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Theorem 13.5 An odd prime number \( p \) can be represented by the quadratic form \( {x}^{2} + 2{y}^{2} \) if and only if \( p \equiv 1 \) or 3 \( \left( {\;\operatorname{mod}\;8}\right) \) .
Proof. Since every square is congruent to 0,1 , or 4 modulo 8 , it follows that an odd integer \( n \) is of the form \( {a}^{2} + 2{b}^{2} \) only if \( n \equiv 1 \) or 3 \( \left( {\;\operatorname{mod}\;8}\right) \) .
Yes
Lemma 13.4 Let \( \mathcal{S} \) and \( {\mathcal{S}}^{\prime } \) be finite sets, and let \( \vartheta : \mathcal{S} \rightarrow {\mathcal{S}}^{\prime } \) be a bijection with inverse \( {\vartheta }^{-1} : {\mathcal{S}}^{\prime } \rightarrow \mathcal{S} \) . If \( G\left( s\right) \) is a function defined for all \( ...
Proof. This follows instantly from the fact that \( {\vartheta }^{-1}\left( {\mathcal{S}}^{\prime }\right) = \mathcal{S} \) .
Yes
Theorem 13.6 If \( F\\left( {x, y, z}\\right) \) is a function that is odd in each of the variables \( x, y \), and \( z \), and if \( F\\left( {x, y, z}\\right) = 0 \) for every even integer \( x \), then\n\n\[ \n\\mathop{\\sum }\\limits_{\\substack{{\\left( {u, d,\\delta }\\right) \\in S\\left( n\\right) } \\ {\\delt...
Proof. Since the function \( F\\left( {x, y, z}\\right) \) is odd in the variable \( y \), we have \( F\\left( {x,0, z}\\right) = 0 \) for all \( x \) and \( z \), and\n\n\[ \n\\mathop{\\sum }\\limits_{{\\left( {u, d,\\delta }\\right) \\in \\mathcal{S}\\left( n\\right) }}F\\left( {d + \\delta, u, d - \\delta }\\right)\...
Yes
Theorem 13.7 Let \( f\left( {x, y}\right) \) be a function that is odd in each of the variables \( x \) and \( y \) . For every positive integer \( n \) ,\n\n\[ \n\mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\de...
Proof. We define the function \( F\left( {x, y, z}\right) \) as follows:\n\n\[ \nF\left( {x, y, z}\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x\text{ or }z\text{ is even,} \\ {\left( -1\right) }^{y + \frac{z + 1}{2}}f\left( {x, y}\right) & \text{ if }x\text{ and }z\text{ are odd. } \end{array}\right.\n\]\n\nThen...
Yes
Theorem 14.1 For all positive integers \( s \) and \( n \) ,
\[ \mathop{\sum }\limits_{{\left| u\right| \leq \sqrt{n}}}\left( {n - \left( {s + 1}\right) {u}^{2}}\right) {R}_{s}\left( {n - {u}^{2}}\right) = 0. \] Proof. If \[ n = {x}_{1}^{2} + \cdots + {x}_{s}^{2} + {x}_{s + 1}^{2}, \] then \( {x}_{s + 1}^{2} \leq n \) and so \[ \left| {x}_{s + 1}\right| \leq \sqrt{n} \] For \( j...
Yes
Theorem 14.2 Let \( \Phi \left( n\right) \) be a function defined for all nonnegative integers \( n \) such that\n\n\[ \Phi \left( 0\right) = 1 \]\n\nand\n\n\[ \mathop{\sum }\limits_{{\left| u\right| \leq \sqrt{n}}}\left( {n - \left( {s + 1}\right) {u}^{2}}\right) \Phi \left( {n - {u}^{2}}\right) = 0 \]\n\nfor \( n \ge...
Proof. This follows immediately from Theorem 14.1.
Yes
Theorem 14.5 Let \( n \) be a positive integer,\n\n\[ n = {2}^{a}m \]\n\nwhere \( a \geq 0 \) and \( m \) is odd. Then\n\n\[ {R}_{6}\left( n\right) = 4\left( {{4}^{a + 1} - {\left( -1\right) }^{\left( {m - 1}\right) /2}}\right) \mathop{\sum }\limits_{{m = {d\delta }}}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}{...
Proof. The function \( f\left( {x, y}\right) = {x}^{3}y \) is odd in each of the variables \( x \) and \( y \), and so we can apply (14.3) with \( {k}_{1} = 3 \) and \( {k}_{2} = 1 \) . The left side of this identity is\n\n\[ \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\op...
Yes
Theorem 14.6 For all positive integers \( n \) , \n\n\[ \n\frac{3{n}^{2}}{2} < {R}_{6}\left( n\right) < {40}{n}^{2} \n\]
Proof. Let \( n = {2}^{a}m \), where \( a \geq 0 \) and \( m \) is odd. The infinite series \( \zeta \left( 2\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }{k}^{-2} \) converges, and \( \zeta \left( 2\right) < 2 \) by Exercise 5 . Then \n\n\[ \n\mathop{\sum }\limits_{{{d\delta } = m}}{\left( -1\right) }^{\left( {\d...
Yes
Theorem 14.8 Let \( n \) be a positive integer,\n\n\[ n = {2}^{a}m \]\n\nwhere \( a \geq 0 \) and \( m \) is odd. Then\n\n\[ {R}_{10}\left( n\right) = \frac{4}{5}\left( {{16}^{a + 1} + {\left( -1\right) }^{\left( {m - 1}\right) /2}}\right) \mathop{\sum }\limits_{{m = {d\delta }}}{\left( -1\right) }^{\left( {\delta - 1}...
Proof. By Theorem 14.2, it suffices to find a function \( \Phi \left( n\right) \) such that \( \Phi \left( 0\right) = 1 \) and\n\n\[ \mathop{\sum }\limits_{{\left| x\right| \leq \sqrt{n}}}\left( {n - {11}{x}^{2}}\right) \Phi \left( {n - {x}^{2}}\right) = 0 \]\n\nfor every positive integer \( n \) .\n\nWe begin by apply...
No
Theorem 15.1 For every positive integer \( n \) , \n\n\[ \n{np}\left( n\right) = \mathop{\sum }\limits_{\substack{{{kv} \leq n} \\ {k, v \geq 1} }}{vp}\left( {n - {kv}}\right) \n\]
Proof. The parts in a partition of \( n \) are positive integers \( v \) not exceeding \( n \) . The number of partitions of \( n \) with at least one part equal to \( v \) is \( p\left( {n - v}\right) \) . For any positive integer \( k \), the number of partitions of \( n \) with at least \( k \) parts equal to \( v \...
Yes
Corollary 15.1 Let \( {p}_{k}\left( n\right) \) denote the number of partitions of \( n \) into at most \( k \) parts. Then\n\n\[ \n{p}_{k}\left( n\right) \sim \frac{{n}^{k - 1}}{k!\left( {k - 1}\right) !} + O\left( {n}^{k - 2}\right) .\n\]
Proof. We know that \( {p}_{k}\left( n\right) \) is also equal to the number of partitions of \( n \) into parts no greater than \( k \) . The result follows from Theorem 15.2 applied to the set \( A = \{ 1,2,\ldots, k\} \) .
Yes
Corollary 15.2 Let \( A \) be an infinite set of positive integers with \( \gcd \left( A\right) = \). Then \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\log {p}_{A}\left( n\right) }{\log n} = \infty \]
Proof. For every sufficiently large integer \( k \) there exists a subset \( {F}_{k} \) of \( A \) of cardinality \( k \) such that \( \gcd \left( {F}_{k}\right) = 1 \) . By Theorem 15.2, \[ {p}_{A}\left( n\right) \geq {p}_{{F}_{k}}\left( n\right) = \frac{{n}^{k - 1}}{\left( {k - 1}\right) !\mathop{\prod }\limits_{{a \...
Yes
Lemma 15.1 If \( 0 < \ell \leq n \), then\n\n\[ \sqrt{n} - \frac{\ell }{2\sqrt{n}} - \frac{{\ell }^{2}}{2{n}^{3/2}} \leq \sqrt{n - \ell } < \sqrt{n} - \frac{\ell }{2\sqrt{n}}. \]
Proof. If \( 0 < x \leq 1 \), then\n\n\[ 1 - \frac{x}{2} - \frac{{x}^{2}}{2} \leq {\left( 1 - x\right) }^{1/2} < 1 - \frac{x}{2}. \]\n\nThe result follows by letting \( x = \ell /n \) .
