Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Lemma 16.7. (Dedekind’s Modular Law) Let \( A, B \), and \( C \) be three subgroups of \( G \). If \( A \) is a subset of \( C \), then \( A\left( {B \cap C}\right) = \left( {AB}\right) \cap C \). | Proof. \( \subseteq \) : If \( a \in A, x \in B \cap C \), then \( {ax} \in {AB} \cap {AC} \subset \left( {AB}\right) \cap C \).\n\n\( \supseteq : \) If \( a \in A, b \in B \) satisfy \( {ab} \in C \), then \( b = {a}^{-1}C \subset C \). | No |
Lemma 16.8. (Zassenhaus Butterfly Lemma) Let \( G \) be a group containing four subgroups \( h, H, k \), and \( K \) satisfying \( h \vartriangleleft H \) and \( k \vartriangleleft K \) . Then we have the following:\n\n(1) \( \left( {H \cap k}\right) \left( {h \cap K}\right) \vartriangleleft H \cap K \) .\n\n(2) \( k\l... | Proof. We use the Second and Third Isomorphism Theorems. As \( k \vartriangleleft K \) in the lemma and \( H \cap K \subset K \) is a subgroup, we have \( k\left( {H \cap K}\right) \) is a group satisfying \( k \vartriangleleft k\left( {H \cap K}\right) \), and \( H \cap k = \left( {H \cap K}\right) \cap k \vartriangle... | Yes |
Theorem 16.10. (Schreier Refinement Theorem) Any two proper subnormal series for a group \( G \) have equivalent refinements. | Proof. Suppose\n\n\[ 1 = {N}_{0} \vartriangleleft {N}_{1} \vartriangleleft \cdots \vartriangleleft {N}_{r} = G\text{ and } \]\n\n\[ 1 = {H}_{0} \vartriangleleft {H}_{1} \vartriangleleft \cdots \vartriangleleft {H}_{s} = G \]\n\nare two proper subnormal series for \( G \) . We have groups:\n\n\[ {N}_{i, j} = {N}_{i}\lef... | Yes |
Theorem 16.11. (Jordan-Holder Theorem) Let \( G \) be a group having a composition series of length \( r \) . Then every composition series for \( G \) has length \( r \) and any proper subnormal series for \( G \) can be refined to a composition series for \( G \) . Moreover, all composition series of \( G \) are equi... | Proof. Let \( 1 = {N}_{0} < \cdots < {N}_{n} < G \) be a proper subnormal series for \( G \) . By the Schreier Refinement Theorem, this subnormal series and a composition series for \( G \) have a common proper refinement. The result follows by the above remark. | No |
Theorem 17.4. Let \( X \) be a non-empty set. Then there exists a free group \( G \) on \( X \) . | Proof. Choose a set \( {X}^{\prime } \) disjoint from \( X \) of the same cardinality, so we have a bijection \( X \rightarrow {X}^{\prime } \) which we denote by \( x \mapsto {x}^{-1} \) . Let \( {X}^{\prime \prime } \) be a set disjoint from \( X \cup {X}^{\prime } \) having precisely one element. Let \( {X}^{\prime ... | Yes |
Every group has a presentation, unique up to isomorphism. | Proof. Let \( G \) be a group and \( X \) a subset of \( G \) that generates \( G \) (e.g., we can take \( G \) itself). Let \( E \) be the free group on basis \( X \) . By the universal property of free groups, there exists a unique group homomorphism \( \varphi : E \rightarrow G \) such that\n\n\n\nby the universal property of free groups, there exist unique group homomorphisms\n\n![c74f12f4-5660... | Yes |
Lemma 18.3. Let \( S \) be a \( G \) -set under \( * \), then \( { \sim }_{G} \) is an equivalence relation on \( S \) . | Proof. Reflexitivity. For all \( s \in S \), we have \( s = e * s \), so \( s{ \sim }_{G}s \) . Symmetry. If \( {s}_{1}{ \sim }_{G}{s}_{2} \), then there exists a \( g \in G \) such that \( {s}_{1} = g * {s}_{2} \) , hence \[ {g}^{-1} * {s}_{1} = {g}^{-1} * \left( {g * {s}_{2}}\right) = \left( {{g}^{-1} \cdot g}\right)... | Yes |
Lemma 18.5. Let \( S \) be a \( G \) -set, \( s \in S \) . Then \( {G}_{s} \) is a subgroup of \( G \) . | Proof. By definition of a \( G \) -action, \( e \in {G}_{s} \), so \( {G}_{s} \) is non-empty. If \( x, y \in G \), then \( \left( {xy}\right) * s = x * \left( {y * s}\right) = x * s = s \) and if \( y * s = s \), then \( s = {y}^{-1} * s \), so \( {G}_{s} \) is a subgroup. | No |
Proposition 18.6. Let \( S \) be a \( G \) -set, \( s \in S \) . Define\n\n\[ \n{f}_{s} : G/{G}_{s} \rightarrow G * s\text{ by }x{G}_{s} \mapsto x * s.\n\]\n\nThen \( {f}_{s} \) is a well-defined bijection. In particular, if \( \left\lbrack {G : {G}_{s}}\right\rbrack \) is finite, then\n\n\[ \n\left| {G * s}\right| = \... | Proof. Let \( x, y \in G \), then\n\n\[ \nx * s = y * s\text{ if and only if }{y}^{-1} * \left( {x * s}\right) = s\n\]\n\n\[ \n\text{if and only if}\left( {{y}^{-1}x}\right) * s = s\n\]\n\n\[ \n\text{if and only if}{y}^{-1}x \in {G}_{s}\n\]\n\n\[ \n\text{if and only if}x{G}_{s} = y{G}_{s}\text{.\n\]\n\nThis shows that ... | Yes |
Let \( S \) be the faces of a cube and \( G \) be the group of rotations of \( S \) . So \( G \) acts on \( S \) . Given any two faces \( {s}_{1},{s}_{2} \), there is an element of \( G \) taking \( {s}_{1} \) to \( {s}_{2} \) . So there is one orbit under this action. We say that \( G \) acts transitively on \( S \) .... | By the proposition, we have \[ \left| G\right| /\left| {C}_{4}\right| = \left\lbrack {G : {G}_{s}}\right\rbrack = \left| {G * s}\right| = \left| S\right| = 6, \] so \( \left| G\right| = {24} \) . | Yes |
Theorem 18.9. (Orbit Decomposition Theorem) Let \( S \) be a G-set. Then\n\n\[ S = {F}_{G}\left( S\right) \vee \mathop{\bigvee }\limits_{{\mathcal{O}}^{ * }}G * s \]\n\nIn particular, if \( S \) is a finite set, then\n\n\[ \left| S\right| = \mathop{\sum }\limits_{\mathcal{O}}\left| {G * s}\right| = \left| {{F}_{G}\left... | Proof. This follows from the Mantra as \( \left| {G * s}\right| = \left\lbrack {G : {G}_{s}}\right\rbrack \) . | No |
Lemma 19.1. Let \( G \) be a finite subgroup of \( {\mathrm{{SO}}}_{3}\left( \mathbb{R}\right) \) of order \( n \), and \( P \) the set of poles of \( G \) . Then \( P \) is a \( G \) -set under the natural action. In particular, if \( p \in P,\left| G\right| = \left\lbrack {G : {G}_{p}}\right\rbrack \left| {G}_{p}\rig... | Proof. Let \( p \) be a pole of \( A \) in \( G \) . If \( B \) is a matrix in \( G \), then \( \left( {{BA}{B}^{-1}}\right) {Bp} = {Bp} \), so \( {Bp} \) is a pole of \( {BA}{B}^{-1} \) . | Yes |
Theorem 20.6. Let \( G \) be a group of order \( {p}^{2} \) with \( p \) a prime. Then \( G \) is abelian. | Proof. Suppose that \( G \) is not abelian, then \( Z\left( G\right) < G \) by Exercise 20.26(1). By Application 20.4, we have \( 1 < Z\left( G\right) \) ; so by Lagrange’s Theorem, we must have \( \left| {Z\left( G\right) }\right| = p \) . Let \( a \in G \smallsetminus Z\left( G\right) \) . Then we have \( Z\left( G\r... | No |
