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Proposition 7. A finite group is nilpotent if and only if every maximal subgroup is normal. | Proof: Let \( G \) be a finite nilpotent group and let \( M \) be a maximal subgroup of \( G \) . As in the proof of Theorem 1, since \( M < {N}_{G}\left( M\right) \) (by Theorem 3(2)) maximality of \( M \) forces \( {N}_{G}\left( M\right) = G \), i.e., \( M \trianglelefteq G \) .\n\nConversely, assume every maximal su... | Yes |
Theorem 8. A group \( G \) is nilpotent if and only if \( {G}^{n} = 1 \) for some \( n \geq 0 \) . More precisely, \( G \) is nilpotent of class \( c \) if and only if \( c \) is the smallest nonnegative integer such that \( {G}^{c} = 1 \) . If \( G \) is nilpotent of class \( c \) then \[ {Z}_{i}\left( G\right) \leq {... | Proof: This is proved by a straightforward induction on the length of either the upper or lower central series. | No |
Theorem 9. A group \( G \) is solvable if and only if \( {G}^{\left( n\right) } = 1 \) for some \( n \geq 0 \) . | Proof: Assume first that \( G \) is solvable and so possesses a series\n\n\[ 1 = {H}_{0} \trianglelefteq {H}_{1} \trianglelefteq \cdots \trianglelefteq {H}_{s} = G \]\n\nsuch that each factor \( {H}_{i + 1}/{H}_{i} \) is abelian. We prove by induction that \( {G}^{\left( i\right) } \leq {H}_{s - i} \) . This is true fo... | Yes |
Proposition 10. Let \( G \) and \( K \) be groups, let \( H \) be a subgroup of \( G \) and let \( \varphi : G \rightarrow K \) be a surjective homomorphism.\n\n(1) \( {H}^{\left( i\right) } \leq {G}^{\left( i\right) } \) for all \( i \geq 0 \) . In particular, if \( G \) is solvable, then so is \( H \), i.e., subgroup... | Proof: Part 1 follows from the observation that since \( H \leq G \), by definition of commutator subgroups, \( \left\lbrack {H, H}\right\rbrack \leq \left\lbrack {G, G}\right\rbrack \), i.e., \( {H}^{\left( 1\right) } \leq {G}^{\left( 1\right) } \) . Then, by induction,\n\n\[{H}^{\left( i\right) } \leq {G}^{\left( i\r... | Yes |
Theorem 11. Let \( G \) be a finite group.\n\n(1) (Burnside) If \( \left| G\right| = {p}^{a}{q}^{b} \) for some primes \( p \) and \( q \), then \( G \) is solvable. | We shall prove Burnside’s Theorem in Chapter 19 and deduce Philip Hall’s generalization of it. | No |
Lemma 13. In a finite group \( G \) if \( {n}_{p} ≢ 1\left( {\;\operatorname{mod}\;{p}^{2}}\right) \), then there are distinct Sylow \( p \) -subgroups \( P \) and \( R \) of \( G \) such that \( P \cap R \) is of index \( p \) in both \( P \) and \( R \) (hence is normal in each). | Proof: The argument is an easy refinement of the proof of the congruence part of Sylow’s Theorem (cf. the exercises at the end of Section 4.5). Let \( P \) act by conjugation on the set \( {\operatorname{Syl}}_{p}\left( G\right) \) . Let \( {\mathcal{O}}_{1},\ldots ,{\mathcal{O}}_{s} \) be the orbits under this action ... | Yes |
Theorem 17. Let \( G \) be a group, \( S \) a set and \( \varphi : S \rightarrow G \) a set map. Then there is a unique group homomorphism \( \Phi : F\left( S\right) \rightarrow G \) such that the following diagram commutes: | Proof: Such a map \(\Phi\) must satisfy \(\Phi \left( {{s}_{1}^{{\epsilon }_{1}}{s}_{2}^{{\epsilon }_{2}}\ldots {s}_{n}^{{\epsilon }_{n}}}\right) = \varphi {\left( {s}_{1}\right) }^{{\epsilon }_{1}}\varphi {\left( {s}_{2}\right) }^{{\epsilon }_{2}}\ldots \varphi {\left( {s}_{n}\right) }^{{\epsilon }_{n}}\) if it is to ... | Yes |
Corollary 18. \( F\left( S\right) \) is unique up to a unique isomorphism which is the identity map on the set \( S \) . | Proof: This follows from the universal property. Suppose \( F\left( S\right) \) and \( {F}^{\prime }\left( S\right) \) are two free groups generated by \( S \) . Since \( S \) is contained in both \( F\left( S\right) \) and \( {F}^{\prime }\left( S\right) \), we have natural injections \( S \hookrightarrow {F}^{\prime ... | Yes |
Proposition 1. Let \( R \) be a ring. Then\n\n(1) \( {0a} = {a0} = 0 \) for all \( a \in R \) . | Proof: These all follow from the distributive laws and cancellation in the additive group \( R \) . For example,(1) follows from \( {0a} = \left( {0 + 0}\right) a = {0a} + {0a} \) . | No |
Proposition 2. Assume \( a, b \) and \( c \) are elements of any ring with \( a \) not a zero divisor. If \( {ab} = {ac} \), then either \( a = 0 \) or \( b = c \) (i.e., if \( a \neq 0 \) we can cancel the \( a \) ’s). In particular, if \( a, b, c \) are any elements in an integral domain and \( {ab} = {ac} \), then e... | Proof: If \( {ab} = {ac} \) then \( a\left( {b - c}\right) = 0 \) so either \( a = 0 \) or \( b - c = 0 \) . The second statement follows from the first and the definition of an integral domain. | Yes |
Corollary 3. Any finite integral domain is a field. | Proof: Let \( R \) be a finite integral domain and let \( a \) be a nonzero element of \( R \) . By the cancellation law the map \( x \mapsto {ax} \) is an injective function. Since \( R \) is finite this map is also surjective. In particular, there is some \( b \in R \) such that \( {ab} = 1 \), i.e., \( a \) is a uni... | Yes |
Proposition 4. Let \( R \) be an integral domain and let \( p\left( x\right), q\left( x\right) \) be nonzero elements of \( R\left\lbrack x\right\rbrack \) . Then\n\n(1) degree \( p\left( x\right) q\left( x\right) = \) degree \( p\left( x\right) + \) degree \( q\left( x\right) \),\n\n(2) the units of \( R\left\lbrack x... | Proof: If \( R \) has no zero divisors then neither does \( R\left\lbrack x\right\rbrack \) ; if \( p\left( x\right) \) and \( q\left( x\right) \) are polynomials with leading terms \( {a}_{n}{x}^{n} \) and \( {b}_{m}{x}^{m} \), respectively, then the leading term of \( p\left( x\right) q\left( x\right) \) is \( {a}_{n... | Yes |
Proposition 5. Let \( R \) and \( S \) be rings and let \( \varphi : R \rightarrow S \) be a homomorphism.\n\n(1) The image of \( \varphi \) is a subring of \( S \).\n\n(2) The kernel of \( \varphi \) is a subring of \( R \). Furthermore, if \( \alpha \in \ker \varphi \) then \( {r\alpha } \) and \( {\alpha r} \in \ker... | Proof: (1) If \( {s}_{1},{s}_{2} \in \operatorname{im}\varphi \) then \( {s}_{1} = \varphi \left( {r}_{1}\right) \) and \( {s}_{2} = \varphi \left( {r}_{2}\right) \) for some \( {r}_{1},{r}_{2} \in R \) . Then \( \varphi \left( {{r}_{1} - {r}_{2}}\right) = {s}_{1} - {s}_{2} \) and \( \varphi \left( {{r}_{1}{r}_{2}}\rig... | Yes |
(1) (The First Isomorphism Theorem for Rings) If \( \varphi : R \rightarrow S \) is a homomorphism of rings, then the kernel of \( \varphi \) is an ideal of \( R \), the image of \( \varphi \) is a subring of \( S \) and \( R/\ker \varphi \) is isomorphic as a ring to \( \varphi \left( R\right) \) . | Proof: This is just a matter of collecting previous calculations. If \( I \) is the kernel of \( \varphi \), then the cosets (under addition) of \( I \) are precisely the fibers of \( \varphi \) . In particular, the cosets \( r + I, s + I \) and \( {rs} + I \) are the fibers of \( \varphi \) over \( \varphi \left( r\ri... | Yes |
