proposition 12.4.2. Every proper (two-sided) ideal \( I \varsubsetneq R \) is contained in a maximal ideal of \( R \) . | Proof. Put \( \mathcal{S} \mathrel{\text{:=}} \{ \) proper ideals of \( R \) containing \( I\} \) . I claim that it is a partially ordered set for inclusion. For this, we need to check that every increasing chain \( {J}_{i} \subseteq \cdots \) of ideals has an upper bound. Indeed, \( J = \mathop{\bigcup }\limits_{{i \i... | Yes |
Proposition 12.4.3. Let \( R \) be a commutative ring. An ideal \( \mathfrak{m} \subseteq R \) is maximal if and only if the quotient \( R/\mathfrak{m} \) is a field. | Proof. By lattice isomorphism theorem, \( \mathfrak{m} \subseteq R \) if and only if \( \bar{R} \mathrel{\text{:=}} R/\mathfrak{m} \) has only two ideals (0) and (1). We claim that the latter statement is equivalent to that \( \bar{R} \) is a field.\n\nIf \( \bar{R} \) is a field, then clearly it has only two ideals \(... | Yes |
Proposition 12.5.3. An ideal \( \mathfrak{p} \subset R \) is a prime ideal if and only if \( R/\mathfrak{p} \) is an integral domain. | Proof. Consider the natural quotient\n\n\[ \pi : R \rightarrow R/\mathfrak{p} \]\n\n\[ a \mapsto \bar{a}\text{.} \]\n\nIf \( R/\mathfrak{p} \) is an integral domain, then for \( a, b \in R \) with \( {ab} \in \mathfrak{p} \), we must have \( \overline{ab} = 0 \), or equivalently \( \bar{a}\bar{b} = 0 \) . This means th... | Yes |
Proposition 12.6.3. Every nonzero prime ideal in a PID is a maximal ideal. | Proof. Let \( \left( p\right) \) be a prime ideal in a PID \( R \) . If \( \mathfrak{m} = \left( m\right) \supseteq \left( p\right) \) is a maximal ideal containing \( \left( p\right) \) . Then \( p = {mn} \) for some \( n \in R \) . Using the property of prime ideals, we see that either \( m \) or \( n \) belongs to \... | Yes |
Example 13.1.3. (3) The ring of Gaussian integers \( R = \mathbb{Z}\left\lbrack i\right\rbrack \) admits a norm\n\n\[ \operatorname{Nm}\left( {x + {yi}}\right) = {x}^{2} + {y}^{2} = {\left| x + yi\right| }^{2}. \] | When \( a, b \in \mathbb{Z}\left\lbrack i\right\rbrack \) with \( b \neq 0 \), let \( q \in \mathbb{Z}\left\lbrack i\right\rbrack \) is taken so that \( \left| {\operatorname{Re}\left( {q - \frac{a}{b}}\right) }\right| \leq \frac{1}{2} \) and \( \left| {\operatorname{Im}\left( {q - \frac{a}{b}}\right) }\right| \leq \fr... | Yes |
Proposition 13.1.4. A Euclidean domain \( R \) is a PID. | Proof. Let \( I \subseteq R \) be a nonzero ideal. Let \( b \) be an element of \( I \smallsetminus \{ 0\} \) with minimal possible norm. We claim that \( I = \left( b\right) \) . It is clear that \( \left( b\right) \subseteq I \) . Now we focus on the other inclusion.\n\nIf \( a \in I \), by Euclidean algorithm, \( a ... | Yes |
Proposition 13.2.2. (1) Prime elements are always irreducible. | Proof. (1) Let \( p \in R \) be a prime element. Then if \( p = {uv} \) for some \( u, v \in R \), then\n\n\[ {uv} \in \left( p\right) \Rightarrow u \in \left( p\right) \text{ or }v \in \left( p\right) . \]\n\nWLOG, \( u = {ps} \) for some \( s \in R \), then \( p = {uv} = {psv} \) . Thus \( 1 = {sv} \) and thus \( v \... | Yes |
Proposition 13.2.6. In a UFD \( R \), for a nonzero element \( p \in R, p \) is a prime element \( \Leftrightarrow \) \( p \) is an irreducible element. | Proof. \ | No |
Proposition 13.2.7. In a UFD, the gcd of nonzero element exists. Namely, for two nonzero element \( a, b \in R \), there exists an element \( d = \gcd \left( {a, b}\right) \in R \) such that if \( {d}^{\prime } \in R \) satisfies \( {d}^{\prime } \mid a \) and \( {d}^{\prime } \mid b \), then \( {d}^{\prime } \mid d \)... | Explicitly, if \( a \) and \( b \) factor as\n\n\[ a = u{p}_{1}^{{c}_{1}}\cdots {p}_{r}^{{c}_{r}}\;\text{ and }\;b = v{p}_{1}^{{d}_{1}}\cdots {p}_{r}^{{d}_{r}} \]\n\nwith \( {p}_{1},\ldots ,{p}_{r} \) irreducible and pairwise non-associate, and \( u, v \in {R}^{ \times },{c}_{i},{d}_{i} \in {\mathbb{Z}}_{ \geq 0} \), t... | Yes |
Theorem 13.3.1. If \( R \) is a PID, then \( R \) is a UFD. | Proof. Existence of factorization. Let \( r \in R \) be a nonzero element that is not a unit.\n\n- If \( r \) is irreducible, then we are done.\n\n- Otherwise, \( r = {a}_{1} \cdot {b}_{1} \) for \( {a}_{1} \) and \( {b}_{1} \) non-unit.\n\nWe may continue this process for \( {a}_{1} \) and \( {b}_{1} \), respectively.... | Yes |
Lemma 14.2.1. (1) If \( F \) is a field, a polynomial \( f \in F\left\lbrack x\right\rbrack \) of degree 2 or 3 is irreducible if and only if it has a root in \( F \) . | Proof. (1) is clear. | No |
