Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Lemma 4.5. Let \( \tau \in {S}_{n} \), and let \( \left( {{a}_{1}\ldots {a}_{r}}\right) \) be a cycle. Then\n\n\[ \tau \left( {{a}_{1}\ldots {a}_{r}}\right) {\tau }^{-1} = \left( {{a}_{1}{\tau }^{-1}\ldots {a}_{r}{\tau }^{-1}}\right) . \] | Proof. This is verified by checking that both sides act in the same way on \( \{ \mathbf{1},\ldots ,\mathbf{n}\} \) . For example, for \( 1 \leq i < r \)\n\n\[ \left( {{a}_{i}{\tau }^{-1}}\right) \left( {\tau \left( {{a}_{1}\ldots {a}_{r}}\right) {\tau }^{-1}}\right) = {a}_{i}\left( {{a}_{1}\ldots {a}_{r}}\right) {\tau... | No |
Proposition 4.6. Two elements of \( {S}_{n} \) are conjugate in \( {S}_{n} \) if and only if they have the same type. | Proof. The 'only if' part of this statement follows immediately from the preceding considerations: conjugating a permutation yields a permutation of the same type. As for the 'if' part, suppose\n\n\[ \n{\sigma }_{1} = \left( {{a}_{1}\ldots {a}_{r}}\right) \left( {{b}_{1}\ldots {b}_{s}}\right) \cdots \left( {{c}_{1}\ldo... | Yes |
In \( {S}_{8} \), \(\left( {18632}\right) \left( {47}\right) \text{ and }\left( {12345}\right) \left( {67}\right)\) must be conjugate, since they have the same type. | The proof of Proposition 4.6 tells us that \(\tau \left( {18632}\right) \left( {47}\right) {\tau }^{-1} = \left( {12345}\right) \left( {67}\right)\) for \(\tau = \left( \begin{array}{llllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 1 & 8 & 6 & 3 & 2 & 4 & 7 & 5 \end{array}\right)\) and of course this may be checked by hand i... | Yes |
Corollary 4.8. The number of conjugacy classes in \( {S}_{n} \) equals the number of partitions of \( n \) . | For example, there are 7 conjugacy classes in \( {S}_{5} \), indexed by the Tetris look-alikes drawn above. It is also reasonably straightforward to compute the number of elements in each conjugacy class, in terms of the type. For example, in order to count the number of permutations of type \( \left\lbrack {2,2,1}\rig... | No |
There are no normal subgroups of size 30 in \( {S}_{5} \) . | Indeed, normal subgroups are unions of conjugacy classes (4.3); since the identity is in every subgroup and \( {30} - 1 = {29} \) cannot be written as a sum of the numbers appearing in the class formula for \( {S}_{5} \), there is no such subgroup. | Yes |
Lemma 4.11. Transpositions generate \( {S}_{n} \) . | Proof. Indeed, by Lemma 4.3 it suffices to show that every cycle is a product of transpositions, and indeed\n\n\[ \left( {{a}_{1}\ldots {a}_{r}}\right) = \left( {{a}_{1}{a}_{2}}\right) \left( {{a}_{1}{a}_{3}}\right) \cdots \left( {{a}_{1}{a}_{r}}\right) ,\]\n\nas may be checked by applying \( {}^{18} \) both sides to e... | No |
Lemma 4.12. Let \( \sigma = {\tau }_{1}\cdots {\tau }_{r} \) be a product of transpositions. Then \( \sigma \) is even, resp., odd, according to whether \( r \) is even, resp., odd. | Proof. This follows immediately from the facts that \( \epsilon \) is a homomorphism and the sign of a transposition is -1 : indeed, \( \left( {ij}\right) \) acts on \( {\Delta }_{n} \) by permuting its factors and changing the sign of an odd number of factors (for \( i < j \), the factor \( \left( {{x}_{i} - {x}_{j}}\... | Yes |
Lemma 4.14. Let \( n \geq 2 \), and let \( \sigma \in {A}_{n} \) . Then \( {\left\lbrack \sigma \right\rbrack }_{{A}_{n}} = {\left\lbrack \sigma \right\rbrack }_{{S}_{n}} \) or the size of \( {\left\lbrack \sigma \right\rbrack }_{{A}_{n}} \) is half the size of \( {\left\lbrack \sigma \right\rbrack }_{{S}_{n}} \), acco... | Proof. (Cf. Exercise 1.16.) Note that\n\n\[ \n{Z}_{{A}_{n}}\left( \sigma \right) = {A}_{n} \cap {Z}_{{S}_{n}}\left( \sigma \right) :\n\]\n\nthis follows immediately from the definition of centralizer (Definition 1.6). Now recall that the centralizer of \( \sigma \) is its stabilizer under conjugation, and therefore the... | No |
Looking again at \( {A}_{5} \), we have noted in \( \$ \underline{4.3} \) that the types of the even permutations in \( {S}_{5} \) are \( \left\lbrack {1,1,1,1,1}\right\rbrack ,\left\lbrack {2,2,1}\right\rbrack ,\left\lbrack {3,1,1}\right\rbrack \), and \( \left\lbrack 5\right\rbrack \) . By Proposition 4.15 the conjug... | Therefore there are exactly 5 conjugacy classes in \( {A}_{5} \), and the class formula for \( {A}_{5} \) is\n\n\[ \n{60} = 1 + {15} + {20} + {12} + {12}.\n\] | Yes |
Corollary 4.17. The alternating group \( {A}_{5} \) is a simple noncommutative group of order 60. | Proof. A normal subgroup of \( {A}_{5} \) is necessarily the union of conjugacy classes, contains the identity, and has order equal to a divisor of 60 (by Lagrange's theorem). The divisors of 60 other than 1 and 60 are\n\n\[ 2,3,4,5,6,{10},{12},{15},{20},{30} \]\n\ncounting the elements other than the identity would gi... | Yes |
Lemma 4.18. The alternating group \( {A}_{n} \) is generated by 3-cycles. | Proof. Since every even permutation is a product of an even number of 2-cycles, it suffices to show that every product of two 2-cycles may be written as product of 3-cycles. Therefore, consider a product\n\n\[ \left( {ab}\right) \left( {cd}\right) \]\n\nwith \( a \neq b, c \neq d \) . If \( \left( {ab}\right) = \left( ... | Yes |
Theorem 4.20. The alternating group \( {A}_{n} \) is simple for \( n \geq 5 \) . | Proof. We have already checked this for \( n = 5 \), and the reader has checked it for \( n = 6 \). For \( n > 6 \), let \( N \) be a nontrivial normal subgroup of \( {A}_{n} \); we will show that necessarily \( N = {A}_{n} \), by proving that \( N \) contains 3-cycles.\n\nLet \( \tau \in N,\tau \neq \left( 1\right) \)... | No |
Corollary 4.21. For \( n \geq 5 \), the group \( {S}_{n} \) is not solvable. | Proof. Since \( {A}_{n} \) is simple, the sequence\n\n\[ \n{S}_{n} \supsetneq {A}_{n} \supsetneq \{ \left( 1\right) \}\n\]\n\n is a composition series for \( {S}_{n} \) . It follows that the composition factors of \( {S}_{n} \) are \( \mathbb{Z}/2\mathbb{Z} \) and \( {A}_{n} \) . By Proposition 3.11, \( {S}_{n} \) is n... | Yes |