Proposition 9. Let \( I \) be an ideal of \( R \). (1) \( I = R \) if and only if \( I \) contains a unit. | Proof: (1) If \( I = R \) then \( I \) contains the unit 1. Conversely, if \( u \) is a unit in \( I \) with inverse \( v \), then for any \( r \in R \)\n\n\[ r = r \cdot 1 = r\left( {vu}\right) = \left( {rv}\right) u \in I \]\n\nhence \( R = I \). | Yes |
Corollary 10. If \( R \) is a field then any nonzero ring homomorphism from \( R \) into another ring is an injection. | Proof: The kernel of a ring homomorphism is an ideal. The kernel of a nonzero homomorphism is a proper ideal hence is 0 by the proposition. | Yes |
Proposition 11. In a ring with identity every proper ideal is contained in a maximal ideal. | Proof: Let \( R \) be a ring with identity and let \( I \) be a proper ideal (so \( R \) cannot be the zero ring, i.e., \( 1 \neq 0 \) ). Let \( \mathcal{S} \) be the set of all proper ideals of \( R \) which contain \( I \) . Then \( \mathcal{S} \) is nonempty \( \left( {I \in \mathcal{S}}\right) \) and is partially o... | Yes |
Proposition 12. Assume \( R \) is commutative. The ideal \( M \) is a maximal ideal if and only if the quotient ring \( R/M \) is a field. | Proof: This follows from the Lattice Isomorphism Theorem together with Proposition 9(2). The ideal \( M \) is maximal if and only if there are no ideals \( I \) with \( M \subset I \subset R \) . By the Lattice Isomorphism Theorem the ideals of \( R \) containing \( M \) correspond bijectively with the ideals of \( R/M... | Yes |
Proposition 13. Assume \( R \) is commutative. Then the ideal \( P \) is a prime ideal in \( R \) if and only if the quotient ring \( R/P \) is an integral domain. | Proof: This proof is simply a matter of translating the definition of a prime ideal into the language of quotients. The ideal \( P \) is prime if and only if \( P \neq R \) and whenever \( {ab} \in P \), then either \( a \in P \) or \( b \in P \) . Use the bar notation for elements of \( R/P \) : \( \bar{r} = r + P \) ... | Yes |
Corollary 14. Assume \( R \) is commutative. Every maximal ideal of \( R \) is a prime ideal. | Proof: If \( M \) is a maximal ideal then \( R/M \) is a field by Proposition 12. A field is an integral domain so the corollary follows from Proposition 13. | Yes |
Corollary 16. Let \( R \) be an integral domain and let \( Q \) be the field of fractions of \( R \). If a field \( F \) contains a subring \( {R}^{\prime } \) isomorphic to \( R \) then the subfield of \( F \) generated by \( {R}^{\prime } \) is isomorphic to \( Q \). | Proof: Let \( \varphi : R \cong {R}^{\prime } \subseteq F \) be a (ring) isomorphism of \( R \) to \( {R}^{\prime } \). In particular, \( \varphi : R \rightarrow F \) is an injective homomorphism from \( R \) into the field \( F \). Let \( \Phi : Q \rightarrow F \) be the extension of \( \varphi \) to \( Q \) as in the... | Yes |
Corollary 18. Let \( n \) be a positive integer and let \( {p}_{1}{}^{{\alpha }_{1}}{p}_{2}{}^{{\alpha }_{2}}\ldots {p}_{k}{}^{{\alpha }_{k}} \) be its factorization into powers of distinct primes. Then\n\n\[ \n\mathbb{Z}/n\mathbb{Z} \cong \left( {\mathbb{Z}/{p}_{1}{}^{{\alpha }_{1}}\mathbb{Z}}\right) \times \left( {\m... | If we compare orders on the two sides of this last isomorphism, we obtain the formula\n\n\[ \n\varphi \left( n\right) = \varphi \left( {{p}_{1}{}^{{\alpha }_{1}}}\right) \varphi \left( {{p}_{2}{}^{{\alpha }_{2}}}\right) \ldots \varphi \left( {{p}_{k}{}^{{\alpha }_{k}}}\right)\n\]\n\nfor the Euler \( \varphi \) -functio... | Yes |
Proposition 1. Every ideal in a Euclidean Domain is principal. More precisely, if \( I \) is any nonzero ideal in the Euclidean Domain \( R \) then \( I = \left( d\right) \), where \( d \) is any nonzero element of \( I \) of minimum norm. | Proof: If \( I \) is the zero ideal, there is nothing to prove. Otherwise let \( d \) be any nonzero element of \( I \) of minimum norm (such a \( d \) exists since the set \( \{ N\left( a\right) \mid a \in I\} \) has a minimum element by the Well Ordering of \( \mathbb{Z} \) ). Clearly \( \left( d\right) \subseteq I \... | Yes |
Proposition 3. Let \( R \) be an integral domain. If two elements \( d \) and \( {d}^{\prime } \) of \( R \) generate the same principal ideal, i.e., \( \left( d\right) = \left( {d}^{\prime }\right) \), then \( {d}^{\prime } = {ud} \) for some unit \( u \) in \( R \) . In particular, if \( d \) and \( {d}^{\prime } \) ... | Proof: This is clear if either \( d \) or \( {d}^{\prime } \) is zero so we may assume \( d \) and \( {d}^{\prime } \) are nonzero. Since \( d \in \left( {d}^{\prime }\right) \) there is some \( x \in R \) such that \( d = x{d}^{\prime } \) . Since \( {d}^{\prime } \in \left( d\right) \) there is some \( y \in R \) suc... | Yes |
Theorem 4. Let \( R \) be a Euclidean Domain and let \( a \) and \( b \) be nonzero elements of \( R \). Let \( d = {r}_{n} \) be the last nonzero remainder in the Euclidean Algorithm for \( a \) and \( b \) described at the beginning of this chapter. Then\n\n(1) \( d \) is a greatest common divisor of \( a \) and \( b... | Proof: By Proposition 1, the ideal generated by \( a \) and \( b \) is principal so \( a, b \) do have a greatest common divisor, namely any element which generates the (principal) ideal \( \left( {a, b}\right) \). Both parts of the theorem will follow therefore once we show \( d = {r}_{n} \) generates this ideal, i.e.... | Yes |
Proposition 5. Let \( R \) be an integral domain that is not a field. If \( R \) is a Euclidean Domain then there are universal side divisors in \( R \) . | Proof: Suppose \( R \) is Euclidean with respect to some norm \( N \) and let \( u \) be an element of \( R - \widetilde{R} \) (which is nonempty since \( R \) is not a field) of minimal norm. For any \( x \in R \) , write \( x = {qu} + r \) where \( r \) is either 0 or \( N\left( r\right) < N\left( u\right) \) . In ei... | Yes |
Proposition 6. Let \( R \) be a Principal Ideal Domain and let \( a \) and \( b \) be nonzero elements of \( R \) . Let \( d \) be a generator for the principal ideal generated by \( a \) and \( b \) . Then\n\n(1) \( d \) is a greatest common divisor of \( a \) and \( b \)\n\n(2) \( d \) can be written as an \( R \) -l... | Proof: This is just Propositions 2 and 3. | No |
Proposition 7. Every nonzero prime ideal in a Principal Ideal Domain is a maximal ideal. | Proof: Let \( \left( p\right) \) be a nonzero prime ideal in the Principal Ideal Domain \( R \) and let \( I = \left( m\right) \) be any ideal containing \( \left( p\right) \) . We must show that \( I = \left( p\right) \) or \( I = R \) . Now \( p \in \left( m\right) \) so \( p = {rm} \) for some \( r \in R \) . Since ... | Yes |
