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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 \( \left( {\;\operatorname{mod}\;8}\right) \) .
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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 \( \left( {\;\operatorname{mod}\;8}\right) \) .
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Yes
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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 \( s \in \mathcal{S} \), then\n\n\[ \mathop{\sum }\limits_{{s \in \mathcal{S}}}G\left( s\right) = \mathop{\sum }\limits_{{{s}^{\prime } \in {\mathcal{S}}^{\prime }}}G\left( {{\vartheta }^{-1}\left( {s}^{\prime }\right) }\right) . \]\n
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Proof. This follows instantly from the fact that \( {\vartheta }^{-1}\left( {\mathcal{S}}^{\prime }\right) = \mathcal{S} \) .
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Yes
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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\[ \n\\mathop{\\sum }\\limits_{\\substack{{\\left( {u, d,\\delta }\\right) \\in S\\left( n\\right) } \\ {\\delta \\equiv 1\\;\\left( {\\;\\operatorname{mod}\\;2}\\right) } }}F\\left( {\\delta - {2u}, u + d,{2u} + {2d} - \\delta }\\right) = {\\left\{ {T}_{0}\\left( \\ell \\right) \\right\} }_{n = {\\ell }^{2}},\n\]\n\nwhere\n\n\[ \n{T}_{0}\\left( \\ell \\right) = \\mathop{\\sum }\\limits_{{j = 1}}^{\\ell }F\\left( {{2j} - 1,\\ell ,{2j} - 1}\\right)\n\]
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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\[ \n\\mathop{\\sum }\\limits_{{\\left( {u, d,\\delta }\\right) \\in \\mathcal{S}\\left( n\\right) }}F\\left( {d + \\delta, u, d - \\delta }\\right)\n\]\n\n\[ \n= \\mathop{\\sum }\\limits_{\\substack{{\\left( {u, d,\\delta }\\right) \\in \\mathcal{S}\\left( n\\right) } \\ {u \\geq 1} }}F\\left( {d + \\delta, u, d - \\delta }\\right) + \\mathop{\\sum }\\limits_{\\substack{{\\left( {u, d,\\delta }\\right) \\in \\mathcal{S}\\left( n\\right) } \\ {u \\leq - 1} }}F\\left( {d + \\delta, u, d - \\delta }\\right)\n\]\n\n\[ \n= \\mathop{\\sum }\\limits_{\\substack{{\\left( {u, d,\\delta }\\right) \\in \\mathcal{S}\\left( n\\right) } \\ {u \\geq 1} }}F\\left( {d + \\delta, u, d - \\delta }\\right) + \\mathop{\\sum }\\limits_{\\substack{{\\left( {u, d,\\delta }\\right) \\in \\mathcal{S}\\left( n\\right) } \\ {u \\geq 1} }}F\\left( {d + \\delta , - u, d - \\delta }\\right)\n\]\n\n\[ \n= \\mathop{\\sum }\\limits_{\\substack{{\\left( {u, d,\\delta }\\right) \\in \\mathcal{S}\\left( n\\right) } \\ {u \\geq 1} }}F\\left( {d + \\delta, u, d - \\delta }\\right) - \\mathop{\\sum }\\limits_{\\substack{{\\left( {u, d,\\delta }\\right) \\in \\mathcal{S}\\left( n\\right) } \\ {u \\geq 1} }}F\\left( {d + \\delta, u, d - \\delta }\\right)\n\]\n\n\[ \n= 0\\text{.}\n\]\n\nSince \( F\\left( {x, y, z}\\right) = 0 \) for all even integers \( x \), we have\n\n\[ \n\\mathop{\\sum }\\limits_{{\\left( {u, d,\\delta }\\right) \\in \\mathcal{S}\\left( n\\right) }}F\\left( {\\delta - {2u}, u + d,{2u} + {2d} - \\delta }\\right) = \\mathop{\\sum }\\limits_{\\substack{{\\left( {u, d,\\delta }\\right) \\in \\mathcal{S}\\left( n\\right) } \\ {\\delta \\equiv 1\\;\\left( {\\;\\operatorname{mod}\\;2}\\right) } }}F\\left( {\\delta - {2u}, u + d,{2u} + {2d} - \\delta }\\right) .\n\]
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Yes
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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\[ \n\mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}f\left( {\delta - {2u}, u + d}\right) = {\left\{ {T}_{0}\left( \ell \right) \right\} }_{n = {\ell }^{2}}, \n\]\n\nwhere\n\n\[ \n{T}_{0}\left( \ell \right) = \mathop{\sum }\limits_{{j = 1}}^{\ell }{\left( -1\right) }^{j + \ell }f\left( {{2j} - 1,\ell }\right) .\n\]
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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 \( F\left( {x, y, z}\right) \) is a function that is odd in each of the variables \( x, y \), and \( z \), and \( F\left( {x, y, z}\right) = 0 \) for every even integer \( x \) . By Theorem 13.6, we have\n\n\[ \n\mathop{\sum }\limits_{\substack{{{u}^{2} + {\delta \delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}F\left( {\delta - {2u}, u + d,{2u} + {2d} - \delta }\right)\n\]\n\n\[ \n= \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}f\left( {\delta - {2u}, u + d}\right)\n\]\n\n\[ \n= {\left\{ {T}_{0}\left( \ell \right) \right\} }_{n = {\ell }^{2}}\n\]\n\nwhere\n\n\[ \n{T}_{0}\left( \ell \right) = \mathop{\sum }\limits_{{j = 1}}^{\ell }F\left( {{2j} - 1,\ell ,{2j} - 1}\right)\n\]\n\n\[ \n= \mathop{\sum }\limits_{{j = 1}}^{\ell }{\left( -1\right) }^{j + \ell }f\left( {{2j} - 1,\ell }\right) .\n\]\n\nThis completes the proof.
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Yes
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Theorem 14.1 For all positive integers \( s \) and \( n \) ,
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\[ \mathop{\sum }\limits_{{\left| u\right| \leq \sqrt{n}}}\left( {n - \left( {s + 1}\right) {u}^{2}}\right) {R}_{s}\left( {n - {u}^{2}}\right) = 0. \] 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 \( j = 1,\ldots ,{R}_{s + 1}\left( n\right) \), let \[ n = \mathop{\sum }\limits_{{i = 1}}^{{s + 1}}{x}_{i, j}^{2} \] denote the \( {R}_{s + 1}\left( n\right) \) representations of \( n \) as a sum of \( s + 1 \) squares. For \( i = 1,\ldots, s \), we define the map \( {\tau }_{i} \) on the set of \( \left( {s + 1}\right) \) -tuples by \[ {\tau }_{i}\left( {{x}_{1},\ldots ,{x}_{i - 1},{x}_{i},{x}_{i + 1},\ldots ,{x}_{s},{x}_{s + 1}}\right) = \left( {{x}_{1},\ldots ,{x}_{i - 1},{x}_{s + 1},{x}_{i + 1},\ldots ,{x}_{s},{x}_{i}}\right) . \] This is an involution on the set of the \( {R}_{s + 1}\left( n\right) \) representations of \( n \) as a sum of \( s + 1 \) squares, and so \[ \mathop{\sum }\limits_{{j = 1}}^{{{R}_{s + 1}\left( n\right) }}{x}_{s + 1, j}^{2} = \mathop{\sum }\limits_{{j = 1}}^{{{R}_{s + 1}\left( n\right) }}{x}_{i, j}^{2}\;\text{ for }i = 1,\ldots, s. \] Summing over all representations of \( n \), we obtain \[ n{R}_{s + 1}\left( n\right) = \mathop{\sum }\limits_{{j = 1}}^{{{R}_{s + 1}\left( n\right) }}\mathop{\sum }\limits_{{i = 1}}^{{s + 1}}{x}_{i, j}^{2} \] \[ = \mathop{\sum }\limits_{{i = 1}}^{{s + 1}}\mathop{\sum }\limits_{{j = 1}}^{{{R}_{s + 1}\left( n\right) }}{x}_{i, j}^{2} \] \[ = \left( {s + 1}\right) \mathop{\sum }\limits_{{j = 1}}^{{{R}_{s + 1}\left( n\right) }}{x}_{s + 1, j}^{2} \] \[ = \left( {s + 1}\right) \mathop{\sum }\limits_{{\left| u\right| \leq \sqrt{n}}}{u}^{2}{R}_{s}\left( {n - {u}^{2}}\right) \] since for every integer \( u \) with \( \left| u\right| \leq \sqrt{n} \) there are \( {R}_{s}\left( {n - {u}^{2}}\right) \) representations \( n = \mathop{\sum }\limits_{{i = 1}}^{{s + 1}}{x}_{i, j}^{2} \) with \( {x}_{s + 1, j} = u \) . This also implies that \[ {R}_{s + 1}\left( n\right) = \mathop{\sum }\limits_{{\left| u\right| \leq \sqrt{n}}}{R}_{s}\left( {n - {u}^{2}}\right) . \] Then \[ n{R}_{s + 1}\left( n\right) = n\mathop{\sum }\limits_{{\left| u\right| \leq \sqrt{n}}}{R}_{s}\left( {n - {u}^{2}}\right) \] and \[ \mathop{\sum }\limits_{{\left| u\right| \leq \sqrt{n}}}\left( {n - \left( {s + 1}\right) {u}^{2}}\right) {R}_{s}\left( {n - {u}^{2}}\right) = 0. \] This completes the proof.