Proposition 14.2.3 (Eisenstein’s criterion). Let \( \mathfrak{p} \) be a prime ideal of an integral domain \( R \), and let \( f\left( x\right) = {x}^{n} + {c}_{n - 1}{x}^{n - 1} + \cdots + {c}_{0} \in R\left\lbrack x\right\rbrack \) . Suppose that\n\n(1) \( {c}_{0},{c}_{1},\ldots ,{c}_{n - 1} \in \mathfrak{p} \), and\... | Proof. We may assume that \( \deg f\left( x\right) \geq 2 \) . Suppose that \( f\left( x\right) = a\left( x\right) b\left( x\right) \) with \( \deg a \geq 1 \) and \( \deg b \geq 1 \) . Then the leading coefficients of \( a\left( x\right) \) and \( b\left( x\right) \) are both units. So we may rescale \( a\left( x\righ... | Yes |
The typical application of Eisenstein's criterion is to the cyclotomic polynomial. Let \( p \) be a prime number, the \( p \) th cyclotomic polynomial is\n\n\[ \n{\Phi }_{p}\left( x\right) = \frac{{x}^{p} - 1}{x - 1} = {x}^{p - 1} + {x}^{p - 2} + \cdots + x + 1.\n\]\n\nWe claim that \( {\Phi }_{p}\left( x\right) \) is ... | This is because we consider \( {\Phi }_{p}\left( {x + 1}\right) \) instead:\n\n\[ \n{\Phi }_{p}\left( {x + 1}\right) = \frac{{\left( x + 1\right) }^{p} - 1}{\left( {x + 1}\right) - 1} = {x}^{p} + \underset{\text{all terms divisible by }p}{\underbrace{p{x}^{p - 1} + \left( \begin{array}{l} p \\ 2 \end{array}\right) {x}^... | Yes |
Lemma 14.3.1. Let \( F \) be a field. A polynomial \( f\left( x\right) \) is irreducible if and only if \( F\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\right) \) is a field. | Proof. This follows from the following equivalences.\n\n\( f\left( x\right) \) is an irreducible polynomial \( \overset{F\left\lbrack x\right\rbrack \text{ UFD }}{ \Leftrightarrow }f\left( x\right) \) is a prime element in \( F\left\lbrack x\right\rbrack \)\n\n\( \Leftrightarrow \;\left( {f\left( x\right) }\right) \) i... | Yes |
Lemma 14.3.3. Let \( F \) be a field. If \( f\left( x\right) = {p}_{1}{\left( x\right) }^{{n}_{1}}\cdots {p}_{r}{\left( x\right) }^{{n}_{r}} \) is the factorization of \( f\left( x\right) \) in \( F\left\lbrack x\right\rbrack \), then | \[ \frac{F\left\lbrack x\right\rbrack }{\left( f\left( x\right) \right) } \cong \frac{F\left\lbrack x\right\rbrack }{\left( {p}_{1}{\left( x\right) }^{{n}_{1}}\right) } \times \cdots \times \frac{F\left\lbrack x\right\rbrack }{\left( {p}_{r}{\left( x\right) }^{{n}_{r}}\right) }. \] Proof. As the \( {p}_{i}{\left( x\rig... | Yes |
Corollary 14.3.5. If \( F \) is a field and \( G \) a finite subgroup of \( F \), then \( G \) is cyclic. | Proof. Assume that \( \\left| G\\right| = n \) . By classification of finite abelian groups, we may write\n\n\[ G = {\\mathbf{Z}}_{{n}_{1}} \\times {\\mathbf{Z}}_{{n}_{2}} \\times \\cdots \\times {\\mathbf{Z}}_{{n}_{r}}\\;\\text{with integers}\\{n}_{1}\\left| {n}_{2}\\right| \\cdots \\mid {n}_{r}\\text{and}n = {n}_{1}\... | Yes |
We explain a key example of the theorem we prove later for classification of finitely generated modules over a PID. | Let \( F \) be a field and \( V \) a vector space, equipped with an action by an \( F \) -linear operator \( T \) . We define an \( F\left\lbrack x\right\rbrack \) -module structure on \( V \) by\n\n\( \left( {{a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n}}\right) \cdot v = {a}_{0} \cdot v + {a}_{1}x \cdot v + \cdots + {... | Yes |
Lemma 15.3.1. Let \( R \) be an integral domain and let \( N \) be a free \( R \) -module of rank \( n \) . Then any \( n + 1 \) elements \( {x}_{1},\ldots ,{x}_{n + 1} \in N \) are linearly dependent, i.e. there exist \( {a}_{1},\ldots ,{a}_{n + 1} \in R \) , not all zero, such that \( {a}_{1}{x}_{1} + \cdots + {a}_{n... | Proof. We may identify \( N \) with \( {R}^{\oplus n} \) . Let \( F \) denote the fraction field of \( R \) . Viewing each \( {x}_{i} \) as an element of \( {R}^{\oplus n} \subseteq {F}^{\oplus n} \) or even a column vector, it is well known that any \( n + 1 \) such column vectors are linearly dependent over \( F \), ... | Yes |
Theorem 15.3.2. Let \( R \) be a PID. Let \( N \) be a free \( R \) -module of rank \( n \) and \( L \) a submodule of \( N \) . Then\n\n(1) \( L \) is free of rank \( \ell \) (with \( \ell \leq n \) ).\n\n(2) There exists a basis \( {y}_{1},{y}_{2},\ldots ,{y}_{n} \) of \( N \) so that \( {a}_{1}{y}_{1},\ldots ,{a}_{\... | Proof. If \( L = 0 \), the theorem is trivial. Now we assume that \( L \neq 0 \) . Our intention is to run an induction on the rank of \( L \), but the actual argument is slightly more complicated. See later.\n\nWe first determine the value of \( {a}_{1} \) . The basic idea is that we identify \( N \) with \( {R}^{\opl... | No |