Yes
Lemma 15.2 If \( x > 0 \), then\n\n\[ \frac{{e}^{-x}}{{\left( 1 - {e}^{-x}\right) }^{2}} < \frac{1}{{x}^{2}} \]\n\nIf \( 0 < x \leq 1 \), then\n\n\[ \frac{{e}^{-x}}{{\left( 1 - {e}^{-x}\right) }^{2}} > \frac{1}{{x}^{2}} - 2 \]
Proof. The power series expansion for \( {e}^{x} \) gives\n\n\[ {e}^{x/2} - {e}^{-x/2} = 2\mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{1}{\left( {{2k} + 1}\right) !}{\left( \frac{x}{2}\right) }^{{2k} + 1} \]\n\n\[ = x + {x}^{3}\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{x}^{{2k} - 2}}{\left( {{2k} + 1}\right) !{2...
Yes
Lemma 15.3 Let \( c \) be a positive real number and let \( n \) be a positive integer. Then\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{e}^{-\frac{ck}{2\sqrt{n}}}}{{\left( 1 - {e}^{-\frac{ck}{2\sqrt{n}}}\right) }^{2}} < \frac{2{\pi }^{2}n}{3{c}^{2}} \]\n\nIf \( n \geq {c}^{2}/4 \), then\n\n\[ \mathop{\sum }\...
Proof. Let \( k \) be a positive integer and\n\n\[ x = \frac{ck}{2\sqrt{n}} \]\n\nBy Lemma 15.2,\n\n\[ \frac{{e}^{-\frac{ck}{2\sqrt{n}}}}{{\left( 1 - {e}^{-\frac{ck}{2\sqrt{n}}}\right) }^{2}} = \frac{{e}^{-x}}{{\left( 1 - {e}^{-x}\right) }^{2}} < \frac{1}{{x}^{2}} = \frac{4n}{{c}^{2}{k}^{2}}, \]\n\nand so\n\n\[ \mathop...
Yes
Lemma 16.1 Let \( A \) be a cofinite set of positive integers. Then\n\n\[ \log {p}_{A}\left( n\right) \sim {c}_{0}\sqrt{n} \]
Proof. If \( A \) is cofinite, then \( A \) contains all sufficiently large integers. Choose a positive integer \( \ell > 1 \) such that \( A \) contains all integers greater than \( \ell \), that is,\n\n\[ B = \{ n \geq \ell + 1\} \subseteq A. \]\n\nThen\n\n\[ {p}_{B}\left( n\right) \leq {p}_{A}\left( n\right) \leq p\...
Yes
Note that from the definition, \( \operatorname{Mor}\left( {A, A}\right) \) is always a monoid, that is, a semigroup with identity.
This is quite general; if \( S \) is a monoid, define a category as follows: obj \( \mathbf{C} = \{ S\} \), and set \( \operatorname{Mor}\left( {S, S}\right) = S \) . Composition is the semigroup multiplication. Note further that the singleton obj \( \mathbf{C} \) can, in fact, be replaced by any other singleton \( \{ ...
Yes
Theorem 1.3 If \( X \) is a set, \( \mathbf{C} \) is a concrete category, and \( F,{F}^{\prime } \) are free on \( X \) (with \( \varphi : X \rightarrow \sigma \left( F\right) ,{\varphi }^{\prime } : X \rightarrow \sigma \left( {F}^{\prime }\right) \) ), then \( F \) and \( {F}^{\prime } \) are isomorphic.
Proof: \( F \) being free, \( \exists f \in \operatorname{Mor}\left( {F,{F}^{\prime }}\right) \) with \( {\varphi }^{\prime } = f \circ \varphi .{F}^{\prime } \) being free, \( \exists g \in \operatorname{Mor}\left( {{F}^{\prime }, F}\right) \) with \( \varphi = g \circ {\varphi }^{\prime } \) . Then \( {gf} \in \opera...
Yes
Theorem 1.4 Suppose \( \mathbf{C} \) is a concrete, uniform category. Suppose \( A, B \in \) obj \( \mathbf{C} \), and \( f \in \operatorname{Mor}\left( {A, B}\right) \) . Suppose that, as a map from \( \sigma \left( A\right) \) to \( \sigma \left( B\right) \) , \( f \) is one-to-one. Then there exists \( C \in \operat...
Proof: Let \( {S}^{\prime } \) be a set which is disjoint from \( \sigma \left( A\right) \), and \( {\varphi }^{\prime } \) a bijection from \( \sigma \left( B\right) \) to \( {S}^{\prime } \) . (Such a pair \( \left( {{S}^{\prime },{\varphi }^{\prime }}\right) \) exists with \( {S}^{\prime } \) inside the power set of...
Yes
Proposition 2.1 Suppose \( {A}_{1},{A}_{2}, A \in {}_{R}\mathbf{M} \), and suppose\n\n\[ \n{A}_{1}\overset{{\pi }_{1}}{ \leftarrow }A\overset{{\pi }_{2}}{ \rightarrow }{A}_{2}\;{A}_{1}\overset{{\varphi }_{1}}{ \rightarrow }A\overset{{\varphi }_{2}}{ \leftarrow }{A}_{2} \]\n\nare morphisms satisfying \( {\pi }_{1}{\varp...
Proof: \( {\varphi }_{1} = {i}_{A}{\varphi }_{1} = \left( {{\varphi }_{1}{\pi }_{1} + {\varphi }_{2}{\pi }_{2}}\right) {\varphi }_{1} = {\varphi }_{1}{\pi }_{1}{\varphi }_{1} + {\varphi }_{2}{\pi }_{2}{\varphi }_{1} \)\n\n\[ \n= {\varphi }_{1}{i}_{{A}_{1}} + {\varphi }_{2}{\pi }_{2}{\varphi }_{1} = {\varphi }_{1} + {\v...
Yes
Proposition 2.2 Suppose \( B \in {}_{R}\mathbf{M} \). a) \( R \otimes B \approx B \) as Abelian groups.
Proof: (a) We show \( B \) is a tensor product of \( R \) with \( B \). Set \( \varphi \left( {r, b}\right) = {rb}.\varphi \) is bilinear from \( R \times B \) to \( B \). Suppose \( \psi \) is bilinear from \( R \times B \) to \( G \in \mathbf{{Ab}} \). Then \( \psi \left( {r, b}\right) = \psi \left( {1,{rb}}\right) \...
Yes
Theorem 2.4 (Fundamental Theorem of Tensor Products) Suppose \( A \in {\mathbf{M}}_{R}, B \in {}_{R}\mathbf{M} \), and \( G \in \mathbf{{Ab}} \). Then, as Abelian groups,\n\n\[ \n{\operatorname{Hom}}_{\mathbb{Z}}\left( {A \otimes B, G}\right) \approx {\operatorname{Hom}}_{R}\left( {B,{\operatorname{Hom}}_{\mathbb{Z}}\l...
Proof: We show that each side is isomorphic (in a natural way) to \( \operatorname{Bil}\left( {A, B;G}\right) \) the group of bilinear maps from \( A \times B \) to \( G \) . Note first that \( {\operatorname{Hom}}_{\mathbb{Z}}\left( {A \otimes B, G}\right) \approx \operatorname{Bil}\left( {A, B;G}\right) \) practicall...
Yes
Proposition 2.7 If \( A \in {}_{R}\mathbf{M} \), then \( \operatorname{Hom}\left( {\bullet, A}\right) \) is left exact.
The proof is very similar to the proof of Proposition 2.6(a), except the compositions are on the other side. Details are left to the reader.
No
Proposition 2.10 (Injective Test Lemma) Suppose \( E \in {}_{R}\mathbf{M} \) . Then \( E \) is injective if and only if a filler \( g \) exists for every diagram\n\n![9de6df55-e4f3-42f3-808c-b3f170e8f76c_40_1.jpg](images/9de6df55-e4f3-42f3-808c-b3f170e8f76c_40_1.jpg)\n\nwhere \( I \) is a left ideal in \( R \) .
Proof: The \
No
Corollary 2.11 Suppose \( R \) is a PID, and suppose \( E \in {}_{R}\mathbf{M} \) has the property that \( {rE} = E \) for all \( r \in R, r \neq 0 \) . Then \( E \) is injective.
Proof: Suppose we are given\n\n![9de6df55-e4f3-42f3-808c-b3f170e8f76c_41_3.jpg](images/9de6df55-e4f3-42f3-808c-b3f170e8f76c_41_3.jpg)\n\nwith \( I = {Rr} \) . If \( r = 0 \), then \( g \equiv 0 \) is a filler. If \( r \neq 0 \), then \( f\left( r\right) \in E = {rE} \), so \( \exists a \in E \) with \( {ra} = f\left( r...