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Yes
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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 \( n \geq 1 \) . Then\n\n\[ \Phi \left( n\right) = {R}_{s}\left( n\right) \]\n\nfor all \( n \geq 0 \) .
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Proof. This follows immediately from Theorem 14.1.
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Yes
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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\[ {R}_{6}\left( n\right) = 4\left( {{4}^{a + 1} - {\left( -1\right) }^{\left( {m - 1}\right) /2}}\right) \mathop{\sum }\limits_{{m = {d\delta }}}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}{d}^{2}. \]
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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\[ \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}{\left( \delta - 2u\right) }^{3}\left( {u + d}\right) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}\left( {u{\delta }^{3} - 6{u}^{2}{\delta }^{2} + {12}{u}^{3}\delta - 8{u}^{4} + d{\delta }^{3} - {6ud}{\delta }^{2}}\right. \]\n\n\[ \left. {+{12}{u}^{2}{d\delta } - 8{u}^{3}d}\right) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}\left( {d{\delta }^{3} - 6{u}^{2}{\delta }^{2} + {12}{u}^{2}{d\delta } - 8{u}^{4}}\right) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}\left( {{\delta }^{2}\left( {n - 7{u}^{2}}\right) + 4{u}^{2}\left( {{3n} - 5{u}^{2}}\right) }\right) . \]\n\nIf \( n = {\ell }^{2} \), then (by Exercise 3) the right side of the identity is\n\n\[ {T}_{0}\left( \ell \right) = {\left( -1\right) }^{\ell - 1}\ell \mathop{\sum }\limits_{{k = 1}}^{\ell }{\left( -1\right) }^{k - 1}{\left( 2k - 1\right) }^{3} \]\n\n\[ = {\left( -1\right) }^{\ell - 1}\ell {\left( -1\right) }^{\ell - 1}\left( {4{\ell }^{3} - 3\ell }\right) \]\n\n\[ = 4{\ell }^{4} - 3{\ell }^{2} \]\n\n\[ = 4{n}^{2} - {3n}\text{. } \]\n\nTherefore,\n\n\[ \mathop{\sum }\limits_{\substack{{{u}^{2} + \delta = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}\left( {{\delta }^{2}\left( {n - 7{u}^{2}}\right) + 4{u}^{2}\left( {{3n} - 5{u}^{2}}\right) }\right) = {\left\{ 4{n}^{2} - 3n\right\} }_{n = {\ell }^{2}}. \]\n\n(14.10)\n\nNext we apply (14.3) to the function \( f\left( {x, y}\right) = x{y}^{3} \) . The left side of the identity is\n\n\[ \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}\left( {\delta - {2u}}\right) {\left( u + d\right) }^{3} \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}\left( {{u}^{3}\delta + 3{u}^{2}{d\delta } + {3u}{d}^{2}\delta + {d}^{3}\delta - 2{u}^{4} - 6{u}^{3}d}\right. \]\n\n\[ \left. {-6{u}^{2}{d}^{2} - {2u}{d}^{3}}\right)
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Yes
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Theorem 14.6 For all positive integers \( n \) , \n\n\[ \n\frac{3{n}^{2}}{2} < {R}_{6}\left( n\right) < {40}{n}^{2} \n\]
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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\[ \n\mathop{\sum }\limits_{{{d\delta } = m}}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}{d}^{2} = {m}^{2}\mathop{\sum }\limits_{{{d\delta } = m}}\frac{{\left( -1\right) }^{\left( {\delta - 1}\right) /2}}{{\delta }^{2}} \n\] \n\n\[ \n\leq {m}^{2}\mathop{\sum }\limits_{{{d\delta } = m}}\frac{1}{{\delta }^{2}} \n\] \n\n\[ \n< {m}^{2}\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{{k}^{2}} \n\] \n\n\[ \n< 2{m}^{2} \n\] \n\nand \n\n\[ \n{4}^{a + 1} - {\left( -1\right) }^{\left( {m - 1}\right) /2} \leq 4 \cdot {4}^{a} + 1 \n\] \n\n\[ \n\leq 5{\left( {2}^{a}\right) }^{2}\text{.} \n\] \n\nTherefore, \n\n\[ \n{R}_{6}\left( n\right) = 4\left( {{4}^{a + 1} - {\left( -1\right) }^{\left( {m - 1}\right) /2}}\right) \mathop{\sum }\limits_{{{d\delta } = m}}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}{d}^{2} \n\] \n\n\[ \n\leq \;4 \cdot 5{\left( {2}^{a}\right) }^{2}2{m}^{2} \n\] \n\n\[ \n= {40}{n}^{2}\text{.} \n\] \n\nThis gives the upper bound. \n\nTo obtain a lower bound, we have \n\n\[ \n\mathop{\sum }\limits_{{{d\delta } = m}}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}{d}^{2} = {m}^{2}\mathop{\sum }\limits_{{{d\delta } = m}}\frac{{\left( -1\right) }^{\left( {\delta - 1}\right) /2}}{{\delta }^{2}} \n\] \n\n\[ \n\geq {m}^{2}\left( {1 - \mathop{\sum }\limits_{\substack{{\delta \mid m} \\ {\delta > 1} }}\frac{1}{{\delta }^{2}}}\right) \n\] \n\n\[ \n> {m}^{2}\left( {1 - \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{{\left( 2k + 1\right) }^{2}}}\right) \n\] \n\n\[ \n> \frac{{m}^{2}}{2} \n\] \n\nby Exercise 6. Also, \n\n\[ \n{4}^{a + 1} - {\left( -1\right) }^{\left( {m - 1}\right) /2} \geq 4 \cdot {4}^{a} - 1 \n\] \n\n\[ \n\geq 3{\left( {2}^{a}\right) }^{2}\text{.} \n\] \n\nTherefore, \n\n\[ \n{R}_{6}\left( n\right) = 4\left( {{4}^{a + 1} - {\left( -1\right) }^{\left( {m - 1}\right) /2}}\right) \mathop{\sum }\limits_{{{d\delta } = m}}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}{d}^{2} \n\] \n\n\[ \n\geq 3{\left( {2}^{a}\right) }^{2}\frac{{m}^{2}}{2} \n\] \n\n\[ \n= \frac{3{n}^{2}}{2}\text{. } \n\] \n\nThis completes the proof.