Proof of Theorem 15.3.3. As \( L \) is finitely generated, say by elements \( {x}_{1},\ldots ,{x}_{n} \), there exists a surjective homomorphism | \[ \phi : {R}^{\oplus n} \rightarrow M \] \[ \left( {{a}_{1},\ldots ,{a}_{n}}\right) \mapsto {a}_{1}{x}_{1} + \cdots + {a}_{n}{x}_{n} \] Then \( \ker \phi \subseteq {R}^{\oplus n} \) is a submodule. Applying Theorem 15.3.2 to the submodule \( \ker \phi \) of the free module \( {R}^{\oplus n} \), we see that there exist... | No |
Lemma 15.3.5. Let \( p \) and \( q \) be prime elements such that \( \left( p\right) \neq \left( q\right) \) and \( m, n \in \mathbb{N} \) . (1) If \( M = R \), then \( M/{p}^{m}M \simeq R/\left( {p}^{m}\right) \) and \( {p}^{m}M/{p}^{m + 1}M \simeq R/\left( p\right) \) . (2) If \( M = R/\left( {p}^{n}\right) \), then ... | Proof. We leave this as an exercise. For (3), we note that \( \left( {{p}^{m},{q}^{n}}\right) = \left( 1\right) \) . Indeed, as \( \left( {p, q}\right) = \left( 1\right) \), we have \( 1 = {ap} + {bq} \) and thus \( 1 = {\left( ap + bq\right) }^{m + n - 1} = {p}^{m} \cdot \left( *\right) + {q}^{n} \cdot \left( *\right)... | No |
Theorem 16.2.3. Let \( F \subseteq E \subseteq K \) be field extensions. Then \( \left\lbrack {K : F}\right\rbrack = \left\lbrack {K : E}\right\rbrack \left\lbrack {E : F}\right\rbrack \) . | Set \( \left\lbrack {K : E}\right\rbrack = m \) and \( \left\lbrack {E : F}\right\rbrack = n \) . Let \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{m}}\right\} \) be a \( E \) -basis of \( K \) and \( \left\{ {{\beta }_{1},\ldots ,{\beta }_{n}}\right\} \) be an \( F \) -basis of \( E \) . Then every element \( x \) of \... | Yes |
Example 16.2.4. |  | No |
Lemma 16.3.1. Let \( F \) and \( E \) be fields. A homomorphism \( \phi : F \rightarrow E \) must be injective. This then realizes \( E \) as an extension of \( \phi \left( F\right) \simeq F \) . | Proof. As \( \ker \phi \) is an ideal of \( F \), namely \( \{ 0\} \) or \( F \) . But our convention of ring homomorphisms require to send \( {1}_{F} \) to \( {1}_{E} \) . So \( \ker \phi \neq F \), and thus \( \ker \phi = \{ 0\} \), i.e. \( \phi \) is injective. | Yes |
Lemma 16.3.3. Equation \( p\left( x\right) = 0 \) has a zero in \( K \) . | Proof. Assume that \( p\left( x\right) = {p}_{0} + {p}_{1}x + \cdots + {p}_{n}{x}^{n} \) . Then\n\n\[ p\left( \theta \right) = {p}_{0} + {p}_{1}\theta + \cdots + {p}_{n}{\theta }^{n} = {p}_{0} + {p}_{1}x + \cdots + {p}_{n}{x}^{n} + \left( {p\left( x\right) }\right) = 0 + \left( {p\left( x\right) }\right) .\n\]\n\nSo \(... | No |
Theorem 16.3.6. Let \( K \) be a field extension of \( F \) and let \( \alpha \in K \) . We have a dichotomy:\n\n(1) either \( 1,\alpha ,{\alpha }^{2},\ldots \) are linearly independent over \( F \), in which case \( F\left( \alpha \right) \simeq F\left( x\right) = \) \( \operatorname{Frac}\left( {F\left\lbrack x\right... | Proof. Consider case (1): the condition implies that\n\n\[ \phi : F\left\lbrack x\right\rbrack \hookrightarrow K \]\n\n\[ f\left( x\right) \mapsto f\left( \alpha \right) \]\n\nis an injective homomorphism. This clearly extends to a homomorphism\n\n\[ \phi : F\left( x\right) \hookrightarrow K \]\n\n\[ f\left( x\right) /... | Yes |
Theorem 16.4.2. The following are equivalent for a field extension \( K \) of \( F \) :\n\n(1) \( K \) is a finite extension of \( F \) .\n\n(2) \( K \) is finitely generated and algebraic over \( F \) . | Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) . If \( K \) is a finite extension of \( F, K \) is generated over \( F \) by the basis element (of \( K \) over \( F \) ). For any \( \alpha \in K,\left\lbrack {F\left( \alpha \right) : F}\right\rbrack \leq \left\lbrack {K : F}\right\rbrack \) is finite; so \( \... | No |
Lemma 16.4.4. Given field extensions of \( K/E/F \) and \( \alpha \in K \), then\n\n\[ \n{m}_{\alpha, E}\left( x\right) \mid {m}_{\alpha, F}\left( x\right) \n\]\n\nas polynomials in \( E\left\lbrack x\right\rbrack \) . In particular,\n\n\[ \n\deg \left( {{m}_{\alpha, E}\left( x\right) }\right) \leq \deg \left( {{m}_{\a... | Proof. This is because \( {m}_{\alpha, F}\left( \alpha \right) = 0 \) . So viewing this in the polynomial ring \( E\left\lbrack x\right\rbrack \), we have \( {m}_{\alpha, F}\left( x\right) \in \left( {{m}_{\alpha, E}\left( x\right) }\right) \) . This implies that \( {m}_{\alpha, E}\left( x\right) \mid {m}_{\alpha, F}\l... | Yes |