Lemma 5.1. Let \( N, H \) be normal subgroups of a group \( G \) . Then\n\n\[ \left\lbrack {N, H}\right\rbrack \subseteq N \cap H \] | Proof. It suffices to verify this on generators; that is, it suffices to check that\n\n\[ \left\lbrack {n, h}\right\rbrack = n\left( {h{n}^{-1}{h}^{-1}}\right) = \left( {{nh}{n}^{-1}}\right) {h}^{-1} \in N \cap H \]\n\nfor all \( n \in N, h \in H \) . But the first expression and the normality of \( N \) show that \( \... | Yes |
Corollary 5.2. Let \( N, H \) be normal subgroups of a group \( G \) . Assume \( N \cap H = \{ e\} \) . Then \( N, H \) commute with each other:\n\n\[ \left( {\forall n \in N}\right) \left( {\forall h \in H}\right) \;{nh} = {hn}. \] | Proof. By Lemma 5.1, \( \left\lbrack {N, H}\right\rbrack = \{ e\} \) if \( N \cap H = \{ e\} \) ; the result follows immediately. | Yes |
Proposition 5.3. Let \( N, H \) be normal subgroups of a group \( G \), such that \( N \cap H = \) \( \{ e\} \) . Then \( {NH} \cong N \times H \) . | Proof. Consider the function\n\n\[ \varphi : N \times H \rightarrow {NH} \]\n\ndefined by \( \varphi \left( {n, h}\right) = {nh} \) . Under the stated hypothesis, \( \varphi \) is a group homomorphism: indeed\n\n\[ \varphi \left( {\left( {{n}_{1},{h}_{1}}\right) \cdot \left( {{n}_{2},{h}_{2}}\right) }\right) = \varphi ... | Yes |
Lemma 5.7. Let \( N \) be a normal subgroup of a group \( G \), and let \( H \) be a subgroup of \( G \) such that \( G = {NH} \) and \( N \cap H = \{ e\} \) . Then \( G \) is a split extension of \( H \) by \( N \) . | Proof. We have to construct an exact sequence\n\n\[ 1 \rightarrow N \rightarrow G \rightarrow H \rightarrow 1 \]\n\nwe let \( N \rightarrow G \) be the inclusion map, and we prove that \( G/N \cong H \) . For this, consider the composition\n\n\[ \alpha : H \hookrightarrow G \rightarrow G/N. \]\n\nThen \( \alpha \) is s... | Yes |
Lemma 5.8. The resulting structure \( \left( {N \times H,{ \bullet }_{\theta }}\right) \) is a group, with identity element \( \left( {{e}_{N},{e}_{H}}\right) \) . | Proof. The reader should carefully verify this. For example, inverses exist because\n\n\[ \left( {{n}_{1},{h}_{1}}\right) { \bullet }_{\theta }\left( {{\theta }_{{h}_{1}^{-1}}\left( {n}_{1}^{-1}\right) ,{h}_{1}^{-1}}\right) = \left( {{n}_{1}{\theta }_{{h}_{1}}\left( {{\theta }_{{h}_{1}^{-1}}\left( {n}_{1}^{-1}\right) }... | No |
Proposition 5.10. Let \( N, H \) be groups, and let \( \theta : H \rightarrow {\operatorname{Aut}}_{\mathrm{{Grp}}}\left( N\right) \) be a homomorphism; let \( G = N{ \rtimes }_{\theta }H \) be the corresponding semidirect product. Then\n\n- \( G \) contains isomorphic copies of \( N \) and \( H \) ;\n\n- the natural p... | Proof. The functions \( N \rightarrow G, H \rightarrow G \) defined for \( n \in N, h \in H \) by\n\n\[ n \mapsto \left( {n,{e}_{H}}\right) ,\;h \mapsto \left( {{e}_{N}, h}\right) \]\n\nare manifestly injective homomorphisms, allowing us to identify \( N, H \) with the corresponding subgroups of \( G \) . It is clear t... | Yes |
Proposition 5.11. Let \( N, H \) be subgroups of a group \( G \), with \( N \) normal in \( G \) . Assume that \( N \cap H = \{ e\} \), and \( G = {NH} \) . Let \( \gamma : H \rightarrow {\operatorname{Aut}}_{\mathrm{{Grp}}}\left( N\right) \) be defined by conjugation: for \( h \in H, n \in N \) ,\n\n\[{\gamma }_{h}\le... | Proof. Define a function\n\n\[ \varphi : N{ \rtimes }_{\gamma }H \rightarrow G \]\n\nby \( \varphi \left( {n, h}\right) = {nh} \) ; this is clearly a bijection. We need to verify that \( \varphi \) is a homomorphism, and indeed \( \left( {\forall {n}_{1},{n}_{2} \in N}\right) ,\left( {\forall {h}_{1},{h}_{2} \in H}\rig... | Yes |
The automorphism group of \( {C}_{3} \) is isomorphic to the cyclic group \( {C}_{2} \) : if \( {C}_{3} = \left\{ {e, y,{y}^{2}}\right\} \), then the two automorphisms of \( {C}_{3} \) are | \[ \text{ id : }\left\{ {\begin{aligned} e & \mapsto e, \\ y & \mapsto y, \\ {y}^{2} & \mapsto {y}^{2}, \end{aligned}\;\sigma : \begin{cases} e & \mapsto e, \\ y & \mapsto {y}^{2}, \\ {y}^{2} & \mapsto y. \end{cases}}\right. \] | Yes |
Lemma 6.1. Let \( G \) be an abelian group, and let \( H, K \) be subgroups such that \( \left| H\right| \) , \( \left| K\right| \) are relatively prime. Then \( H + K \cong H \oplus K \) . | Proof. By Lagrange’s theorem (Corollary 118.14), \( H \cap K = \{ 0\} \) . Since subgroups of abelian groups are automatically normal, the statement follows from Proposition 5.3 | No |
Corollary 6.2. Every finite abelian group is the direct sum of its nontrivial Sylow subgroups. | (The diligent reader knew already that this had to be the case, since abelian groups are nilpotent; cf. Exercise 5.1) Thus, we already know that every finite abelian group is a direct sum of \( p \) -groups, and our main task amounts to classifying abelian \( p \) -groups for a fixed prime \( p \) . This is somewhat te... | No |
Lemma 6.3. Let \( G \) be an abelian p-group, and let \( g \in G \) be an element of maximal order. Then the exact sequence \[ 0 \rightarrow \langle g\rangle \rightarrow G \rightarrow G/\langle g\rangle \rightarrow 0 \] splits. | Put otherwise, there is a subgroup \( L \) of \( G \) such that \( L \) maps isomorphically to \( G/\langle g\rangle \) via the canonical projection, that is, such that \( \langle g\rangle \cap L = \{ 0\} \) and \( \langle g\rangle + L = G \) . Note that it will follow that \( G \cong \langle g\rangle \oplus L \), by P... | No |
Lemma 6.4. Let \( p \) be a prime integer and \( r \geq 1 \). Let \( G \) be a noncyclic abelian group of order \( {p}^{r + 1} \), and let \( g \in G \) be an element of order \( {p}^{r} \). Then there exists an element \( h \in G, h \notin \langle g\rangle \), such that \( \left| h\right| = p \). | Proof of Lemma 6.4. Denote \( \langle g\rangle \) by \( K \), and let \( {h}^{\prime } be any element of \( G,{h}^{\prime } \notin K \). The subgroup \( K \) is normal in \( G \) since \( G \) is abelian; the quotient group \( G/K \) has order \( p \). Since \( {h}^{\prime } \notin K \), the coset \( {h}^{\prime } + K ... | Yes |