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Yes
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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\[ {R}_{10}\left( n\right) = \frac{4}{5}\left( {{16}^{a + 1} + {\left( -1\right) }^{\left( {m - 1}\right) /2}}\right) \mathop{\sum }\limits_{{m = {d\delta }}}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}{d}^{4} \]\n\n\[ + \frac{16}{5}\mathop{\sum }\limits_{{n = {v}^{2} + {w}^{2}}}\left( {{v}^{4} - 3{v}^{2}{w}^{2}}\right) \]
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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 applying identity (14.3) to each of the monomials \( {x}^{5}y,{x}^{3}{y}^{3} \) , and \( x{y}^{5} \) . With \( f\left( {x, y}\right) = {x}^{5}y \), we obtain\n\n\[ \mathop{\sum }\limits_{\substack{{{u}^{2} + {\delta \delta } \equiv n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}{\left( \delta - 2u\right) }^{5}\left( {u + d}\right) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2} \]\n\n\[ \times \left( {\mathop{\sum }\limits_{\substack{{0 \leq k \leq 5} \\ {k \equiv 1{\;(\operatorname{mod}\;2)}} }}\left( \begin{array}{l} 5 \\ k \end{array}\right) {\left( -2\right) }^{k}{\delta }^{5 - k}{u}^{k + 1} + \mathop{\sum }\limits_{\substack{{0 \leq k \leq 5} \\ {k \equiv 0{\;(\operatorname{mod}\;2)}} }}\left( \begin{array}{l} 5 \\ k \end{array}\right) {\left( -2\right) }^{k}d{\delta }^{5 - k}{u}^{k}}\right) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{u}^{2} + {ds} = n} \\ {\delta \equiv 1\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}\left( {d{\delta }^{5} - {10}{\delta }^{4}{u}^{2} + {40d}{\delta }^{3}{u}^{2} - {80}{\delta }^{2}{u}^{4}}\right. \]\n\n\[ \left. {+{80d\delta }{u}^{4} - {32}{u}^{6}}\right) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}\left( {{\delta }^{4}\left( {n - {u}^{2}}\right) - {10}{\delta }^{4}{u}^{2} + {40}{\delta }^{2}{u}^{2}\left( {n - {u}^{2}}\right) }\right. \]\n\n\[ \; - {80}{\delta }^{2}{u}^{4} + {16}{u}^{4}\left( {{5n} - 5{u}^{2}}\right) - {32}{u}^{6}) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{\left( {\delta - 1}\right) /2}\left( {{\delta }^{4}\left( {n - {11}{u}^{2}}\right) + {40}{\delta }^{2}{u}^{2}\left( {n - 3{u}^{2}}\right) }\right. \]\n\n\[ \; + {16}{u}^{4}\left( {{5n} - 7{u}^{2}}\right) ) \]\n\n\[ = {\left\{ \ell \mathop{\sum }\limits_{{j = 1}}^{\ell }{\left( -1\right)
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No
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Theorem 15.1 For every positive integer \( n \) , \n\n\[ \n{np}\left( n\right) = \mathop{\sum }\limits_{\substack{{{kv} \leq n} \\ {k, v \geq 1} }}{vp}\left( {n - {kv}}\right) \n\]
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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 \( v \) is \( p\left( {n - {kv}}\right) \), and so the number of partitions of \( n \) with exactly \( k \) parts equal to \( v \) is \( p\left( {n - {kv}}\right) - p\left( {n - \left( {k + 1}\right) v}\right) \) . Therefore, the number of parts equal to \( v \) that occur in all partitions of \( n \) is \n\n\[ \n\mathop{\sum }\limits_{{k \geq 1}}k\left( {p\left( {n - {kv}}\right) - p\left( {n - \left( {k + 1}\right) v}\right) }\right) = \mathop{\sum }\limits_{{k \geq 1}}p\left( {n - {kv}}\right) . \n\] \n\nWe list the \( p\left( n\right) \) partitions of \( n \) as follows: \n\n\[ \nn = {a}_{1,1} + {a}_{1,2} + \cdots + {a}_{1,{k}_{1}} \n\] \n\n\[ \nn = {a}_{2,1} + {a}_{2,2} + \cdots + {a}_{2,{k}_{2}} \n\] \n\n\[ \nn = {a}_{3,1} + {a}_{3,2} + \cdots + {a}_{3,{k}_{3}} \n\] \n\n\[ \n\vdots \n\] \n\n\[ \nn = {a}_{p\left( n\right) ,1} + {a}_{p\left( n\right) ,2} + \cdots + {a}_{p\left( n\right) ,{k}_{p\left( n\right) }}. \n\] \n\nAdding the \( p\left( n\right) \) rows of this array, we obtain \n\n\[ \n{np}\left( n\right) = \mathop{\sum }\limits_{{i = 1}}^{{p\left( n\right) }}\mathop{\sum }\limits_{{j = 1}}^{{k}_{i}}{a}_{i, j} \n\] \n\n\[ \n= \mathop{\sum }\limits_{{v = 1}}^{n}v\mathop{\sum }\limits_{{{a}_{i, j} = v}}1 \n\] \n\n\[ \n= \mathop{\sum }\limits_{{v = 1}}^{n}v\mathop{\sum }\limits_{{k \geq 1}}p\left( {n - {kv}}\right) \n\] \n\n\[ \n= \mathop{\sum }\limits_{\substack{{{kv} \leq n} \\ {k, v \geq 1} }}{vp}\left( {n - {kv}}\right) \n\] \n\nThis completes the proof.
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Yes
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Corollary 15.1 Let \( {p}_{k}\left( n\right) \) denote the number of partitions of \( n \) into at most \( k \) parts. Then\n\n\[ \n{p}_{k}\left( n\right) \sim \frac{{n}^{k - 1}}{k!\left( {k - 1}\right) !} + O\left( {n}^{k - 2}\right) .\n\]
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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 \( A = \{ 1,2,\ldots, k\} \) .
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Yes
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Corollary 15.2 Let \( A \) be an infinite set of positive integers with \( \gcd \left( A\right) = \). Then \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\log {p}_{A}\left( n\right) }{\log n} = \infty \]
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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, \[ {p}_{A}\left( n\right) \geq {p}_{{F}_{k}}\left( n\right) = \frac{{n}^{k - 1}}{\left( {k - 1}\right) !\mathop{\prod }\limits_{{a \in {F}_{k}}}a} + O\left( {n}^{k - 2}\right) , \] and so there exists a positive constant \( {c}_{k} \) such that \[ {p}_{A}\left( n\right) \geq {c}_{k}{n}^{k - 1} \] for all sufficiently large integers \( n \) . Then \[ \log {p}_{A}\left( n\right) \geq \log {p}_{{F}_{k}}\left( n\right) \geq \left( {k - 1}\right) \log n + \log {c}_{k}. \] Dividing by \( \log n \), we obtain \[ \mathop{\liminf }\limits_{{n \rightarrow \infty }}\frac{\log {p}_{A}\left( n\right) }{\log n} \geq k - 1 \] This is true for all sufficiently large \( k \), and so \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\log {p}_{A}\left( n\right) }{\log n} = \infty \] This completes the proof.
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Yes
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Lemma 15.1 If \( 0 < \ell \leq n \), then\n\n\[ \sqrt{n} - \frac{\ell }{2\sqrt{n}} - \frac{{\ell }^{2}}{2{n}^{3/2}} \leq \sqrt{n - \ell } < \sqrt{n} - \frac{\ell }{2\sqrt{n}}. \]
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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 \( x = \ell /n \) .
