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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 II.8.14), \( H \cap K = \{ 0\} \) . Since subgroups of abelian groups are automatically normal, the statement follows from Proposition 5.3. | Yes |
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 t... | 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\n\n\[ 0 \rightarrow \langle g\rangle \rightarrow G \rightarrow G/\langle g\rangle \rightarrow 0 \]\n\nsplits. | 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 } + ... | 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). We 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. We 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.... | 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}}{360\mathbb{Z}}, \frac{\mathbb{Z}}{2\mat... | Yes |
Lemma 6.9. Let \( G \) be a finite abelian group, and assume that for every integer \( n \) 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\[ G \cong \frac{\mathbb{Z}}{\left( {d}_{1}\right) } \oplus \cdots \oplus \frac{\mathbb{Z}}{\left( {d}_{s}\right) } \]\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... | 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 |
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\[ R\text{ Noetherian } \Rightarrow R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \text{ Noetherian; } \]\n\nand an immediate induction reduces the statement to the following particular case, which carries a distingui... | 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 is, \(... | 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. Since PIDs are Noetherian, the a.c.c. holds in \( R \) (for all ideals, hence in particular for principal ideals); 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, and assume \( {bc... | 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 III.4.4) and \( k\left\lbrack x\right\rbrack \) (which the reader has hopefully worked out in Exercise III.4.4).\n\nProof. Let \( I \) be an ideal of \( R \) ; we have to prove that \( I \) is principal. If \( I = \{ 0\} \) , there i... | 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 |
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\nWe claim that \( U \) is a proper ideal containing \( I \) ; hen... | 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\righ... | 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( {\mathrm{{cont}}}_{f}\right) \left( \underline{f}\right) }\right) \left( {\left( {\mathrm{{cont}}}_{g}\right) \left( \underline{g}\right) }\right) = \left( {\mathrm{{cont}}}_{f}\right) \left( {\mathrm{{cont}}}_{g}\righ... | 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 we 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 |
Example 5.4. 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... | \[ \frac{{\mathbb{F}}_{2}\left\lbrack t\right\rbrack }{\left( {t}^{2} + t + 1\right) } \] | 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\n\n\[ 3 - {2x} + 3{x}^{2} - 2{x}^{3} + 3{x}^{4} - 2{x}^{5} \] | 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 |
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 \( i : k \hookrightarrow F \) (obtained as ... | 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. | Yes |
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 proo \( {\mathrm{f}}^{21} \) of this fundamental theorem (which is indeed known as the fundamental theorem of algebra.)\n\n'Algebraic' proofs of the fundamental theorem of algebra require more than we know at this point (we will encounter one in §VII.6, after we have seen a li... | 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 comple... | 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- \( {a}_{i} \in \mathfrak{p} \) for \( ... | 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 §VII.5.... | Claim 5.20. For \( p \) prime and \( k = 1,\ldots, p - 1, p \) divides \( \left( \begin{array}{l} p \\ k \end{array}\right) \) .\n\nThere is nothing to this, because\n\n\[ \left( \begin{array}{l} p \\ k \end{array}\right) = \frac{p!}{k!\left( {p - k}\right) !} \]\n\nand \( p \) divides the numerator and does not divide... | 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\[ \n\varphi : R \rightarrow \frac{R}{{I}_{1}} \times \cdots \times \frac{R}{{I}_{k}}\n\]\n\nis surjective and induces an isomorphism\n\n\[ \n... | 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 \... | Yes |