No
Theorem 2.12 Suppose \( A \in {\mathbf{M}}_{R} \) is flat, and suppose \( G \in {}_{\mathbb{Z}}\mathbf{M} \) is injective. Then \( {\operatorname{Hom}}_{\mathbb{Z}}\left( {A, G}\right) \) is injective in \( {}_{R}\mathbf{M} \) .
Proof: Suppose\n\n\[ 0 \rightarrow B \rightarrow C \rightarrow D \rightarrow 0 \]\n\nis exact in \( {}_{R}\mathbf{M} \) . Since \( A \) is flat,\n\n\[ 0 \rightarrow A \otimes B \rightarrow A \otimes C \rightarrow A \otimes D \rightarrow 0 \]\n\nis exact. Since \( G \) is injective in \( {}_{\mathbb{Z}}\mathbf{M} \), by...
Yes
Corollary 2.13 (Enough Injectives) If \( A \in {}_{R}\mathbf{M} \), then there exists an injective \( E \in {}_{R}\mathbf{M} \) and a one-to-one homomorphism: \( A \rightarrow E \) .
Proof: As an Abelian group, there exists a divisible Abelian group \( G \) and an injection \( \varphi : A \rightarrow G \) . \( G \) is injective since \( \mathbb{Z} \) is a PID. Hence, as \( R \) - modules, recalling that \( {\operatorname{Hom}}_{\mathbb{Z}} \) denotes homomorphisms of Abelian groups:\n\n\[ A \approx...
No
Proposition 2.14 R is left Noetherian if and only if every direct sum of injectives in \( {}_{R}\mathbf{M} \) is injective.
Proof: Suppose \( R \) is left Noetherian, that is, every left ideal is finitely generated. We use Proposition 2.10, the injective test lemma. Let \( I \) be a left ideal, \( I = \left\langle {{a}_{1},\ldots ,{a}_{n}}\right\rangle \), and suppose \( {E}_{i} \) are injective. If \( \varphi : I \rightarrow \oplus {E}_{i}...
Yes
Proposition 3.1 Suppose \( B,{B}^{\prime } \in {}_{R}\mathbf{M} \), and \( \varphi \in \operatorname{Hom}\left( {B,{B}^{\prime }}\right) \) . Suppose \( \left\langle {{P}_{n},{d}_{n}}\right\rangle \) is a projective resolution of \( B \), and \( \left\langle {{P}_{n}^{\prime },{d}_{n}^{\prime }}\right\rangle \) is a pr...
Proof: Define \( {\varphi }_{0} \) as a filler for\n\n![9de6df55-e4f3-42f3-808c-b3f170e8f76c_52_0.jpg](images/9de6df55-e4f3-42f3-808c-b3f170e8f76c_52_0.jpg)\n\nSuppose \( {\varphi }_{0},\ldots ,{\varphi }_{n} \) have been defined; we define \( {\varphi }_{n + 1} \) recursively. Note first that \( x \in \operatorname{im...
Yes
Example 10 Let \( B = {\mathbb{Z}}_{p}, R = \mathbb{Z} \). As a projective resolution, use \( \cdots \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow {\mathbb{Z}}_{p} \rightarrow 0 \). The map from \( \mathbb{Z} \) to \( \mathbb{Z} \) is multiplication by \( p \). Tensoring with \( A \) and delet...
\[ \cdots \rightarrow 0 \rightarrow 0 \rightarrow \cdots \rightarrow 0 \rightarrow A\overset{\times p}{ \rightarrow }A\overset{\times p}{ \rightarrow }A \rightarrow 0. \] Hence, \( {\operatorname{Tor}}_{0}^{\mathbb{Z}}\left( {A,{\mathbb{Z}}_{p}}\right) \approx A/{pA} \), while \( {\operatorname{Tor}}_{1}^{\mathbb{Z}}\l...
Yes
Proposition 3.2 If \( A \in {\mathbf{M}}_{R}, B \in {}_{R}\mathbf{M}, C \in {}_{R}\mathbf{M} \), then\n\n\[ \text{a)}{\operatorname{Tor}}_{0}\left( {A, B}\right) \approx A \otimes B\text{.} \]\n\n\[ \text{b)}{\operatorname{Ext}}^{0}\left( {B, C}\right) \approx \operatorname{Hom}\left( {B, C}\right) \text{.} \]
Proof: First (a). Since \( A \otimes \) is right exact, the sequence\n\n\[ A \otimes {P}_{1}\xrightarrow[]{A \otimes {d}_{1}}A \otimes {P}_{0}\xrightarrow[]{A \otimes \pi }A \otimes B \rightarrow 0 \]\n\nis exact. Hence, \( A \otimes B \approx A \otimes {P}_{0}/\operatorname{im}\left( {A \otimes {d}_{1}}\right) = {\ope...
Yes
Corollary 3.5 Suppose \( 0 \rightarrow A \rightarrow F \rightarrow {A}^{\prime } \rightarrow 0 \) is short exact in \( {\mathbf{M}}_{R} \), with \( F \) flat. Then \( {\operatorname{Tor}}_{n}\left( {A, B}\right) \approx {\operatorname{Tor}}_{n + 1}\left( {{A}^{\prime }, B}\right) \) for all \( B \in {}_{R}\mathbf{M} \)...
Proof: \( 0 = {\operatorname{Tor}}_{n + 1}\left( {F, B}\right) \rightarrow {\operatorname{Tor}}_{n + 1}\left( {{A}^{\prime }, B}\right) \rightarrow {\operatorname{Tor}}_{n}\left( {A, B}\right) \rightarrow {\operatorname{Tor}}_{n}\left( {F, B}\right) \) \( = 0 \) is exact.
Yes
Corollary 3.6 Suppose \( 0 \rightarrow C \rightarrow E \rightarrow {C}^{\prime } \rightarrow 0 \) is short exact in \( {}_{R}\mathbf{M} \), with \( E \) injective. Then \( {\operatorname{Ext}}^{n}\left( {B,{C}^{\prime }}\right) \approx {\operatorname{Ext}}^{n + 1}\left( {B, C}\right) \) for all \( B \in {}_{R}\mathbf{M...
Proof: \( 0 = {\operatorname{Ext}}^{n}\left( {B, E}\right) \rightarrow {\operatorname{Ext}}^{n}\left( {B,{C}^{\prime }}\right) \rightarrow {\operatorname{Ext}}^{n + 1}\left( {B, C}\right) \rightarrow {\operatorname{Ext}}^{n + 1}\left( {B, E}\right) \) \( = 0 \) is exact.
Yes
Corollary 3.7 Suppose \( B \in {}_{R}\mathbf{M} \), and suppose \( {\operatorname{Tor}}_{1}\left( {R/I, B}\right) = 0 \) for every finitely generated right ideal \( I \) . Then \( B \) is flat.
Proof: Applying Theorem 3.4(a) to \( 0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0 \) yields, in part,\n\n\[ 0 = {\operatorname{Tor}}_{1}\left( {R/I, B}\right) \rightarrow I \otimes B \rightarrow R \otimes B \approx B. \]\n\nHence \( I \otimes B \rightarrow {IB} \) is one-to-one. By the flat test lemma (C...
Yes
Corollary 3.8 Suppose \( B \in {}_{R}\mathbf{M} \) . The following are equivalent:\n\n(i) \( B \) is projective.\n\n(ii) For all \( C \in {}_{R}\mathbf{M} \) and \( n \geq 1,{\operatorname{Ext}}^{n}\left( {B, C}\right) = 0 \) .\n\n(iii) For all \( C \in {}_{R}\mathbf{M},{\operatorname{Ext}}^{1}\left( {B, C}\right) = 0 ...
Proof: (i) \( \Rightarrow \) (ii) is Proposition 3.2(d). (ii) \( \Rightarrow \) (iii) is trivial. Given (iii), if \( 0 \rightarrow C \rightarrow {C}^{\prime } \rightarrow {C}^{\prime \prime } \rightarrow 0 \) is exact in \( {}_{R}\mathbf{M} \), then Theorem 3.4(b) says, in part,\n\n\[ 0 \rightarrow \operatorname{Hom}\l...
Yes
Corollary 3.10 \( {\operatorname{Tor}}_{n}\left( {A, B}\right) \) is isomorphic to the nth homology of a flat resolution of \( A \), tensored with \( B \) (and \( A \otimes B \) deleted). Furthermore, this isomorphism is natural in that if \( \varphi \in \operatorname{Hom}\left( {B,{B}^{\prime }}\right) \), and if \( {...