Corollary 16.4.5. Given field extensions of \( K/E/F \) and \( \alpha \in K \), then\n\n\[ \left\lbrack {E\left( \alpha \right) : E}\right\rbrack \leq \left\lbrack {F\left( \alpha \right) : F}\right\rbrack \] | Proof. This is because \( \left\lbrack {E\left( \alpha \right) : E}\right\rbrack = \deg {m}_{\alpha, E}\left( x\right) \) and \( \left\lbrack {F\left( \alpha \right) : F}\right\rbrack = \deg {m}_{\alpha, F}\left( x\right) \) . | Yes |
Inside \( \mathbb{C} \), the composite of \( \mathbb{Q}\left( \sqrt{2}\right) \) and \( \mathbb{Q}\left( \sqrt{3}\right) \) is | \[ \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) = \{ a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} \mid a, b, c, d \in \mathbb{Q}\} . \] | Yes |
Corollary 16.4.8. Let \( {K}_{1} \) and \( {K}_{2} \) be two intermediate fields in the field extension \( K \) over \( F \) such that \( \left\lbrack {{K}_{i} : F}\right\rbrack < + \infty \) . Then\n\n\[ \left\lbrack {{K}_{1}{K}_{2} : F}\right\rbrack \leq \left\lbrack {{K}_{1} : F}\right\rbrack \cdot \left\lbrack {{K}... | Proof. As \( {K}_{1} \) is a finite extension of \( F \), we may write \( {K}_{1} = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) . We consider the following tower\n\n\n\nApplying Corollary 16.4.5 to each p... | Yes |
Corollary 16.4.10. Let \( K \) be a field extension of \( F \) and let \( \alpha ,\beta \in K \) be elements algebraic over \( F \) . Then \( \alpha \pm \beta ,{\alpha \beta } \), and \( \alpha /\beta \) (when \( \beta \neq 0 \) ) are all algebraic over \( F \) . | Proof. This is because \( \alpha \pm \beta ,{\alpha \beta } \), and \( \alpha /\beta \) all belong to the field \( F\left( {\alpha ,\beta }\right) \) which is a finite extension of \( F \) . | Yes |
Theorem 16.4.13. If \( L/K \) and \( K/F \) are both algebraic extensions, then \( L/F \) is algebraic. | Proof. Let \( \alpha \in L \) . Its minimal polynomial \( {m}_{\alpha }\left( x\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} \) over \( K \) involves only finitely many elements of \( K \), each of them being algebraic over \( F \) . So we see that \( F\left( \alpha \right) \) is contained in the field ... | Yes |
Theorem 17.1.3. For any field \( F \) and \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) of degree \( n \), a splitting field \( K \) of \( F \) exists. Moreover, \( \left\lbrack {K : F}\right\rbrack \leq n! \) . | Proof. We use induction on \( \deg f\left( x\right) = n \) . When \( n = 1, F \) itself is the splitting field of \( f\left( x\right) \) over \( F \) . Suppose that the statement is proved for polynomials of strictly smaller degrees (over any field).\n\nLet \( p\left( x\right) \) be an irreducible factor of \( f\left( ... | Yes |
Lemma 17.2.1. If \( \eta : F \cong {F}^{\prime } \) is an isomorphism of fields and \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) is irreducible, then \( {p}^{\prime }\left( x\right) \mathrel{\text{:=}} \eta \left( {p\left( x\right) }\right) \in {F}^{\prime }\left\lbrack x\right\rbrack \) is irreducible, and ... | \[ \eta : F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \overset{ \simeq }{ \rightarrow }{F}^{\prime }\left\lbrack x\right\rbrack /\left( {{p}^{\prime }\left( x\right) }\right) . \] | Yes |
Proposition 17.2.3. Let \( \eta : F\overset{ \simeq }{ \rightarrow }{F}^{\prime } \) be an isomorphism of fields and \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) . Put \( {f}^{\prime }\left( x\right) \mathrel{\text{:=}} \eta \left( {f\left( x\right) }\right) \in {F}^{\prime }\left\lbrack x\right\rbrack \) . ... | Proof. We will prove that for the splitting field \( K \) of \( f\left( x\right) \) over \( F \) constructed in the previous lemma, we have the following commutative diagram with top row isomorphisms.\n\n\[\\begin{array}{l} E\\overset{{\\sigma }_{1}}{ \\leftarrow }K\\overset{{\\sigma }^{\\prime }}{ \\rightarrow }{E}^{\... | Yes |
Lemma 17.2.4. If we have the following tower of field extensions such that both \( E \) and \( {E}^{\prime } \) are splitting fields of some polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) . Then \( E = {E}^{\prime } \) (as equality of subfields of \( K \) ). | Proof. By the definition of splitting field, we may - split \( f\left( x\right) \) in \( E \) as \( c\left( {x - {\alpha }_{1}}\right) \cdots \left( {x - {\alpha }_{n}}\right) \) with \( c \in {E}^{ \times } \) and \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in E \), and - split \( f\left( x\right) \) in \( {E}^{\prime } \... | Yes |