Proposition 20.8. Let \( G \) be a p-group and \( N \) a non-trivial normal subgroup of \( G \) . Then there exists an non-identity element \( x \in N \) such that \( {xy} = {yx} \) for all \( y \in G \), i.e., \( 1 < N \cap Z\left( G\right) \) . | Proof. Let \( G \) act on \( N \) by conjugation. Every subgroup of \( G \) is either 1 or a \( p \) -group, hence\n\n\[ 0 \equiv \left| N\right| = \left| {{F}_{G}\left( N\right) }\right| + \mathop{\sum }\limits_{{\mathcal{O}}^{ * }}\left\lbrack {G : {Z}_{G}\left( a\right) }\right\rbrack \equiv \left| {{F}_{G}\left( N\... | Yes |
Theorem 20.23. (Cauchy's Theorem) Let \( p \) be a (positive) prime dividing the order of the finite group \( G \) . Then there exists an element of \( G \) of order \( p \) . | Proof. Let \( S \) be the \( \mathbb{Z}/p\mathbb{Z} \) -group under the shift action of Example 20.22. Apply the Orbit Decomposition Theorem 18.9 to get\n\n\[ \left| {G}^{p}\right| = \left| S\right| = \left| {{F}_{\mathbb{Z}/p\mathbb{Z}}\left( S\right) }\right| + \mathop{\sum }\limits_{{\mathcal{O}}^{ * }}\left\lbrack ... | Yes |
Corollary 20.25. Let \( p \) and \( q \) be primes and \( G \) of order \( {pq} \) . Then \( G \) is not simple. | Proof. If \( p = q \), then \( G \) is abelian hence not simple as \( G \) is not of prime order. If \( p < q \), then \( G \) contains a normal group of order \( q \) by Useful Counting 12.7. | No |
Theorem 21.2. (Generalized First Sylow Theorem) Let \( p \) be a (positive) prime such that \( {p}^{s} \), with \( s \geq 0 \), divides the order of a finite group \( G \) . Then there exists a subgroup of \( G \) of order \( {p}^{s} \) . | Proof. We induct on the order of \( G \) . If \( \left| G\right| \leq p \), the result is trivial, so we may assume that \( s > 1 \) . We make the following:\n\nInduction Hypothesis. If \( T \) is any finite group satisfying \( \left| T\right| < \left| G\right| \) and \( {p}^{s}\left| \right| T \mid \), then \( T \) co... | Yes |
Lemma 21.3. Let \( H \) and \( K \) be subgroups of a finite group \( G \) . Then \( \left| {HK}\right| = \left| H\right| \left| K\right| /\left| {H \cap K}\right| \) . | Proof. This is Exercise 13.6(3). | No |
Lemma 21.4. Let \( G \) be a finite group, \( P \) a Sylow p-subgroup, and \( H \) a subgroup of \( G \) that is also a p-group. If \( H \) is a subgroup of \( {N}_{G}\left( P\right) \), i.e., \( {hP}{h}^{-1} = P \) for all \( h \in H \), then \( H \) is a subgroup of \( P \) . In particular, if, in addition, \( H \) i... | Proof. The key to the proof of this lemma is the Second Isomorphism Theorem together with the counting version of the Second Isomorphism Theorem of Lemma 21.3. By definition, \( P \vartriangleleft {N}_{G}\left( P\right) \) and by hypothesis \( H \subset {N}_{G}\left( P\right) \) . Hence, by the Second Isomorphism Theor... | Yes |
Lemma 21.5. Let \( P \) be a Sylow p-subgroup of a finite group \( G \) and \( H \) a subgroup of \( G \) that is also a p-group. Then the following hold:\n\n(1) \( C\left( P\right) \) consists of Sylow p-subgroups of \( G \) .\n\n(2) The conjugacy class \( C\left( P\right) \) is an \( H \) -set by conjugation, i.e., v... | Proof. (1): As \( \left| P\right| = \left| {{xP}{x}^{-1}}\right| \) for all \( x \) in \( G \), this is clear.\n\n(2) is immediate.\n\n(3): As \( T \) is an \( H \) -fixed point, the orbit \( H * T = \{ T\} \), so \( {xT}{x}^{-1} = T \) for all \( x \in H \) . By definition, this means that \( H \subset {N}_{G}\left( T... | Yes |
Proposition 21.7. Let \( G \) be a finite group, \( p \) a (positive) prime dividing the order of \( G \), and \( P \) a Sylow p-subgroup. Then the following are equivalent:\n\n(1) \( {\operatorname{Syl}}_{p}\left( G\right) = \{ P\} \) .\n\n(2) \( C\left( P\right) = \{ P\} \) .\n\n(3) \( P \vartriangleleft G \) .\n\n(4... | Proof. The equivalence of (1), (2), and (3) follows from the Second Sylow Theorem. Clearly, (1) implies (4) as all isomorphic images of a group have the same cardinality. Finally (4) implies (3) by the comment above. | No |
Corollary 21.8. Let \( G \) be a finite group, \( p \) a (positive) prime dividing the order of \( G \), and \( P \) a Sylow p-subgroup. Then \( {N}_{G}\left( P\right) = \) \( {N}_{G}\left( {{N}_{G}\left( P\right) }\right) \) . | Proof. As \( P \) is a Sylow \( p \) -subgroup of \( {N}_{G}\left( P\right) \) and normal in it, \( P \) is a characteristic subgroup of \( {N}_{G}\left( P\right) \) by the proposition. As \( {N}_{G}\left( P\right) \) is a normal subgroup of \( {N}_{G}\left( {{N}_{G}\left( P\right) }\right) \), we conclude by Exercise ... | No |
Theorem 21.10. Let \( p \) and \( q \) be primes, \( G \) a group of order \( {p}^{s}q \) with \( s \geq 1 \) . Then \( G \) is not simple. | Proof. We may assume that \( p \neq q \) and there exist no normal Sylow subgroups. In particular, if \( {n}_{p} = \left| {{\operatorname{Syl}}_{p}\left( G\right) }\right| \), then \( 1 < {n}_{p} = \) \( 1 + {kp} \mid q \) with \( k \geq 1 \), an integer. In particular, \( {n}_{p} \leq q \) . Among all the Sylow \( p \... | Yes |
Lemma 22.4. If \( \alpha \) and \( \beta \) are disjoint cycles, then \( \alpha \) and \( \beta \) commute, i.e., \( {\alpha \beta } = {\beta \alpha } \) . | Proof. \( \beta \left( {a}_{i}\right) = {a}_{i} \) for all \( i \) and \( \alpha \left( {b}_{j}\right) = {b}_{j} \) for all \( j \) . | No |
Proposition 22.6. Let \( n > 1 \), then every element in \( {S}_{n} \) is a product of the transpositions \( \left( {12}\right) ,\left( {13}\right) ,\ldots ,\left( {1n}\right) \) . | Proof. We first show that \( {S}_{n} \) is generated by transpositions. If \( \sigma = \left( {{a}_{1}\ldots {a}_{r}}\right) \), then\n\n(22.7)\n\n\[ \sigma = \left( {{a}_{r - 1}{a}_{r}}\right) \cdots \left( {{a}_{2}{a}_{r}}\right) \left( {{a}_{1}{a}_{r}}\right) \]\n\nis a product of \( r - 1 \) transpositions.\n\nTo s... | Yes |
The alternating group \( {A}_{n} \) is a normal subgroup of index two in \( {S}_{n} \), and, if \( n \geq 3 \), then \( {A}_{n} \) is generated by the 3-cycles: | Proof. We need only show the last statement. We first show that \( {A}_{n} \) is generated by 3-cycles. We have shown that every element in \( {A}_{n} \) is a product of an even number of transpositions, so it suffices to show a product of two distinct transpositions is a product of 3-cycles. Suppose that \( i, j, k, l... | Yes |