Corollary 8. If \( R \) is any commutative ring such that the polynomial ring \( R\left\lbrack x\right\rbrack \) is a Principal Ideal Domain (or a Euclidean Domain), then \( R \) is necessarily a field. | Proof: Assume \( R\left\lbrack x\right\rbrack \) is a Principal Ideal Domain. Since \( R \) is a subring of \( R\left\lbrack x\right\rbrack \) then \( R \) must be an integral domain (recall that \( R\left\lbrack x\right\rbrack \) has an identity if and only if \( R \) does). The ideal \( \left( x\right) \) is a nonzer... | Yes |
Proposition 9. The integral domain \( R \) is a P.I.D. if and only if \( R \) has a Dedekind-Hasse norm. | Proof: Let \( I \) be any nonzero ideal in \( R \) and let \( b \) be a nonzero element of \( I \) with \( N\left( b\right) \) minimal. Suppose \( a \) is any nonzero element in \( I \), so that the ideal \( \left( {a, b}\right) \) is contained in \( I \) . Then the Dedekind-Hasse condition on \( N \) and the minimalit... | No |
Proposition 10. In an integral domain a prime element is always irreducible. | Proof: Suppose \( \left( p\right) \) is a nonzero prime ideal and \( p = {ab} \) . Then \( {ab} = p \in \left( p\right) \), so by definition of prime ideal one of \( a \) or \( b \), say \( a \), is in \( \left( p\right) \) . Thus \( a = {pr} \) for some \( r \) . This implies \( p = {ab} = {prb} \) so \( {rb} = 1 \) a... | Yes |
Proposition 11. In a Principal Ideal Domain a nonzero element is a prime if and only if it is irreducible. | Proof: We have shown above that prime implies irreducible. We must show conversely that if \( p \) is irreducible, then \( p \) is a prime, i.e., the ideal \( \left( p\right) \) is a prime ideal. If \( M \) is any ideal containing \( \left( p\right) \) then by hypothesis \( M = \left( m\right) \) is a principal ideal. ... | Yes |
Proposition 12. In a Unique Factorization Domain a nonzero element is a prime if and only if it is irreducible. | Proof: Let \( R \) be a Unique Factorization Domain. Since by Proposition 10, primes of \( R \) are irreducible it remains to prove that each irreducible element is a prime. Let \( p \) be an irreducible in \( R \) and assume \( p \mid {ab} \) for some \( a, b \in R \) ; we must show that \( p \) divides either \( a \)... | Yes |
Proposition 13. Let \( a \) and \( b \) be two nonzero elements of the Unique Factorization Domain \( R \) and suppose\n\n\[ a = u{p}_{1}{}^{{e}_{1}}{p}_{2}{}^{{e}_{2}}\cdots {p}_{n}{}^{{e}_{n}}\;\text{ and }\;b = v{p}_{1}{}^{{f}_{1}}{p}_{2}{}^{{f}_{2}}\cdots {p}_{n}{}^{{f}_{n}} \]\n\nare prime factorizations for \( a ... | Proof: Since the exponents of each of the primes occurring in \( d \) are no larger than the exponents occurring in the factorizations of both \( a \) and \( b, d \) divides both \( a \) and \( b \) . To show that \( d \) is a greatest common divisor, let \( c \) be any common divisor of \( a \) and \( b \) and let \( ... | Yes |
Corollary 16. Let \( R \) be a P.I.D. Then there exists a multiplicative Dedekind-Hasse norm on \( R \) . | Proof: If \( R \) is a P.I.D. then \( R \) is a U.F.D. Define the norm \( N \) by setting \( N\left( 0\right) = 0 \) , \( N\left( u\right) = 1 \) if \( u \) is a unit, and \( N\left( a\right) = {2}^{n} \) if \( a = {p}_{1}{p}_{2}\cdots {p}_{n} \) where the \( {p}_{i} \) are irreducibles in \( R \) (well defined since t... | Yes |
Lemma 17. The prime number \( p \in \mathbb{Z} \) divides an integer of the form \( {n}^{2} + 1 \) if and only if \( p \) is either 2 or is an odd prime congruent to 1 modulo 4. | Proof: The statement for \( p = 2 \) is trivial since \( 2 \mid {1}^{2} + 1 \) . If \( p \) is an odd prime, note that \( p \mid {n}^{2} + 1 \) is equivalent to \( {n}^{2} = - 1 \) in \( \mathbb{Z}/p\mathbb{Z} \) . This in turn is equivalent to saying the residue class of \( n \) is of order 4 in the multiplicative gro... | Yes |
Let \( n \) be a positive integer and write\n\n\[ n = {2}^{k}{p}_{1}^{{a}_{1}}\ldots {p}_{r}^{{a}_{r}}{q}_{1}^{{b}_{1}}\ldots {q}_{s}^{{b}_{s}} \]\n\nwhere \( {p}_{1},\ldots ,{p}_{r} \) are distinct primes congruent to 1 modulo 4 and \( {q}_{1},\ldots ,{q}_{s} \) are distinct primes congruent to 3 modulo 4 . Then \( n ... | The first statement in the corollary was proved above. Assume now that \( {b}_{1},\ldots ,{b}_{s} \) are all even. For each prime \( {p}_{i} \) congruent to 1 modulo 4 write \( {p}_{i} = {\pi }_{i}\overline{{\pi }_{i}} \) for \( i = 1,2,\ldots, r \), where \( {\pi }_{i} \) and \( \overline{{\pi }_{i}} \) are irreducibl... | Yes |
Proposition 2. Let \( I \) be an ideal of the ring \( R \) and let \( \left( I\right) = I\left\lbrack x\right\rbrack \) denote the ideal of \( R\left\lbrack x\right\rbrack \) generated by \( I \) (the set of polynomials with coefficients in \( I \) ). Then\n\n\[ R\left\lbrack x\right\rbrack /\left( I\right) \cong \left... | Proof: There is a natural map \( \varphi : R\left\lbrack x\right\rbrack \rightarrow \left( {R/I}\right) \left\lbrack x\right\rbrack \) given by reducing each of the coefficients of a polynomial modulo \( I \) . The definition of addition and multiplication in these two rings shows that \( \varphi \) is a ring homomorph... | Yes |
Theorem 3. Let \( F \) be a field. The polynomial ring \( F\left\lbrack x\right\rbrack \) is a Euclidean Domain. Specifically, if \( a\left( x\right) \) and \( b\left( x\right) \) are two polynomials in \( F\left\lbrack x\right\rbrack \) with \( b\left( x\right) \) nonzero, then there are unique \( q\left( x\right) \) ... | Proof: If \( a\left( x\right) \) is the zero polynomial then take \( q\left( x\right) = r\left( x\right) = 0 \) . We may therefore assume \( a\left( x\right) \neq 0 \) and prove the existence of \( q\left( x\right) \) and \( r\left( x\right) \) by induction on \( n = \operatorname{degree}a\left( x\right) \) . Let \( b\... | Yes |
Corollary 4. If \( F \) is a field, then \( F\left\lbrack x\right\rbrack \) is a Principal Ideal Domain and a Unique Factorization Domain. | Proof: This is immediate from the results of the last chapter. | No |
Proposition 5. (Gauss’ Lemma) Let \( R \) be a Unique Factorization Domain with field of fractions \( F \) and let \( p\left( x\right) \in R\left\lbrack x\right\rbrack \) . If \( p\left( x\right) \) is reducible in \( F\left\lbrack x\right\rbrack \) then \( p\left( x\right) \) is reducible in \( R\left\lbrack x\right\r... | Proof: The coefficients of the polynomials on the right hand side of the equation \( p\left( x\right) = A\left( x\right) B\left( x\right) \) are elements in the field \( F \), hence are quotients of elements from the Unique Factorization Domain \( R \) . Multiplying through by a common denominator for all these coeffic... | Yes |