Proof: The isomorphism was just proved. To see naturality, take a diagram ![9de6df55-e4f3-42f3-808c-b3f170e8f76c_71_1.jpg](images/9de6df55-e4f3-42f3-808c-b3f170e8f76c_71_1.jpg) and tensor it with the flat resolution \( \left\langle {{F}_{k},{d}_{k}}\right\rangle \) of \( A \). The result is a two-layer, three dimension...
Yes
Proposition 3.13 (Second Long Exact Sequence for Ext). Suppose \( 0 \rightarrow B \rightarrow {B}^{\prime } \rightarrow {B}^{\prime \prime } \rightarrow 0 \) is a short exact sequence in \( {}_{R}\mathbf{M} \), and suppose \( C \in {}_{R}\mathbf{M} \) . Then there is a long exact sequence:\n\n\[ \n{\operatorname{Ext}}^...
Proof: Take an injective resolution \( \left\langle {{E}_{i},{d}_{i}}\right\rangle \) of \( C \) and apply \( \operatorname{Hom}\left( {B, \bullet }\right) \) , \( \operatorname{Hom}\left( {{B}^{\prime }, \bullet }\right) \), and \( \operatorname{Hom}\left( {{B}^{\prime \prime }, \bullet }\right) \) to it, deleting the...
Yes
Corollary 3.14 Suppose \( 0 \rightarrow B \rightarrow P \rightarrow {B}^{\prime } \rightarrow 0 \) is short exact in \( {}_{R}\mathbf{M} \) , with \( P \) projective. Then \( {\operatorname{Ext}}^{n}\left( {B, C}\right) \approx {\operatorname{Ext}}^{n + 1}\left( {{B}^{\prime }, C}\right) \) for all \( C \in {}_{R}\math...
Proof: \( 0 = {\operatorname{Ext}}^{n}\left( {P, C}\right) \rightarrow {\operatorname{Ext}}^{n}\left( {B, C}\right) \rightarrow {\operatorname{Ext}}^{n + 1}\left( {{B}^{\prime }, C}\right) \rightarrow {\operatorname{Ext}}^{n + 1}\left( {P, C}\right) \) \( = 0 \) is exact.
Yes
Corollary 3.15 Suppose \( C \in {}_{R}\mathbf{M} \) . The following are equivalent:\n\ni) \( C \) is injective.\n\nii) \( {\operatorname{Ext}}^{n}\left( {B, C}\right) = 0 \) for all \( B \in {}_{R}\mathbf{M} \) and \( n \geq 1 \) .\n\niii) \( {\operatorname{Ext}}^{1}\left( {R/I, C}\right) = 0 \) for all left ideals \( ...
Proof: (i) \( \Rightarrow \) (ii) is Proposition 3.2(d). (ii) \( \Rightarrow \) (iii) is trivial. Given (iii), one has, as part of the long exact sequence,\n\n\[ 0 \rightarrow \operatorname{Hom}\left( {R/I, C}\right) \rightarrow \operatorname{Hom}\left( {R, C}\right) \rightarrow \operatorname{Hom}\left( {I, C}\right) \...
Yes
Example 11 \( {\operatorname{Ext}}_{\mathbb{Z}}^{1}\left( {\mathbb{Q},\mathbb{Z}}\right) \approx \mathbb{R} \) (as groups).
Use an injective resolution of \( \mathbb{Z} \) :\n\n\[ 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow 0 \rightarrow 0 \rightarrow \cdots \]\n\nThe complex is \( 0 \rightarrow \operatorname{Hom}\left( {\mathbb{Q},\mathbb{Q}}\right) \rightarrow \operatorname{Hom}\left( {\ma...
Yes
Proposition 3.16 \( {\operatorname{Tor}}_{n}^{R}\left( {A, B}\right) \approx {\operatorname{Tor}}_{n}^{{R}^{\text{op }}}\left( {B, A}\right) \) .
Proof: \( {\operatorname{Tor}}_{n}^{{R}^{\text{op }}}\left( {B, A}\right) \) is computed using a projective resolution of \( A \) . Since a projective resolution of \( A \) is a flat resolution, this follows from Corollary 3.10.
No
Corollary 3.17 (Second Long Exact Sequence for Tor) If \( 0 \rightarrow B \rightarrow \) \( {B}^{\prime } \rightarrow {B}^{\prime \prime } \rightarrow 0 \) is short exact, then there is a long exact sequence\n\n\[ \rightarrow {\operatorname{Tor}}_{n}^{R}\left( {A, B}\right) \rightarrow {\operatorname{Tor}}_{n}^{R}\left...
Proof: Apply the corollary to Proposition 3 to \( {\operatorname{Tor}}^{{R}^{\mathrm{{op}}}}\left( {\bullet, A}\right) \), then use Proposition 3.16 to put \( A \) on the left.
No
Corollary 3.18 Suppose \( A \in {\mathbf{M}}_{R} \) . The following are equivalent:\n\ni) \( A \) is flat.\n\nii) \( {\operatorname{Tor}}_{n}^{R}\left( {A, B}\right) = 0 \) for all \( B \in {}_{R}\mathbf{M}, n \geq 1 \) .\n\niii) \( {\operatorname{Tor}}_{1}^{R}\left( {A, R/I}\right) = 0 \) for every finitely generated ...
Proof: (i) \( \Rightarrow \) (ii) is Proposition 3.2(c). (ii) \( \Rightarrow \) (iii) is trivial. (iii) \( \Rightarrow \) (i) follows from Proposition 3.16 and Corollary 3.7.
Yes
Corollary 3.19 Suppose \( B \in {}_{R}\mathbf{M} \) . The following are equivalent:\n\ni) \( B \) is flat.\n\nii) \( {\operatorname{Tor}}_{n}^{R}\left( {A, B}\right) = 0 \) for all \( A \in {\mathbf{M}}_{R}, n \geq 1 \) .\n\niii) \( {\operatorname{Tor}}_{1}^{R}\left( {R/J, B}\right) = 0 \) for every finitely generated ...
Proof: Corollary 3.18 plus Proposition 3.16.
No
Corollary 3.20 Suppose \( 0 \rightarrow B \rightarrow F \rightarrow {B}^{\prime } \rightarrow 0 \) is short exact in \( {}_{R}\mathbf{M} \) , with \( F \) flat. Then \( {\operatorname{Tor}}_{n}\left( {A, B}\right) \approx {\operatorname{Tor}}_{n + 1}\left( {A,{B}^{\prime }}\right) \) for all \( A \in {\mathbf{M}}_{R} \...
Proof: \( 0 = {\operatorname{Tor}}_{n + 1}\left( {A, F}\right) \rightarrow {\operatorname{Tor}}_{n + 1}\left( {A,{B}^{\prime }}\right) \rightarrow {\operatorname{Tor}}_{n}\left( {A, B}\right) \rightarrow {\operatorname{Tor}}_{n}\left( {A, F}\right) \) \( = 0 \) is exact.
Yes
Proposition 4.2 Suppose \( 0 \rightarrow D \rightarrow {L}_{1} \rightarrow {L}_{2} \rightarrow \cdots \rightarrow {L}_{n} \rightarrow {D}^{\prime } \rightarrow 0 \) is exact in \( {}_{R}\mathbf{M} \), and \( d \geq 0 \) . a) If \( \mathrm{P} - \dim {L}_{j} \leq d \) for all \( j \), then \( {\operatorname{Ext}}^{k}\lef...
Proof: They all work essentially the same way, so only (a) will be proved. The proof is by induction on \( n \) ; the discussion of the \( n = 1 \) case also carries out the induction step.\n\nGiven \( n = 1 \), we have \( 0 \rightarrow D \rightarrow {L}_{1} \rightarrow {D}^{\prime } \rightarrow 0 \) short exact, and a...
Yes
Corollary 4.3 Suppose \( 0 \rightarrow {Q}_{n} \rightarrow {Q}_{n - 1} \rightarrow \cdots \rightarrow {Q}_{1} \rightarrow {Q}_{0} \rightarrow B \rightarrow 0 \) is exact in \( {}_{R}\mathbf{M} \) . a) If \( \mathrm{P} - \dim {Q}_{j} \leq d \), for all \( j \), then \( \mathrm{P} - \dim B \leq d + n \) . b) If \( \mathr...