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Yes
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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\[ \frac{{e}^{-x}}{{\left( 1 - {e}^{-x}\right) }^{2}} > \frac{1}{{x}^{2}} - 2 \]
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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\[ = x + {x}^{3}\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{x}^{{2k} - 2}}{\left( {{2k} + 1}\right) !{2}^{2k}}. \]\n\nIf \( x > 0 \), then\n\n\[ {e}^{x/2} - {e}^{-x/2} > x \]\n\nand so\n\n\[ \frac{{e}^{-x}}{{\left( 1 - {e}^{-x}\right) }^{2}} = \frac{1}{{\left( {e}^{x/2} - {e}^{-x/2}\right) }^{2}} < \frac{1}{{x}^{2}}. \]\n\nIf \( 0 < x \leq 1 \), then\n\n\[ {e}^{x/2} - {e}^{-x/2} < x + {x}^{3}\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{{2}^{2k}} < x + {x}^{3} < \frac{x}{1 - {x}^{2}}, \]\n\nand so\n\n\[ \frac{{e}^{-x}}{{\left( 1 - {e}^{-x}\right) }^{2}} = \frac{1}{{\left( {e}^{x/2} - {e}^{-x/2}\right) }^{2}} > {\left( \frac{1}{x} - x\right) }^{2} > \frac{1}{{x}^{2}} - 2. \]
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Yes
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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\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{e}^{-\frac{ck}{2\sqrt{n}}}}{{\left( 1 - {e}^{-\frac{ck}{2\sqrt{n}}}\right) }^{2}} > \frac{2{\pi }^{2}n}{3{c}^{2}} - \frac{8\sqrt{n}}{c}. \]
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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\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{e}^{-\frac{ck}{2\sqrt{n}}}}{{\left( 1 - {e}^{-\frac{ck}{2\sqrt{n}}}\right) }^{2}} < \frac{4n}{{c}^{2}}\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{{k}^{2}} = \frac{4{\pi }^{2}n}{6{c}^{2}} = \frac{2{\pi }^{2}n}{3{c}^{2}}. \]\n\nIf \( \sqrt{n} \geq c/2 \) and \( 1 \leq k \leq 2\sqrt{n}/c \), then \( 0 < x \leq 1 \) and, by Lemma 15.2,\n\n\[ \frac{{e}^{-\frac{ck}{2\sqrt{n}}}}{{\left( 1 - {e}^{-\frac{ck}{2\sqrt{n}}}\right) }^{2}} > \frac{1}{{x}^{2}} - 2 = \frac{4n}{{c}^{2}{k}^{2}} - 2. \]\n\nTherefore,\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{e}^{-\frac{ck}{2\sqrt{n}}}}{{\left( 1 - {e}^{-\frac{ck}{2\sqrt{n}}}\right) }^{2}} > \mathop{\sum }\limits_{{k \leq 2\sqrt{n}/c}}\frac{{e}^{-\frac{ck}{2\sqrt{n}}}}{{\left( 1 - {e}^{-\frac{ck}{2\sqrt{n}}}\right) }^{2}} \]\n\n\[ > \mathop{\sum }\limits_{{k \leq 2\sqrt{n}/c}}\left( {\frac{4n}{{c}^{2}{k}^{2}} - 2}\right) \]\n\n\[ \geq \frac{4n}{{c}^{2}}\left( {\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{{k}^{2}} - \mathop{\sum }\limits_{{k > 2\sqrt{n}/c}}\frac{1}{{k}^{2}}}\right) - \frac{4\sqrt{n}}{c} \]\n\n\[ = \frac{2{\pi }^{2}n}{3{c}^{2}} - \frac{4n}{{c}^{2}}\mathop{\sum }\limits_{{k = \left\lbrack {2\sqrt{n}/c}\right\rbrack + 1}}^{\infty }\frac{1}{{k}^{2}} - \frac{4\sqrt{n}}{c}. \]\n\nFor \( k \geq 1 \) we have\n\n\[ \frac{1}{{k}^{2}} < \frac{1}{{k}^{2} - 1/4} = {\int }_{k - 1/2}^{k + 1/2}\frac{dt}{{t}^{2}} \]\n\nand so\n\n\[ \frac{4n}{{c}^{2}}\mathop{\sum }\limits_{{k = \left\lbrack {2\sqrt{n}/c}\right\rbrack + 1}}^{\infty }\frac{1}{{k}^{2}} < \frac{4n}{{c}^{2}}{\int }_{\left\lbrack {2\sqrt{n}/c}\right\rbrack + 1/2}^{\infty }\frac{dt}{{t}^{2}} = \frac{4n}{{c}^{2}}\frac{1}{\left\lbrack {2\sqrt{n}/c}\right\rbrack + 1/2} \]\n\n\[ < \frac{4n}{{c}^{2}}\frac{1}{2\sqrt{n}/c - 1/2} \leq \frac{4\sqrt{n}}{c}. \]\n\nIn the last inequality we used the fact that \( \sqrt{n} \geq c/2 \) . Therefore,\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{e}^{-\frac{ck}{2\sqrt{n}}}}{{\left( 1 - {e}^{-\frac{ck}{2\sqrt{n}}}\right) }^{2}} > \frac{2{\pi }^{2}n}{3{c}^{2}} - \frac{8\sqrt{n}}{c}. \]
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Yes
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Lemma 16.1 Let \( A \) be a cofinite set of positive integers. Then\n\n\[ \log {p}_{A}\left( n\right) \sim {c}_{0}\sqrt{n} \]
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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\[ {p}_{B}\left( n\right) \leq {p}_{A}\left( n\right) \leq p\left( n\right) \]\n\nSince \( \log p\left( n\right) \sim {c}_{0}\sqrt{n} \), it suffices to prove that \( \log {p}_{B}\left( n\right) \sim {c}_{0}\sqrt{n} \) .\n\nConsider the finite set \( F = \{ 1,2,\ldots ,\ell \} \) . Since \( \gcd \left( F\right) = 1 \), Theorem 15.2 implies that there exists a constant \( c \geq 1 \) such that \( {p}_{F}\left( n\right) \leq c{n}^{\ell - 1} \) for all positive integers \( n \) . Each part of an unrestricted partition of \( n \) belongs to \( F \) or to \( B \), and so every partition of \( n \) is uniquely of the form \( n = \left( {n - m}\right) + m \) , where \( n - m \) is a sum of elements of \( F \) and \( m \) is a sum of elements of \( B \) . By Exercise 4, the partition function \( {p}_{B}\left( n\right) \) is increasing for \( n \geq 1 \), and so\n\n\[ p\left( n\right) = \mathop{\sum }\limits_{{m = 0}}^{n}{p}_{F}\left( {n - m}\right) {p}_{B}\left( m\right) \]\n\n\[ \leq c{n}^{\ell - 1}\mathop{\sum }\limits_{{m = 0}}^{n}{p}_{B}\left( m\right) \]\n\n\[ \leq {2c}{n}^{\ell }{p}_{B}\left( n\right) \]\n\n\[ \leq \;{2c}{n}^{\ell }p\left( n\right) . \]\n\nTaking logarithms and dividing by \( {c}_{0}\sqrt{n} \), we have\n\n\[ \frac{\log p\left( n\right) }{{c}_{0}\sqrt{n}} \leq \frac{\log {2c} + \ell \log n}{{c}_{0}\sqrt{n}} + \frac{\log {p}_{B}\left( n\right) }{{c}_{0}\sqrt{n}} \]\n\n\[ \leq \frac{\log {2c} + \left( {\ell - 1}\right) \log n}{{c}_{0}\sqrt{n}} + \frac{\log p\left( n\right) }{{c}_{0}\sqrt{n}}. \]\n\nLetting \( n \) go to infinity, we obtain \( \log {p}_{B}\left( n\right) \sim {c}_{0}\sqrt{n} \) . This completes the proof.
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Yes
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Note that from the definition, \( \operatorname{Mor}\left( {A, A}\right) \) is always a monoid, that is, a semigroup with identity.