Lemma 6.2. 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. A simple induction reduces the general statement to the case \( k = 2 \) . Therefore, assume \( I \) and \( J \) are ideals of \( R \), such that \( I + J = \left( 1\right) \) . The inclusion \( {IJ} \subseteq I \cap J \) holds for all ideals \( I, J \), so the task amounts to proving \( I \cap J \subseteq {IJ} ... | Yes |
Lemma 6.3. 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 \( 1 - {a}_{k} \in {I}_{i} \) . Then\n\n\[ \left( {1 - {a}_{1}}\right) \cdots \left( {1 - {a}_{k - 1}}\right) \in {I}_{1}\cdots {I}_{k - 1} \]\n\nand\n\n\[ 1 - \left( {1 - {a}_{1}}\right) \cdots \left( {1 - {a}_{k - 1}}... | Yes |
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. This is not the case for arbitrary UFDs, and indeed the natural map\n\n\[ \mathbb{Z}\left\lbrack x\right\rbrack \rightarrow \frac{\mathbb{Z}\left\... | 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 ther... | 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 \( \mathbb{Z} \) . | Proof. 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 \... | 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 identifie... | 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 \hookrighta... | No |
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 |
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 \( \det \left( {A}^{\prime }\right) = - \det \left( A\right) \). | 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 the thir... | 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 \( \det A \) 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 generall... | Proof for \( R = \) 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 \( \det A = 0... | 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 we leave to the diligent reader. | No |
Corollary 3.5. Let \( R \) be a commutative ring and \( A \in {\mathcal{M}}_{n}\left( R\right) \) . Then\n\n\[ A \cdot \left( \begin{matrix} {A}^{\left( {11}\right) } & & & {A}^{\left( n1\right) } \\ \vdots & & & \vdots \\ {A}^{\left( 1n\right) } & & & {A}^{\left( nn\right) } \end{matrix}\right) = \left( \begin{matrix}... | Proof. Along the diagonal of the right-hand side, this is a restatement of Lemma 3.4. Off the diagonal, one is evaluating (for example)\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{{i}^{\prime }j}{A}^{\left( ij\right) } \]\n\nfor \( {i}^{\prime } \neq i \) . By Lemma 3.4 this is the same as the determinant of the matr... | Yes |
Proposition 3.6 (Cramer’s rule). Assume \( \det \left( A\right) \) is a unit, and let \( {A}^{\left( j\right) } \) be the matrix obtained by replacing the \( j \) -th column of \( A \) by the column vector \( b \) . Then\n\n\[ \n{x}_{j} = \det {\left( A\right) }^{-1}\det \left( {A}^{\left( j\right) }\right) .\n\] | Proof. Using Lemma 3.4, expand \( \det \left( {A}^{\left( j\right) }\right) \) with respect to the \( j \) -th column:\n\n\[ \n\det \left( {A}^{\left( j\right) }\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{A}^{\left( ij\right) }{b}_{i} \n\]\n\nTherefore\n\n\[ \n\left( \begin{matrix} {x}_{1} \\ \vdots \\ {x}_{n} \end{m... | Yes |
Proposition 3.7. The row rank of a matrix over a field \( k \) equals its column rank. | Proof. Equivalent matrices have the same ranks. Indeed, let \( P \in {\mathcal{M}}_{m, n}\left( k\right) \) ; the row space of \( P \) consists of all row vectors\n\n\[ \left( \begin{array}{ll} {a}_{1} & {a}_{m} \end{array}\right) = \left( \begin{array}{ll} {v}_{1} & {v}_{m} \end{array}\right) \cdot P \]\n\nobtained as... | No |