Corollary 6.5. Let \( G \) be a finite abelian group. Then \( G \) is a direct sum of cyclic groups, which may be assumed to be cyclic p-groups. | Proof. As noted in Corollary 6.2, \( G \) is a direct sum of \( p \) -groups (as a consequence of the Sylow theorems). I claim that every abelian \( p \) -group \( P \) is a direct sum of cyclic \( p \) -groups.\n\nTo establish this, argue by induction on \( \left| P\right| \) . There is nothing to prove if \( P \) is ... | Yes |
Theorem 6.6. Let \( G \) be a finite nontrivial abelian group. Then\n\n- there exist prime integers \( {p}_{1},\ldots ,{p}_{r} \) and positive integers \( {n}_{ij} \) such that \( \left| G\right| = \) \( \mathop{\prod }\limits_{{i, j}}{p}_{i}^{{n}_{i, j}} \) and\n\n\[ G \cong {\bigoplus }_{i, j}\frac{\mathbb{Z}}{{p}_{i... | The first form is nothing but a more explicit version of the statement of Corollary 6.5, so it has already been proven. I will explain how to obtain the second form from the first. The uniqueness statement \( {}^{29} \) is left to the reader (Exercise 6.1).\n\nThe prime powers appearing in the first form of Theorem 6.6... | No |
There are exactly 6 isomorphism classes of abelian groups of order 360. | Indeed, \( 360 = 2^3 \cdot 3^2 \cdot 5 \) ; the six possible tables of elementary divisors are shown below. In terms of invariant factors, the six distinct abelian groups of order 360 (up to isomorphism, by the uniqueness part of Theorem 6.6) are therefore\n\n\[ \frac{\mathbb{Z}}{3\mathbb{Z}} \oplus \frac{\mathbb{Z}}{{... | Yes |
Lemma 6.9. Let \( G \) be a finite abelian group, and assume that for every integer \( n > 0 \) the number of elements \( g \in G \) such that \( {ng} = 0 \) is at most \( n \) . Then \( G \) is cyclic. | Indeed, by Theorem 6.6\n\n\[ G \cong \frac{\mathbb{Z}}{{d}_{1}\mathbb{Z}} \oplus \cdots \oplus \frac{\mathbb{Z}}{{d}_{s}\mathbb{Z}} \]\n\nfor some positive integers \( 1 < {d}_{1}\left| \cdots \right| {d}_{s} \) . But if \( s > 1 \), then \( \left| G\right| > {d}_{s} \) and \( {d}_{s}g = 0 \) for all \( g \in G \) (so ... | Yes |
Theorem 6.10. Let \( F \) be a field, and let \( G \) be a finite subgroup of the multiplicative group \( \left( {{F}^{ * }, \cdot }\right) \) . Then \( G \) is cyclic. | Proof. By the considerations preceding the statement, for every \( n \) there are at most \( n \) elements \( a \in F \) such that \( {a}^{n} - 1 = 0 \), that is, at most \( n \) elements \( a \in G \) such that \( {a}^{n} = 1 \) . Lemma 6.9 implies then that \( G \) is cyclic. | No |
Proposition 1.1. Let \( R \) be a commutative ring, and let \( M \) be an \( R \) -module. Then the following are equivalent:\n\n(1) \( M \) is Noetherian; that is, every submodule of \( M \) is finitely generated.\n\n(2) Every ascending chain of submodules of \( M \) stabilizes; that is, if\n\n\[ \n{N}_{1} \subseteq {... | Proof. (1) \( \Rightarrow \) (2): Assume that \( M \) is Noetherian, and let\n\n\[ \n{N}_{1} \subseteq {N}_{2} \subseteq {N}_{3} \subseteq \cdots \n\]\n\nbe a chain of submodules of \( M \) . Consider the union\n\n\[ \nN = \mathop{\bigcup }\limits_{i}{N}_{i} \n\]\n\nthe reader will verify that \( N \) is a submodule of... | No |
Theorem 1.2. Let \( R \) be a Noetherian ring, and let \( J \) be an ideal of the polynomial ring \( R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then the ring \( R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /J \) is Noetherian. | The proof of this deep fact is surprisingly easy. By Exercise 1.1 it suffices to prove that\n\n\[ R\text{Noetherian} \Rightarrow R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \text{Noetherian;} \] | No |
Lemma 1.5. Let \( a, b \) be nonzero elements of an integral domain \( R \) . Then \( a \) and \( b \) are associates if and only if \( a = {ub} \), for \( u \) a unit in \( R \) . | Proof. Assume \( a \) and \( b \) are associates. Then \( \exists c, d \in R \) such that\n\n\[ b = {ac},\;a = {bd}; \]\n\ntherefore \( a = {bd} = {acd} \), i.e.,\n\n\[ a\left( {1 - {cd}}\right) = 0. \]\n\nSince cancellation by nonzero elements hold in integral domains, this implies \( {cd} = 1 \) . Thus \( c \) is a u... | No |
Lemma 1.7. Let \( R \) be an integral domain, and let \( a \in R \) be a nonzero prime element. Then a is irreducible. | Proof. Since \( \left( a\right) \) is prime, \( \left( a\right) \neq \left( 1\right) \) ; hence \( a \) is not a unit. If \( a = {bc} \), then \( {bc} = a \in \left( a\right) \) ; therefore \( b \in \left( a\right) \) or \( c \in \left( a\right) \) since \( \left( a\right) \) is prime. Assuming without loss of generali... | Yes |
Proposition 1.11. Let \( R \) be an integral domain, and let \( r \) be a nonzero, nonunit element of \( R \) . Assume that every ascending chain of principal ideals\n\n\[ \left( r\right) \subseteq \left( {r}_{1}\right) \subseteq \left( {r}_{2}\right) \subseteq \left( {r}_{3}\right) \subseteq \cdots \]\n\nstabilizes. T... | Proof. Assume that \( r \) does not have a factorization into irreducible elements. In particular, \( r \) is itself not irreducible; thus \( \exists {r}_{1},{s}_{1} \in R \) such that \( r = {r}_{1}{s}_{1} \) and \( \left( r\right) \varsubsetneq \left( {r}_{1}\right) ,\left( r\right) \varsubsetneq \left( {s}_{1}\right... | Yes |
Corollary 1.12. Let \( R \) be a Noetherian domain. Then factorizations exist in \( R \) . | Proof. By Proposition 1.1, Noetherian domains satisfy the ascending chain condition for all ideals. | No |
Lemma 2.1. Let \( R \) be a UFD, and let \( a, b, c \) be nonzero elements of \( R \) . Then\n\n- \( \left( a\right) \subseteq \left( b\right) \Leftrightarrow \) the multiset of irreducible factors of \( b \) is contained in the multiset of irreducible factors of \( a \) ;\n\n- \( a \) and \( b \) are associates (that ... | The proof is left to the reader (Exercise 2.1). | No |