Lemma 22.10. Let \( K \) be a normal subgroup of \( {A}_{n} \) . If \( K \) contains a 3-cycle, then \( K = {A}_{n} \) . | Proof. As \( K \) contains a 3-cycle, we have \( n \geq 3 \) . By the proposition, it suffices to show that \( \left( {12k}\right) \) lies in \( K \) for \( k = 3,\ldots, n \) . Changing notation, we may assume that (123) lies in \( K \), hence (213) \( = {\left( {123}\right) }^{-1} \) lies in \( K \) . Let \( \sigma =... | Yes |
Lemma 22.13. Let \( n \geq 5 \), then \( {A}_{n} \) is the only subgroup of \( {S}_{n} \) of index two. | Proof. Suppose that \( K \) is a normal subgroup of \( G \) of index two and \( K \neq {A}_{n} \) . By the Second Isomorphism Theorem, we conclude \( {A}_{n}K \) is a group with \( K < {A}_{n}K \) normal and \( {A}_{n}K/K \cong {A}_{n}/\left( {{A}_{n} \cap K}\right) \) . As \( {A}_{n} < {A}_{n}K \subset {S}_{n} \), we ... | Yes |
Lemma 22.18. Let \( G \) be a group of order \( {mn} \) and \( \lambda : G \rightarrow \sum \left( G\right) \) the regular representation with \( m > 1 \) . If the element \( a \) in \( G \) has order \( m \) , then\n\n(1) \( {\lambda }_{a} \) is a product of \( n \) disjoint \( m \) -cycles.\n\n(2) \( {\lambda }_{a} \... | Proof. (1): By the review, we know if \( 1 \leq r < m \), then the permutation \( {\lambda }_{{a}^{r}} = {\left( {\lambda }_{a}\right) }^{r} \) has no fixed points. Let \( {\lambda }_{a} = {\gamma }_{1}\cdots {\gamma }_{s} \) be a full cycle decomposition of \( {\lambda }_{a} \) . As \( {\lambda }_{{a}^{r}} \) has no f... | Yes |
Lemma 22.19. Let \( G \) be a finite group and \( \lambda : G \rightarrow \sum \left( G\right) \) the regular representation. If \( \lambda \left( G\right) \) contains an odd permutation, then there exists a normal subgroup of \( G \) of index two. | Proof. Let \( \left| G\right| = n \) and identify \( \sum \left( G\right) \) and \( {S}_{n} \) . As \( \lambda \left( G\right) \) contains an odd permutation, \( H \mathrel{\text{:=}} {A}_{n} \cap \lambda \left( G\right) < \lambda \left( G\right) \) and \( \lambda \left( G\right) {A}_{n} = \sum \left( G\right) \) . The... | Yes |
Theorem 22.21. Let \( G \) be a finite group of order \( {2}^{r}m \) with \( m \) odd. If \( G \) contains a cyclic Sylow 2-subgroup then there exists a normal subgroup of \( G \) of index \( {2}^{r} \) . In particular, if \( m > 1 \) or \( r > 1 \), then \( G \) is not simple. | Proof. We may assume that \( m > 1 \) and \( r \geq 1 \) . Let \( \lambda : G \rightarrow \sum \left( G\right) \) be the regular representation. By Lemma 22.18, the group \( \lambda \left( G\right) \) contains an odd permutation, so by Lemma 22.19, there exists a normal subgroup \( N \) of \( G \) of index two. Hence \... | Yes |
Lemma 23.20. Let \( R \) be a commutative ring and \( \mathfrak{p} < R \) an ideal. Then \( \mathfrak{p} \) is a prime ideal if and only if for all ideals \( \mathfrak{A} \) and \( \mathfrak{B} \) in \( R \) satisfying \( \mathfrak{A}\mathfrak{B} \subset \mathfrak{p} \), either \( \mathfrak{A} \subset \mathfrak{p} \) o... | Proof. ( \( \Leftarrow \) ): If \( {xy} \in \mathfrak{p} \) with \( x, y \in R \) then \( \left( x\right) \left( y\right) \subset \mathfrak{p} \), so either \( x \in \left( x\right) \subset \mathfrak{p} \) or \( y \in \left( y\right) \subset \mathfrak{p} \) .\n\n\( \left( \Rightarrow \right) \) : Suppose that the ideal... | Yes |
Proposition 24.2. Let \( R \) be a ring and \( \mathfrak{A} \) an ideal in \( R \) . Then \( \equiv \) \( {\;\operatorname{mod}\;\mathfrak{A}} \) is an equivalence relation. Suppose that the elements \( a,{a}^{\prime }, b,{b}^{\prime } \) in \( R \) satisfy:\n\n\[ a \equiv {a}^{\prime }{\;\operatorname{mod}\;\mathfrak{... | Proofs analogous to those in Group Theory show: | No |
Proposition 24.5. Let \( R \) be a nontrivial commutative ring and \( \mathfrak{A} \) an ideal of \( R \) . Then (1) \( \mathfrak{A} \) is a prime ideal of \( R \) if and only if \( R/\mathfrak{A} \) is a domain. | Proof. Let \( - : R \rightarrow R/\mathfrak{A} \) be the canonical epimorphism, so \( {\ker }^{ - } = \mathfrak{A} \) . (1): \( \bar{R} \) is a domain if and only if \( \left( \overline{0}\right) \) is a prime ideal in \( \bar{R} \) if and only if whenever \( a, b \in R \) satisfy \( \bar{a} \cdot \bar{b} = \overline{0... | Yes |
Proposition 24.6. Let \( R \) be a ring and \( \mathfrak{A} \subset \mathfrak{B} \subset R \) ideals. Then\n\n\[ \mathfrak{B}/\mathfrak{A} \mathrel{\text{:=}} \{ b + \mathfrak{A} \mid b \in \mathfrak{B}\} \subset R/\mathfrak{A} \]\n\nis an ideal in \( R/\mathfrak{A} \) and\n\n\[ R/\mathfrak{B} \cong \left( {R/\mathfrak... | Proof. It is easy to check that \( R/\mathfrak{A} \rightarrow R/\mathfrak{B} \) given by \( r + \mathfrak{A} \mapsto \) \( r + \mathfrak{B} \) is a well-defined epimorphism with kernel \( \mathfrak{B}/\mathfrak{A} \) . | No |
Proposition 24.7. Let \( \varphi : R \rightarrow S \) be a ring epimorphism of commutative rings with kernel \( \mathfrak{A} \) . Then the map\n\n\( \{ \mathfrak{p} \mid \mathfrak{p} < R \) a prime ideal with \( \mathfrak{A} \subset \mathfrak{p}\} \rightarrow \{ \mathfrak{P} \mid \mathfrak{P} < S \) a prime ideal \( \}... | Proof. We may assume that \( S \) is not the trivial ring for if not then both sets of prime ideals would be empty. By the First Isomorphism Theorem, we may assume that \( S = R/\mathfrak{A} \) and \( \varphi \) is the canonical epimorphism \( {}^{ - } : R \rightarrow R/\mathfrak{A} \) . Let \( \mathfrak{A} \subset \ma... | Yes |
Proposition 24.12. Let \( R \) be a domain. Then either\n\n(1) \( \operatorname{char}\left( R\right) = 0 \) and there exists a ring monomorphism \( \mathbb{Z} \rightarrow R \) or\n\n(2) \( \operatorname{char}\left( R\right) = p, p \) a prime, and there exists a ring monomorphism\n\n\( \mathbb{Z}/p\mathbb{Z} \rightarrow... | Proof. Let \( \iota : \mathbb{Z} \rightarrow R \) be the unique ring homomorphism, then \( \ker \varphi = \left( n\right) \) for some non-negative integer \( n \), so \( \operatorname{char}\left( R\right) = n \) . As \( R \) is a domain so is \( \operatorname{im}\iota \cong \mathbb{Z}/n\mathbb{Z} \) . Therefore, \( n =... | Yes |
Lemma 25.5. (Zorn’s Lemma) Let \( S \) be a non-empty inductive poset. Then \( S \) contains a maximal element. | This is indeed an extension of finite induction, for if \( S \) is a subset of \( {\mathbb{Z}}^{ + } \), then \( \left( {S, \geq }\right) \) has a maximal element, which is exactly the Well-Ordering Principle. As mentioned, we cannot prove this lemma; we accept it as an axiom. | No |