Lemma 17.2.5. Consider a tower of extensions \( K/E/F \) . If \( E \) is a splitting field over \( F \) of some polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \), then for any automorphism \( \sigma : K\overset{ \simeq }{ \rightarrow }K \) such that \( {\left. \sigma \right| }_{F} = \mathrm{{id}} \) , ... | Proof. This is because \( \sigma \left( E\right) \) is a splitting field of \( \sigma \left( f\right) = f \) . By the above lemma, we deduce that \( \sigma \left( E\right) = E \) . | No |
Theorem 17.3.2. A finite extension \( K \) of \( F \) is normal if and only if it is the splitting field of some \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) . | Proof. We first prove that a finite normal extension \( K \) of \( F \) is a splitting field. Indeed, write \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right. \) for \( {\alpha }_{1},\ldots ,{\alpha }_{r} \in K \) . For each \( {\alpha }_{i} \), the minimal polynomial \( {m}_{{\alpha }_{i}}\left( x\right) \in ... | Yes |
Corollary 17.3.3. If \( K \) is a finite and normal extension of \( F \), for any intermediate field \( E \) , the field \( K \) is a normal extension of \( E \) . (Note that \( E \) need not be a normal extension of \( F \) .) | Proof. By the theorem above, \( K \) is a splitting field of some \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) over \( F \) . Hence \( K \) is a splitting field of \( f\left( x\right) \) over \( E \) . The reverse implication of the above theorem implies that \( K \) is normal over \( E \) . | Yes |
Lemma 17.3.5. A normal closure of a finite extension \( K \) over \( F \) exists and is unique up to (some) isomorphism. | Proof. Existence: Assume that \( K = F\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) . Put\n\n\[ f\left( x\right) \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 1}}^{r}{m}_{{\alpha }_{i}, F}\left( x\right) \in F\left\lbrack x\right\rbrack \]\n\nTake \( L \) to be a splitting field of \( f\left( x\right) \) ov... | Yes |
(2) The splitting field of \( {x}^{3} - 2 \) is |  | No |
Example 17.4.2. (1) \( F = {\mathbb{F}}_{p} \) is a perfect field (so is any finite field). | This is because \( \sigma : F \rightarrow F \) is always injective, so also surjective by counting elements. | Yes |
Proposition 17.4.3. Algebraic extensions of perfect fields are still perfect. | Proof. Let \( K \) be an algebraic extension of \( F \) with \( F \) a perfect field of characteristic \( p \) . It suffices to show that each \( \alpha \in K \) admits a \( p \) th root in \( E \mathrel{\text{:=}} F\left( \alpha \right) (E \) is a finite extension of \( F) \) .\n\nConsider the Frobenius endomorphism \... | Yes |
Theorem 18.1.2. Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be a polynomial of degree \( \geq 1 \) . Then \( f\left( x\right) \) has no repeated roots in its splitting field \( K \) if and only if \( \left( {f\left( x\right), D\left( f\right) \left( x\right) }\right) = \left( 1\right) \) . | Proof. \ | No |
Corollary 18.1.3. If \( f\left( x\right) \) is an irreducible polynomial in \( F\left\lbrack x\right\rbrack \), then \( f\left( x\right) \) has repeated roots in its splitting field if and only if \( D\left( f\right) \left( x\right) = 0 \) . | Proof. By Theorem 18.1.2, the polynomial \( f\left( x\right) \) has repeated zero if and only if \( \left( {f\left( x\right), D\left( f\right) \left( x\right) }\right) = \) (1). As \( f\left( x\right) \) is irreducible, this is further equivalent to \( f\left( x\right) \mid D\left( f\right) \left( x\right) \), which is... | Yes |
Corollary 18.1.5. If \( \operatorname{char}\left( F\right) = 0 \), all irreducible polynomials are separable. | Proof. This is because when \( f\left( x\right) \neq 0 \) and \( \deg f \geq 1 \), we have \( D\left( f\right) \left( x\right) \neq 0 \) . | Yes |
Corollary 18.1.6. If \( \operatorname{char}\left( F\right) = p > 0 \), and if \( f\left( x\right) \) is inseparable, then\n\n\[ f\left( x\right) = g\left( {x}^{p}\right) \;\text{ for some }g \in F\left\lbrack x\right\rbrack \text{ irreducible. } \]\n\nMoreover, this can only happen when \( F \) is imperfect. | Proof. Let \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \) be an irreducible polynomial with repeated roots. Corollary 18.1.3 implies that\n\n\[ D\left( f\right) \left( x\right) = {a}_{1} + 2{a}_{2}x + \cdots + n{a}_{n}{x}^{n - 1} = 0. \]\n\nThis means that \( i{a}_{i} = 0 \) and thus \( {a}_{i} =... | Yes |