Lemma 2.3. Let \( R \) be a UFD, and let \( a, b \) be nonzero elements of \( R \). Then \( a, b \) have a greatest common divisor. | Proof. We can write\n\n\[ a = u{q}_{1}^{{\alpha }_{1}}\cdots {q}_{r}^{{\alpha }_{r}},\;b = v{q}_{1}^{{\beta }_{1}}\cdots {q}_{r}^{{\beta }_{r}} \]\n\nwhere \( u \) and \( v \) are units, the elements \( {q}_{i} \) are irreducible, \( {q}_{i} \) is not an associate of \( {q}_{j} \) for \( i \neq j \), and \( {\alpha }_{... | Yes |
Lemma 2.4. Let \( R \) be a UFD, and let a be an irreducible element of \( R \) . Then a is prime. | Proof. The element \( a \) is not a unit, by definition of irreducible. Assume \( {bc} \in \left( a\right) \) : thus \( \left( {bc}\right) \subseteq \left( a\right) \), and by Lemma 2.1 the irreducible factors of \( a \), that is, \( a \) itself, must be among the factors of \( b \) or of \( c \) . We have \( b \in \le... | Yes |
Theorem 2.5. An integral domain \( R \) is a UFD if and only if\n\n- the a.c.c. for principal ideals holds in \( R \) and\n\n- every irreducible element of \( R \) is prime. | Proof. ( \( \Rightarrow \) ) Assume that \( R \) is a UFD. Lemma 2.4 shows that irreducible elements of \( R \) are prime. To prove that the a.c.c. for principal ideals holds, consider an ascending chain\n\n\[ \left( {r}_{1}\right) \varsubsetneq \left( {r}_{2}\right) \varsubsetneq \left( {r}_{3}\right) \varsubsetneq \c... | Yes |
Proposition 2.6. If \( R \) is a PID, then it is a UFD. | Proof. Let \( R \) be a PID. The a.c.c. (for principal ideals, as all ideals in \( R \) are principal!) holds in \( R \) since PIDs are Noetherian. We verify that irreducible elements are prime in \( R \), which implies that \( R \) is a UFD by Theorem 2.5\n\nLet \( a \in R \) be an irreducible element. Ideals generate... | No |
Proposition 2.8. Let \( R \) be a Euclidean domain. Then \( R \) is a PID. | The proof is modeled after the instances encountered for \( \mathbb{Z} \) (Proposition 1111.4) and \( k\left\lbrack x\right\rbrack \) (which the reader has hopefully worked out in Exercise 11114.4).\n\nProof. Let \( I \) be an ideal of \( R \) ; we have to prove that \( I \) is principal. If \( I = \{ 0\} \) , there is... | Yes |
Lemma 2.9. Let \( a = {bq} + r \) in \( a \) ring \( R \) . Then \( \left( {a, b}\right) = \left( {b, r}\right) \) . | Proof. Indeed, \( r = a - {bq} \in \left( {a, b}\right) \), proving \( \left( {b, r}\right) \subseteq \left( {a, b}\right) \) ; and \( a = {bq} + r \in \left( {b, r}\right) \) , proving \( \left( {a, b}\right) \subseteq \left( {b, r}\right) \) . | Yes |
Corollary 2.10. Assume \( a = {bq} + r \) . Then \( a, b \) have a gcd if and only if \( b, r \) have \( {agcd} \), and in this case \( \gcd \left( {a, b}\right) = \gcd \left( {b, r}\right) \) . | Of course ’ \( \gcd \left( {a, b}\right) = \gcd \left( {b, r}\right) \) ’ means that the two classes of associate elements coincide. | No |
Proposition 2.12. With notation as above, \( {r}_{N - 1} \) is a gcd of \( a, b \) . | Proof. By Corollary 2.10,\n\n\[ \gcd \left( {a, b}\right) = \gcd \left( {b,{r}_{1}}\right) = \gcd \left( {{r}_{1},{r}_{2}}\right) = \cdots = \gcd \left( {{r}_{N - 2},{r}_{N - 1}}\right) .\n\]\n\nBut \( {r}_{N - 2} = {r}_{N - 1}{q}_{N - 1} \) gives \( {r}_{N - 2} \in \left( {r}_{N - 1}\right) \) ; hence \( \left( {{r}_{... | Yes |
Theorem 3.3 (Well-ordering theorem). Every set admits a well-ordering. | Well-ordering theorem \( \Rightarrow \) Zorn’s lemma. Let \( \left( {Z, \leq }\right) \) be a nonempty poset such that every chain in \( Z \) has an upper bound in \( Z \) . By the well-ordering theorem, there is a well-ordering \( {}^{13} \preccurlyeq \) on \( Z \) . Define a function \( f \) from \( Z \) to the power... | Yes |
Proposition 3.5. Let \( I \neq \left( 1\right) \) be a proper ideal of a commutative ring \( R \) . Then there exists a maximal ideal \( \mathfrak{m} \) of \( R \) containing \( I \) . | Proof. The set \( \mathcal{I} \) of proper ideals of \( R \) containing \( I \) is ordered by inclusion. Then let \( \mathcal{C} \) be a chain of proper ideals, and consider\n\n\[ U \mathrel{\text{:=}} \mathop{\bigcup }\limits_{{J \in \mathcal{C}}}J \]\n\nI claim that \( U \) is a proper ideal containing \( I \) ; henc... | Yes |
Lemma 4.1. Let \( R \) be a ring, and let \( I \) be an ideal of \( R \) . Then\n\n\[ \frac{R\left\lbrack x\right\rbrack }{{IR}\left\lbrack x\right\rbrack } \cong \frac{R}{I}\left\lbrack x\right\rbrack \]\n | The proof of this lemma is a standard application of the first isomorphism theorem and is left to the reader (Exercise 4.1). | No |
Corollary 4.2. If \( I \) is a prime ideal of \( R \), then \( {IR}\left\lbrack x\right\rbrack \) is prime in \( R\left\lbrack x\right\rbrack \) . | Proof. If \( I \) is prime in \( R \), then \( R/I \) is an integral domain; hence so is \( R\left\lbrack x\right\rbrack /{IR}\left\lbrack x\right\rbrack \cong \) \( \left( {R/I}\right) \left\lbrack x\right\rbrack \), and therefore \( {IR}\left\lbrack x\right\rbrack \) is prime in \( R\left\lbrack x\right\rbrack \) . | Yes |
Lemma 4.4. Let \( R \) be a commutative ring. Then for \( f, g \in R\left\lbrack x\right\rbrack \)\n\n\( {fg} \) is primitive \( \Leftrightarrow \) both \( f \) and \( g \) are primitive. | Proof. This is an easy consequence of Corollary 4.2\n\n\( {fg} \) primitive \( \Leftrightarrow \forall \mathfrak{p} \) prime and principal in \( R,{fg} \notin \mathfrak{p}R\left\lbrack x\right\rbrack \)\n\n\( \Leftrightarrow \forall \mathfrak{p} \) prime and principal in \( R, f \notin \mathfrak{p}R\left\lbrack x\right... | Yes |
Lemma 4.5. Let \( R \) be a commutative ring and \( f = {a}_{0} + {a}_{1}x + \cdots + {a}_{d}{x}^{d} \in R\left\lbrack x\right\rbrack \) as above.\n\n- \( f \) is very primitive if and only if \( \left( {{a}_{0},\ldots ,{a}_{d}}\right) = \left( 1\right) \).\n\n- If \( R \) is a UFD, then \( f \) is primitive if and onl... | Proof. If \( \left( {{a}_{0},\ldots ,{a}_{d}}\right) = \left( 1\right) \), then no prime ideal can contain all coefficients \( {a}_{i} \) , and it follows that \( f \) is very primitive. Conversely, if \( f \) is very primitive, then the coefficients of \( f \) are not all contained in any one prime ideal, and in parti... | Yes |