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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 \( \{ A\} \), with \( \operatorname{Mor}\left( {A, A}\right) = S \) .
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Yes
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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.
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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 \( {gf} \in \operatorname{Mor}\left( {F, F}\right) \) . Also, \( \varphi = g{\varphi }^{\prime } = \) \( g\left( {f\varphi }\right) = \left( {gf}\right) \varphi \) . The uniqueness of the map (namely \( {i}_{F} \) ) satisfying \( \varphi = {h\varphi } \) implies that \( {gf} = {i}_{F} \) . Similarly, \( {fg} = {i}_{{F}^{\prime }} \) .
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Yes
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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 \( C \in \operatorname{obj}\mathbf{C} \), as well as \( g \in \operatorname{Mor}\left( {A, C}\right) \) , \( h \in \operatorname{Mor}\left( {C, B}\right) \), such that \( f = {hg} \), and\ni) \( h \) is an isomorphism of \( C \) with \( B \) .\nii) \( \sigma \left( A\right) \subset \sigma \left( C\right) \), and \( g\left( x\right) = x \) for all \( x \in \sigma \left( A\right) \) (that is, \( g : A \rightarrow C \) is set inclusion).
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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 \( \sigma \left( A\right) \cup \sigma \left( B\right) \) for reasons of cardinality.) Define a set \( S \) and \( \varphi : \sigma \left( B\right) \rightarrow S \) , as follows:\n\n\[ S = \left\lbrack {{S}^{\prime } \sim {\varphi }^{\prime }\left( {f\left( {\sigma \left( A\right) }\right) }\right) }\right\rbrack \cup \sigma \left( A\right) \]\n\n\[ = {\varphi }^{\prime }\left( {\sigma \left( B\right) \sim f\left( {\sigma \left( A\right) }\right) }\right) \cup \sigma \left( A\right) \]\n\n\[ \varphi \left( y\right) = \left\{ \begin{array}{l} x,\;\text{ if }y = f\left( x\right) \text{ for some }x \in \sigma \left( A\right) \\ {\varphi }^{\prime }\left( y\right) ,\text{ if }y \notin f\left( {\sigma \left( A\right) }\right) \end{array}\right. \]\n\n(Roughly speaking, we \
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Yes
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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 \( {\pi }_{1}{\varphi }_{1} = {i}_{{A}_{1}},{\pi }_{2}{\varphi }_{2} = {i}_{{A}_{2}} \), and \( {\varphi }_{1}{\pi }_{1} + {\varphi }_{2}{\pi }_{2} = {i}_{A} \) . Then \( {\varphi }_{2}{\pi }_{1} = 0,{\varphi }_{1}{\pi }_{2} = 0 \), and \( A \) is both a product and a coproduct of \( {A}_{1} \) and \( {A}_{2} \) .
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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\[ \n= {\varphi }_{1}{i}_{{A}_{1}} + {\varphi }_{2}{\pi }_{2}{\varphi }_{1} = {\varphi }_{1} + {\varphi }_{2}{\pi }_{2}{\varphi }_{1}. \]\n\nHence \( {\varphi }_{2}{\pi }_{2}{\varphi }_{1} = 0 \), so \( 0 = {\pi }_{2}{\varphi }_{2}{\pi }_{2}{\varphi }_{1} = {i}_{{A}_{2}}{\pi }_{2}{\varphi }_{1} = {\pi }_{2}{\varphi }_{1}.{\pi }_{1}{\varphi }_{2} = 0 \) by a similar argument.\n\nNow suppose \( B \in {}_{R}\mathbf{M} \), and \( {\psi }_{i} : {A}_{i} \rightarrow B \) are given. If \( \theta : A \rightarrow B \) makes\n\n\n\ncommute, then\n\n\[ \n\theta = \theta {i}_{A} = \theta \left( {{\varphi }_{1}{\pi }_{1} + {\varphi }_{2}{\pi }_{2}}\right) = \theta {\varphi }_{1}{\pi }_{1} + \theta {\varphi }_{2}{\pi }_{2} = {\psi }_{1}{\pi }_{1} + {\psi }_{2}{\pi }_{2}. \]\n\nBut, in fact, this works. Setting \( \theta = {\psi }_{1}{\pi }_{1} + {\psi }_{2}{\pi }_{2} \) gives \( \theta {\varphi }_{1} = \left( {{\psi }_{1}{\pi }_{1} + }\right. \) \( {\psi }_{2}{\pi }_{2}){\varphi }_{1} = {\psi }_{1}{\pi }_{1}{\varphi }_{1} + {\psi }_{2}{\pi }_{2}{\varphi }_{1} = {\psi }_{1}{i}_{{A}_{1}} + 0 = {\psi }_{1} \), and (similarly) \( \theta {\varphi }_{2} = {\psi }_{2} \) . This shows that \( A \) is a coproduct (with unique filler \( {\psi }_{1}{\pi }_{1} + {\psi }_{2}{\pi }_{2} \) ).\n\nFinally, suppose \( B \in {}_{R}\mathbf{M} \), and \( {\rho }_{i} : B \rightarrow {A}_{i} \) are given. If \( \eta : B \rightarrow A \) makes\n\n\n\ncommute, then\n\n\[ \n\eta = {i}_{A}\eta = \left( {{\varphi }_{1}{\pi }_{1} + {\varphi }_{2}{\pi }_{2}}\right) \eta = {\varphi }_{1}{\pi }_{1}\eta + {\varphi }_{2}{\pi }_{2}\eta = {\varphi }_{1}{\rho }_{1} + {\varphi }_{2}{\rho }_{2}. \]\n\nAs before, setting \( \eta = {\varphi }_{1}{\rho }_{1} + {\varphi }_{2}{\rho }_{2} \) works; details are left to the reader.
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Yes
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Proposition 2.2 Suppose \( B \in {}_{R}\mathbf{M} \). a) \( R \otimes B \approx B \) as Abelian groups.
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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 \( \psi \left( {r, b}\right) = \psi \left( {1,{rb}}\right) \) so \( \theta = \psi \left( {1, \bullet }\right) \) is the unique element of \( {\operatorname{Hom}}_{\mathbb{Z}}\left( {B, G}\right) \) making commutative (i.e., \( \theta \left( {rb}\right) = \psi \left( {r, b}\right) \) ).