Proposition 25.6. Let \( V \) be a nonzero vector space over a field \( F \) and \( S \) a linearly independent subset of \( V \) . Then \( S \) extends to a basis of \( V \) , i.e., is part of a basis for \( V \) . | Proof. Let\n\n\[ \mathcal{S} = \{ T \mid S \subset T \subset V\text{with}T\text{linearly independent}\} \text{,} \]\n\n a nonempty poset under \( \subset \) . Let \( \mathcal{C} \) be a chain in \( \mathcal{S} \) .\n\nClaim. \( A = \mathop{\bigcup }\limits_{\mathcal{C}}T \) is an upper bound for \( \mathcal{C} \) in \(... | Yes |
Proposition 25.7. Let \( V \) be a nonzero vector space over a field \( F \) and \( S \) a spanning set for \( V \) . Then a subset of \( S \) is a basis of \( V \) . | We leave a proof of this as an exercise. You should try the obvious proof. It does not work. Can you figure out what this means? Rather you will have to modify the proof above. | No |
Proposition 25.9. Let \( R \) be a nontrivial ring and \( \mathfrak{A} < R \) an ideal. Then there exists a maximal ideal \( \mathfrak{m} \) in \( R \) such that \( \mathfrak{A} \subset \mathfrak{m} \) . In particular, any nontrivial ring contains maximal ideals. | Proof. Let\n\n\[ S = \{ \mathfrak{B} \mid \mathfrak{A} \subset \mathfrak{B} < R\text{ an ideal }\} .\n\]\n\nThe set \( S \) is a poset under \( \subset \), and \( S \) is non-empty as \( \mathfrak{A} \in S \) . Let \( \mathcal{C} \) be a chain in \( S \), so\n\n\[ \text{for all}{\mathfrak{B}}^{\prime }\text{and}{\mathf... | Yes |
Theorem 25.13. (Krull) Let \( R \) be a commutative ring and \( S \) a multiplicative set. Suppose that \( \mathfrak{A} < R \) is an ideal in \( R \) excluding \( S \) . Then there exists an ideal \( \mathfrak{B} \subset R \) containing \( \mathfrak{A} \) excluding \( S \) and maximal with respect to these two properti... | Proof. Let\n\n\[ \mathcal{I} = \{ \mathfrak{B} \mid \mathfrak{A} \subset \mathfrak{B} < R\text{ an ideal excluding }S\} \]\n\nWe know that \( \mathcal{I} \) is not empty as \( \mathfrak{A} \in \mathcal{I} \) . Partially order \( \mathcal{I} \) by \( \subset \) . Let \( \mathcal{C} \) be a chain in \( \mathcal{I} \) .\n... | Yes |
Corollary 25.16. Let \( R \) be a non-trivial commutative ring. Then \( \operatorname{nil}\left( R\right) \) is an ideal in \( R \) and\n\n\[ \operatorname{nil}\left( R\right) = \mathop{\bigcap }\limits_{\substack{{\mathfrak{p} < R} \\ {\mathfrak{p}\text{ a prime ideal }} }}\mathfrak{p}. \] | Proof. We leave it as an exercise to prove \( \operatorname{nil}\left( R\right) \) is a ideal.\n\n\( \operatorname{nil}\left( R\right) \subset \mathop{\bigcap }\limits_{{\mathfrak{p} < R}}\mathfrak{p} \) : If \( a \in \operatorname{nil}\left( R\right) \), there exists an \( n \in {\mathbb{Z}}^{ + } \) such that \( {a}^... | No |
Corollary 27.6. (Euclid’s Lemma) Let \( R \) be a PID and \( r \in R \) . Then \( r \) is irreducible if and only if \( r \) is a prime element. | Proof. We have already shown that prime elements are irreducible. Conversely, assume that \( r \) is an irreducible element in \( R \) and satisfies \( r \mid {ab} \) with \( a, b \in R \) . As \( R \) is a PID, there exists a \( c \in R \) satisfying \( \left( c\right) = \left( {r, a}\right) \) . Write \( r = {xc} \) ... | Yes |
Proposition 27.9. (Euclid's Argument) Let \( R \) be a domain. Suppose that \( {p}_{1}\cdots {p}_{n} = {f}_{1}\cdots {f}_{m} \) with \( {p}_{1},\ldots ,{p}_{n} \) prime elements in \( R \) and \( {f}_{1},\ldots ,{f}_{m} \) irreducible elements in \( R \) for some \( n, m \in {\mathbb{Z}}^{ + } \) . Then \( n = m \) and... | Proof. As \( {p}_{1} \mid {f}_{1}\cdots {f}_{m} \) in \( R \), there exists an \( i \) such that \( {f}_{i} = u{p}_{1} \) for some element \( u \) in \( R \) . Since \( {f}_{i} \) is irreducible and \( {p}_{1} \) not a unit, \( u \) is a unit in \( R \) . Now \( {p}_{1}\cdots {p}_{n} = u{p}_{1}{f}_{2}\cdots {f}_{m} \) ... | Yes |
Theorem 27.14. Let \( R \) be a Noetherian domain and \( r \) a nonzero non-unit in \( R \). Then \( r \) is a product of (finitely many) irreducible elements. | Proof. Let\n\n\[ S = \{ \left( a\right) \mid \left( 0\right) < \left( a\right) < R \]\n\nwith \( a \) not a product of irreducible elements \( \} \).\n\nSuppose that \( S \) is non-empty, i.e., the result is false. Then there exists a principal ideal \( \left( a\right) \in S \), a maximal element by the Maximal Princip... | No |
Theorem 27.15. Let \( R \) be a PID. Then \( R \) is a UFD. | Proof. As \( R \) is a PID, it is a Noetherian domain. Hence every nonzero nonunit factors into a product of irreducible elements. As every irreducible element in a PID is a prime, factorization is (essentially) unique by Euclid's Argument. | Yes |
Theorem 27.20. If \( R \) is a euclidean domain, then \( R \) is a PID, hence \( a ⫫ {UFD} \) . | Proof. Let \( 0 < \mathfrak{A} \subset R \) be an ideal and\n\n\[ \varnothing \neq \mathcal{S} = \{ \partial a \mid 0 \neq a \in \mathfrak{A}\} \subset {\mathbb{Z}}^{ + } \cup \{ 0\} \]\n\n(as \( \mathfrak{A} \neq 0 \) ), By the Well-ordering Principle, there exists an element \( a \) in \( \mathfrak{A} \) with \( \par... | Yes |
Theorem 28.1. (Kaplansky) Let \( R \) be a domain. Then \( R \) is a UFD if and only if every nonzero prime ideal contains a prime element. | Proof. \( \left( \Rightarrow \right) \) : Let \( 0 < \mathfrak{p} < R \) be a prime ideal and \( a \) a nonzero element in \( \mathfrak{p} \) . Then \( a = {f}_{1}\cdots {f}_{r} \), with \( {f}_{1},\ldots ,{f}_{r} \) irreducible in \( R \) . Since \( \mathfrak{p} \) is a prime ideal, there exist an \( i \) such that \(... | Yes |
Lemma 29.2. \( \mathbb{Z}\left\lbrack \sqrt{-1}\right\rbrack \) is a domain and satisfies the following:\n\n(1) \( {\left. N\right| }_{\mathbb{Z}\left\lbrack \sqrt{-1}\right\rbrack } : \mathbb{Z}\left\lbrack \sqrt{-1}\right\rbrack \rightarrow {\mathbb{Z}}^{ + } \cup \{ 0\} \) and satisfies \( {\left. N\right| }_{\mathb... | Proof. We only prove (2) leaving the rest as an easy check. If \( {uv} = 1 \) with \( u, v \in \mathbb{Z}\left\lbrack t\right\rbrack \), then \( 1 = N\left( 1\right) = N\left( u\right) N\left( v\right) \), so we have \( N\left( u\right) \in {\mathbb{Z}}^{ \times } \cap {\mathbb{Z}}^{ + } = \{ 1\} \) . As \( {x}^{2} + {... | No |
Theorem 29.3. The domain \( \mathbb{Z}\left\lbrack \sqrt{-1}\right\rbrack \) is a (strong) euclidean domain, hence a PID and so a UFD. | Proof. Claim. \( N \) is a euclidean function on \( \mathbb{Z}\left\lbrack \sqrt{-1}\right\rbrack \) . (Consequently, \( \mathbb{Z}\left\lbrack \sqrt{-1}\right\rbrack \) is a strong euclidean domain.):\n\nLet \( \alpha \) and \( \beta \) be elements in \( \mathbb{Z}\left\lbrack \sqrt{-1}\right\rbrack \) with \( \beta \... | Yes |