Proposition 4.8 (Gauss’s lemma). Let \( R \) be a UFD, and let \( f, g \in R\left\lbrack x\right\rbrack \) . Then\n\n\[ \left( {\operatorname{cont}}_{fg}\right) = \left( {\operatorname{cont}}_{f}\right) \left( {\operatorname{cont}}_{g}\right) \] | Proof. This follows easily from our preparatory work. Write\n\n\[ \left( {fg}\right) = \left( {\left( {\operatorname{cont}}_{f}\right) \left( \underline{f}\right) }\right) \left( {\left( {\operatorname{cont}}_{g}\right) \left( \underline{g}\right) }\right) = \left( {\operatorname{cont}}_{f}\right) \left( {\operatorname... | Yes |
Example 4.12. With the notation introduced above, \( K\left( \mathbb{Z}\right) = \mathbb{Q} \) . | The universal property implies immediately that \( F \hookrightarrow K\left( F\right) \) is an isomorphism if \( F \) is itself a field. Thus, the construction adds nothing to \( \mathbb{Q},\mathbb{R},\mathbb{C},\mathbb{Z}/p\mathbb{Z} \), etc. | No |
Theorem 4.14. Let \( R \) be a UFD; then \( R\left\lbrack x\right\rbrack \) is a UFD. | By Theorem 2.5, in order to prove Theorem 4.14, we have to verify that \( R\left\lbrack x\right\rbrack \) satisfies the a.c.c. for principal ideals and that every irreducible element in \( R\left\lbrack x\right\rbrack \) is prime, provided that \( R \) is itself a UFD. The general idea is to reduce these questions to m... | Yes |
Lemma 4.15. Let \( R \) be a UFD, and let \( K = K\left( R\right) \) be its field of fractions. For nonzero \( f, g \in R\left\lbrack x\right\rbrack \), denote by \( \left( f\right) ,\left( g\right) \) the principal ideals \( {fR}\left\lbrack x\right\rbrack ,{gR}\left\lbrack x\right\rbrack \) in \( R\left\lbrack x\righ... | Proof. Since \( {\left( g\right) }_{K} \subseteq {\left( f\right) }_{K} \), we have \( g = {fh} \), where \( h \in K\left\lbrack x\right\rbrack \) . Write \( h = \frac{a}{b}\underline{h} \), where \( a, b \in R \) and \( \underline{h} \in R\left\lbrack x\right\rbrack \) is a primitive polynomial: this can be done by co... | Yes |
Proposition 4.16. Let \( R \) be a UFD, and let \( K \) be its field of fractions. Let \( f \in \) \( R\left\lbrack x\right\rbrack \) be a nonconstant, irreducible polynomial. Then \( f \) is irreducible as an element of \( K\left\lbrack x\right\rbrack \) . | Proof. First note that \( f \) is primitive: otherwise we could factor out its content, and \( f \) would not be irreducible.\n\nNext, assume \( f = {gh} \), with \( g, h \in K\left\lbrack x\right\rbrack \) ; we have to prove that either \( g \) or \( h \) is a unit in \( K\left\lbrack x\right\rbrack \) . Let \( c, d \... | Yes |
Corollary 4.17. Let \( R \) be a UFD and \( K \) the field of fractions of \( R \) . Let \( f \in R\left\lbrack x\right\rbrack \) be a nonconstant polynomial. Then \( f \) is irreducible in \( R\left\lbrack x\right\rbrack \) if and only if it is irreducible in \( K\left\lbrack x\right\rbrack \) and primitive. | The proof amounts to tying up loose ends, and I leave it to the reader (Exercise 4.21). | No |
Lemma 5.1. Let \( R \) be an integral domain, and let \( f \in R\left\lbrack x\right\rbrack \) be a polynomial of degree \( n \) . Then the number of roots of \( f \), counted with multiplicity, is at most \( n \) . | Proof. The number of roots of \( f \) in \( R \) is less than or equal to the number of roots of \( f \) viewed as a polynomial over the field of fractions \( K \) of \( R \) ; so we may replace \( R \) by \( K \) .\n\nNow, \( K\left\lbrack x\right\rbrack \) is a UFD, and the roots of \( f \) correspond to the irreduci... | Yes |
Corollary 5.2. Let \( R \) be an infinite integral domain, and let \( f, g \in R\left\lbrack x\right\rbrack \) be polynomials. Then \( f = g \) if and only if the evaluation functions \( r \mapsto f\left( r\right), r \mapsto g\left( r\right) \) agree. | Proof. Indeed, the two functions agree if and only if every \( a \in R \) is a root of \( f - g \) ; but a nonzero polynomial over \( R \) cannot have infinitely many roots, by Lemma 5.1 | Yes |
Proposition 5.3. Let \( k \) be a field. A polynomial \( f \in k\left\lbrack x\right\rbrack \) of degree 2 or 3 is irreducible if and only if it has no roots. | Proof. Exercise 5.5. | No |
Let \( {\mathbb{F}}_{2} \) be the field \( \mathbb{Z}/2\mathbb{Z} \). The polynomial \( f\left( t\right) = {t}^{2} + t + 1 \in {\mathbb{F}}_{2}\left\lbrack t\right\rbrack \) is irreducible, since it has no roots: \( f\left( 0\right) = f\left( 1\right) = 1 \) . Therefore the ideal \( \left( {{t}^{2} + t + 1}\right) \) i... | \[ \frac{{\mathbb{F}}_{2}\left\lbrack t\right\rbrack }{\left( {t}^{2} + t + 1\right) }.\] | Yes |
Proposition 5.5. Let \( R \) be a UFD, and let \( K \) be its field of fractions. Let\n\n\[ f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \in R\left\lbrack x\right\rbrack ,\]\n\nand let \( c = \frac{p}{q} \in K \) be a root of \( f \), with \( p, q \in R,\gcd \left( {p, q}\right) = 1 \) . Then \( p \m... | Proof. By hypothesis,\n\n\[ {a}_{0} + {a}_{1}\frac{p}{q} + \cdots + {a}_{n}\frac{{p}^{n}}{{q}^{n}} = 0 \]\n\nthat is,\n\n\[ {a}_{0}{q}^{n} + {a}_{1}p{q}^{n - 1} + \cdots + {a}_{n}{p}^{n} = 0. \]\n\nTherefore\n\n\[ {a}_{0}{q}^{n} = - p\left( {{a}_{1}{q}^{n - 1} + \cdots + {a}_{n}{p}^{n - 1}}\right) ,\]\n\nproving that \... | Yes |
Looking for rational roots of the polynomial \[ 3 - {2x} + 3{x}^{2} - 2{x}^{3} + 3{x}^{4} - 2{x}^{5} \] | is therefore reduced to trying fractions \( \frac{p}{q} \) with \( q = \pm 1, \pm 2, p = \pm 1, \pm 3 \) . As it happens, \( \frac{3}{2} \) is the only root found among these possibilities, and it follows that it is the only rational root of the polynomial. | Yes |