Corollary 6. Let \( R \) be a Unique Factorization Domain, let \( F \) be its field of fractions and let \( p\left( x\right) \in R\left\lbrack x\right\rbrack \) . Suppose the greatest common divisor of the coefficients of \( p\left( x\right) \) is 1 . Then \( p\left( x\right) \) is irreducible in \( R\left\lbrack x\rig... | Proof: By Gauss’ Lemma above, if \( p\left( x\right) \) is reducible in \( F\left\lbrack x\right\rbrack \), then it is reducible in \( R\left\lbrack x\right\rbrack \) . Conversely, the assumption on the greatest common divisor of the coefficients of \( p\left( x\right) \) implies that if it is reducible in \( R\left\lb... | Yes |
Corollary 8. If \( R \) is a Unique Factorization Domain, then a polynomial ring in an arbitrary number of variables with coefficients in \( R \) is also a Unique Factorization Domain. | Proof: For finitely many variables, this follows by induction from Theorem 7, since a polynomial ring in \( n \) variables can be considered as a polynomial ring in one variable with coefficients in a polynomial ring in \( n - 1 \) variables. The general case follows from the definition of a polynomial ring in an arbit... | Yes |
Proposition 9. Let \( F \) be a field and let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) . Then \( p\left( x\right) \) has a factor of degree one if and only if \( p\left( x\right) \) has a root in \( F \), i.e., there is an \( \alpha \in F \) with \( p\left( \alpha \right) = 0 \) . | Proof: If \( p\left( x\right) \) has a factor of degree one, then since \( F \) is a field, we may assume the factor is monic, i.e., is of the form \( \left( {x - \alpha }\right) \) for some \( \alpha \in F \) . But then \( p\left( \alpha \right) = 0 \) . Conversely, suppose \( p\left( \alpha \right) = 0 \) . By the Di... | Yes |
Proposition 10. A polynomial of degree two or three over a field \( F \) is reducible if and only if it has a root in \( F \) . | Proof: This follows immediately from the previous proposition, since a polynomial of degree two or three is reducible if and only if it has at least one linear factor. | Yes |
Proposition 11. Let \( p\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} \) be a polynomial of degree \( n \) with integer coefficients. If \( r/s \in \mathbb{Q} \) is in lowest terms (i.e., \( r \) and \( s \) are relatively prime integers) and \( r/s \) is a root of \( p\left( x\right) \),... | Proof: By hypothesis, \( p\left( {r/s}\right) = 0 = {a}_{n}{\left( r/s\right) }^{n} + {a}_{n - 1}{\left( r/s\right) }^{n - 1} + \cdots + {a}_{0} \) . Multiplying through by \( {s}^{n} \) gives\n\n\[ 0 = {a}_{n}{r}^{n} + {a}_{n - 1}{r}^{n - 1}s + \cdots + {a}_{0}{s}^{n}. \]\n\nThus \( {a}_{n}{r}^{n} = s\left( {-{a}_{n -... | Yes |
Proposition 12. Let \( I \) be a proper ideal in the integral domain \( R \) and let \( p\left( x\right) \) be a nonconstant monic polynomial in \( R\left\lbrack x\right\rbrack \) . If the image of \( p\left( x\right) \) in \( \left( {R/I}\right) \left\lbrack x\right\rbrack \) cannot be factored in \( \left( {R/I}\righ... | Proof: Suppose \( p\left( x\right) \) cannot be factored in \( \left( {R/I}\right) \left\lbrack x\right\rbrack \) but that \( p\left( x\right) \) is reducible in \( R\left\lbrack x\right\rbrack \) . As noted at the end of the preceding section this means there are monic, nonconstant polynomials \( a\left( x\right) \) a... | Yes |
Proposition 13. (Eisenstein’s Criterion) Let \( P \) be a prime ideal of the integral domain \( R \) and let \( f\left( x\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} \) be a polynomial in \( R\left\lbrack x\right\rbrack \) (here \( n \geq 1 \) ). Suppose \( {a}_{n - 1},\ldots ,{a}_{1},{a}_{0... | Proof: Suppose \( f\left( x\right) \) were reducible, say \( f\left( x\right) = a\left( x\right) b\left( x\right) \) in \( R\left\lbrack x\right\rbrack \), where \( a\left( x\right) \) and \( b\left( x\right) \) are nonconstant polynomials. Reducing this equation modulo \( P \) and using the assumptions on the coeffici... | Yes |
Corollary 14. (Eisenstein’s Criterion for \( \mathbb{Z}\left\lbrack x\right\rbrack \) ) Let \( p \) be a prime in \( \mathbb{Z} \) and let \( f\left( x\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} \in \mathbb{Z}\left\lbrack x\right\rbrack, n \geq 1 \) . Suppose \( p \) divides \( {a}_{i} \) f... | Proof: This is simply a restatement of Proposition 13 in the case of the prime ideal \( \left( p\right) \) in \( \mathbb{Z} \) together with Corollary 6. | Yes |
Proposition 15. The maximal ideals in \( F\left\lbrack x\right\rbrack \) are the ideals \( \left( {f\left( x\right) }\right) \) generated by irreducible polynomials \( f\left( x\right) \) . In particular, \( F\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\right) \) is a field if and only if \( f\left( x\right)... | Proof: This follows from Proposition 7 of Section 8.2 applied to the Principal Ideal Domain \( F\left\lbrack x\right\rbrack \) . | Yes |
Proposition 16. Let \( g\left( x\right) \) be a nonconstant element of \( F\left\lbrack x\right\rbrack \) and let\n\n\[ g\left( x\right) = {f}_{1}{\left( x\right) }^{{n}_{1}}{f}_{2}{\left( x\right) }^{{n}_{2}}\cdots {f}_{k}{\left( x\right) }^{{n}_{k}} \]\n\nbe its factorization into irreducibles, where the \( {f}_{i}\l... | Proof: This follows from the Chinese Remainder Theorem (Theorem 7.17), since the ideals \( \left( {{f}_{i}{\left( x\right) }^{{n}_{i}}}\right) \) and \( \left( {{f}_{j}{\left( x\right) }^{{n}_{j}}}\right) \) are comaximal if \( {f}_{i}\left( x\right) \) and \( {f}_{j}\left( x\right) \) are distinct (they are relatively... | Yes |
Proposition 17. If the polynomial \( f\left( x\right) \) has roots \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{k} \) in \( F \) (not necessarily distinct), then \( f\left( x\right) \) has \( \left( {x - {\alpha }_{1}}\right) \cdots \left( {x - {\alpha }_{k}}\right) \) as a factor. In particular, a polynomial of d... | Proof: The first statement follows easily by induction from Proposition 9. Since linear factors are irreducible, the second statement follows since \( F\left\lbrack x\right\rbrack \) is a Unique Factorization Domain. | No |
Proposition 18. A finite subgroup of the multiplicative group of a field is cyclic. In particular, if \( F \) is a finite field, then the multiplicative group \( {F}^{ \times } \) of nonzero elements of \( F \) is a cyclic group. | Proof: We give a proof of this result using the Fundamental Theorem of Finitely Generated Abelian Groups (Theorem 3 in Section 5.2). A more number-theoretic proof is outlined in the exercises, or Proposition 5 in Section 6.1 may be used in place of the Fundamental Theorem. By the Fundamental Theorem, the finite subgrou... | No |
Corollary 19. Let \( p \) be a prime. The multiplicative group \( {\left( \mathbb{Z}/p\mathbb{Z}\right) }^{ \times } \) of nonzero residue classes \( {\;\operatorname{mod}\;p} \) is cyclic. | Proof: This is the multiplicative group of the finite field \( \mathbb{Z}/p\mathbb{Z} \) . | No |