Proof: For (a) \( {\mathrm{{Ext}}}^{d + n + 1}\left( {B, C}\right) \approx {\mathrm{{Ext}}}^{d + 1}\left( {{Q}_{n}, C}\right) = 0 \) for all \( C \in {}_{R}\mathbf{M} \) . For (b) \( {\operatorname{Tor}}_{d + n + 1}\left( {A, B}\right) \approx {\operatorname{Tor}}_{d + 1}\left( {A,{Q}_{n}}\right) = 0 \) for all \( A \i...
Yes
Proposition 4.4 (Projective Dimension Theorem) Suppose \( B \in {}_{R}\mathbf{M} \) . The following are equivalent:\n\ni) \( \mathrm{P} - \dim B \leq n \) .\n\nii) The nth kernel of any projective resolution of \( B \) is projective.\n\niii) There exists a projective resolution of \( B \) whose nth kernel is projective...
Proof: (i) \( \Rightarrow \) (ii) follows from Corollary 3.8, since \( {\operatorname{Ext}}^{1}\left( {{K}_{n}, C}\right) \approx \) \( {\operatorname{Ext}}^{n + 1}\left( {B, C}\right) \) if \( {K}_{n} \) is the \( n \) th kernel of a projective resolution of \( B \) . (ii) \( \Rightarrow \) (iii) is trivial. (iii) \( ...
Yes
Proposition 4.5 (Flat Dimension Theorem) Suppose \( B \in {}_{R}\mathbf{M} \) . The following are equivalent:\n\ni) \( \mathrm{F} - \dim B \leq n \) .\n\nii) \( {\operatorname{Tor}}_{n + 1}\left( {R/I, B}\right) = 0 \) for all finitely generated right ideals \( I \) .\n\niii) The nth kernel of any flat resolution of \(...
Proof: Pretty much like Proposition 4.4, except (ii) \( \Rightarrow \) (iii) uses Corollary 3.19.
No
Corollary 4.6 For all \( B \in {}_{R}\mathbf{M},\mathrm{F} \) -dim \( B \leq \mathrm{P} \) -dim \( B \) .
Proof: If P-dim \( B = \infty \), this is immediate. If P-dim \( B = n < \infty \), then the \( n \) th kernel of a projective resolution of \( B \) is projective, hence flat. Thus, F-dim \( B \leq n \) .
Yes
Corollary 4.7 LG-dim \( R \geq \mathrm{W} \) -dim \( R \) and RG-dim \( R \geq \mathrm{W} \) -dim \( R \) .
Proof: Take the supremum; see also Proposition 3.16.
No
Proposition 4.8 (Injective Dimension Theorem) Suppose \( C \in {}_{R}\mathbf{M} \) . The following are equivalent:\n\ni) I-dim \( C \leq n \) .\n\nii) \( {\operatorname{Ext}}^{n + 1}\left( {R/I, C}\right) = 0 \) for all left ideals \( I \) .\n\niii) The nth cokernel of any injective resolution of \( C \) is injective.\...
Proof: Again, like Proposition 4.4, with arrows reversed. (ii) \( \Rightarrow \) (iii) uses Corollary 3.15.
No
Proposition 4.9 (Global Dimension Theorem)\n\n\\[ \n\\text{LG-dim}R = \\sup \\{ \\mathrm{P}\\text{-dim}\\left( {R/I}\\right) : I\\text{a left ideal}\\} \\text{.} \n\\]
Proof: Set \\( n = \\sup \\{ \\mathrm{P} - \\dim \\left( {R/I}\\right) : I \\) a left ideal \\( \\} .n \\leq \\mathrm{{LG}} - \\dim R \\) by definition, so suppose \\( n < \\mathrm{{LG}} \\) -dim \\( R \\) . Then \\( n < \\infty \\), so by Proposition 4.1(a), there exists a \\( C \\in {}_{R}\\mathbf{M} \\) with I-dim \...
No
Corollary 4.10 If LG-dim \( R > 0 \), then LG-dim \( R = 1 + \sup \{ \) P-dim \( I \) : \( I \) a left ideal \( \} \) .
Proof: From \( 0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0 \), for all \( n \geq 1,{\operatorname{Ext}}^{n}\left( {I, C}\right) \approx \) \( {\operatorname{Ext}}^{n + 1}\left( {R/I, C}\right) \) . Hence, if \( R/I \) is not projective, then \( n + 1 > \) P-dim \( \left( {R/I}\right) \) if and only if \...
No
Corollary 4.12 If \( R \) is a PID, then LG-dim \( R \leq 1 \) .
Proof: If \( I = {Ra} \neq 0 \), then \( I \) is isomorphic to \( R \) in \( {}_{R}\mathbf{M} \) since \( R \) is an integral domain \( \left( {I\text{is free on}\{ a\} }\right) \) .
No
Corollary 4.12 If \( R \) is a PID, then LG-dim \( R \leq 1 \) .
Proof: If \( I = {Ra} \neq 0 \), then \( I \) is isomorphic to \( R \) in \( {}_{R}\mathbf{M} \) since \( R \) is an integral domain \( \left( {I\text{is free on}\{ a\} }\right) \) .
No
Proposition 4.16 (Projective Basis Theorem) Suppose \( P \in {}_{R}\mathbf{M} \) . The following are equivalent:\n\ni) \( P \) is projective.\n\nii) If \( P \) is generated by \( \\left\\{ {{s}_{i} : i \\in \\mathcal{I}}\\right\\} \), then there exist \( {\\varphi }_{i} \\in {P}^{ * }, i \\in \\mathcal{I} \) such that ...
Proof: (i) \( \\Rightarrow \) (ii). Suppose \( P \) is generated by \( \\left\\{ {{s}_{i} : i \\in \\mathcal{I}}\\right\\} \) . Let \( F \) be the free module on \( \\mathcal{I},\\pi : F \\rightarrow P \) defined via \( i \\mapsto {s}_{i} \) . Then \( F \\rightarrow P \\rightarrow 0 \) is exact, hence splits since \( P...
Yes
Corollary 4.17 Suppose \( P \) is finitely generated. Then \( P \) is projective if and only if the image of the natural map: \( {P}^{ * } \otimes P \rightarrow \operatorname{Hom}\left( {P, P}\right) \) contains \( {i}_{P} \) .
Proof: \( x = \sum {\varphi }_{i}\left( x\right) \cdot {s}_{i} \) for all \( x \Leftrightarrow \sum {\varphi }_{i} \otimes {s}_{i} \mapsto {i}_{P} \) .
No
Proposition 4.18 Suppose \( B \in {}_{R}\mathbf{M} \) is flat, and suppose \( C \in {}_{R}\mathbf{M} \) is finitely presented. Then \( {C}^{ * } \otimes B \rightarrow \operatorname{Hom}\left( {C, B}\right) \) is an isomorphism.
Proof: We may suppose that we have finitely generated free modules \( {F}_{0} \) and \( {F}_{1} \), and an exact sequence \( {F}_{1} \rightarrow {F}_{0} \rightarrow C \rightarrow 0 \) . \( \operatorname{Hom}\left( {\bullet, R}\right) \) is left exact, so \( 0 \rightarrow {C}^{ * } \rightarrow {F}_{0}^{ * } \rightarrow ...
Yes
Corollary 4.21 Suppose \( R \) is left Noetherian. Then LG-dim \( R = \) W-dim \( R \) .
Proof: The global and weak dimension theorems imply that only dimensions of quotients \( R/I \) need to be examined, and P-dim \( R/I = \mathrm{F} - \dim R/I \) by Proposition 4.20.
No
Corollary 4.22 (Auslander) Suppose \( R \) is both right and left Noetherian. Then LG-dim \( R = \mathrm{{RG}} \)-dim \( R \) .
Proof: Both equal W-dim \( R \) .
No
Lemma 4.27 Every submodule of a semisimple module is semisimple.
Proof: This follows from the modular law (the term is from lattice theory, and its proof is left as an exercise): If \( A, B \), and \( C \) are submodules of \( D \) , with \( A \subset C \), then \( A + \left( {B \cap C}\right) = \left( {A + B}\right) \cap C \) . Taking \( C \subset D \) with \( D \) semisimple and \...
No
Proposition 4.28 Suppose \( R \) is a ring, and \( B \in {}_{R}\mathbf{M} \) . Suppose \( B \) is generated by a set \( S \) together with an element \( x \), but is not generated by \( S \) alone. Then any submodule of \( B \) that contains \( S \), and is maximal (under set inclusion) with respect to the property of ...
Proof: If \( {B}^{\prime } \) is such a submodule, then \( S \subset {B}^{\prime }, x \notin {B}^{\prime } \), and \( \left( {{B}^{\prime } \subset {B}^{\prime \prime } \subset }\right. \n\n\( B,{B}^{\prime \prime } \) a sub-module \( ) \Rightarrow x \in {B}^{\prime \prime } \) . But then \( x \in {B}^{\prime \prime } ...