Proposition 5.7. Let \( k \) be a field, and let \( f\left( t\right) \in k\left\lbrack t\right\rbrack \) be a nonzero irreducible polynomial. Then \[ F \mathrel{\text{:=}} \frac{k\left\lbrack t\right\rbrack }{\left( f\left( t\right) \right) } \] is a field, endowed with a natural homomorphism \( i : k \hookrightarrow F... | Proof. Since \( k \) is a field, \( k\left\lbrack t\right\rbrack \) is a PID; hence \( \left( {f\left( t\right) }\right) \) is a maximal ideal of \( k\left\lbrack t\right\rbrack \), by Proposition III 4.13 Therefore \( F \) is indeed a field. Denoting cosets in \( k\left\lbrack t\right\rbrack /\left( {f\left( t\right) ... | Yes |
For \( k = \mathbb{R} \) and \( f\left( x\right) = {x}^{2} + 1 \), the field constructed in Proposition 5.7 is (isomorphic to) \( \mathbb{C} \) | this was checked carefully in Example III 4.8 | No |
Proposition 5.11. Let \( k \) be an algebraically closed field. Then \( k \) is infinite. | Proof. By contradiction, assume that \( k \) is algebraically closed and finite; let the elements of \( k \) be \( {c}_{1},\ldots ,{c}_{N} \) . Then there are exactly \( N \) irreducible monic polynomials in \( k\left\lbrack x\right\rbrack \), namely \( (x - \) \( \left. {c}_{1}\right) ,\ldots ,\left( {x - {c}_{N}}\rig... | Yes |
Theorem 5.12. \( \mathbb{C} \) is algebraically closed. | Gauss is credited with providing the first proof 21 of this fundamental theorem (which is indeed known as the fundamental theorem of algebra.) 'Algebraic' proofs of the fundamental theorem of algebra require more than we know at this point (we will encounter one in SVII 7.1 after we have seen a little Galois theory); c... | No |
Proposition 5.13. Every polynomial \( f \in \mathbb{R}\left\lbrack x\right\rbrack \) of degree \( \geq 3 \) is reducible. | Proof. Let \( f \in \mathbb{R}\left\lbrack x\right\rbrack \) be a nonconstant polynomial:\n\n\[ f = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n}, \]\n\nwith all \( {a}_{i} \in \mathbb{R} \). By Theorem 5.12 f has a complex root \( z \):\n\n\[ {a}_{0} + {a}_{1}z + \cdots + {a}_{n}{z}^{n} = 0. \]\n\nApplying complex conj... | Yes |
Proposition 5.15. Let \( f \in \mathbb{Z}\left\lbrack x\right\rbrack \) be a primitive polynomial, and let \( p \) be a prime integer. Assume \( f{\;\operatorname{mod}\;p} \) has the same degree as \( f \) and is irreducible in \( \mathbb{Z}/p\mathbb{Z}\left\lbrack x\right\rbrack \) . Then \( f \) is irreducible in \( ... | Proof. Argue contrapositively: if \( f \) is primitive and reducible in \( \mathbb{Z}\left\lbrack x\right\rbrack \) and \( \deg f = \) \( n \), then \( f = {gh} \) with \( \deg g = d,\deg h = e, d + e = n \), and both \( d, e \), positive. But then the same can be said of \( f{\;\operatorname{mod}\;p} \), so \( f{\;\op... | Yes |
Corollary 5.16. There are irreducible polynomials in \( \mathbb{Z}\left\lbrack x\right\rbrack \) and \( \mathbb{Q}\left\lbrack x\right\rbrack \) of arbitrarily large degree. | Proof. By Proposition 4.16, the statement for \( \mathbb{Z}\left\lbrack x\right\rbrack \) implies the one for \( \mathbb{Q}\left\lbrack x\right\rbrack \) . By Proposition 5.15, it suffices to verify that there are irreducible polynomials in \( \mathbb{Z}/p\mathbb{Z}\left\lbrack x\right\rbrack \) of arbitrarily large de... | No |
Proposition 5.17. Let \( R \) be a (commutative) ring, and let \( \mathfrak{p} \) be a prime ideal of \( R \). Let\n\n\[ f = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \in R\left\lbrack x\right\rbrack \]\n\nbe a polynomial, and assume that\n\n- \( {a}_{n} \notin \mathfrak{p} \);\n\n- \( {a}_{i} \in \mathfrak{p} \) fo... | Proof. Argue by contradiction. Assume \( f = {gh} \) in \( R\left\lbrack x\right\rbrack \), with both \( d = \deg g \) and \( e = \deg h \) less than \( n = \deg f \); write\n\n\[ g = {b}_{0} + {b}_{1}x + \cdots + {b}_{d}{x}^{d},\;h = {c}_{0} + {c}_{1}x + \cdots + {c}_{e}{x}^{e}, \]\n\nand note that necessarily \( d > ... | Yes |
For all \( n \) and all primes \( p \), the polynomial \( {x}^{n} - p \) is irreducible in \( \mathbb{Z}\left\lbrack x\right\rbrack \) . | This follows immediately from Eisenstein’s criterion and gives an alternative proof of Corollary 5.16. | No |
Example 5.19. This is probably the most famous application of Eisenstein's criterion. Let \( p \) be a prime integer, and let\n\n\[ f\left( x\right) = 1 + x + {x}^{2} + \cdots + {x}^{p - 1} \in \mathbb{Z}\left\lbrack x\right\rbrack . \]\n\nThese polynomials are called cyclotomic; we will encounter them again in SVII 5.... | \[ f\left( {x + 1}\right) = \frac{{\left( x + 1\right) }^{p} - 1}{\left( {x + 1}\right) - 1} = {x}^{p - 1} + \left( \begin{matrix} p \\ p - 1 \end{matrix}\right) {x}^{p - 2} + \cdots + \left( \begin{array}{l} p \\ 3 \end{array}\right) {x}^{2} + \left( \begin{array}{l} p \\ 2 \end{array}\right) x + \left( \begin{array}{... | Yes |
Theorem 6.1. Let \( {I}_{1},\ldots ,{I}_{k} \) be ideals of \( R \) such that \( {I}_{i} + {I}_{j} = \left( 1\right) \) for all \( i \neq j \) . Then the natural homomorphism\n\n\[ \varphi : R \rightarrow \frac{R}{{I}_{1}} \times \cdots \times \frac{R}{{I}_{k}} \]\n\nis surjective and induces an isomorphism\n\n\[ \wide... | The ’natural’ homomorphism \( \varphi \) is determined by the canonical projections \( R \rightarrow R/{I}_{j} \) and the universal property of products; the homomorphism \( \widetilde{\varphi } \) is induced by virtue of the universal property of quotients, since \( {I}_{1}\cdots {I}_{k} \subseteq {I}_{j} \) for all \... | No |
Lemma 6.2. Let \( {I}_{1},\ldots ,{I}_{k} \) be ideals of \( R \) such that \( {I}_{i} + {I}_{k} = \left( 1\right) \) for all \( i = \) \( 1,\ldots, k - 1 \) . Then \( \left( {{I}_{1}\cdots {I}_{k - 1}}\right) + {I}_{k} = \left( 1\right) \) . | Proof. By hypothesis, for \( i = 1,\ldots, k - 1 \) there exists \( {a}_{i} \in {I}_{k} \) such that \( 1 - {a}_{i} \in {I}_{i} \) . Then \[ \left( {1 - {a}_{1}}\right) \cdots \left( {1 - {a}_{k - 1}}\right) \in {I}_{1}\cdots {I}_{k - 1} \] and \[ 1 - \left( {1 - {a}_{1}}\right) \cdots \left( {1 - {a}_{k - 1}}\right) \... | Yes |