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Yes
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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\[ \n{\operatorname{Hom}}_{\mathbb{Z}}\left( {A \otimes B, G}\right) \approx {\operatorname{Hom}}_{R}\left( {B,{\operatorname{Hom}}_{\mathbb{Z}}\left( {A, G}\right) }\right) .\n\]\n\nThis isomorphism is natural in \( A, B \), and \( G \) ; that is,\n\na) If \( \varphi \in {\operatorname{Hom}}_{R}\left( {B,{B}^{\prime }}\right) \), inducing \( {\varphi }_{ * } \in {\operatorname{Hom}}_{\mathbb{Z}}\left( {A \otimes B, A \otimes {B}^{\prime }}\right) \), then\n\n\[ \n{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A \otimes B, G}\right) \approx {\mathrm{{Hom}}}_{R}\left( {B,{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A, G}\right) }\right) \n\]\n\n\[ \n{ \uparrow }_{{\operatorname{Hom}}_{\mathbb{Z}}\left( {{\varphi }_{ * }, G}\right) }\;{ \uparrow }_{{\operatorname{Hom}}_{R}\left( {\varphi ,{\operatorname{Hom}}_{\mathbb{Z}}\left( {A, G}\right) }\right) } \n\]\n\n\[ \n{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A \otimes {B}^{\prime }, G}\right) \approx {\mathrm{{Hom}}}_{R}\left( {{B}^{\prime },{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A, G}\right) }\right) \n\]\ncommutes.\n\nb) If \( \varphi \in {\operatorname{Hom}}_{R}\left( {A,{A}^{\prime }}\right) \), inducing \( {\varphi }_{ * } \in \operatorname{Hom}\left( {A \otimes B,{A}^{\prime } \otimes B}\right) \) and \( {\varphi }^{ * } : \) \( {\operatorname{Hom}}_{\mathbb{Z}}\left( {{A}^{\prime }, G}\right) \rightarrow {\operatorname{Hom}}_{\mathbb{Z}}\left( {A, G}\right) \), then\n\n\[ \n{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A \otimes B, G}\right) \approx {\mathrm{{Hom}}}_{R}\left( {B,{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A, G}\right) }\right) \n\]\n\n\[ \n{ \uparrow }_{{\operatorname{Hom}}_{\mathbb{Z}}\left( {{\varphi }_{ * }, G}\right) }\;{ \uparrow }_{{\operatorname{Hom}}_{R}\left( {B,{\varphi }^{ * }}\right) } \n\]\n\n\[ \n{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {{A}^{\prime } \otimes B, G}\right) \approx {\mathrm{{Hom}}}_{R}\left( {B,{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {{A}^{\prime }, G}\right) }\right) \n\]\n\ncommutes.\n\nc) If \( \varphi \in \operatorname{Hom}\left( {G,{G}^{\prime }}\right) \), then\n\n\[ \n{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A \otimes B, G}\right) \approx {\mathrm{{Hom}}}_{R}\left( {B,{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A, G}\right) }\right) \n\]\n\n\[ \n\downarrow {\operatorname{Hom}}_{\mathbb{Z}}\left( {A \otimes B,\varphi }\right) \; \downarrow {\operatorname{Hom}}_{R}\left( {B,{\operatorname{Hom}}_{\mathbb{Z}}\left( {A,\varphi }\right) }\right) \n\]\n\n\[ \n{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A \otimes B,{G}^{\prime }}\right) \approx {\mathrm{{Hom}}}_{R}\left( {B,{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A,{G}^{\prime }}\right) }\right) \n\]\n\ncommutes.
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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 \( {\operatorname{Hom}}_{\mathbb{Z}}\left( {A \otimes B, G}\right) \approx \operatorname{Bil}\left( {A, B;G}\right) \) practically by definition, for the map \( \eta : A \times B \rightarrow A \otimes B \) gives, via post-composition, a map from \( {\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A \otimes B, G}\right) \) to \( \operatorname{Bil}\left( {A, B;G}\right) \) . However, the universal mapping property of \( A \otimes B \) just says that post-composition is onto (existence of a filler) and one-to-one (uniqueness of fillers). It preserves the group structure, too, since that is inherited from \( G \) . It remains to show that\n\n\[ \n\operatorname{Bil}\left( {A, B;G}\right) \approx {\operatorname{Hom}}_{R}\left( {B,{\operatorname{Hom}}_{\mathbb{Z}}\left( {A, G}\right) }\right) .\n\]\n\nThe correspondence is the usual one:\n\n\[ \n\{ \text{maps} : A \times B \rightarrow G\} \leftrightarrow \{ \text{maps} : B \rightarrow \left( {\text{maps} : A \rightarrow G}\right) \} \n\]\n\n\[ \nf \leftrightarrow g \n\]\n\n\[ \nf\left( {a, b}\right) = \left\lbrack {g\left( b\right) }\right\
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Yes
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Proposition 2.7 If \( A \in {}_{R}\mathbf{M} \), then \( \operatorname{Hom}\left( {\bullet, A}\right) \) is left exact.
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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.
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No
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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\n\nwhere \( I \) is a left ideal in \( R \) .
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Proof: The \
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No
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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.
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Proof: Suppose we are given\n\n\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 \( {ra} = f\left( r\right) \) . Set \( g\left( x\right) = {xa} \) . Then \( g\left( {xr}\right) = {xra} = {xf}\left( r\right) = f\left( {xr}\right) , \) so \( g \) is a filler.
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No
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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 \( {}_{R}\mathbf{M} \) .
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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 Theorem 2.4,\n\n\[ 0 \rightarrow {\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A \otimes D, G}\right) \rightarrow {\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A \otimes C, G}\right) \rightarrow {\mathrm{{Hom}}}_{\mathbb{Z}}\left( {A \otimes B, G}\right) \rightarrow 0 \]\n\nis exact. Using the fundamental theorem of tensor products,\n\n\[ 0 \rightarrow {\operatorname{Hom}}_{R}\left( {D,{\operatorname{Hom}}_{\mathbb{Z}}\left( {A, G}\right) }\right) \rightarrow {\operatorname{Hom}}_{R}\left( {C,{\operatorname{Hom}}_{\mathbb{Z}}\left( {A, G}\right) }\right) \]\n\n\[ \rightarrow {\operatorname{Hom}}_{R}\left( {B,{\operatorname{Hom}}_{\mathbb{Z}}\left( {A, G}\right) }\right) \rightarrow 0 \]\n\nis exact. That is, \( {\operatorname{Hom}}_{R}\left( {\bullet ,{\operatorname{Hom}}_{\mathbb{Z}}\left( {A, G}\right) }\right) \) is an exact functor, so that \( {\operatorname{Hom}}_{\mathbb{Z}}(A \) , \( G) \) is injective.
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Yes
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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: \( A \rightarrow E \) .
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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\[ A \approx {\operatorname{Hom}}_{R}\left( {R, A}\right) \subset {\operatorname{Hom}}_{\mathbb{Z}}\left( {R, A}\right) \approx {\operatorname{Hom}}_{\mathbb{Z}}\left( {R,\varphi \left( A\right) }\right) \subset {\operatorname{Hom}}_{\mathbb{Z}}\left( {R, G}\right) . \]\n\nFurthermore, the injection may be taken to be set inclusion by the pulltab theorem. If \( E \) is injective, and \( E \subset {E}^{\prime } \), then as with projectives (but backwards), a filler for\n\n\n\nturns \( E \) into a direct summand of \( {E}^{\prime } \) . Conversely, if \( E \) is a direct summand of \( {E}^{\prime } \) and \( {E}^{\prime } \) is injective, then \( E \) is injective by Proposition 2.9(b). Thus:\n\n\( E \) is injective if and only if \( E \) is an absolute direct summand, that is,\n\n\( E \) is a direct summand of any module having \( E \) as a submodule.
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No
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Proposition 2.14 R is left Noetherian if and only if every direct sum of injectives in \( {}_{R}\mathbf{M} \) is injective.
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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 \( \varphi : I \rightarrow \oplus {E}_{i} \) , with \( \varphi = \oplus {\varphi }_{i} \), then \( \left\{ {i : {\varphi }_{i}\left( {a}_{j}\right) \neq 0}\right\} \) is finite for \( j = 1,\ldots, n \), so that\n\n\n\nfactors as\n\n\n\nwhere \( { \oplus }^{\prime }{E}_{i} \) is the direct sum (and hence direct product) of finitely many \( {E}_{i} \) . Hence, one has a filler \( g \) for\n\n\n\nby Proposition 2.9(b).\n\nNow suppose \( R \) is not left Noetherian. Then there exists a chain of left ideals \( {I}_{1} \subset {I}_{2} \subset {I}_{3} \subset \cdots \) . Let \( I = \bigcup {I}_{n} \) . Choose injectives \( {E}_{n} \) and injections\n\n\( {\varphi }_{n} : I/{I}_{n} \rightarrow {E}_{n} \) . Define \( \varphi \left( x\right) = \oplus {\varphi }_{n}\left( {x + {I}_{n}}\right) \) if \( x \in I \) . Since any \( x \in I \) is in \( {I}_{n} \) for large \( n,{\varphi }_{n}\left( {x + {I}_{n}}\right) = 0 \) for all but finitely many \( n \) . That is, \( \varphi \) takes values in \( \oplus {E}_{n} \) . If \( g \) is a filler for\n\n\n\nthen \( {\varphi }_{n}\left( {x + {I}_{n}}\right) = {g}_{n}\left( x\right) \) where \( g\left( x\right) = \oplus {g}_{n}\left( x\right) \) . But now \( {\varphi }_{n}\left( {x + {I}_{n}}\right) = \) \( {g}_{n}\left( x\right) = x{g}_{n}\left( 1\right) \) for \( x \notin {I}_{n} \) implies that \( {g}_{n}\left( 1\right) \neq 0 \) for all \( n \) . Thus, \( g\left( 1\right) \) does not take values in \( \oplus {E}_{n} \), a contradiction.