Lemma 29.4. Let \( p \) be a positive prime in \( \mathbb{Z} \) satisfying \( p \equiv 1{\;\operatorname{mod}\;4} \) . Then there exists an integer \( x \) satisfying \( p \mid {x}^{2} + 1 \) in \( \mathbb{Z} \) . In particular, -1 is a square modulo \( p \) if \( p \equiv 1{\;\operatorname{mod}\;4} \) . | Proof. Let \( x = 1 \cdot 2 \cdot 3 \cdot \cdots \cdot \frac{p - 1}{2} \) in \( \mathbb{Z} \) . By Wilson’s Theorem (Exercise 10.16(11)), we have\n\n\[ - 1 \equiv \left( {p - 1}\right) ! = \left( {1 \cdot 2 \cdot 3\cdots \frac{p - 1}{2}}\right) \left( {p - \frac{p - 1}{2}\cdots p - 3 \cdot p - 2 \cdot p - 1}\right) \]\... | Yes |
Theorem 29.5. (Fermat) Let \( p \) be a positive prime congruent to 1 modulo 4. Then there exist integers \( x \) and \( y \) such that \( p = {x}^{2} + {y}^{2} \) . | Proof. By the lemma there exists an integer \( x \) such that \( p \mid {x}^{2} + 1 \) . Hence \( p \mid \left( {x + \sqrt{-1}}\right) \left( {x - \sqrt{-1}}\right) \) in the UFD \( \mathbb{Z}\left\lbrack \sqrt{-1}\right\rbrack \) .\n\nClaim. \( p \) is not irreducible (i.e., it is reducible) in \( \mathbb{Z}\left\lbra... | Yes |
Lemma 29.6. Let \( a \) and \( b \) be two nonzero relatively prime integers and \( {pa} \) (positive) odd prime such that \( p \mid {a}^{2} + {b}^{2} \) . Then \( p \equiv 1{\;\operatorname{mod}\;4} \) . | Proof. As \( p \mid {a}^{2} + {b}^{2} \), we have \( p \mid a \) if and only if \( p \mid b \) . Since \( a \) and \( b \) are relatively prime, \( p \) cannot divide \( a \) or \( b \) . By Fermat’s Little Theorem 10.14, we know that \( {b}^{p - 3}{b}^{2} = {b}^{p - 1} \equiv 1{\;\operatorname{mod}\;p} \) . Consequent... | Yes |
Example 29.9. \( \left( \frac{0}{7}\right) = 0,\left( \frac{1}{7}\right) = \left( \frac{2}{7}\right) = \left( \frac{4}{7}\right) = 1 \), and \( \left( \frac{3}{7}\right) = \left( \frac{5}{7}\right) = \) \( \left( \frac{6}{7}\right) = - 1 \) | We have proven:\n\nProposition 29.10. (Eul | No |
The set \( \{ {4n} + 1 \mid n \in \mathbb{Z}\} \) contains infinitely many primes. | We prove the stronger result that the set \( \{ {8n} + 5 \mid n \in \mathbb{Z}\} \) contains infinitely many primes.\n\nCheck. If \( n \in \mathbb{Z} \), then \( {n}^{2} \equiv 0,1,4{\;\operatorname{mod}\;8} \) . In particular, if \( n \) is odd, then \( {n}^{2} \equiv 1{\;\operatorname{mod}\;8} \) .\n\nLet \( p \) be ... | Yes |
Lemma 30.3. \( 1 \leq m < p \) and \( {mp} \) is a sum of four integer squares. | Proof. Let\n\n\[ \n{S}_{1} \mathrel{\text{:=}} \left\{ {{0}^{2},{1}^{2},\ldots ,\frac{{\left( p - 1\right) }^{2}}{2}}\right\} \n\] \n\n\[ \n{S}_{2} \mathrel{\text{:=}} \left\{ {-{0}^{2} - 1, - {1}^{2} - 1,\ldots , - \frac{{\left( p - 1\right) }^{2}}{2} - 1}\right\} . \n\] \n\nIf \( {x}^{2} = {y}^{2}{\;\operatorname{mod... | No |
Lemma 31.3. Let \( R \) be a domain, then \( R{\left\lbrack t\right\rbrack }^{ \times } = {R}^{ \times } \) . | Proof. Clearly, units in \( R \) remain units in \( R\left\lbrack t\right\rbrack \) . Conversely, let \( f \) , \( g \) be polynomials in \( R\left\lbrack t\right\rbrack \) satisfying \( {fg} = 1 \) . Then \( \deg \left( {fg}\right) = \deg f + \) \( \deg g = 0 \), so \( \deg f = \deg g = 0 \) and \( f, g \) lie in \( R... | Yes |
Let \( F \) be a field. Then \( F\left\lbrack t\right\rbrack \) is a euclidean domain. | In particular, \( F\left\lbrack t\right\rbrack \) is a PID, hence also a UFD. | No |
Corollary 31.6. (Remainder Theorem) Let \( R \) be a commutative ring, \( \alpha \) an element in \( R \), and \( f \) a nonzero polynomial in \( R\left\lbrack t\right\rbrack \) . Then there exists a polynomial \( q \) in \( R\left\lbrack t\right\rbrack \) satisfying \( f = \left( {t - \alpha }\right) q + f\left( \alph... | Proof. Apply the General Division Algorithm with \( g = t - \alpha \) to get \( f = \left( {t - \alpha }\right) q + r \) in \( R\left\lbrack t\right\rbrack \) for some \( q, r \in R\left\lbrack t\right\rbrack \) with \( r = 0 \) or \( \deg r < \deg \left( {t - \alpha }\right) = 1 \) . It follows that \( r \in R \), hen... | Yes |
Let \( R \) be a domain, \( f \) a nonzero polynomial in \( R\left\lbrack t\right\rbrack \) , and \( {x}_{1},\ldots ,{x}_{n} \) distinct roots of \( f \) in \( R \) . Then \( \mathop{\prod }\limits_{{i = 1}}^{n}\left( {t - {x}_{i}}\right) \mid f \) in \( R\left\lbrack t\right\rbrack \) . In particular, \( \deg f \geq n... | Proof. If \( n = 1 \), the result follows by the Remainder Theorem, so assume that \( n > 1 \) . By iterating the Remainder Theorem, we can write \( f = {\left( t - {x}_{1}\right) }^{r}h \) in \( R\left\lbrack t\right\rbrack \) for some \( r \) in \( {\mathbb{Z}}^{ + } \) and \( h \) in \( R\left\lbrack t\right\rbrack ... | Yes |
Corollary 31.12. (Kronecker’s Theorem) Let \( F \) be a field and \( f \) an irreducible element in \( F\left\lbrack t\right\rbrack \) . Set \( K = F\left\lbrack t\right\rbrack /\left( f\right) \) . Then \( K \) is a field. Viewing \( F \subset K \), i.e., as a subfield of \( K \) as above, we have \( \bar{t} \) is a r... | Proof. As \( f \) is irreducible in the UFD \( F\left\lbrack t\right\rbrack \), it is a prime element. As \( F\left\lbrack t\right\rbrack \) is a PID, the nonzero prime ideal \( \left( f\right) \) is maximal, so \( K \) is a field. As \( 0 = \bar{f} = f\left( \bar{t}\right) \), by the above, the result follows. | Yes |
Corollary 31.13. Let \( F \) be a field and \( f \) a non-constant polynomial in \( F\left\lbrack t\right\rbrack \) . Then there exists a field \( K \) with \( F \subset K \) a subfield such that \( f \) has a root in \( K \) . | Proof. We can write \( f = {f}_{1}g \) with \( {f}_{1} \) and \( g \) polynomials in \( F\left\lbrack t\right\rbrack \) and \( {f}_{1} \) irreducible. By Kronecker’s Theorem, \( {f}_{1} \) hence \( f \) has a root in \( F\left\lbrack t\right\rbrack /\left( {f}_{1}\right) \), a field containing \( F \) . | Yes |
Theorem 31.15. Let \( F \) be a field and \( G \) a finite (multiplicative) subgroup of \( {F}^{ \times } \) . Then \( G \) is cyclic. In particular, if \( F \) is a finite field, the group \( {F}^{ \times } \) is cyclic. | Proof. Let \( p \) be a prime dividing \( \left| G\right| \) and \( P \) a Sylow \( p \) -subgroup of \( G \) (so the only one as \( G \) is abelian). Choose \( {x}_{P} \) in \( P \) such that the order of the cyclic subgroup \( \left\langle {x}_{P}\right\rangle \) generated by \( {x}_{P} \) in \( P \) is maximal. Let ... | No |