Lemma 6.3. Let \( {I}_{1},\ldots ,{I}_{k} \) be ideals of \( R \) such that \( {I}_{i} + {I}_{j} = \left( 1\right) \) for all \( i \neq j \) . Then \( {I}_{1}\cdots {I}_{k} = {I}_{1} \cap \cdots \cap {I}_{k} \) . | Proof. By Lemma 6.2, under the stated hypotheses we have that \( {I}_{1}\cdots {I}_{k - 1} + {I}_{k} = \) (1) for \( k \geq 3 \) . Thus, the general statement is reduced by induction to the case \( k = 2 \) . (By the way, this case is Exercise III 4.5) Assume \( I \) and \( J \) are ideals of \( R \) such that \( I + J... | No |
Corollary 6.4. Let \( R \) be a PID, and let \( {a}_{1},\ldots ,{a}_{k} \in R \) be elements such that \( \gcd \left( {{a}_{i},{a}_{j}}\right) = 1 \) for all \( i \neq j \) . Let \( a = {a}_{1}\cdots {a}_{k} \) . Then the function\n\n\[ \varphi : \frac{R}{\left( a\right) } \rightarrow \frac{R}{\left( {a}_{1}\right) } \... | This is an immediate consequence of Theorem 6.1, since (in a PID!) \( \gcd \left( {a, b}\right) = \) 1 if and only if \( \left( {a, b}\right) = \left( 1\right) \) as ideals. | Yes |
Lemma 6.5. The function \( N \) is a Euclidean valuation on \( \mathbb{Z}\left\lbrack i\right\rbrack \) ; further, \( N \) is multiplicative in the sense that \( \forall z, w \in \mathbb{Z}\left\lbrack i\right\rbrack \)\n\n\[ N\left( {zw}\right) = N\left( z\right) N\left( w\right) \] | Proof. The multiplicativity is an immediate consequence of the elementary properties of complex conjugation:\n\n\[ N\left( {zw}\right) = \left( {zw}\right) \left( \overline{zw}\right) = \left( {z\bar{z}}\right) \left( {w\bar{w}}\right) = N\left( z\right) N\left( w\right) . \] | Yes |
Lemma 6.6. The units of \( \mathbb{Z}\left\lbrack i\right\rbrack \) are \( \pm 1, \pm i \) . | Proof. If \( u \) is a unit in \( \mathbb{Z}\left\lbrack i\right\rbrack \), then there exists \( v \in \mathbb{Z}\left\lbrack i\right\rbrack \) such that \( {uv} = 1 \) . But then \( N\left( u\right) N\left( v\right) = N\left( {uv}\right) = N\left( 1\right) = 1 \) by multiplicativity, so \( N\left( u\right) \) is a uni... | Yes |
Lemma 6.7. Let \( q \in \mathbb{Z}\left\lbrack i\right\rbrack \) be a prime element. Then there is a prime integer \( p \in \mathbb{Z} \) such that \( N\left( q\right) = p \) or \( N\left( q\right) = {p}^{2} \) . | Proof. Since \( q \) is not a unit, \( N\left( q\right) \neq 1 \) (by Lemma 6.6). Thus \( N\left( q\right) \) is a nontrivial product of (integer) primes, and since \( q \) is prime in \( \mathbb{Z}\left\lbrack i\right\rbrack \supseteq \mathbb{Z}, q \) must divide one of the prime integer factors of \( N\left( q\right)... | Yes |
The prime integer 3 is a prime element of \( \mathbb{Z}\left\lbrack i\right\rbrack \) ; this can be verified by proving that 3 is irreducible in \( \mathbb{Z}\left\lbrack i\right\rbrack \) (since \( \mathbb{Z}\left\lbrack i\right\rbrack \) is a UFD). | For this purpose, note that since \( N\left( 3\right) = 9 \), the norm of a factor of 3 would have to be a divisor of 9, that is, 1, 3, or 9. Gaussian integers with norm 1 are units, and those with norm 9 are associates of 3 (Exercise 6.10); thus a nontrivial factor of 3 would necessarily have norm equal to 3 . But the... | No |
A positive integer prime \( p \in \mathbb{Z} \) splits in \( \mathbb{Z}\left\lbrack i\right\rbrack \) if and only if it is the sum of two squares in \( \mathbb{Z} \). | First assume that \( p = {a}^{2} + {b}^{2} \), with \( a, b \in \mathbb{Z} \). Then\n\n\[ p = \left( {a + {bi}}\right) \left( {a - {bi}}\right) \]\n\nin \( \mathbb{Z}\left\lbrack i\right\rbrack \), and \( N\left( {a \pm {bi}}\right) = {a}^{2} + {b}^{2} = p \neq 1 \), so neither of the two factors is a unit in \( \mathb... | Yes |
Lemma 1.2. Let \( M \) be an \( R \) -module, and let \( S \subseteq M \) be a linearly independent subset. Then there exists a maximal linearly independent subset of \( M \) containing \( S \) . | Proof. Consider the family \( \mathcal{S} \) of linearly independent subsets of \( M \) containing \( S \) , ordered by inclusion. Since \( S \) is linearly independent, \( \mathcal{S} \neq \varnothing \) . By Zorn’s lemma, it suffices to verify that every chain in \( \mathcal{S} \) has an upper bound. Indeed, the unio... | Yes |
Lemma 1.5. An R-module \( M \) is free if and only if it admits a basis. In fact, \( B \subseteq M \) is a basis if and only if the natural homomorphism \( {R}^{\oplus B} \rightarrow M \) is an isomorphism. | Proof. This is immediate from Definition 1.1 if \( B \subseteq M \) is linearly independent and generates \( M \), then the corresponding homomorphism \( {R}^{\oplus B} \rightarrow M \) is injective and surjective. Conversely, if \( \varphi : {R}^{\oplus B} \rightarrow M \) is an isomorphism, then \( B \) is identified... | Yes |
Lemma 1.6. Let \( R = k \) be a field, and let \( V \) be a \( k \) -vector space. Let \( B \) be a maximal linearly independent subset of \( V \) ; then \( B \) is a basis of \( V \). | Proof. Let \( v \in V, v \notin B \) . Then \( B \cup \{ v\} \) is not linearly independent, by the maximality of \( B \) ; therefore, there exist \( {c}_{0},\ldots ,{c}_{t} \in k \) and (distinct) \( {b}_{1},\ldots ,{b}_{t} \in B \) such that\n\n\[ \n{c}_{0}v + {c}_{1}{b}_{1} + \cdots + {c}_{t}{b}_{t} = 0 \n\]\n\nwith... | Yes |
Proposition 1.7. Let \( R = k \) be a field, and let \( V \) be a \( k \) -vector space. Let \( S \) be a linearly independent set of vectors of \( V \) . Then there exists a basis \( B \) of \( V \) containing \( S \) . | Proof. Put Lemma 1.2, Lemma 1.5, and Lemma 1.6 together. | No |
Lemma 1.8. Let \( R = k \) be a field, and let \( V \) be a \( k \) -vector space. Let \( B \) be a minimal generating set for \( V \) ; then \( B \) is a basis of \( V \) . | ## Proof. Exercise 1.6. | No |
Proposition 1.9. Let \( R \) be an integral domain, and let \( M \) be a free \( R \) -module. Let \( B \) be a maximal linearly independent subset of \( M \), and let \( S \) be a linearly independent subset. Then 4 \( \left| S\right| \leq \left| B\right| \) . | Proof. By taking fields of fractions, the general case over an integral domain is easily reduced to the case of vector spaces over a field; see Exercise 1.7 We may then assume that \( R = k \) is a field and \( M = V \) is a \( k \) -vector space.\n\nWe have to prove that there is an injective map \( j : S \hookrightar... | No |