Corollary 18.1.7. If \( \operatorname{char}\left( F\right) = p > 0 \), an irreducible polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) is of the form \( f\left( x\right) = g\left( {x}^{{p}^{e}}\right) \) with \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) is an irreducible and separable polyn... | Proof. The first statement follows from the previous corollary. For the second statement, we note that, if \( g\left( x\right) = \mathop{\prod }\limits_{i}\left( {x - {\alpha }_{i}}\right) \) in a splitting field of \( F \), then we have\n\n\[ f\left( x\right) = \mathop{\prod }\limits_{i}\left( {{x}^{{p}^{e}} - {\alpha... | Yes |
Example 18.2.5. We consider the case when \( F = \mathbb{Q}, K = \mathbb{Q}\left( \sqrt[3]{2}\right) \) and \( M = \mathbb{Q}\left( {\sqrt[3]{2},{\zeta }_{3}}\right) \) . | The set \( {\operatorname{Hom}}_{\mathbb{Q}}\left( {K, M}\right) = {\operatorname{Hom}}_{\mathbb{Q}}\left( {\mathbb{Q}\left\lbrack x\right\rbrack /\left( {{x}^{3} - 2}\right), M}\right) \) is given by\n\n\[ K = \mathbb{Q}\left( \sqrt[3]{2}\right) \rightarrow M = \mathbb{Q}\left( {\sqrt[3]{2},{\zeta }_{3}}\right) \]\n\n... | Yes |
Lemma 18.2.6. If \( K = F\left( \alpha \right) \) with \( {m}_{\alpha, F}\left( x\right) = g\left( {x}^{{p}^{e}}\right) \) for some \( g \in F\left\lbrack x\right\rbrack \) irreducible and separable, then \[ \# {\operatorname{Hom}}_{F}\left( {F\left( \alpha \right), M}\right) = \deg g\left( x\right) \leq \left\lbrack {... | Proof. Such a \( \phi \in {\operatorname{Hom}}_{F}\left( {F\left( \alpha \right), M}\right) \) is determined by where \( \alpha \) goes. The constraint on \( \phi \left( \alpha \right) \) is that, it must be a zero of \( {m}_{\alpha, F}\left( x\right) \) in \( M \). There are precisely \( \deg \left( g\right) \) of the... | Yes |
Corollary 18.2.8. Let \( K \) be a finite extension of \( F \) and \( M \) a normal extension of \( F \) containing \( K \) . Then\n\n(18.2.8.1) \( \# {\operatorname{Hom}}_{F}\left( {K, M}\right) \leq \left\lbrack {K : F}\right\rbrack . \)\n\nMoreover, the following are equivalent:\n\n(1) \( K = F\left( {{\alpha }_{1},... | Proof. We consider the following tower of extensions and their embeddings into \( M \) .\n\n\n\nBy Corollary 18.2.8, we have\n\n\[ \# {\operatorname{Hom}}_{F}\left( {F\left( {\alpha }_{1}\right), M}\right) \leq \left... | Yes |
Theorem 18.3.1. A finite separable extension of fields is generated by one element. | In fact, we have a stronger statement: if \( K = F\left( {\alpha ,\beta }\right) \) with \( \alpha ,\beta \) algebraic over \( F \) and \( \beta \) separable over \( F \) . Then \( K = F\left( \gamma \right) \) for some \( \gamma \in K \) . | No |
Theorem 18.4.1. (1) If \( F \) is a finite field, then \( \operatorname{char}\left( F\right) = p > 0 \) for a prime \( p \), and \( \# F = {p}^{n} \) for \( n = \left\lbrack {F : {\mathbb{F}}_{p}}\right\rbrack \) . | Proof. (1) is clear. | No |
Lemma 18.4.3. (1) The field \( {\mathbb{F}}_{{p}^{m}} \) can be viewed as a subfield of \( {\mathbb{F}}_{{p}^{n}} \) if and only if \( m \mid n \) . In this case, as a subset, \( {\mathbb{F}}_{{p}^{m}} \subseteq {\mathbb{F}}_{{p}^{n}} \) is unique. | Proof. (1) If \( {\mathbb{F}}_{{p}^{m}} \) is a subfield of \( {\mathbb{F}}_{{p}^{n}} \), then \( {\mathbb{F}}_{{p}^{n}} \) is a vector space over \( {\mathbb{F}}_{{p}^{m}} \) . Thus, \( {p}^{n} \) is a power of \( {p}^{m} \) . So \( m \mid n \) .\n\nConversely, if \( m \mid n,{\mathbb{F}}_{{p}^{m}} \) is a splitting f... | Yes |
Lemma 19.1.2. If \( K \) is a finite Galois extension of \( F,\# \operatorname{Gal}\left( {K/F}\right) = \left\lbrack {K : F}\right\rbrack \) . | Proof. Consider the above situation we recalled with \( K = M \) Galois over \( F \), then each \( \phi \in {\operatorname{Hom}}_{F}\left( {K, K}\right) \) is an automorphism of \( K \) fixing \( F \) . This is because \( \phi : K \rightarrow K \) is an injective \( F \) -linear map between the same finite dimensional ... | Yes |
Consider \( K = \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) \) as an extension of \( \mathbb{Q} \). The Galois group is equal to \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) = \{ \mathrm{{id}},\sigma ,\tau ,{\sigma \tau }\} \), where\n\n- id is the identity map;\n\n- \( \sigma \left( {a + b\sqrt{2} + c\sqrt{3} + d... | We can compute the subfields that are invariant under these automorphisms:\n\n\[ {K}^{\sigma = 1} = \{ x \in K \mid \sigma \left( x\right) = x\} = \{ a + c\sqrt{3} \mid a, c \in \mathbb{Q}\} = \mathbb{Q}\left( \sqrt{3}\right) ,\]\n\n\[ {K}^{\tau = 1} = \{ x \in K \mid \sigma \left( x\right) = x\} = \{ a + b\sqrt{2} \mi... | No |
Example 19.2.1. Corresponding to Example 19.1.3, we have the following corresponding diagram | Note that \( \langle \left( {123}\right) \rangle \) is a normal subgroup; this corresponds to that \( \mathbb{Q}\left( {\zeta }_{3}\right) \) is a Galois extension of \( \mathbb{Q} \) . | No |