Corollary 20. Let \( n \geq 2 \) be an integer with factorization \( n = {p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}\cdots {p}_{r}^{{\alpha }_{r}} \) in \( \mathbb{Z} \), where \( {p}_{1},\ldots ,{p}_{r} \) are distinct primes. We have the following isomorphisms of (multiplicative) groups:\n\n\( \left( \widehat{1}\r... | Proof: This is mainly a matter of collecting previous results. The isomorphism in (1) follows from the Chinese Remainder Theorem (see Corollary 18, Section 7.6). | Yes |
Corollary 22. Every ideal in the polynomial ring \( F\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) with coefficients from a field \( F \) is finitely generated. | If \( I \) is an ideal in \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) generated by a (possibly infinite) set \( \mathcal{S} \) of polynomials, Corollary 22 shows that \( I \) is finitely generated, and in fact \( I \) is generated by a finite number of the polynomials from the set \( \mathcal{S} \) (cf. ... | No |
Theorem 23. Fix a monomial ordering on \( \;R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \;\mathrm{{and}}\;\mathrm{{suppose}}\;\{ {g}_{1},\ldots ,{g}_{m}\} \) is a Gröbner basis for the nonzero ideal \( I \) in \( R \) . Then\n\n(1) Every polynomial \( f \in R \) can be written uniquely in the form\n\n\[ f ... | Proof: Letting \( {f}_{I} = \mathop{\sum }\limits_{{i = 1}}^{m}{q}_{i}{g}_{i} \in I \) in the general polynomial division of \( f \) by \( {g}_{1},\ldots ,{g}_{m} \) immediately gives a decomposition \( f = {f}_{l} + r \) for any generators \( {g}_{1},\ldots ,{g}_{m} \) . Suppose now that \( \left\{ {{g}_{1},\ldots ,{g... | Yes |
Proposition 24. Fix a monomial ordering on \( R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) and let \( I \) be a nonzero ideal in \( R \) . (1) If \( {g}_{1},\ldots ,{g}_{m} \) are any elements of \( I \) such that \( {LT}\left( I\right) = \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\righ... | Proof: Suppose \( {g}_{1},\ldots ,{g}_{m} \in I \) with \( {LT}\left( I\right) = \left( {{LT}\left( {g}_{1}\right) ,\ldots ,{LT}\left( {g}_{m}\right) }\right) \) . We need to see that \( {g}_{1},\ldots ,{g}_{m} \) generate the ideal \( I \) . If \( f \in I \), use general polynomial division to write \( f = \mathop{\su... | Yes |
Lemma 25. Suppose \( {f}_{1},\ldots ,{f}_{m} \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) are polynomials with the same multidegree \( \alpha \) and that the linear combination \( h = {a}_{1}{f}_{1} + \cdots + {a}_{m}{f}_{m} \) with constants \( {a}_{i} \in F \) has strictly smaller multidegree. Then\n\n... | Proof: Write \( {f}_{i} = {c}_{i}{f}_{i}^{\prime } \) where \( {c}_{i} \in F \) and \( {f}_{i}^{\prime } \) is a monic polynomial of multidegree \( \alpha \) . We have\n\n\[ h = \sum {a}_{i}{c}_{i}{f}_{i}^{\prime } = {a}_{1}{c}_{1}\left( {{f}_{1}^{\prime } - {f}_{2}^{\prime }}\right) + \left( {{a}_{1}{c}_{1} + {a}_{2}{... | Yes |
Theorem 27. Fix a monomial ordering on \( R = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then there is a unique reduced Gröbner basis for every nonzero ideal \( I \) in \( R \) . | Proof: By Exercise 15, two reduced bases have the same number of elements and the same leading terms since reduced bases are also minimal bases. If \( G = \{ {g}_{1},\ldots ,{g}_{m}\} \) and \( {G}^{\prime } = \left\{ {{g}_{1}^{\prime },\ldots ,{g}_{m}^{\prime }}\right\} \) are two reduced bases for the same nonzero id... | Yes |
Proposition 29. (Elimination) Suppose \( G = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) is a Gröbner basis for the nonzero ideal \( I \) in \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) with respect to the lexicographic monomial ordering \( {x}_{1} > \cdots > {x}_{n} \) . Then \( G \cap F\left\lbrack {{x... | Proof: Denote \( {G}_{i} = G \cap F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack \) . Then \( {G}_{i} \subseteq {I}_{i} \), so by Proposition 24, to see that \( {G}_{i} \) is a Gröbner basis of \( {I}_{i} \) it suffices to see that \( {LT}\left( {G}_{i}\right) \), the leading terms of the elements in \( {G}_{... | Yes |
Proposition 30. If \( I \) and \( J \) are any two ideals in \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) then \( {tI} + \left( {1 - t}\right) J \) is an ideal in \( F\left\lbrack {t,{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) and \( I \cap J = \left( {{tI} + \left( {1 - t}\right) J}\right) \cap F\left\lbrac... | Proof: First, \( {tI} \) and \( \left( {1 - t}\right) J \) are clearly ideals in \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}, t}\right\rbrack \), so also their sum \( {tI} + \left( {1 - t}\right) J \) is an ideal in \( F\left\lbrack {{x}_{1},\ldots ,{x}_{n}, t}\right\rbrack \) . If \( f \in I \cap J \), then \( f = {tf} ... | Yes |
Proposition 1. (The Submodule Criterion) Let \( R \) be a ring and let \( M \) be an \( R \)-module. A subset \( N \) of \( M \) is a submodule of \( M \) if and only if\n\n(1) \( N \neq \varnothing \), and\n\n(2) \( x + {ry} \in N \) for all \( r \in R \) and for all \( x, y \in N \). | Proof: If \( N \) is a submodule, then \( 0 \in N \) so \( N \neq \varnothing \) . Also \( N \) is closed under addition and is sent to itself under the action of elements of \( R \) . Conversely, suppose (1) and (2) hold. Let \( r = - 1 \) and apply the subgroup criterion (in additive form) to see that \( N \) is a su... | Yes |
A map \( \varphi : M \rightarrow N \) is an \( R \) -module homomorphism if and only if \( \varphi \left( {{rx} + y}\right) = {r\varphi }\left( x\right) + \varphi \left( y\right) \) for all \( x, y \in M \) and all \( r \in R. \) | \( \textit{Proof:}\;\left( 1\right) \;\textit{Certainly}\;\varphi \left( {{rx} + y}\right) = {r\varphi }\left( x\right) + \varphi \left( y\right) \;\textit{if}\;\varphi \;\textit{is an}\;R\textit{-module homomorphism.} \) Conversely, if \( \varphi \left( {{rx} + y}\right) = {r\varphi }\left( x\right) + \varphi \left( y... | Yes |
Proposition 3. Let \( R \) be a ring, let \( M \) be an \( R \) -module and let \( N \) be a submodule of \( M \) . The (additive, abelian) quotient group \( M/N \) can be made into an \( R \) -module by defining an action of elements of \( R \) by \[ r\left( {x + N}\right) = \left( {rx}\right) + N,\;\text{ for all }r ... | Proof: Since \( M \) is an abelian group under + the quotient group \( M/N \) is defined and is an abelian group. To see that the action of the ring element \( r \) on the coset \( x + N \) is well defined, suppose \( x + N = y + N \), i.e., \( x - y \in N \) . Since \( N \) is a (left) \( R \) -submodule, \( r\left( {... | Yes |