Lemma 4.2. Submodules and direct sums of torsion-free modules are torsion-free. Free modules over an integral domain are torsion-free. | Proof. The first statement is immediate; the second follows from the first, since an integral domain is torsion-free as a module over itself. | No |
Let \( R = \mathbb{Z}\left\lbrack x\right\rbrack \), and let \( I = \left( {2, x}\right) \) . Then \( I \) is not a free \( R \) -module. More generally, let \( I \) be any nonprincipal ideal of an integral domain \( R \) ; then \( I \) is a torsion-free module which is not free. | Indeed, if \( I \) were free, then its rank would have to be 1 at most, by Proposition 1.9 (a basis for \( I \) would be a linearly independent subset of \( R \), and \( R \) has rank 1 over itself); thus one element would suffice to generate \( I \), and \( I \) would be principal. | Yes |
Lemma 4.5. Let \( R \) be an integral domain. Assume that every cyclic \( R \) -module is torsion-free. Then \( R \) is a field. | Proof. Let \( c \in R, c \neq 0 \) ; then \( M = R/\left( c\right) \) is a cyclic module. Note that \( \operatorname{Tor}\left( M\right) = \) \( M \) : indeed, the class of 1 generates \( R/\left( c\right) \) and belongs to \( \operatorname{Tor}\left( M\right) \) since \( c \cdot 1 \) is 0 \( {\;\operatorname{mod}\;\le... | Yes |
Lemma 4.8. If \( R \) is a Noetherian ring, then every finitely generated \( R \)-module is finitely presented. | Proof. If \( M \) is a finitely generated module, there is an exact sequence\n\n\[ \n{R}^{m}\overset{\pi }{ \rightarrow }M \rightarrow 0 \n\]\n\nfor some \( m \) . Since \( R \) is Noetherian, \( {R}^{m} \) is Noetherian as an \( R \)-module (Corollary III.6.8). Thus \( \ker \pi \) is finitely generated; that is, there... | Yes |
Proposition 4.10. Let \( R \) be an integral domain. Then \( R \) is a field if and only if every finitely generated \( R \) -module is free. | Proof. If \( R \) is a field, then every \( R \) -module is free, by Proposition 1.7. For the converse, assume that every finitely generated \( R \) -module is free; in particular, every cyclic module is free; in particular, every cyclic module is torsion-free. But then \( R \) is a field, by Lemma 4.5. | Yes |
Lemma 4.12. Let \( A, B \) be matrices with entries in an integral domain \( R \), and let \( M \) , \( N \) denote the corresponding \( R \) -modules. Then \( M \oplus N \) corresponds to the block matrix \[ \left( \begin{matrix} A & 0 \\ 0 & B \end{matrix}\right) \] | Proof. This follows immediately from Exercise 4.16. | No |
Proposition 4.13. Let \( A \) be a matrix with entries in an integral domain \( R \), and let \( B \) be obtained from \( A \) by any sequence of the following operations:\n- switch two rows or two columns;\n- add to one row (resp., column) a multiple of another row (resp., column);\n- multiply all entries in one row (... | Proof. The first three operations are the ’elementary operations’ of \( §{2.3} \), and they transform a matrix into an equivalent one (by Proposition 2.7); as observed above, this does not affect the corresponding module, up to isomorphism.\n\nAs for the fourth operation, if \( u \) is a unit and the only nonzero entry... | Yes |
Example 4.14. The matrix with integer entries\n\n\[ \n\\left( \\begin{array}{ll} 1 & 3 \\\\ 2 & 3 \\\\ 5 & 9 \\end{array}\\right) \n\]\n\ndetermines an abelian group \( G \) . | Subtract three times the first column from the second column, obtaining\n\n\[ \n\\left( \\begin{matrix} 1 & 0 \\\\ 2 & - 3 \\\\ 5 & - 6 \\end{matrix}\\right) \n\]\n\nthe \( \\left( {1,1}\\right) \) entry is a unit and the only nonzero entry in the first row, so we can remove the first row and column:\n\n\[ \n\\left( \\... | Yes |