Yes
Lemma 4.29 Every nonzero semisimple module contains a simple submodule.
Proof: Suppose \( B \) is semisimple, and \( 0 \neq x \in B \) . Let \( {B}^{\prime } \) be the submodule generated by \( x \), and let \( S = \varnothing \) . Using Proposition 4.28, there exists a \( {B}^{\prime \prime } \) which is a maximal submodule of \( {B}^{\prime } \), so that \( {B}^{\prime }/{B}^{\prime \pri...
Yes
Lemma 4.30 Every semisimple module is the sum of its simple submodules.
Proof: Let \( B \) be semisimple, and let \( {B}^{\prime } \) denote the sum of the simple submodules of \( B \) . If \( B \neq {B}^{\prime } \), then \( B = {B}^{\prime } \oplus {B}^{\prime \prime } \) for a nonzero submodule \( {B}^{\prime \prime } \) since \( B \) is semisimple. But \( {B}^{\prime \prime } \) is sem...
Yes
Lemma 4.31 Suppose \( B \) is an \( R \) -module, \( \mathcal{I} \) is an index set, and \( {B}_{i} \) is a simple submodule of \( B \) for each \( i \in \mathcal{I} \) . Also suppose \( B = {\sum }_{\mathcal{I}}{B}_{i} \), that is, \( B \) is the sum (possibly not direct) of the \( {B}_{i} \) . Then for any submodule ...
Proof: Consider all subsets \( \mathcal{J} \) of \( \mathcal{I} \) such that \( {B}^{\prime } + \left( {{\sum }_{\mathcal{J}}{B}_{i}}\right) \) is a direct sum, \( {B}^{\prime } \oplus \left( {{ \oplus }_{\mathcal{J}}{B}_{i}}\right) \), and partially order by set inclusion. (The empty set is such a subset of \( \mathca...
Yes
Proposition 4.32 Suppose \( B \) is an \( R \) -module. The following are equivalent:\n\ni) \( B \) is semisimple.\n\nii) \( B \) is a sum of simple submodules.\n\niii) \( B \) is a direct sum of simple submodules.
Proof: (i) \( \Rightarrow \) (ii) is Lemma 4.30. (ii) \( \Rightarrow \) (iii) is Lemma 4.31, with \( {B}^{\prime } = 0 \) . (iii) \( \Rightarrow \) (ii) is trivial. (ii) \( \Rightarrow \) (i) follows from Lemma 4.31, since Lemma 4.31 says that any submodule \( {B}^{\prime } \) is a direct summand, the other factor bein...
Yes
Proposition 4.33 If \( B \) and \( {B}^{\prime } \) are simple \( R \) -modules, then every nonzero element of \( \operatorname{Hom}\left( {B,{B}^{\prime }}\right) \) is an isomorphism.
Proof: If \( 0 \neq \varphi \in \operatorname{Hom}\left( {B,{B}^{\prime }}\right) \), then \( \varphi \neq 0 \Rightarrow \ker \varphi \neq B \), so \( \ker \varphi = 0 \) since \( B \) is simple. Also, \( \varphi \neq 0 \Rightarrow \operatorname{im}\varphi \neq 0 \), so \( \operatorname{im}\varphi = {B}^{\prime } \) si...
Yes
Lemma 4.35 \( \operatorname{End}\left( {B}^{n}\right) \approx {M}_{n}\left( {\operatorname{End}\left( B\right) }\right) \) .
Note that \( \operatorname{Hom}\left( {{\bigoplus }_{i = 1}^{n}B,{\bigoplus }_{j = 1}^{n}B}\right) \approx {\bigoplus }_{i = 1}^{n}{\bigoplus }_{j = 1}^{n}\operatorname{Hom}\left( {B, B}\right) \) . One need only verify that matrix multiplication gives functional composition on the direct sum.
No
Lemma 4.36 Suppose \( {B}_{1},\ldots ,{B}_{N} \) are pairwise nonisomorphic simple \( R \) - modules. Then\n\n\[ \operatorname{End}\left( {{B}_{1}^{{n}_{1}} \oplus \cdots \oplus {B}_{N}^{{n}_{N}}}\right) \approx {M}_{{n}_{1}}\left( {\operatorname{End}\left( {B}_{1}\right) }\right) \oplus \cdots \oplus {M}_{{n}_{N}}\lef...
Proof:\n\n\[ \operatorname{End}\left( {{B}_{1}^{{n}_{1}} \oplus \cdots \oplus {B}_{N}^{{n}_{N}}}\right) \approx {\bigoplus }_{i = 1}^{N}{\bigoplus }_{j = 1}^{N}\operatorname{Hom}\left( {{B}_{i}^{{n}_{i}},{B}_{j}^{{n}_{j}}}\right) \]\n\n\[ \approx {\bigoplus }_{i = 1}^{N}\operatorname{Hom}\left( {{B}_{i}^{{n}_{i}},{B}_{...
Yes
Lemma 4.37 Suppose \( B \) is a finitely generated semisimple \( R \) -module. Then \( B \) is a finite direct sum of simple modules.
Proof: If \( B = \sum {B}_{i} \) over some index set \( \mathcal{I} \), and \( B \) is generated by \( {x}_{1},\ldots ,{x}_{n} \) , then for each \( j \), there exists a finite subset \( {\mathcal{F}}_{j} \) of \( \mathcal{I} \) such that \( {x}_{j} \in {\sum }_{{\mathcal{F}}_{j}}{B}_{i} \) . Set \( \mathcal{F} = \cup ...
Yes
Lemma 4.39 \( \operatorname{End}\left( {{}_{R}R}\right) \approx {R}^{\mathrm{{op}}} \) .
Proof: Send \( R \) to \( \operatorname{End}\left( {{}_{R}R}\right) \) by sending \( r \) to \( {\varphi }_{r} \), where \( {\varphi }_{r}\left( x\right) = {xr} \) . That pesky problem of writing \( r \) on the right guarantees that if we define \( \Phi \left( r\right) = \) \( {\varphi }_{r} \), then \( \Phi \left( {rs...
No
Theorem 4.40 (Artin–Wedderburn Structure Theorem) Suppose \( R \) is a ring. The following are equivalent:\n\ni) \( \\mathrm{{LG}} - \\dim R = 0 \).\n\nii) Every left \( R \) -module is projective.\n\niii) Every left \( R \) -module is injective.\n\niv) Every left \( R \) -module is semisimple.\n\nv) Every short exact ...
Proof: It's hard to follow the logic without a scorecard. The implications are proved as follows:\n\n![9de6df55-e4f3-42f3-808c-b3f170e8f76c_98_0.jpg](images/9de6df55-e4f3-42f3-808c-b3f170e8f76c_98_0.jpg)\n\nThe implications marked with a check are trivial as statements. Most of the rest are quick. If \( B \\in {}_{R}\\...
Yes
Lemma 4.42 Suppose \( R \) is a ring, and \( I \) is a left ideal. Then \( I \) is a direct summand of \( R \) if and only if \( I \) is principal and generated by an idempotent.
Proof: If \( I = {Re} \), with \( e = {e}^{2} \), set \( f = 1 - e \) . Then \( 1 = e + \left( {1 - e}\right) \in I + {Rf} \) , while if \( r, s \in R \) and \( {re} = s\left( {1 - e}\right) \in I \cap {Rf} \), then \( {re} = s - {se} \Rightarrow s = {se} + {re} \Rightarrow \) \( {se} = \left( {{se} + {re}}\right) e = ...
Yes
Lemma 4.43 Suppose \( R \) is a ring, and suppose \( e \) and \( f \) are idempotents in \( R \) such that \( {ef} = 0 = {fe} \) . Then \( e + f \) is idempotent and \( {Re} + {Rf} = R\left( {e + f}\right) \) .
Proof: \( {\left( e + f\right) }^{2} = {e}^{2} + {ef} + {fe} + {f}^{2} = {e}^{2} + {f}^{2} = e + f \) . Further, \( e + f \in \) \( {Re} + {Rf} \), so \( R\left( {e + f}\right) \subset {Re} + {Rf} \) . Finally, \( e = e\left( {e + f}\right) \in R\left( {e + f}\right) \) and \( f = f\left( {e + f}\right) \in R\left( {e ...
Yes
Lemma 4.44 Suppose \( R \) is a ring. Then \( {Ra} + {Rb} = {Ra} + {Rb}\left( {1 - a}\right) \) .