An uncountable subset of \( \mathbb{C}\left\lbrack x\right\rbrack \) is necessarily linearly dependent. | Indeed, \( \mathbb{C}\left\lbrack x\right\rbrack \) has a countable basis over \( \mathbb{C} \) : for example, \( \left\{ {1, x,{x}^{2},{x}^{3},\ldots }\right\} \). | Yes |
Corollary 1.11. Let \( R \) be an integral domain, and let \( A, B \) be sets. Then\n\n\[ \n{F}^{R}\left( A\right) \cong {F}^{R}\left( B\right) \Leftrightarrow \text{ there is a bijection }A \cong B.\n\] | Proof. Exercise 1.8. | No |
Proposition 1.15. Let \( R \) be an integral domain, and let \( M \) be a free \( R \) -module; assume that \( M \) is generated by \( S : M = \langle S\rangle \) . Then \( S \) contains a maximal linearly independent subset of \( M \) . | Proof. By Exercise 1.7 we may assume that \( R \) is a field and \( M = V \) is a vector space. Use Zorn’s lemma to obtain a linearly independent subset \( B \subseteq S \) which is maximal among subsets of \( S \) . Arguing as in the proof of Lemma 1.6 shows that \( S \) is in the span of \( B \), and it follows that ... | No |
Lemma 2.1. For all \( m \times n \) matrices \( A \) with entries in \( R \) :\n\n- The function \( \varphi : {R}^{n} \rightarrow {R}^{m} \) defined by \( \varphi \left( \mathbf{v}\right) = A \cdot \mathbf{v} \) is a homomorphism of \( R \) -modules.\n\n- Every \( R \) -module homomorphism \( {R}^{n} \rightarrow {R}^{m... | Proof. The first point follows immediately from the elementary properties of matrix multiplication recalled above: \( \forall r, s \in R,\forall \mathbf{v},\mathbf{w} \in {R}^{n} \)\n\n\[ \varphi \left( {r\mathbf{v} + s\mathbf{w}}\right) = A \cdot \left( {r\mathbf{v} + s\mathbf{w}}\right) = {rA} \cdot \mathbf{v} + {sA}... | Yes |
Corollary 2.2. The correspondence introduced in Lemma 2.1 gives an isomorphism of \( R \) -modules\n\n\[ \n{\mathcal{M}}_{m, n}\left( R\right) \cong {\operatorname{Hom}}_{R}\left( {{R}^{n},{R}^{m}}\right) .\n\] | Proof. The reader will check that the correspondence is a bijective homomorphism of \( R \) -modules; this is enough, by Exercise III 5.12 | No |
Lemma 2.3. This diagram commutes. That is, the matrix corresponding to a composition \( \varphi \circ \psi \) is the product of the matrices corresponding to \( \varphi \) and \( \psi \) . | Proof. This follows immediately from the associativity of matrix multiplication: for \( \mathbf{v} \in {R}^{n} \) and \( A \in {\mathcal{M}}_{m, p}\left( R\right), B \in {\mathcal{M}}_{p, n}\left( R\right) \) ,\n\n\[ A \cdot \left( {B \cdot \mathbf{v}}\right) = \left( {A \cdot B}\right) \cdot \mathbf{v} \]\n\nthat is, ... | Yes |
Proposition 2.7. Two matrices \( P, Q \in {\mathcal{M}}_{m, n}\left( R\right) \) are equivalent if \( Q \) may be obtained from \( P \) by a sequence of elementary operations. | Proof. To see that elementary operations produce equivalent matrices, it suffices (by Proposition 2.5) to express them as multiplications on the left or right \( {}^{12} \) by invertible matrices. Indeed, these operations may be performed by suitably multiplying by the matrices obtained from the identity matrix by perf... | No |
Proposition 2.9. Let \( R = k \) be a field, and let \( n \geq 0 \) be an integer. Then \( {\mathrm{{GL}}}_{n}\left( k\right) \) is generated by elementary matrices. | Proof. Let \( A = \left( {a}_{ij}\right) \) be an \( n \times n \) invertible matrix. In particular, some entry in the first column of \( A \) is nonzero; by performing a row switch if necessary, we may assume that \( {a}_{11} \) is nonzero. Multiplying the first row by \( {a}_{11}^{-1} \), we may assume that \( {a}_{1... | Yes |
Over a field, every \( m \times n \) matrix is equivalent to a matrix of the form\n\n\[ \left( \begin{matrix} {I}_{r} & 0 \\ 0 & 0 \end{matrix}\right) \]\n\n(where \( r \leq \min \left( {m, n}\right) \) and ’ 0 ’ stands for null matrices of appropriate sizes). | Different matrices of the type displayed in Proposition 2.10 are inequivalent (for example by rank considerations; cf. (3.3). Thus, Proposition 2.10 describes all equivalence classes of matrices over a field and shows that for any given \( m, n \) there are in fact only finitely many such classes (over a field!). | No |
Let \( A \) be a square matrix with entries in an integral domain \( R \). Let \( {A}^{\prime } \) be obtained from \( A \) by switching two rows or two columns. Then \( \det \left( {A}^{\prime }\right) = - \det \left( A\right) . | Proof. These are all essentially immediate from Definition 3.1 For example, switching two columns amounts to correcting each \( \sigma \) in the definition by a fixed transposition, changing the sign of all contributions to the \( \sum \) in the definition. The third point is immediate from distributivity. Combining th... | No |
Let \( R \) be a commutative ring.\n\n- A square matrix \( A \in {\mathcal{M}}_{n}\left( R\right) \) is invertible if and only if \( \det \left( A\right) \) is a unit in \( R \) .\n\n- The determinant is a homomorphism \( {}^{19}{\mathrm{{GL}}}_{n}\left( R\right) \rightarrow \left( {{R}^{ * }, \cdot }\right) \) . More ... | Proof for \( R = \mathbf{a} \) field. If \( R = k \) is a field, we can use the considerations immediately preceding the statement. The first point is reduced to the case of a block matrix\n\n\[ \left( \begin{matrix} {I}_{r} & 0 \\ 0 & 0 \end{matrix}\right) \]\n\nfor which it is immediate. In fact, this shows that \( \... | No |
Lemma 3.4. With notation as above,\n\n\[ \n\\text{- for all}i = 1,\\ldots, n,\\det \\left( A\\right) = \\mathop{\\sum }\\limits_{{j = 1}}^{n}{a}_{ij}{A}^{\\left( ij\\right) }\\text{,}\n\]\n\n\[ \n\\text{- for all}j = 1,\\ldots, n,\\det \\left( A\\right) = \\mathop{\\sum }\\limits_{{i = 1}}^{n}{a}_{ij}{A}^{\\left( ij\\r... | Proof. This is a simple (if slightly messy) induction on \( n \), which I leave to the diligent reader. | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.