We give a case of Galois group being \( {D}_{8} \), with \( \theta = \sqrt[4]{2} \). The normal closure of \( \mathbb{Q}\left( \sqrt[4]{2}\right) \) is \( \mathbb{Q}\left( {\sqrt[4]{2},\mathbf{i}}\right) \). There are obvious automorphisms of \( \mathbb{Q}\left( \sqrt[4]{2}\right) \) given by | \[ s\left( \mathbf{i}\right) = - \mathbf{i},\;s\left( \sqrt[4]{2}\right) = \sqrt[4]{2}, \] \[ r\left( \sqrt[4]{2}\right) = \mathbf{i}\sqrt[4]{2},\;r\left( \mathbf{i}\right) = \mathbf{i}. \] One can check that \[ \begin{matrix} \text{ rsrs } : \sqrt[4]{2} \mapsto \sqrt[s]{2} \mapsto \mathbf{i}\sqrt[4]{2} \mapsto \mathbf... | Yes |
Corollary 19.2.6. If \( K \) is a Galois extension of \( F \) with Galois group \( G = \operatorname{Gal}\left( {K/F}\right) \) being an abelian group, then any intermediate field \( E \) is Galois over \( F \) . | Proof. This is because any subgroup of an abelian group is normal. | No |
Lemma 19.3.2. The Galois group of \( {\mathbb{F}}_{{q}^{n}} \) over \( {\mathbb{F}}_{q} \) is isomorphic to \( {\mathbf{Z}}_{n} \), generated by \( {\phi }_{q} \) . | Proof. For the \( q \) -Frobenius automorphism \( {\phi }_{q} \) of \( {\mathbb{F}}_{{q}^{n}},{\phi }_{q}^{n}\left( b\right) = b \) for any \( b \in {\mathbb{F}}_{{q}^{n}} \) . So \( {\phi }_{q}^{n} = \mathrm{{id}} \) . The rest is clear. | No |
We give an example of the Galois diagram for extensions of finite fields. |  | No |
Lemma 19.4.3. We have\n\n\[ \n{x}^{n} - 1 = \mathop{\prod }\limits_{{d \mid n}}{\Phi }_{d}\left( x\right) \n\]\n\nEach \( {\Phi }_{n}\left( x\right) \) is a polynomial of degree \( \varphi \left( n\right) \) with coefficients in \( \mathbb{Z} \) . | Proof. The first equality is easy:\n\n(19.4.3.1)\n\n\[ \n{x}^{n} - 1 = \mathop{\prod }\limits_{{b \in {\mathbf{Z}}_{n}}}\left( {x - {\zeta }_{n}^{b}}\right) = \mathop{\prod }\limits_{{d \mid n}}\mathop{\prod }\limits_{{i \in {\mathbf{Z}}_{d}^{ \times }}}\left( {x - {\zeta }_{n}^{di}}\right) = \mathop{\prod }\limits_{{d... | No |
Theorem 19.4.4. The polynomial \( {\Phi }_{n}\left( x\right) \) is irreducible in \( \mathbb{Q}\left\lbrack x\right\rbrack \) . So \( \left\lbrack {\mathbb{Q}\left( {\zeta }_{n}\right) : \mathbb{Q}}\right\rbrack = \varphi \left( n\right) \) . | Proof. It suffices to show that \( {\Phi }_{n}\left( x\right) \) is irreducible in \( \mathbb{Z}\left\lbrack x\right\rbrack \) . Let \( \zeta \) be a primitive \( n \) th root of unity in a splitting field of \( {\Phi }_{n}\left( x\right) \).\n\nWe need to show that the minimal polynomial \( f\left( x\right) \mathrel{\... | Yes |
Corollary 19.4.5. The Galois group of \( \mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q} \) is \( {\mathbf{Z}}_{n}^{ \times } \) . Explicitly, for \( a \in {\mathbf{Z}}_{n}^{ \times } \), the associated automorphism is | \[ \mathbb{Q}\left( {\zeta }_{n}\right) \overset{ \cong }{ \leftarrow }\mathbb{Q}\left\lbrack x\right\rbrack /\left( {{\Phi }_{n}\left( x\right) }\right) \overset{ \cong }{ \rightarrow }\mathbb{Q}\left( {\zeta }_{n}\right) \] \[ {\zeta }_{n} \leftarrow x + {\Phi }_{n}\left( x\right) , \mapsto {\zeta }_{n}^{a}. \] | Yes |
Corollary 19.4.6. For every finite abelian group \( G \), there exists a finite Galois extension \( K \) of \( \mathbb{Q} \) with Galois group \( G \) . | Proof. Write \( G = {\mathbf{Z}}_{{n}_{1}} \times {\mathbf{Z}}_{{n}_{r}} \) . For each \( {n}_{i} \), find distinct odd prime number \( {p}_{i} \) such that \( {p}_{i} \equiv 1{\;\operatorname{mod}\;{n}_{i}} \) . Then \( G \) is a quotient of\n\n\[{\mathbf{Z}}_{{p}_{1}}^{ \times } \times {\mathbf{Z}}_{{p}_{r}}^{ \times... | Yes |
We illustrate the above proof by constructing a cyclic extension of \( \mathbb{Q} \) of degree 3. Write \( \zeta = {\zeta }_{7} \). Then we have | But we have to translate \( \ker \phi \) in terms of \( {\mathbf{Z}}_{7}^{ \times } \), it is the subset \( \{ 1, - 1\} \) (namely \( \left\{ {x \in {\mathbf{Z}}_{7}^{ \times } \mid {x}^{2} = }\right. \) \( 1{\;\operatorname{mod}\;7}\} \) ). Taking trace, we have\n\n\[ \theta \mathrel{\text{:=}} \zeta + {\zeta }^{-1} \... | Yes |
Lemma 19.4.10. If \( K \) is a finite extension of \( F \) and \( E \) an intermediate field, we have for \( \alpha \in K \) ,\n\n\[ \n{\operatorname{Tr}}_{K/F}\left( \alpha \right) = {\operatorname{Tr}}_{E/F}\left( {{\operatorname{Tr}}_{K/E}\left( \alpha \right) }\right) \;\text{ and }\;{\operatorname{Nm}}_{K/F}\left(... | Proof. We leave this as an exercise. | No |