Proposition 5. Let \( {N}_{1},{N}_{2},\ldots ,{N}_{k} \) be submodules of the \( R \) -module \( M \) . Then the following are equivalent:\n\n(1) The map \( \pi : {N}_{1} \times {N}_{2} \times \cdots \times {N}_{k} \rightarrow {N}_{1} + {N}_{2} + \cdots + {N}_{k} \) defined by\n\n\[ \pi \left( {{a}_{1},{a}_{2},\ldots ,... | Proof: To prove (1) implies (2), suppose for some \( j \) that (2) fails to hold and let \( {a}_{j} \in \left( {{N}_{1} + \cdots + {N}_{j - 1} + {N}_{j + 1} + \cdots + {N}_{k}}\right) \cap {N}_{j} \), with \( {a}_{j} \neq 0 \) . Then\n\n\[ {a}_{j} = {a}_{1} + \cdots + {a}_{j - 1} + {a}_{j + 1} + \cdots + {a}_{k} \]\n\n... | Yes |
Theorem 6. For any set \( A \) there is a free \( R \) -module \( F\left( A\right) \) on the set \( A \) and \( F\left( A\right) \) satisfies the following universal property: if \( M \) is any \( R \) -module and \( \varphi : A \rightarrow M \) is any map of sets, then there is a unique \( R \) -module homomorphism \(... | Proof: Let \( F\left( A\right) = \{ 0\} \) if \( A = \varnothing \) . If \( A \) is nonempty let \( F\left( A\right) \) be the collection of all set functions \( f : A \rightarrow R \) such that \( f\left( a\right) = 0 \) for all but finitely many \( a \in A \) . Make\n\n\( F\left( A\right) \) into an \( R \) -module b... | No |
Theorem 8. Let \( R \) be a subring of \( S \), let \( N \) be a left \( R \) -module and let \( \iota : N \rightarrow S{ \otimes }_{R}N \) be the \( R \) -module homomorphism defined by \( \iota \left( n\right) = 1 \otimes n \) . Suppose that \( L \) is any left \( S \) - module (hence also an \( R \) -module) and tha... | Proof: Suppose \( \varphi : N \rightarrow L \) is an \( R \) -module homomorphism to the \( S \) -module \( L \) . By the universal property of free modules (Theorem 6 in Section 3) there is a \( \mathbb{Z} \) -module homomorphism from the free \( \mathbb{Z} \) -module \( F \) on the set \( S \times N \) to \( L \) tha... | Yes |
Let \( \iota : N \rightarrow S{ \otimes }_{R}N \) be the \( R \) -module homomorphism in Theorem 8. Then \( N/\ker \iota \) is the unique largest quotient of \( N \) that can be embedded in any \( S \) -module. In particular, \( N \) can be embedded as an \( R \) -submodule of some left \( S \) -module if and only if \... | Proof: The quotient \( N/\ker \iota \) is mapped injectively (by \( \iota \) ) into the \( S \) -module \( S{ \otimes }_{R}N \) . Suppose now that \( \varphi \) is an \( R \) -module homomorphism injecting the quotient \( N/\ker \varphi \) of \( N \) into an \( S \) -module \( L \) . Then, by Theorem 8, \( \ker \iota \... | Yes |
Theorem 10. Suppose \( R \) is a ring with \( 1, M \) is a right \( R \) -module, and \( N \) is a left \( R \) -module. Let \( M{ \otimes }_{R}N \) be the tensor product of \( M \) and \( N \) over \( R \) and let \( \iota : M \times N \rightarrow \) \( M{ \otimes }_{R}N \) be the \( R \) -balanced map defined above.\... | Proof: The proof of (1) is immediate from the properties of \( \iota \) above. For (2), the map \( \varphi \) defines a unique \( \mathbb{Z} \) -module homomorphism \( \widetilde{\varphi } \) from the free group on \( M \times N \) to \( L \) (Theorem 6 in Section 3) such that \( \widetilde{\varphi }\left( {m, n}\right... | Yes |
Corollary 11. Suppose \( D \) is an abelian group and \( {\iota }^{\prime } : M \times N \rightarrow D \) is an \( R \) -balanced map such that\n\n(i) the image of \( {\iota }^{\prime } \) generates \( D \) as an abelian group, and\n\n(ii) every \( R \) -balanced map defined on \( M \times N \) factors through \( {\iot... | Proof: Since \( {\iota }^{\prime } : M \times N \rightarrow D \) is a balanced map, the universal property in (2) of Theorem 10 implies there is a (unique) homomorphism \( f : M{ \otimes }_{R}N \rightarrow D \) with \( {\iota }^{\prime } = f \circ \iota \) . In particular \( {\iota }^{\prime }\left( {m, n}\right) = f\l... | Yes |
Suppose \( R \) is a commutative ring. Let \( M \) and \( N \) be two left \( R \) -modules and let \( M{ \otimes }_{R}N \) be the tensor product of \( M \) and \( N \) over \( R \), where \( M \) is given the standard \( R \) -module structure. Then \( M{ \otimes }_{R}N \) is a left \( R \) -module with\n\n\[ r\left( ... | Proof: We have shown \( M{ \otimes }_{R}N \) is an \( R \) -module and that \( \iota \) is bilinear. It remains only to check that in the bijective correspondence in Theorem 10 the bilinear maps correspond with the \( R \) -module homomorphisms. If \( \varphi : M \times N \rightarrow L \) is bilinear then it is an \( R... | Yes |
Theorem 14. (Associativity of the Tensor Product) Suppose \( M \) is a right \( R \) -module, \( N \) is an \( \left( {R, T}\right) \) -bimodule, and \( L \) is a left \( T \) -module. Then there is a unique isomorphism\n\n\[ \left( {M{ \otimes }_{R}N}\right) { \otimes }_{T}L \cong M{ \otimes }_{R}\left( {N{ \otimes }_... | Proof: Note first that the \( \left( {R, T}\right) \) -bimodule structure on \( N \) makes \( M{ \otimes }_{R}N \) into a right \( T \) -module and \( N{ \otimes }_{T}L \) into a left \( R \) -module, so both sides of the isomorphism are well defined. For each fixed \( l \in L \), the mapping \( \left( {m, n}\right) \m... | Yes |
Theorem 17. (Tensor Products of Direct Sums) Let \( M,{M}^{\prime } \) be right \( R \) -modules and let \( N,{N}^{\prime } \) be left \( R \) -modules. Then there are unique group isomorphisms\n\n\[ \left( {M \oplus {M}^{\prime }}\right) { \otimes }_{R}N \cong \left( {M{ \otimes }_{R}N}\right) \oplus \left( {{M}^{\pri... | Proof: The map \(\left( {M \oplus {M}^{\prime }}\right) \underset{x}{ \times }N \rightarrow \left( {M{ \otimes }_{R}N}\right) \oplus \left( {{M}^{\prime }{ \otimes }_{R}N}\right) \) defined by \(\left( {\left( {m,{m}^{\prime }}\right), n}\right) \mapsto \left( {m \otimes n,{m}^{\prime } \otimes n}\right) \) is well def... | Yes |
Corollary 18. (Extension of Scalars for Free Modules) The module obtained from the free \( R \) -module \( N \cong {R}^{n} \) by extension of scalars from \( R \) to \( S \) is the free \( S \) -module \( {S}^{n} \) , i.e., \[ S{ \otimes }_{R}{R}^{n} \cong {S}^{n} \] as left \( S \) -modules. | Proof: This follows immediately from Theorem 17 and the isomorphism \( S{ \otimes }_{R}R \cong \) \( S \) proved in Example 7 previously. | No |
Corollary 19. Let \( R \) be a commutative ring and let \( M \cong {R}^{s} \) and \( N \cong {R}^{t} \) be free \( R \) -modules with bases \( {m}_{1},\ldots ,{m}_{s} \) and \( {n}_{1},\ldots ,{n}_{t} \), respectively. Then \( M{ \otimes }_{R}N \) is a free \( R \) -module of rank \( {st} \), with basis \( {m}_{i} \oti... | Proof: This follows easily from Theorem 17 and the first example following Corollary 9. | No |