Proof: \( b\left( {1 - a}\right) = b - {ba} = - {ba} + b \in {Ra} + {Rb} \), so \( {Rb}\left( {1 - a}\right) \subset {Ra} + {Rb} \) . \( {Ra} \subset {Ra} + {Rb} \) trivially, so \( {Ra} + {Rb}\left( {1 - a}\right) \subset {Ra} + {Rb} \) . On the other hand, \( b = {ba} + b\left( {1 - a}\right) \in {Ra} + {Rb}\left( {1...
Yes
Lemma 4.45 Suppose \( R \) is regular. Then every finitely generated left ideal is principal (and generated by an idempotent).
Proof: Suppose we knew the sum of two principal left ideals was principal. Then the set of principal left ideals would be closed under addition of ideals, and so would include every finitely generated ideal. It thus suffices to show that \( {Re} + {Rf} \) is principal if \( e \) and \( f \) are idempotents. (Recall tha...
Yes
Theorem 4.46 (Weak Dimension Zero Characterization) Suppose \( R \) is a ring. The following conditions are equivalent:\n\n\[ \n\\text{i)}\\mathrm{W} - \\dim R = 0\\text{}.\n\]\n\nii) Every left \( R \) -module is flat.\n\niii) For every finitely generated left ideal \( I, R/I \) is projective.\n\niv) \( {\\operatornam...
\( \\textbf{Proof: We prove that (i)} \\Rightarrow \) (ii) \( \\Rightarrow \) (iv) \( \\Rightarrow \) (v) \( \\Rightarrow \) (vi) \( \\Rightarrow \) (iii) \( \\Rightarrow \) (i). (i) \( \\Rightarrow \) (ii) by definition. (ii) \( \\Rightarrow \) (iv) by Corollary 3.19. (iv) \( \\Rightarrow \) (v) trivially. (v) \( \\Ri...
No
Theorem 5.1 Suppose \( F : {}_{S}\mathbf{M} \rightarrow {}_{R}\mathbf{M} \) is an exact, strongly additive covariant functor. Then for all \( B \in {}_{S}\mathbf{M} \) :\n\n\[ \text{a)}\mathrm{P} - {\dim }_{R}F\left( B\right) \leq \mathrm{P} - {\dim }_{S}B + \mathrm{P} - {\dim }_{R}F\left( S\right) \text{, and} \]\n\n\...
Proof: First case: \( B \) is free. Then \( B = { \oplus }_{\mathcal{I}}S \) for some index set \( \mathcal{I} \), and \( F\left( B\right) = { \oplus }_{\mathcal{I}}F\left( S\right) \) since \( F \) is strongly additive. But \( \mathrm{P} - {\dim }_{R}{ \oplus }_{\mathcal{I}}F\left( S\right) \) is P- \( {\dim }_{R}F\le...
Yes
Proposition 5.2 Suppose \( B \in {}_{R}\mathbf{M},{B}^{\prime } \) is a submodule of \( B \), and \( {B}^{\prime \prime } \) is a submodule of \( {B}^{\prime } \) . Then\n\n\[ \mathrm{{SP}} - \dim \left( {{B}^{\prime \prime }, B}\right) = \max \left\{ {\mathrm{{SP}} - \dim \left( {{B}^{\prime \prime },{B}^{\prime }}\ri...
Proof: Since \( {B}^{\prime \prime } \subset C \subset {B}^{\prime } \) or \( {B}^{\prime } \subset C \subset B \Rightarrow {B}^{\prime \prime } \subset C \subset B \), SP-dim \( \left( {B}^{\prime \prime }\right. \) , \( B) \geq \max \left\{ {\mathrm{{SP}}\text{-dim }\left( {{B}^{\prime \prime },{B}^{\prime }}\right) ...
Yes
Corollary 5.3 If LG-dim \( R > 0 \), and \( 0 = {I}_{0} \subset {I}_{1} \subset \cdots \subset {I}_{n} = R \) is a chain of left ideals in \( R \), then LG-dim \( R = 1 + \max \left\{ {\mathrm{{SP}}\text{-dim }\left( {{I}_{j - 1},{I}_{j}}\right) }\right. \) : \( j = 1,\ldots, n\} \) .
Proof: See comments preceding Proposition 5.2.
No
Proposition 5.4 Suppose \( B, C \in {}_{R}\mathbf{M} \) . Then\n\n\[ \n\text{ SP-dim }\left( {B \oplus C}\right) = \max \{ \text{ SP-dim }B,\text{ SP-dim }C\} .\n\]
Proof: SP-dim \( \left( {B \oplus C}\right) = \max \{ \) SP-dim \( \left( {B \oplus 0}\right) , \) SP-dim \( \left( {B \oplus 0, B \oplus C}\right) \) by Proposition 5.2. But any module between \( B \oplus 0 \) and \( B \oplus C \) corresponds to a submodule of \( C \approx B \oplus C/B \oplus 0 \) by the fundamental i...
Yes
Corollary 5.5 If LG-dim \( R > 0 \), and if \( R = {I}_{1} \oplus \cdots \oplus {I}_{n} \) is a direct sum of left ideals, then LG-dim \( R = 1 + \max \left\{ {\mathrm{{SP}}\text{-dim }{I}_{j} : j = 1,\ldots, n}\right\} \) .
Proof: Again, see the comments preceeding Proposition 5.2.
No
Proposition 5.6 Suppose \( \phi : R \rightarrow \widehat{R} \) is a surjective ring homomorphism, and suppose \( \widehat{R} \) is \( R \) -projective. Then \( \mathrm{P} - {\dim }_{R}\widehat{B} = \mathrm{P} - {\dim }_{\widehat{R}}\widehat{B} \) for all \( \widehat{B} \in \) \( {}_{\widehat{R}} \) M.
Proof: \( \mathrm{P} - {\dim }_{R}\widehat{B} \leq \mathrm{P} - {\dim }_{\widehat{R}}\widehat{B} \) by Theorem 5.1. Hence, all \( \widehat{R} \) -projective modules are \( R \) -projective. Suppose \( \widehat{B} \) is \( R \) -projective. There is an \( \widehat{R} \) -projective module \( \widehat{P} \) and a surject...
Yes
Corollary 5.9 Suppose a is central in \( R \), and a is neither a unit nor a zero divisor. Set \( \widehat{R} = R/{Ra} \), and suppose LG-dim \( \widehat{R} < \infty \) . Then LG-dim \( R \) \( \geq 1 + \operatorname{LG} - \dim \widehat{R} \) .
Proof: Take the supremum over \( {}_{\widehat{R}}\mathbf{M} \) in Proposition 5.8.
No
Corollary 5.10 If LG-dim \( R < \infty \) , then LG-dim \( R\left\lbrack x\right\rbrack \geq 1 + \) LG-dim \( R \) .
Proof: Take \( a = x \in \widehat{R} = R\left\lbrack x\right\rbrack : \widehat{R}/\widehat{R}x \approx R \) .
No
Proposition 5.11 For all \( B \in {}_{R}\mathbf{M},\mathrm{P} - {\dim }_{R\left\lbrack x\right\rbrack }B\left\lbrack x\right\rbrack = \mathrm{P} - {\dim }_{R}B \) .
Proof: This is much like the two-functor argument of Theorem 5.7 in the last section. Let \( \widehat{R} = R\left\lbrack x\right\rbrack \) . P- \( {\dim }_{R}\widehat{R} = 0 \), so by Theorem 5.1 we have that \( \mathrm{P} - {\dim }_{R}\widehat{B} \leq \mathrm{P} - {\dim }_{\widehat{R}}\widehat{B} \) for any \( \wideha...
No
Theorem 5.13 If \( R \) is any ring, then LG-dim \( R\left\lbrack x\right\rbrack = 1 + \) LG-dim \( R \) .
Proof: If LG-dim \( R = \infty \), then LG-dim \( R\left\lbrack x\right\rbrack = \infty \), too, by taking the supremum over \( B \in {}_{R}\mathbf{M} \) in Proposition 5.11. This verifies the theorem if LG-dim \( R = \infty \), so suppose LG-dim \( R < \infty \) . Then LG-dim \( R\left\lbrack x\right\rbrack \geq \) \(...
Yes
Corollary 5.14 (Hilbert’s Syzygy Theorem) If \( K \) is a field, then LG-dim \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack = n \) .
Proof: Induction on \( n : K\left\lbrack {{x}_{1},\ldots ,{x}_{n + 1}}\right\rbrack = K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \left\lbrack {x}_{n + 1}\right\rbrack \) for the induction step.