Lemma 19.4.11. Let \( K \) be a finite extension over \( F \) of degree \( n \), and let \( \alpha \in K \) be an element with minimal polynomial \( {m}_{\alpha, F}\left( x\right) = {x}^{m} + {a}_{1}{x}^{m - 1} + \cdots + {a}_{m} \in F\left\lbrack x\right\rbrack \) with \( m \mid n \) . Then\n\n\[{\operatorname{Tr}}_{K... | Proof. We first treat the case when \( K = F\left( \alpha \right) \) (in this case \( m = n \) ). In this case, \( K \cong \) \( F\left\lbrack x\right\rbrack /\left( {{m}_{\alpha, F}\left( x\right) }\right) \) with basis elements \( 1, x,\ldots ,{x}^{n - 1} \), and multiplication by \( \alpha \) is represented by the m... | Yes |
Proposition 20.1.1. Let \( K \) be a finite Galois (namely normal and separable) extension of \( F \), and let \( L \) be an extension of \( K \) normal over \( F \). | We have the following equalities\n\n\[ \begin{matrix} \left\lbrack {K : F}\right\rbrack = \# {\operatorname{Hom}}_{F}\left( {K, L}\right) \; = \;\# {\operatorname{Hom}}_{F}\left( {K, K}\right) = \# \operatorname{Gal}\left( {K/F}\right) . \\ \uparrow \\ K/F\text{ separable }\;K/F\text{ normal } \\ \text{ so can take }K ... | No |
Theorem 20.1.2 (Galois theory). Let \( K \) be a finite Galois extension with \( G = \operatorname{Gal}\left( {K/F}\right) \). (1) Then there is a one-to-one correspondence between \[ \{ \text{subgroups}\;H \leq G\} \leftarrow \; \rightarrow \{ \text{intermediate fields}\;E\;\text{of}\;K/F\} \] \[ H \mapsto {K}^{H} = \... | Proof. (1) Since \( K \) is a finite normal extension of \( F, K \) is a splitting field for some \( f\left( x\right) \in \) \( F\left\lbrack x\right\rbrack \) . (This implies that \( K \) is also the splitting field of \( f\left( x\right) \) over an intermediate field \( E \) .) It follows from Proposition 20.1.1 that... | Yes |
Consider the following diagram of field extensions.\n\n\n\nAssume that \( K \) is a finite Galois extension and \( E \) is an arbitrary field extension of \( F \) (not even assume to be algebraic). Then \( {KE} \) is... | Proof of Proposition 20.2.1. Since \( K \) is a finite Galois extension of \( F, K \) is the splitting field of some separable polynomial \( f\left( x\right) \) over \( F \) . This further implies that \( {KE} \) is the splitting field of the same polynomial \( f\left( x\right) \) over \( E \) . This implies that \( {K... | Yes |
Proposition 20.2.3. Suppose that we have a tower of extensions below, in which \( {K}_{1} \) and \( {K}_{2} \) are Galois over \( F \) .\n\n\n\nThen we have\n\n(1) \( {K}_{1} \cap {K}_{2} \) is Galois over \( F \) .\... | Proof. (1) We need to show that \( {K}_{1} \cap {K}_{2} \) is normal over \( F \) . Suppose that \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) is a polynomial that has a zero in \( {K}_{1} \cap {K}_{2} \) . Then by normality of \( f\left( x\right) \), all zeros of \( f\left( x\right) \) belong to both \( {K}_... | Yes |
Theorem 20.3.3 (Linearly independence of characters). If \( {\chi }_{1},\ldots ,{\chi }_{n} \) are distinct characters of a group \( H \) with values in \( L \), then they are linearly independent. | Proof of Theorem 20.3.3. Suppose that these characters \( {\chi }_{1},\ldots ,{\chi }_{n} \) are linearly dependent. Then among all (nonzero) linear relations, there is a unique one with minimal number of nonzero \( {a}_{i} \) ’s. Without loss of generality, we assume this is\n\n\[ \n{a}_{1}{\chi }_{1} + {a}_{2}{\chi }... | Yes |
Lemma 20.3.7. Let \( K \) be a finite separable field extension of \( F \) so that \( K = F\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\right) \) and let \( L \) be normal extension of \( F \) containing \( K \) . Then we have an isomorphism of \( L \) -algebras. | \[ L{ \otimes }_{F}K \mathrel{\text{:=}} L\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\right) \overset{\varphi = {\left( {\varphi }_{\sigma }\right) }_{\sigma }}{ \rightarrow }\mathop{\prod }\limits_{{\sigma \in {\operatorname{Hom}}_{F}\left( {K, L}\right) }}L \] \[ a \otimes f\left( x\right) \mapsto {\left(... | Yes |
Example 20.3.8. \[ \mathbb{R}{ \otimes }_{\mathbb{Q}}\mathbb{Q}\left( \sqrt{2}\right) \cong \mathbb{R}\left\lbrack x\right\rbrack /\left( {{x}^{2} - 2}\right) \xrightarrow[]{ \cong }\mathbb{R} \times \mathbb{R} \] | Proof of Lemma 20.3.7. The map \( \varphi \) is clearly a well-defined homomorphism and \( L \) -linear. Both sides are \( L \) -vector spaces of dimension \( \left\lbrack {K : F}\right\rbrack = \# {\operatorname{Hom}}_{F}\left( {K, L}\right) \) . It suffices to show injectivity and this is exactly the linearly indepen... | No |
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