Proposition 5.1. Let \( R \) be a PID, let \( F \) be a finitely generated free module over \( R \) , and let \( M \subseteq F \) be a submodule. Then \( M \) is free. | We will actually prove a more precise result, in view of the full statement of the classification theorem: we will show that there is a basis \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) of \( F \) and elements \( {a}_{1},\ldots ,{a}_{m} \) of \( R \) (with \( m \leq n \) ) such that\n\n\[ \n{y}_{1} = {a}_{1}{x}_{1},\... | Yes |
Lemma 5.2. Let \( R \) be a PID, let \( F \) be a finitely generated free module over \( R \), and let \( M \subseteq F \) be a nonzero submodule. Then there exist \( a \in R, x \in F, y \in M \), and submodules \( {F}^{\prime } \subseteq F \) and \( {M}^{\prime } \subseteq M \), such that \( y = {ax},{M}^{\prime } = {... | Proof. For all \( \varphi \in {\operatorname{Hom}}_{R}\left( {F, R}\right) ,\varphi \left( M\right) \) is a submodule of \( R \), that is, an ideal. The family of all these ideals is nonempty, and PIDs are Noetherian; therefore (by Proposition V.1.1) there exists a maximal element in the family, say \( \alpha \left( M\... | Yes |
Corollary 5.3. Let \( R \) be a PID, let \( F \) be a finitely generated free module over \( R \) , and let \( M \subseteq F \) be a submodule. Then there exist a basis \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) of \( F \) and nonzero elements \( {a}_{1},\ldots ,{a}_{m} \) of \( R\left( {m \leq n}\right) \) such tha... | Proof. Now that we know that submodules of a free module are free, we see that the submodule \( {F}^{\prime } \subseteq F \) produced in Lemma 5.2 is free. The first part of the statement then follows from Lemma 5.2, by an inductive argument analogous to the proof of Proposition 5.1, and is left to the reader.\n\nThe m... | No |
Proposition 5.4. Let \( R \) be an integral domain. Then \( R \) is a PID if and only if for every finitely generated \( R \) -module \( M \) and every epimorphism \[ {R}^{{m}_{0}}\overset{{\pi }_{0}}{ \rightarrow }M \rightarrow 0, \] there exist a free \( R \) -module \( {R}^{{m}_{1}} \) and a homomorphism \( {\pi }_{... | Proof. The fact that the stated condition implies that \( R \) is a PID was proved in Claim 4.11. For the converse, let \( {\pi }_{0} : {R}^{{m}_{0}} \rightarrow M \) be an epimorphism; then \( \ker {\pi }_{0} \) is free by Proposition 5.1; the result follows by choosing any isomorphism \( {\pi }_{1} : {R}^{{m}_{1}} \r... | Yes |
Lemma 5.7. Let \( M \) be a torsion module, expressed as in Theorem 5.6 (with \( \operatorname{rk}M = \) \( 0) \) . Then \( \operatorname{Ann}\left( M\right) = \left( {a}_{m}\right) \) . Further, the prime ideals \( \left( {q}_{i}\right) \) are precisely the prime ideals of \( R \) containing \( \operatorname{Ann}\left... | Proof. By hypothesis\n\n\[ M \cong \frac{R}{\left( {a}_{1}\right) } \oplus \cdots \oplus \frac{R}{\left( {a}_{m}\right) } \]\n\nwith \( {a}_{1}\left| \cdots \right| {a}_{m} \) . If \( r \in \operatorname{Ann}\left( M\right) \), then\n\n\[ 0 = r\left( {1,\ldots ,1}\right) = \left( {r,\ldots, r}\right) \]\n\nIn particula... | Yes |
Proposition 6.2. Two matrices \( A, B \in {\mathcal{M}}_{n}\left( R\right) \) are similar if and only if there exists an invertible matrix \( P \) such that\n\n\[ B = {PA}{P}^{-1}\text{.} \] | The reader who has really understood Proposition 2.5 will not need any detailed proof of this statement: it should be apparent from staring at the butterfly diagram\n\n\n\nwhich we are essentially copying from \( §{2... | No |
Proposition 6.5. Let \( \alpha \) be a linear transformation of a free \( R \) -module \( F \cong {R}^{n} \) . Then \( \det \alpha \neq 0 \) if and only if \( \alpha \) is injective. | Proof. Embed \( R \) in its field of fractions \( K \), and view \( \alpha \) as a linear transformation of \( {K}^{n} \) ; note that the determinant of \( \alpha \) is the same whether it is computed over \( R \) or over \( K \) . Then \( \alpha \) is injective as a linear transformation \( {R}^{n} \rightarrow {R}^{n}... | Yes |