Theorem 31.17. (Fundamental Theorem of Algebra) The field of complex numbers is algebraically closed. | We shall, however, assume here that it has already been established. | No |
Lemma 32.3. Let \( R \) be a UFD, \( K \) the quotient field of \( R \), and \( {fg} \) nonzero polynomial in \( K\left\lbrack t\right\rbrack \) . Then there exist a primitive polynomial \( {f}_{1} \in R\left\lbrack t\right\rbrack \) (of the same degree as \( f \) ) and an element \( \alpha \) in \( K \) satisfying \( ... | Proof. Existence: Write \( f = \mathop{\sum }\limits_{{i = 0}}^{n}\frac{{a}_{i}}{{b}_{i}}{t}^{i} \) with \( {a}_{i},{b}_{i} \) in \( R \) and \( {b}_{i} \neq 0 \) , \( i = 0,\ldots, n \) . We clear denominators. Let \( b = {b}_{0}\cdots {b}_{n} \neq 0 \) in the domain \( R \) . Then \( {bf} \in R\left\lbrack t\right\rb... | Yes |
Corollary 32.4. Let \( R \) be a UFD, \( f, g \) primitive polynomials in \( R\left\lbrack t\right\rbrack \) , \( h \) a polynomial in \( R\left\lbrack t\right\rbrack \), and \( K \) the quotient field of \( R \) .\n\n(1) If \( g = {sf} \) in \( K\left\lbrack t\right\rbrack \) for some \( s \) in \( K \), then \( s \) ... | Proof. (1): We have \( 1 \cdot g = s \cdot f \) in \( K\left\lbrack t\right\rbrack \) with both \( f \) and \( g \) primitive, so \( s = u \cdot 1 \) for some unit \( u \) in \( R \) by Lemma 32.3, hence \( s \) is a unit in \( R \) .\n\n(2): Write \( h = C\left( h\right) {h}_{1} \) with \( {h}_{1} \) primitive in \( R... | Yes |
Lemma 32.5. (Gauss’ Lemma) Let \( R \) be a UFD and \( f \) , \( g \) non-constant polynomials. Then \( C\left( {fg}\right) \approx C\left( f\right) C\left( g\right) \) . In particular, the product of primitive polynomials in \( R\left\lbrack t\right\rbrack \) is primitive. | Proof. Write \( f = C\left( f\right) {f}_{1} \) and \( g = C\left( g\right) {g}_{1} \) with \( {f}_{1} \) and \( {g}_{1} \) primitive polynomials in \( R\left\lbrack t\right\rbrack \) . Then\n\n\[ C\left( {fg}\right) \approx C\left( {C\left( f\right) {f}_{1}C\left( g\right) {g}_{1}}\right) \approx C\left( f\right) C\le... | Yes |
Corollary 32.6. Suppose that \( R \) is a UFD with quotient field \( K \) . Let \( f, g \) be non-constant primitive polynomials in \( R\left\lbrack t\right\rbrack \) and \( h \) a non-constant polynomial in \( K\left\lbrack t\right\rbrack \) . If \( f = {gh} \) in \( K\left\lbrack t\right\rbrack \), then \( h \) lies ... | Proof. By Lemma 32.3, we can write \( h = \alpha {h}_{1} \) with \( \alpha \in K \) and \( {h}_{1} \) a primitive polynomial in \( R\left\lbrack t\right\rbrack \), so \( f = {\alpha g}{h}_{1} \) . By Gauss’ Lemma, \( g{h}_{1} \) is primitive, so \( \alpha \) lies in \( {R}^{ \times } \) by Corollary 32.4. It follows th... | Yes |
Lemma 32.7. Let \( R \) be a UFD with quotient field \( K \) and \( f \) a nonconstant irreducible polynomial in \( R\left\lbrack t\right\rbrack \) . Then \( f \) remains irreducible in \( K\left\lbrack t\right\rbrack \) . | Proof. As \( f \) is a non-constant irreducible polynomial in \( R\left\lbrack t\right\rbrack \), it is primitive. Suppose that \( f \) factors as \( f = {g}_{1}{g}_{2} \) in \( K\left\lbrack t\right\rbrack \), with \( {g}_{i} \in K\left\lbrack t\right\rbrack \) and \( 0 < \deg {g}_{i} < \deg f \), for \( i = 1,2 \) . ... | Yes |
Theorem 32.8. Let \( R \) be a UFD, then \( R\left\lbrack t\right\rbrack \) is a UFD. | Proof. Existence. Every nonzero element in \( R \) is a unit or factors into irreducibles that remain irreducible in \( R\left\lbrack t\right\rbrack \), so it suffices to factor a non-constant polynomial in \( R\left\lbrack t\right\rbrack \) . Let \( f \) be such a polynomial. As \( f = C\left( f\right) {f}_{1} \) with... | Yes |
Lemma 34.2. Let \( g \) be a nonzero element in \( F\left\lbrack \left\lbrack {{t}_{1},{t}_{2}}\right\rbrack \right\rbrack \) . Then there exists a positive integer \( r \) satisfying \( g\left( {{t}_{2}^{r},{t}_{2}}\right) \) is nonzero. | Proof. Order the nonzero monomials \( {a}_{{r}_{1}{r}_{2}}{t}_{1}^{{r}_{1}}{t}_{2}^{{r}_{2}} \) of \( g \) lexicographically, i.e., \( \left( {{r}_{1},{r}_{2}}\right) \geq \left( {{s}_{1},{s}_{2}}\right) \) if \( {r}_{1} > {s}_{1} \) or \( {r}_{1} = {s}_{1} \) and \( {r}_{2} \geq {s}_{2} \) . Let \( {a}_{{s}_{1}{s}_{2}... | Yes |
Lemma 34.3. Let \( {f}_{1},\ldots ,{f}_{m} \) be nonzero elements in \( F\left\lbrack \left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \right\rbrack \) . Then there exists a ring automorphism \( \sigma \) of \( F\left\lbrack \left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \right\rbrack \) fixing \( F\left\lbrack... | Proof. As \( F\left\lbrack \left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \right\rbrack \) is a domain, \( f = {f}_{1}\cdots {f}_{m} \) is nonzero. So we are done if we find a \( \sigma \) that works for \( f \), i.e., we may assume that \( f = {f}_{1} \) . By the previous lemma, there exist \( {r}_{1},\ldots ,{r}... | Yes |
Theorem 34.4. (Algebraic Weierstraß Preparation Theorem) Let \( R = \) \( F\\left\\lbrack \\left\\lbrack {{t}_{1},\\ldots ,{t}_{n}}\\right\\rbrack \\right\\rbrack \) with \( F \) a field and \( f \\in R \) regular in \( {t}_{n} \). Suppose that \( s = {\\operatorname{ord}}_{{t}_{n}}f \) and \( g \) lies in \( R \). The... | Proof. We induct on \( n \).\n\nSuppose that \( n = 1 \) and \( t = {t}_{1} \): Then\n\n\[ f = {c}_{s}{t}^{s} + \\mathop{\\sum }\\limits_{{j > s}}{c}_{j}{t}^{j}\\text{ with }{c}_{s} \\neq 0 \]\n\n\[ g = \\mathop{\\sum }\\limits_{{i = 0}}^{{s - 1}}{a}_{i}{t}^{i} + \\mathop{\\sum }\\limits_{{i \\geq s}}{a}_{i}{t}^{i} = \... | Yes |
Corollary 34.6. Let \( R = F\left\lbrack \left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \right\rbrack \) with \( F \) a field and \( f \) an element of \( R \) regular in \( {t}_{n} \) with \( s = {\operatorname{ord}}_{{t}_{n}}f \) . Then there exists a unique pseudo-polynomial of degree \( s \) that is an associa... | Proof. Apply the Weierstraß Preparation Theorem to \( f \) and \( g = \) \( {t}_{n}^{s} \), to obtain unique elements \( h \in R \) and \( r \in F\left\lbrack \left\lbrack {{t}_{1},\ldots ,{t}_{n - 1}}\right\rbrack \right\rbrack \left\lbrack {t}_{n}\right\rbrack \) with \( {\deg }_{{t}_{n}}r < s \) satisfying \( {t}_{n... | No |