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Yes
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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 projective resolution of \( {B}^{\prime } \) . Then there exist fillers \( {\varphi }_{n} \in \operatorname{Hom}\left( {{P}_{n},{P}_{n}^{\prime }}\right) \) making\n\n\n\ncommutative. Further, if \( {\varphi }_{n}^{\prime } \in \operatorname{Hom}\left( {{P}_{n},{P}_{n}^{\prime }}\right) \) also serve as fillers, then \( {\varphi }_{n} \) and \( {\varphi }_{n}^{\prime } \) are homotopic, that is, there exist \( {D}_{n} : {P}_{n} \rightarrow {P}_{n + 1}^{\prime } \) (with \( {D}_{-1} \equiv 0 \) ) such that \( {\varphi }_{n} - {\varphi }_{n}^{\prime } = {d}_{n + 1}^{\prime }{D}_{n} + {D}_{n - 1}{d}_{n} \) .
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Proof: Define \( {\varphi }_{0} \) as a filler for\n\n\n\nSuppose \( {\varphi }_{0},\ldots ,{\varphi }_{n} \) have been defined; we define \( {\varphi }_{n + 1} \) recursively. Note first that \( x \in \operatorname{im}{d}_{n + 1} \Rightarrow {d}_{n}\left( x\right) = 0 \Rightarrow 0 = {\varphi }_{n - 1}{d}_{n}\left( x\right) = {d}_{n}^{\prime }{\varphi }_{n}\left( x\right) \) . That is, \( {\varphi }_{n}\left( {\operatorname{im}{d}_{n + 1}}\right) \subset \ker {d}_{n}^{\prime } = \operatorname{im}{d}_{n + 1}^{\prime } \) . (This works if \( n > 0 \) ; essentially the same argument works if \( n = 0 \), replacing \( {d}_{0} \) with \( \pi \) .) We can thus find \( {\varphi }_{n + 1} \) as a filler for\n\n\n\nIt remains to show any two fillers are homotopic. The \( {D}_{n} \) are constructed recursively, too. \( {D}_{-1} = 0 \) is given. As above, the recursion step is much like the construction of \( {D}_{0} \) . Note that \( {\pi }^{\prime }{\varphi }_{0} = {\varphi \pi } = {\pi }^{\prime }{\varphi }_{0}^{\prime } \), that is, \( {\pi }^{\prime }\left( {{\varphi }_{0} - {\varphi }_{0}^{\prime }}\right) = \) 0 . Thus \( {\varphi }_{0} - {\varphi }_{0}^{\prime } \) takes values in \( \ker {\pi }^{\prime } = \operatorname{im}{d}_{1}^{\prime } \) . \( {D}_{0} \) is a filler for \n\nWe now suppose we are given \( {D}_{0},\ldots ,{D}_{n} \) . In this case, we know that \( {\varphi }_{n} - {\varphi }_{n}^{\prime } = {d}_{n + 1}^{\prime }{D}_{n} + {D}_{n - 1}{d}_{n} \), so that\n\n\[{d}_{n + 1}^{\prime }\left( {{\varphi }_{n + 1} - {\varphi }_{n + 1}^{\prime } - {D}_{n}{d}_{n + 1}}\right) = {d}_{n + 1}^{\prime }{\varphi }_{n + 1} - {d}_{n + 1}^{\prime }{\varphi }_{n + 1}^{\prime } - {d}_{n + 1}^{\prime }{D}_{n}{d}_{n + 1}\]\n\n\[= {\varphi }_{n}{d}_{n + 1} - {\varphi }_{n}^{\prime }{d}_{n + 1} - {d}_{n + 1}^{\prime }{D}_{n}{d}_{n + 1}\]\n\n\[= \left( {{\varphi }_{n} - {\varphi }_{n}^{\prime } - {d}_{n + 1}^{\prime }{D}_{n}}\right) {d}_{n + 1}\]\n\n\[= {D}_{n - 1}{d}_{n}{d}_{n + 1}\]\n\n\[= 0\text{.}\]\n\nThat is, \( \operatorname{im}\left( {{\varphi }_{n + 1} - {\varphi }_{n + 1}^{\prime } - {D}_{n}{d}_{n + 1}}\right) \subset \ker {d}_{n + 1}^{\prime } = \operatorname{im}{d}_{n + 2}^{\prime } \) . (This works without change if \( n = 0 \) .) Hence, \( {D}_{n + 1} \) can be constructed as a filler for 
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Yes
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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 deleting, we get
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\[ \cdots \rightarrow 0 \rightarrow 0 \rightarrow \cdots \rightarrow 0 \rightarrow A\overset{\times p}{ \rightarrow }A\overset{\times p}{ \rightarrow }A \rightarrow 0. \] Hence, \( {\operatorname{Tor}}_{0}^{\mathbb{Z}}\left( {A,{\mathbb{Z}}_{p}}\right) \approx A/{pA} \), while \( {\operatorname{Tor}}_{1}^{\mathbb{Z}}\left( {A,{\mathbb{Z}}_{p}}\right) \approx \{ x \in A : {px} = 0\} \).
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Yes
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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\[ \text{b)}{\operatorname{Ext}}^{0}\left( {B, C}\right) \approx \operatorname{Hom}\left( {B, C}\right) \text{.} \]
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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, \( A \otimes B \approx A \otimes {P}_{0}/\operatorname{im}\left( {A \otimes {d}_{1}}\right) = {\operatorname{Tor}}_{0}\left( {A, B}\right) \). The proof of (b) is similar, since \( \operatorname{Hom}\left( {\bullet, C}\right) \) is left exact.
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Yes
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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 \( B \in {}_{R}\mathbf{M} \) and \( n \geq 1 \) .
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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.
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Yes
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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 \( B \in {}_{R}\mathbf{M} \) and \( n \geq 1 \) .
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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.
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Yes
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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.
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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 (Chapter 2, Exercise 11), \( B \) is flat.
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Yes
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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 \( C \in {}_{R}\mathbf{M},{\operatorname{Ext}}^{1}\left( {B, C}\right) = 0 \) .
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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\[ 0 \rightarrow \operatorname{Hom}\left( {B, C}\right) \rightarrow \operatorname{Hom}\left( {B,{C}^{\prime }}\right) \rightarrow \operatorname{Hom}\left( {B,{C}^{\prime \prime }}\right) \rightarrow {\operatorname{Ext}}^{1}\left( {B, C}\right) = 0 \]\n\nis exact, i.e. \( \operatorname{Hom}\left( {B, \bullet }\right) \) is an exact functor. Hence \( B \) is projective.
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Yes
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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 \( {H}_{n} \) denotes the nth homology of a chain complex, and \( \left\langle {{F}_{k},{d}_{k}}\right\rangle \) is a flat resolution of \( A \), then  \( \mathit{{commutes}} \).
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Proof: The isomorphism was just proved. To see naturality, take a diagram  and tensor it with the flat resolution \( \left\langle {{F}_{k},{d}_{k}}\right\rangle \) of \( A \). The result is a two-layer, three dimensional diagram whose typical cube looks like:  Now simply observe that a zigzag for \( \left\langle {{F}_{i} \otimes {P}_{j}}\right\rangle \) maps to a zigzag for \( \langle {F}_{i} \otimes {P}_{j}^{\prime }\rangle \) ：  The first homology, \( {Z}_{1}/{B}_{1} \), works even more easily, and these are all that appear here.