No
Corollary 5.16 Suppose \( R \) is a commutative ring and \( S \) is an admissible multiplicative subset of \( R \) . Then the map \( B \mapsto {S}^{-1}B \) is an exact, strongly additive covariant functor.
Proof: \( {S}^{-1}B \approx {S}^{-1}R \otimes B \) . Tensor products are always right exact, strongly additive covariant functors. It is exact in this case since \( {S}^{-1}R \) is flat.
Yes
Proposition 5.17 Suppose \( R \) is a commutative ring, and \( S \) is an admissible multiplicative subset. Set \( \widehat{R} = {S}^{-1}R,\phi : R \rightarrow \widehat{R} \) the associated ring homomorphism.\n\na) If \( B \in {}_{R}\mathbf{M} \), and \( \psi : B \rightarrow {S}^{-1}B \) is the associated module homomo...
Proof: (a) \( b \in \ker \psi \Leftrightarrow b/1 = 0 \Leftrightarrow \exists {s}^{ * } \in S \) for which \( {s}^{ * }b = 0 \Leftrightarrow \exists {s}^{ * } \in S \) for which \( {s}^{ * } \in \operatorname{ann}\left( b\right) .b/s = \psi \left( c\right) \Leftrightarrow b/s = c/1 \Leftrightarrow \exists {s}^{ * } \in...
Yes
Theorem 5.18 Suppose \( R \) is a commutative ring and \( S \) is an admissible multiplicative subset of \( R \) . We have\n\n\[ \n\text{a)}\mathrm{P} - {\dim }_{{S}^{-1}R}{S}^{-1}B \leq \mathrm{P} - {\dim }_{R}B\text{for any}B \in {}_{R}\mathbf{M}\text{.}\n\]\n\n\[ \n\text{b)}\mathrm{F} - {\dim }_{R}\widehat{B} \leq \...
Proof: (a) Apply Theorem 5.1(a) to the functor\n\n\[ \nB \mapsto {S}^{-1}B : \mathrm{P} - {\dim }_{{S}^{-1}R}{S}^{-1}R = 0,\n\]\n\nyielding the inequality.\n\n(b) Apply Theorem 5.1(b) to the functor \( \widehat{B} \) -as- \( {S}^{-1}R \) -module \( \mapsto \widehat{B} \) -as- \( R \) - module. F- \( {\dim }_{R}{S}^{-1}...
Yes
Proposition 5.19 Suppose \( R \) is commutative and \( A \in {}_{R}\mathbf{M} \) . Suppose \( {A}_{M} \) is \( {R}_{M} \) -flat for every maximal ideal \( M \) . Then \( A \) is flat.
Proof: Suppose \( A \) is not flat. We must produce a maximal ideal \( M \) such that \( {A}_{M} \) is not \( {R}_{M} \) -flat.\n\nTo begin with, there is an exact sequence \( 0 \rightarrow B\overset{j}{ \rightarrow }C \) such that \( A \otimes \) \( j : A \otimes B \rightarrow A \otimes C \) is not one-to-one. Choose ...
Yes
Theorem 5.20 Suppose \( R \) is commutative. Then \( \mathrm{{LG}} \) - \( \dim R \geq \mathrm{{LG}} \) - \( \dim {R}_{P} \) for every prime ideal \( P \) . Moreover, if \( R \) is Noetherian, then LG-dim \( R = \) \( \sup \left\{ {\mathrm{{LG}} - \dim {R}_{M} : M\text{ a maximal ideal }}\right\} \) .
Proof: LG-dim \( R \geq \) LG-dim \( {R}_{P} \) by Theorem 5.18. Hence, LG-dim \( R \geq \) \( \sup \left\{ {\mathrm{{LG}} - \dim {R}_{M} : M}\right. \) a maximal ideal \( \} \) . Suppose \( R \) is Noetherian but this inequality is strict. Set \( n = \sup \left\{ {\mathrm{{LG}}\text{-dim }{R}_{M} : M}\right. \) a maxi...
Yes
Proposition 6.1 Let \( F : {}_{R}\mathbf{M} \rightarrow {}_{S}\mathbf{M} \) be a covariant functor. The following are equivalent:\n\ni) \( F \) is additive.\n\nii) \( F\left( {{B}_{1} \oplus {B}_{2}}\right) \approx F\left( {B}_{1}\right) \oplus F\left( {B}_{2}\right) \) for all \( {B}_{1},{B}_{2} \in {}_{R}\mathbf{M} \...
Proof: (i) \( \Rightarrow \) (ii): To show \( F\left( {{B}_{1} \oplus {B}_{2}}\right) \) is a biproduct of \( F\left( {B}_{1}\right) \) with \( F\left( {B}_{2}\right) \), suppose we have the arrows specifying a biproduct:\n\n\[ {B}_{1}\overset{{\pi }_{1}}{ \leftarrow }{B}_{1} \oplus {B}_{2}\overset{{\pi }_{2}}{ \righta...
Yes
Proposition 6.3 Suppose \( F \) is an additive functor from \( {}_{R}\mathbf{M} \) to \( {}_{S}\mathbf{M} \) . a) If \( F \) is covariant and right exact, then \( {\mathcal{L}}_{0}F\left( B\right) \approx F\left( B\right) \) for all \( B \in \) \( {}_{R} \) M.
Proof: (Sketch) The proofs for Proposition 3.2 can be suitably modified for the covariant cases. To see this, consider (a). Since \( F \) is right exact, \[ F\left( {P}_{1}\right) \overset{F\left( {d}_{1}\right) }{ \rightarrow }F\left( {P}_{0}\right) \overset{F\left( \pi \right) }{ \rightarrow }F\left( B\right) \righta...
Yes
Proposition 6.5 Suppose \( 0 \rightarrow B \rightarrow {B}^{\prime } \rightarrow {B}^{\prime \prime } \rightarrow 0 \) is short exact in \( {}_{R}\mathbf{M} \). a) Given projective resolutions of \( B \) and \( {B}^{\prime \prime } \), there exist maps \( {\pi }^{\prime } : {P}_{0} \oplus {P}_{0}^{\prime \prime } \righ...
Proof: We do (a); (b) is similar and is left as an exercise (Exercise 7). First, \( {\pi }^{\prime } \). We find a filler \( f \) using projectivity of \( {P}_{0}^{\prime \prime } \): Set \( {\pi }^{\prime }\left( {{x}_{0},{x}_{0}^{\prime \prime }}\right) = {j\pi }\left( {x}_{0}\right) + f\left( {x}_{0}^{\prime \prime ...
No
Theorem 6.6 Suppose \( F : {}_{R}\mathbf{M} \rightarrow {}_{S}\mathbf{M} \) is a functor, and \( 0 \rightarrow B \rightarrow {B}^{\prime } \rightarrow {B}^{\prime \prime } \rightarrow 0 \) is short exact in \( {}_{R}\mathbf{M} \).
Proof: These all work out like the long exact sequences of Chapter 3; only (d), the right derived contravariant functors, are new, so we prove (d) only.\n\nWe have a commutative diagram by applying \( F \) to the array in Proposition \( {6.5}\left( \mathrm{\;b}\right) \), and deleting \( F\left( B\right), F\left( {B}^{...
No
Theorem 6.6 Suppose \( F : {}_{R}\mathbf{M} \rightarrow {}_{S}\mathbf{M} \) is a functor, and \( 0 \rightarrow B \rightarrow {B}^{\prime } \rightarrow {B}^{\prime \prime } \rightarrow 0 \) is short exact in \( {}_{R}\mathbf{M} \). d) If \( F \) is contravariant, then there is a long exact sequence \[ \text{-}{\mathcal{...
Proof: These all work out like the long exact sequences of Chapter 3; only (d), the right derived contravariant functors, are new, so we prove (d) only.\n\nWe have a commutative diagram by applying \( F \) to the array in Proposition \( {6.5}\left( \mathrm{\;b}\right) \), and deleting \( F\left( B\right), F\left( {B}^{...
Yes
Corollary 6.7 Suppose \( F : {}_{R}\mathbf{M} \rightarrow {}_{S}\mathbf{M} \) is a functor. Then:\n\na) If \( F \) is covariant, then \( {\mathcal{L}}_{0}F \) is right exact, and \( {\mathcal{L}}_{n}F \) is half exact for all \( n \) .\n\nb) If \( F \) is covariant, then \( {\mathcal{R}}_{0}F \) is left exact, and \( {...
Proof: Again they all work out the same way; the first clause after \
No