Proposition 20. Suppose \( R \) is a commutative ring and \( M, N \) are left \( R \) -modules, considered with the standard \( R \) -module structures. Then there is a unique \( R \) -module isomorphism\n\n\[ M{ \otimes }_{R}N \cong N{ \otimes }_{R}M \]\n\nmapping \( m \otimes n \) to \( n \otimes m \) . | Proof: The map \( M \times N \rightarrow N \otimes M \) defined by \( \left( {m, n}\right) \mapsto n \otimes m \) is \( R \) -balanced. Hence it induces a unique homomorphism \( \;f\;\mathrm{{from}}\;M \otimes N\;\mathrm{{to}}\;N \otimes M\;\mathrm{{with}}\;f\left( {m \otimes n}\right) = \) \( n \otimes m \) . Similarl... | Yes |
Proposition 21. Let \( R \) be a commutative ring and let \( A \) and \( B \) be \( R \) -algebras. Then the multiplication \( \left( {a \otimes b}\right) \left( {{a}^{\prime } \otimes {b}^{\prime }}\right) = a{a}^{\prime } \otimes b{b}^{\prime } \) is well defined and makes \( A{ \otimes }_{R}B \) into an \( R \) -alg... | Proof: Note first that the definition of an \( R \) -algebra shows that\n\n\[ \begin{matrix} r\left( {a \otimes b}\right) = r \\ a \otimes b = a \\ r \otimes b = a \otimes r \\ b = a \otimes b \\ r = \left( {a \otimes b}\right) r \end{matrix} \]\n\nfor every \( r \in R, a \in A \) and \( b \in B \) . To show that \( A ... | No |
Proposition 22. Let \( A, B \) and \( C \) be \( R \) -modules over some ring \( R \) . Then\n\n(1) The sequence \( 0 \rightarrow A\overset{\psi }{ \rightarrow }B \) is exact (at \( A \) ) if and only if \( \psi \) is injective.\n\n(2) The sequence \( B\overset{\varphi }{ \rightarrow }C \rightarrow 0 \) is exact (at \(... | Proof: The (uniquely defined) homomorphism \( 0 \rightarrow A \) has image \( 0 \) in A. This will be the kernel of \( \psi \) if and only if \( \psi \) is injective. Similarly, the kernel of the (uniquely defined) zero homomorphism \( C \rightarrow 0 \) is all of \( C \), which is the image of \( \varphi \) if and onl... | Yes |
Proposition 24. (The Short Five Lemma) Let \( \alpha ,\beta ,\gamma \) be a homomorphism of short exact sequences\n\n\n\n(1) If \( \alpha \) and \( \gamma \) are injective then so is \( \beta \) .\n\n(2) If \( \alpha... | Proof: We shall prove (1), leaving the proof of (2) as an exercise (and (3) follows immediately from (1) and (2)). Suppose then that \( \alpha \) and \( \gamma \) are injective and suppose \( b \in B \) with \( \beta \left( b\right) = 0 \) . Let \( \psi : A \rightarrow B \) and \( \varphi : B \rightarrow C \) denote th... | No |
Proposition 25. The short exact sequence \( 0 \rightarrow A\overset{\psi }{ \rightarrow }B\overset{\varphi }{ \rightarrow }C \rightarrow 0 \) of \( R \) -modules is split if and only if there is an \( R \) -module homomorphism \( \mu : C \rightarrow B \) such that \( \varphi \circ \mu \) is the identity map on \( C \) ... | Proof: This follows directly from the definitions: if \( \mu \) is given define \( {C}^{\prime } = \mu \left( C\right) \subseteq \) \( B \) and if \( {C}^{\prime } \) is given define \( \mu = {\varphi }^{-1} : C \cong {C}^{\prime } \subseteq B \) . | No |
Proposition 26. Let \( 0 \rightarrow A\overset{\psi }{ \rightarrow }B\overset{\varphi }{ \rightarrow }C \rightarrow 0 \) be a short exact sequence of modules (respectively, \( 1 \rightarrow A\overset{\psi }{ \rightarrow }B\overset{\varphi }{ \rightarrow }C \rightarrow 1 \) a short exact sequence of groups). Then \( B =... | Proof: This is similar to the proof of Proposition 25. If \( \lambda \) is given, define \( {C}^{\prime } = \) \( \ker \lambda \subseteq B \) and if \( {C}^{\prime } \) is given define \( \lambda : B = \psi \left( A\right) \oplus {C}^{\prime } \rightarrow A \) by \( \lambda \left( {\left( {\psi \left( a\right) ,{c}^{\p... | Yes |
Proposition 27. Let \( D, L \) and \( M \) be \( R \) -modules and let \( \psi : L \rightarrow M \) be an \( R \) -module homomorphism. Then the map\n\n\[ \n{\psi }^{\prime } : {\operatorname{Hom}}_{R}\left( {D, L}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {D, M}\right)\n\]\n\n\[ \nf \mapsto {f}^{\prime } = \ps... | Proof: The fact that \( {\psi }^{\prime } \) is a homomorphism is immediate. If \( \psi \) is injective, then distinct homomorphisms \( f \) and \( g \) from \( D \) into \( L \) give distinct homomorphisms \( \psi \circ f \) and \( \psi \circ g \) from \( D \) into \( M \), which is to say that \( {\psi }^{\prime } \)... | Yes |
Theorem 28. Let \( D, L, M \), and \( N \) be \( R \) -modules. If\n\n\[ 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0\;\text{ is exact,}\]\n\nthen the associated sequence\n\n\[ 0 \rightarrow {\operatorname{Hom}}_{R}\left( {D, L}\right) \overset{{\psi }^{\prime }}{ \righ... | Proof: The only item in the first statement that has not already been proved is the exactness of (10) at \( {\operatorname{Hom}}_{R}\left( {D, M}\right) \), i.e., \( \ker {\varphi }^{\prime } = \operatorname{image}{\psi }^{\prime } \) . Suppose \( F : D \rightarrow M \) is an element of \( {\operatorname{Hom}}_{R}\left... | Yes |
Proposition 29. Let \( D, L \) and \( N \) be \( R \) -modules. Then\n\n(1) \( {\operatorname{Hom}}_{R}\left( {D, L \oplus N}\right) \cong {\operatorname{Hom}}_{R}\left( {D, L}\right) \oplus {\operatorname{Hom}}_{R}\left( {D, N}\right) \), and\n\n(2) \( {\operatorname{Hom}}_{R}\left( {L \oplus N, D}\right) \cong {\oper... | Proof: Let \( {\pi }_{1} : L \oplus N \rightarrow L \) be the natural projection from \( L \oplus N \) to \( L \) and similarly let \( {\pi }_{2} \) be the natural projection to \( N \) . If \( f \in {\operatorname{Hom}}_{R}\left( {D, L \oplus N}\right) \) then the compositions \( {\pi }_{1} \circ f \) and \( {\pi }_{2... | No |
Proposition 30. Let \( P \) be an \( R \) -module. Then the following are equivalent:\n\n(1) For any \( R \) -modules \( L, M \), and \( N \), if\n\n\[ 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0 \]\n\nis a short exact sequence, then\n\n\[ 0 \rightarrow {\operatorname{... | Proof: The equivalence of (1) and (2) is a restatement of a result in Theorem 28. Suppose now that (2) is satisfied, and let \( 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }P \rightarrow 0 \) be exact. By (2), the identity map from \( P \) to \( P \) lifts to a homomorphism \( \mu \) m... | Yes |
If \( D \) is an \( R \) -module, then the functor \( {\operatorname{Hom}}_{R}\left( {D, \bot }\right) \) from the category of \( R \) -modules to the category of abelian groups is left exact. It is exact if and only if \( D \) is a projective \( R \) -module. | Note that if \( {\operatorname{Hom}}_{R}\left( {D,\_ }\right) \) takes short exact sequences to short exact sequences, then it takes exact sequences of any length to exact sequences since any exact sequence can be broken up into a succession of short exact sequences. | No |