Corollary 34.8. Let \( R = F\left\lbrack \left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \right\rbrack \) with \( F \) a field and \( f \) and \( g \) elements of \( R \) both regular in \( {t}_{n} \) . Then \( {\left( fg\right) }^{ * } = {f}^{ * }{g}^{ * } \) . | Proof. Note that \( {fg} \) is regular at \( {t}_{n} \) as \( f \) and \( g \) are. By the previous result, there exist unique units \( u, v, w \in {R}^{ \times } \) satisfying \( {f}^{ * } = \) \( {uf},{g}^{ * } = {vg} \), and \( {\left( fg\right) }^{ * } = {wfg} \) pseudo-polynomials of the appropriate degree. Thus \... | Yes |
Theorem 34.9. Let \( F \) be a field, then \( F\left\lbrack \left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \right\rbrack \) is a UFD. | Proof. We induct on \( n \), the case \( n = 0 \) being immediate. So we may assume that \( n \geq 1 \) . Let \( {R}_{0} = F\left\lbrack \left\lbrack {{t}_{1},\ldots ,{t}_{n - 1}}\right\rbrack \right\rbrack \) and \( R = {R}_{0}\left\lbrack \left\lbrack {t}_{n}\right\rbrack \right\rbrack = \) \( F\left\lbrack \left\lbr... | Yes |
Lemma 35.12. Let \( M \) be an \( R \) -module and \( m,{m}^{\prime } \) elements in \( M \) . Then\n\n(1) \( {\operatorname{ann}}_{R}m \subset R \) is a left ideal.\n\n(2) \( {\rho }_{m} : R \rightarrow M \) given by \( r \mapsto {rm} \) is an \( R \) -homomorphism and satisfies \( \ker {\rho }_{m} = {\operatorname{an... | Proof. (1) and (2) follow immediately.\n\n(3) follows from the First Isomorphism Theorem for modules.\n\n(4): Suppose that we know that \( m = a{m}^{\prime } \) for some \( a \in R \) . If \( r{m}^{\prime } = 0 \) , then \( {ra}{m}^{\prime } = {ar}{m}^{\prime } = 0 \) . | Yes |
Corollary 35.13. Let \( M \) be an \( R \) -module. Then \( M \) is a cyclic \( R \) - module if and only if there exists a left ideal \( \mathfrak{A} \) in \( R \) satisfying \( M \cong \) \( R/\mathfrak{A} \) . | Proof. \( R/\mathfrak{A} = \langle 1 + \mathfrak{A}\rangle = R\left( {1 + \mathfrak{A}}\right) \) is cyclic, so this follows by the lemma. | No |
Proposition 35.14. Let \( R \) be a commutative ring, \( M \) a cyclic \( R \) - module. Then there exists a unique ideal \( \mathfrak{A} \) in \( R \) such that \( M \cong R/\mathfrak{A} \) . Moreover, if \( M = {Rm} \), then \( \mathfrak{A} = {\operatorname{ann}}_{R}m \) . In particular, if \( R \) is a PID, there ex... | Proof. We already know if \( M = {Rm} \), then \( M \cong R/{\operatorname{ann}}_{R}m \) . Suppose that we have an \( R \) -isomorphism \( f : R/\mathfrak{A} \rightarrow M \) for some ideal \( \mathfrak{A} \) of \( R \) . Let \( m = f\left( {1 + \mathfrak{A}}\right) \) . We have\n\n\[ f\left( {r + \mathfrak{A}}\right) ... | Yes |
Lemma 35.17. (Five Lemma) Suppose the following is a commutative diagram of \( R \) -modules and \( R \) -homomorphisms with exact rows: \n\nIf two of \( {h}_{A},{h}_{B},{h}_{C} \) are \( R \) -isomorphisms, then the... | The proof of this lemma is called diagram chasing, and we leave it as an exercise. | No |
Theorem 36.3. (Universal Property of Free Modules) Let \( \mathcal{B} = {\left\{ {x}_{i}\right\} }_{I} \) be a basis for a free \( R \) -module \( M \) . If \( N \) is an \( R \) -module and \( {y}_{i}, i \in I \) , elements in \( N \) (not necessarily distinct), then there exists a unique \( R \) -homomorphism \( f : ... | Proof. If \( z \in M \), there exist unique \( {r}_{i} \in R \), almost all \( {r}_{i} = 0 \) , such that \( z = \mathop{\sum }\limits_{I}{r}_{i}{x}_{i} \) . In particular, the uniqueness of the \( {r}_{i} \) implies that \( f : M \rightarrow \bar{N} \) given by \( z \mapsto \mathop{\sum }\limits_{I}{r}_{i}{y}_{i} \) i... | Yes |
Corollary 36.5. Let \( M \) and \( N \) be free \( R \) -modules on bases \( \mathcal{B} \) and \( \mathcal{C} \) respectively. If there exists a bijection \( g : \mathcal{B} \rightarrow \mathcal{C} \), i.e., \( \left| \mathcal{B}\right| = \left| \mathcal{C}\right| \), then \( M \cong N \) . | Proof. The maps \( g \) and \( {g}^{-1} \) of sets induce inverse \( R \) -isomorphisms \( M \rightarrow N \) and \( N \rightarrow M \) . | Yes |
Corollary 36.6. Let \( M \) be an \( R \) -module (respectively, a finitely generated R-module). Then there exists a free R-module (respectively, a finitely generated free \( R \) -module) \( P \) and an \( R \) -epimorphism \( g : P \rightarrow M \) . In particular, we have a short exact sequence\n\n\[ 0 \rightarrow \... | Proof. Let \( Y = {\left\{ {y}_{i}\right\} }_{I} \) generate \( M \) . If \( M \) is finitely generated, we may assume that \( I \) is finite. Let \( P = \mathop{\coprod }\limits_{I}R \) and \( {\mathcal{S}}_{I} \) the standard basis for \( P \) . (So \( {\mathcal{S}}_{I} \mathrel{\text{:=}} {\left\{ {e}_{i} \mid {e}_{... | Yes |
Proposition 37.1. Let \( R \) be a ring and \( M \) an \( R \) -module. Then the following are equivalent:\n\n(1) Every submodule of \( M \) is finitely generated.\n\n(2) \( M \) satisfies the ascending chain condition, i.e., if \( {M}_{i} \subset M \) are submodules and\n\n\[ \n{M}_{1} \subset {M}_{2} \subset \cdots \... | Proof. \( \\left( 1\\right) \\Rightarrow \\left( 2\\right) \) : Let\n\n\[ \n\\mathcal{C} : \\;{M}_{1} \\subset {M}_{2} \\subset \\cdots \\subset {M}_{n} \\subset \\cdots\n\]\n\nbe a chain of submodules of \( M \) . It follows that the subset \( {M}^{\\prime } \\mathrel{\\text{:=}} \) \( \\mathop{\\bigcup }\\limits_{{i ... | Yes |
Proposition 37.4. Let \( M \) be an \( R \) -module and \( N \) a submodule of \( M \) . Then \( M \) is \( R \) -Noetherian if and only if \( N \) and \( M/N \) are \( R \) -Noetherian. In particular, if\n\n\[ 0 \rightarrow {M}^{\prime } \rightarrow M \rightarrow {M}^{\prime \prime } \rightarrow 0 \]\n\nis an exact se... | Proof. \( \Rightarrow \) : Since \( {N}_{0} \subset N \) is a submodule, then \( {N}_{0} \subset M \) is a submodule hence finitely generated - or any ascending chain in \( N \) is an ascending chain in \( M \) . Thus \( N \) is \( R \) -Noetherian. By the Correspondence Principle, a (countable) chain of submodules in ... | No |
Corollary 37.5. If \( M, N \) are Noetherian R-modules, so is \( {M}_{ \coprod }N \) . | Proof. \( \left( {M \coprod N}\right) /N \cong M \) and \( N \) are Noetherian. | No |
Theorem 37.6. Let \( R \) be a Noetherian ring. If \( M \) is a finitely generated \( R \) -module, then \( M \) is \( R \) -Noetherian. | Proof. Suppose \( M = \mathop{\sum }\limits_{{i = 1}}^{n}R{x}_{i} \) . Let \( f : {R}^{n} \rightarrow M \) be the \( R \) -epimorphism given by \( {e}_{i} \mapsto {x}_{i} \), where \( \left\{ {{e}_{1},\ldots ,{e}_{n}}\right\} \) is the standard basis for \( {R}^{n} \) . Since \( R \) is \( R \) -Noetherian so is \( {R}... | Yes |