Lemma 6.7. Let \( A, B \in {\mathcal{M}}_{n}\left( R\right) \) . Then \( \operatorname{tr}\left( {AB}\right) = \operatorname{tr}\left( {BA}\right) \) . | Proof. Let \( A = \left( {a}_{ij}\right), B = \left( {b}_{ij}\right) \) . Then \( {AB} = \left( {\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik}{b}_{kj}}\right) \) ; hence\n\n\[ \operatorname{tr}\left( {AB}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik}{b}_{ki} \]\n\nThis expression... | Yes |
Proposition 6.9. Let \( F \) be a free \( R \) -module of rank \( n \), and let \( \alpha \in {\operatorname{End}}_{R}\left( F\right) \) .\n\n- The characteristic polynomial \( {P}_{\alpha }\left( t\right) \) is a monic polynomial of degree \( n \) .\n\n- The coefficient of \( {t}^{n - 1} \) in \( {P}_{\alpha }\left( t... | Proof. The first point is immediate, and the third is checked by setting \( t = 0 \) . To verify the second assertion, let \( A = \left( {a}_{ij}\right) \) be a matrix representing \( \alpha \) with respect to any basis for \( F \), so that\n\n\[ \n{P}_{\alpha }\left( t\right) = \det \left( \begin{matrix} t - {a}_{11} ... | Yes |
Lemma 6.10. If \( \alpha \) and \( \beta \) are similar, then \( {\mathcal{I}}_{\alpha } = {\mathcal{I}}_{\beta } \) . | Proof. By hypothesis there exists an invertible \( \pi \) such that \( \beta = \pi \circ \alpha \circ {\pi }^{-1} \) . As\n\n\( {\beta }^{k} = {\left( \pi \circ \alpha \circ {\pi }^{-1}\right) }^{k} = \left( {\pi \circ \alpha \circ {\pi }^{-1}}\right) \circ \left( {\pi \circ \alpha \circ {\pi }^{-1}}\right) \circ \cdot... | Yes |
Theorem 6.11 (Cayley-Hamilton). Let \( {P}_{\alpha }\left( t\right) \) be the characteristic polynomial of the linear transformation \( \alpha \in {\operatorname{End}}_{R}\left( F\right) \) . Then\n\n\[ \n{P}_{\alpha }\left( \alpha \right) = 0.\n\] | This beautiful observation can be proved directly by judicious use of Cramer's \( {\text{rule}}^{30} \), in the form of Corollary 3.5; cf. Exercise 6.9. In any case, the Cayley-Hamilton theorem will become essentially evident once we connect these linear algebra considerations with the classification theorem for finite... | No |
Lemma 6.14. Let \( F \) be a finitely generated \( R \) -module, and let \( \alpha \in {\operatorname{End}}_{R}\left( F\right) \) . Then the set of eigenvalues of \( \alpha \) is precisely the set of roots in \( R \) of the characteristic polynomial \( {P}_{\alpha }\left( t\right) \) . | Proof. This is a straightforward consequence of Proposition 6.5:\n\n\( \lambda \) is an eigenvalue for \( \alpha \Leftrightarrow \exists \mathbf{v} \neq 0 \) such that \( \alpha \left( \mathbf{v}\right) = {\lambda I}\left( \mathbf{v}\right) \)\n\n\[ \Leftrightarrow \exists \mathbf{v} \neq 0\text{ such that }\left( {{\l... | Yes |
Corollary 6.18. The number of eigenvalues of a linear transformation of \( {R}^{n} \) is at most \( n \) . If the base ring \( R \) is an algebraically closed field, then every linear transformation has exactly n eigenvalues (counted with algebraic multiplicity). | Proof. Immediate from Lemmas 6.14, V.5.1, and V.5.10. | No |
Lemma 7.2. Let \( \alpha ,\beta \) be linear transformations of a free \( R \) -module \( F \) . Then the corresponding \( R\left\lbrack t\right\rbrack \) -module structures on \( F \) are isomorphic if and only if \( \alpha \) and \( \beta \) are similar. | Proof. Denote by \( {F}_{\alpha },{F}_{\beta } \) the two \( R\left\lbrack t\right\rbrack \) -modules defined on \( F \) by \( \alpha ,\beta \) as per Claim 7.1.\n\nAssume first that \( \alpha \) and \( \beta \) are similar. Then there exists an invertible \( R \) -linear transformation \( \pi : F \rightarrow F \) such... | No |
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