Theorem 33. Let \( D, L, M \), and \( N \) be \( R \) -modules. If\n\n\[ 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0\;\text{ is exact,}\]\n\nthen the associated sequence\n\n\[ 0 \rightarrow {\operatorname{Hom}}_{R}\left( {N, D}\right) \overset{{\varphi }^{\prime }}{ \r... | Proof: The only item remaining to be proved in the first statement is the exactness of (12) at \( {\operatorname{Hom}}_{R}\left( {M, D}\right) \) . The proof of this statement is very similar to the proof of the corresponding result in Theorem 28 and is left as an exercise. Note also that the injectivity of \( \psi \) ... | No |
Proposition 34. Let \( Q \) be an \( R \) -module. Then the following are equivalent:\n\n(1) For any \( R \) -modules \( L, M \), and \( N \), if\n\n\[ 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0 \]\n\nis a short exact sequence, then\n\n\[ 0 \rightarrow {\operatorname{... | Proof: The equivalence of (1) and (2) is part of Theorem 33. Suppose now that (2) is satisfied and let \( 0 \rightarrow Q\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0 \) be exact. Taking \( L = Q \) and \( f \) the identity map from \( Q \) to itself, it follows by (2) that there is a ... | No |
Proposition 36. Let \( Q \) be an \( R \) -module.\n\n(1) (Baer’s Criterion) The module \( Q \) is injective if and only if for every left ideal \( I \) of \( R \) any \( R \) -module homomorphism \( g : I \rightarrow Q \) can be extended to an \( R \) -module homomorphism \( G : R \rightarrow Q \) . | Proof: If \( Q \) is injective and \( g : I \rightarrow Q \) is an \( R \) -module homomorphism from the nonzero ideal \( I \) of \( R \) into \( Q \), then \( g \) can be extended to an \( R \) -module homomorphism from \( R \) into \( Q \) by Proposition 34(2) applied to the exact sequence \( 0 \rightarrow I \rightar... | No |
Theorem 38. Let \( R \) be a ring with 1 and let \( M \) be an \( R \) -module. Then \( M \) is contained in an injective \( R \) -module. | ## Proof: A proof is outlined in Exercises 15 to 17. | No |
Theorem 39. Suppose that \( D \) is a right \( R \) -module and that \( L, M \) and \( N \) are left \( R \) -modules. If\n\n\[ 0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0\;\text{ is exact,}\]\n\nthen the associated sequence of abelian groups\n\n\[ D{ \otimes }_{R}L\ov... | Proof: For the first statement it remains to prove the exactness of (13) at \( D{ \otimes }_{R}M \) . Since \( \varphi \circ \psi = 0 \), we have\n\n\[ \left( {1 \otimes \varphi }\right) \left( {\sum {d}_{i} \otimes \psi \left( {l}_{i}\right) }\right) = \sum {d}_{i} \otimes \left( {\varphi \circ \psi \left( {l}_{i}\rig... | Yes |
Theorem 43. (Adjoint Associativity) Let \( R \) and \( S \) be rings, let \( A \) be a right \( R \) -module, let \( B \) be an \( \left( {R, S}\right) \) -bimodule and let \( C \) be a right \( S \) -module. Then there is an isomorphism of abelian groups:\n\n\[{\operatorname{Hom}}_{S}\left( {A{ \otimes }_{R}B, C}\righ... | Proof: Suppose \( \varphi : A{ \otimes }_{R}B \rightarrow C \) is a homomorphism. For any fixed \( a \in A \) define the map \( \Phi \left( a\right) \) from \( B \) to \( C \) by \( \Phi \left( a\right) \left( b\right) = \varphi \left( {a \otimes b}\right) \) . It is easy to check that \( \Phi \left( a\right) \) is a h... | Yes |
Corollary 44. If \( R \) is commutative then the tensor product of two projective \( R \) -modules is projective. | Proof: Let \( {P}_{1} \) and \( {P}_{2} \) be projective modules. Then by Corollary 32, \( {\operatorname{Hom}}_{R}\left( {{P}_{2}, \bot }\right) \) is an exact functor from the category of \( R \) -modules to the category of \( R \) -modules. Then the composition \( {\operatorname{Hom}}_{R}\left( {{P}_{1},{\operatorna... | Yes |
Proposition 1. Assume the set \( \mathcal{A} = \left\{ {{v}_{1},{v}_{2},\ldots ,{v}_{n}}\right\} \) spans the vector space \( V \) but no proper subset of \( \mathcal{A} \) spans \( V \) . Then \( \mathcal{A} \) is a basis of \( V \) . In particular, any finitely generated (i.e., finitely spanned) vector space over \( ... | Proof: It is only necessary to prove that \( {v}_{1},{v}_{2},\ldots ,{v}_{n} \) are linearly independent. Suppose \( {\alpha }_{1}{v}_{1} + {\alpha }_{2}{v}_{2} + \cdots + {\alpha }_{n}{v}_{n} = 0 \) where not all of the \( {\alpha }_{i} \) are 0 . By reordering, we may assume that \( {\alpha }_{1} \neq 0 \) and then\n... | Yes |
Corollary 2. Assume the finite set \( \mathcal{A} \) spans the vector space \( V \) . Then \( \mathcal{A} \) contains a basis of \( V \) . | Proof: Any subset \( \mathcal{B} \) of \( \mathcal{A} \) spanning \( V \) such that no proper subset of \( \mathcal{B} \) also spans \( V \) (there clearly exist such subsets) is a basis for \( V \) by Proposition 1. | Yes |
Theorem 3. (A Replacement Theorem) Assume \( \mathcal{A} = \left\{ {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right\} \) is a basis for \( V \) containing \( n \) elements and \( \left\{ {{b}_{1},{b}_{2},\ldots ,{b}_{m}}\right\} \) is a set of linearly independent vectors in \( V \) . Then there is an ordering \( {a}_{1},{a}_{2... | Proof: Proceed by induction on \( k \) . If \( k = 0 \) there is nothing to prove, since \( \mathcal{A} \) is given as a basis for \( V \) . Suppose now that \( \left\{ {{b}_{1},{b}_{2},\ldots ,{b}_{k},{a}_{k + 1},{a}_{k + 2},\ldots ,{a}_{n}}\right\} \) is a basis for \( V \) . Then in particular this is a spanning set... | Yes |
Corollary 5. (Building-Up Lemma) If \( A \) is a set of linearly independent vectors in the finite dimensional space \( V \) then there exists a basis of \( V \) containing \( A \) . | Proof: This is also immediate from Theorem 3, since we can use the elements of \( A \) to successively replace the elements of any given basis for \( V \) (which exists by the assumption that \( V \) is finite dimensional). | No |
Theorem 6. If \( V \) is an \( n \) dimensional vector space over \( F \) , then \( V \cong {F}^{n} \) . In particular, any two finite dimensional vector spaces over \( F \) of the same dimension are isomorphic. | Proof: Let \( {v}_{1},{v}_{2},\ldots ,{v}_{n} \) be a basis for \( V \) . Define the map\n\n\[ \varphi : {F}^{n} \rightarrow V\;\text{by}\;\varphi \left( {{\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n}}\right) = {\alpha }_{1}{v}_{1} + {\alpha }_{2}{v}_{2} + \cdots + {\alpha }_{n}{v}_{n}. \]\n\nThe map \( \varphi \) ... | Yes |
Theorem 7. Let \( V \) be a vector space over \( F \) and let \( W \) be a subspace of \( V \) . Then \( V/W \) is a vector space with \( \dim V = \dim W + \dim V/W \) (where if one side is infinite then both are). | Proof: Suppose \( W \) has dimension \( m \) and \( V \) has dimension \( n \) over \( F \) and let \( {w}_{1},{w}_{2},\ldots ,{w}_{m} \) be a basis for \( W \) . By Corollary 5, these linearly independent elements of \( V \) can be extended to a basis \( {w}_{1},{w}_{2},\ldots ,{w}_{m},{v}_{m + 1},\ldots ,{v}_{n} \) o... | Yes |
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