Corollary 37.7. Let \( R \) be a Noetherian ring and \( M \) a finitely generated R-module. Then there exists positive integers \( m \) and \( n \) and an exact sequence\n\n\[ {R}^{m}\overset{g}{ \rightarrow }{R}^{n}\overset{f}{ \rightarrow }M \rightarrow 0. \] | Proof. As is the proof of the theorem there exists an integer \( n \) and an \( R \) -epimorphism \( f : {R}^{n} \rightarrow M \) . Since \( M \) is finitely generated hence \( R \) -noetherian, \( \ker f \) is also finitely generated. Therefore, there exists an \( R \) -epimorphism \( h : {R}^{m} \rightarrow \ker f \)... | Yes |
Proposition 37.10. Suppose that \( f : R \rightarrow S \) is a ring epimorphism of commutative rings. If \( R \) is a Noetherian ring, then \( S \) is also a Noetherian ring. | Proof. Let \( \mathfrak{A} \subset S \) be an ideal. Then \( {f}^{-1}\left( \mathfrak{A}\right) \subset R \) is an ideal hence finitely generated. Thus \( \mathfrak{A} = f\left( {{f}^{-1}\left( \mathfrak{A}\right) }\right) \) is finitely generated. | Yes |
Theorem 38.1. (Hilbert Basis Theorem) If \( R \) is a Noetherian ring so is the ring \( R\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) . | Proof. By induction on \( n \), it suffices to show that \( R\left\lbrack t\right\rbrack \) is Noetherian. Let \( \mathfrak{B} \subset R\left\lbrack t\right\rbrack \) be an ideal. We must show that \( \mathfrak{B} \) is finitely generated. Let\n\n\[ \mathfrak{A} = \{ r \in R \mid r = \operatorname{lead}f, f \in \mathfr... | Yes |
Proposition 38.5. Let \( R \) be a commutative ring and \( S \) a finitely generated commutative R-algebra. If \( R \) is Noetherian so is \( S \) . | Proof. Let \( S = R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Since \( R\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) is Noetherian and we have a ring epimorphism \( R\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \rightarrow R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) via \( f\l... | Yes |
Lemma 38.7. (Artin-Tate) Let \( T \) be a commutative ring and \( R \subset S \) be subrings of \( T \). Suppose that \( R \) is Noetherian and \( T \) is a finitely generated commutative R-algebra. Suppose that as an S-module \( T \) is finitely generated. Then \( S \) is a finitely generated commutative \( R \) -alge... | Proof. Let \( T = R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) as a commutative \( R \) -algebra for some \( {x}_{i} \in T \), some \( n \) and \( T = \mathop{\sum }\limits_{{i = 1}}^{m}S{y}_{i} \) as an \( S \) -module for some \( {y}_{i} \in \) \( T \), some \( m \). Then for all \( i = 1,\ldots, n \) and ... | Yes |
Theorem 38.11. (Hilbert Nullstellensatz) (Weak Form) Suppose that \( F \) is an algebraically closed field, \( R = F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \), and \( \mathfrak{A} = \left( {{f}_{1},\ldots ,{f}_{r}}\right) \) is an ideal in \( R \) . Then \( {Z}_{F}\left( \mathfrak{A}\right) \) is the empty ... | Proof. Certainly if \( \mathfrak{A} = R \), then there exists no \( \underline{a} \in {F}^{n} \) such that the element 1 in \( R \) evaluated at \( \underline{a} \) takes the value zero, so we need only show if \( \mathfrak{A} < R \), then \( {Z}_{F}\left( \mathfrak{A}\right) \) is not empty.\n\nSo suppose that \( \mat... | Yes |
Theorem 38.13. (Hilbert Nullstellensatz) (Strong Form) Suppose that \( F \) be an algebraically closed field and \( R = F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) . Let \( f,{f}_{1},\ldots ,{f}_{r} \) be elements in \( R \) and \( \mathfrak{A} = \left( {{f}_{1},\ldots ,{f}_{r}}\right) \subset R \) . Suppos... | Proof. (Rabinowitch Trick). We may assume that \( f \) is nonzero. Let \( S = R\left\lbrack t\right\rbrack \) . Define the ideal \( \mathfrak{B} \) in \( S \) by \( \mathfrak{B} = \left( {{f}_{1},\ldots ,{f}_{r},1 - {tf}}\right) \subset S \) . If \( \mathfrak{B} < S \), then there exists a point \( \left( {{a}_{1},\ldo... | Yes |
Let \( F \) be a field and \( \mathfrak{A} = \left( {XY}\right) \) in \( F\left\lbrack {X, Y}\right\rbrack \). Then \( {Z}_{F}\left( \mathfrak{A}\right) = {Z}_{F}\left( X\right) \cup {Z}_{F}\left( Y\right) \) is an irreducible decomposition. | Here \( {Z}_{F}\left( X\right) \) is the \( Y \) -axis defined by \( X = 0 \) and \( {Z}_{F}\left( Y\right) \) is the \( X \) -axis defined by \( Y = 0 \). The origin is the only point in \( {Z}_{F}\left( {X, Y}\right) \). As \( {Z}_{F}\left( {X, Y}\right) \subset {Z}_{F}\left( {XY}\right) \) (cf. Exercise 33.5), we se... | No |
Proposition 39.3. Let \( R \) be a UFD and \( f \) and \( g \) non-constant polynomials in \( R\left\lbrack {X, Y}\right\rbrack \) having no non-constant common factor in \( R\left\lbrack {X, Y}\right\rbrack \) . Then\n\n\[ \n{Z}_{R}\left( f\right) \cap {Z}_{R}\left( g\right) \mathrel{\text{:=}} \left\{ {\left( {a, b}\... | Proof. Let \( F \) be the quotient field of \( R \) and \( K \) the quotient field of \( F\left\lbrack X\right\rbrack \) . We view \( f \) and \( g \) in \( K\left\lbrack Y\right\rbrack \) . By Exercise 32.12(3ii), \( f \) and \( g \) have no common factor in \( K\left\lbrack Y\right\rbrack \), i.e., they are relativel... | Yes |
Proposition 39.8. Let \( F \) be an algebraically closed field of characteristic zero, e.g., \( F = \mathbb{C} \). Suppose \( f \) is an irreducible polynomial in \( F\left\lbrack {X, Y}\right\rbrack \) with the degree \( {\deg }_{Y}f \) of \( f \) in the variable \( Y \) satisfying \( {\deg }_{Y}f = n > 0 \). Then\n\n... | Proof. Write \( f = \mathop{\sum }\limits_{{i = 0}}^{n}{q}_{i}\left( X\right) {Y}^{i} \) with \( {q}_{i}\left( X\right) \) polynomials in \( F\left\lbrack X\right\rbrack \) , \( {q}_{n}\left( X\right) \) nonzero for some \( n > 0 \). If \( a \) lies in \( F \), we have \( f\left( {a, Y}\right) = \) \( \mathop{\sum }\li... | Yes |
Lemma 40.3. Let \( R \) be a PID and \( A = \left( {a}_{ij}\right) \) a nonzero \( m \times n \) matrix in \( {R}^{m \times n} \) . Fix \( i, j \) with \( {a}_{ij} \neq 0 \) and \( j < n \) . Let \( d \) in the ideal \( \left( {{a}_{ij},{a}_{i, j + 1}}\right) \) in \( R \) satisfy \( \left( d\right) = \left( {{a}_{ij},... | Proof. (of the lemma). That \( d \) satisfies the first condition is immediate. We construct \( B \) in (2). For notational convenience, write \( v = {a}_{ij} \) and \( w = {a}_{i, j + 1} \), so \( \left( d\right) = \left( {v, w}\right) \) . Then \( 0 \neq d = {xv} + {yw} = \left( {{xa} + {yb}}\right) d \) in the domai... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.