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Yes
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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\[ \n{\operatorname{Ext}}^{n}\left( {{B}^{\prime \prime }, C}\right) \rightarrow {\operatorname{Ext}}^{n}\left( {{B}^{\prime }, C}\right) \rightarrow {\operatorname{Ext}}^{n}\left( {B, C}\right) \n\]\n\n\( {\operatorname{Ext}}^{1}\left( {{B}^{\prime \prime }, C}\right) \rightarrow {\operatorname{Ext}}^{1}\left( {{B}^{\prime }, C}\right) \rightarrow {\operatorname{Ext}}^{n}\left( {B, C}\right) \)\n\n\( 0 \rightarrow \operatorname{Hom}\left( {{B}^{\prime \prime }, C}\right) \rightarrow \operatorname{Hom}\left( {{B}^{\prime }, C}\right) \rightarrow \operatorname{Hom}\left( {B, C}\right) \)
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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 terms with \( C \) :\n\n\n\nColumns are exact since all \( {E}_{j} \) are injective. The result now follows from Theorem 3.3 and Corollary 3.12.
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Yes
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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 \( C \in {}_{R}\mathbf{M} \) and \( n \geq 1 \) .
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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.
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Yes
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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 \( I \) .
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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\[ 0 \rightarrow \operatorname{Hom}\left( {R/I, C}\right) \rightarrow \operatorname{Hom}\left( {R, C}\right) \rightarrow \operatorname{Hom}\left( {I, C}\right) \rightarrow {\operatorname{Ext}}^{1}\left( {R/I, C}\right) = 0 \]\n\nso that \( \operatorname{Hom}\left( {R, C}\right) \rightarrow \operatorname{Hom}\left( {I, C}\right) \) is onto. This is exactly what gives a filler for any diagram\n\n\n\nso that \( C \) is injective by the injective test lemma.
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Yes
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Example 11 \( {\operatorname{Ext}}_{\mathbb{Z}}^{1}\left( {\mathbb{Q},\mathbb{Z}}\right) \approx \mathbb{R} \) (as groups).
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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 \( 0 \rightarrow \operatorname{Hom}\left( {\mathbb{Q},\mathbb{Q}}\right) \rightarrow \operatorname{Hom}\left( {\mathbb{Q},\mathbb{Q}/\mathbb{Z}}\right) \rightarrow 0 \rightarrow 0 \rightarrow \cdots \) . Now Hom \( \left( {\mathbb{Q},\mathbb{Q}}\right) \approx \mathbb{Q} \), since \( \mathbb{Q} \) is uniquely divisible: If \( f \in \) Hom \( \left( {\mathbb{Q},\mathbb{Q}}\right) \) and \( f\left( 1\right) = q \), then \( q = f\left( {n/n}\right) = {nf}\left( {1/n}\right) \Rightarrow f\left( {1/n}\right) = q/n \), so that \( f\left( {m/n}\right) = {mf}\left( {1/n}\right) = \frac{m}{n} \cdot q \) . Hence \( f \mapsto f\left( 1\right) \) is an isomorphism of \( \operatorname{Hom}\left( {\mathbb{Q},\mathbb{Q}}\right) \) with \( \mathbb{Q} \) . To finish, note that \( \mathbb{Q} \in {}_{\mathbb{Z}}{\mathbf{M}}_{\mathbb{Q}} \), so that \( \operatorname{Hom}\left( {\mathbb{Q},\mathbb{Q}/\mathbb{Z}}\right) \) is a vector space over \( \mathbb{Q} \) . It suffices to show that its dimension over \( \mathbb{Q} \) is the same as the dimension of \( \mathbb{R} \), that is, a continuum. That way the dimension of \( {\operatorname{Ext}}_{\mathbb{Z}}^{1}\left( {\mathbb{Q},\mathbb{Z}}\right) \) will be a continuum. For this purpose, it suffices to show (since \( \mathbb{Q} \) is countable) that \( \operatorname{Hom}\left( {\mathbb{Q},\mathbb{Q}/\mathbb{Z}}\right) \) is a continuum. To see this, note that one can define an \( f \in \operatorname{Hom}\left( {\mathbb{Q},\mathbb{Q}/\mathbb{Z}}\right) \) by recursive choices:\n\n i) Choose \( {r}_{0} \in \mathbb{Q} \cap \lbrack 0,1) \) ; set \( f\left( k\right) = k{r}_{0} + \mathbb{Z} \).\n\n ii) Choose \( {r}_{1} = \frac{1}{2}{r}_{0} \) or \( \frac{1}{2}{r}_{0} + \frac{1}{2} \) ; set \( f\left( \frac{k}{2}\right) = k{r}_{1} + \mathbb{Z} \).\n\n iii) Choose \( {r}_{2} = \frac{1}{3}{r}_{1},\frac{1}{3}{r}_{1} + \frac{1}{3} \) or \( \frac{1}{3}{r}_{1} + \frac{2}{3} \) ; set \( f\left( \frac{k}{6}\right) = k{r}_{2} + \mathbb{Z} \).\n\n Etc. There is a continuum of total choices.
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Yes
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Proposition 3.16 \( {\operatorname{Tor}}_{n}^{R}\left( {A, B}\right) \approx {\operatorname{Tor}}_{n}^{{R}^{\text{op }}}\left( {B, A}\right) \) .
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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.
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No
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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\[ \rightarrow {\operatorname{Tor}}_{n}^{R}\left( {A, B}\right) \rightarrow {\operatorname{Tor}}_{n}^{R}\left( {A,{B}^{\prime }}\right) \rightarrow {\operatorname{Tor}}_{n}^{R}\left( {A,{B}^{\prime \prime }}\right) \]\n\n\[ \cdots \rightarrow {\operatorname{Tor}}_{1}^{R}\left( {A,{B}^{\prime \prime }}\right) \]\n\n- \( A \otimes B \rightarrow A \otimes {B}^{\prime } \rightarrow A \otimes {B}^{\prime \prime } \rightarrow 0. \)
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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.
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No
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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 left ideal \( I \) .
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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.
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Yes
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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 right ideal \( J \) .
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Proof: Corollary 3.18 plus Proposition 3.16.
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No
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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 \( A \in {\mathbf{M}}_{R} \), and \( n \geq 1 \) .
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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.
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Yes
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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 \( {\operatorname{Ext}}^{k}\left( {D, C}\right) \approx {\operatorname{Ext}}^{k + n}\left( {{D}^{\prime }, C}\right) \) for all \( C \in {}_{R}\mathbf{M} \), and \( k > d \) .
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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 piece of the long exact sequence of Proposition 3.13 yields, for \( k > d \) :\n\n\[ 0 = {\operatorname{Ext}}^{k}\left( {{L}_{1}, C}\right) \rightarrow {\operatorname{Ext}}^{k}\left( {D, C}\right) \rightarrow {\operatorname{Ext}}^{k + 1}\left( {{D}^{\prime }, C}\right) \rightarrow {\operatorname{Ext}}^{k + 1}\left( {{L}_{1}, C}\right) = 0. \]\n\nThe induction step (with \( n \geq 2 \) ) comes from defining \( Q \) to be the kernel of \( {L}_{n} \rightarrow {D}^{\prime } \) . We have two sequences, \( 0 \rightarrow D \rightarrow {L}_{1} \rightarrow {L}_{2} \rightarrow \cdots \rightarrow {L}_{n - 1} \rightarrow \) \( Q \rightarrow 0 \) and \( 0 \rightarrow Q \rightarrow {L}_{n} \rightarrow {D}^{\prime } \rightarrow 0 \) . We thus get (by induction)\n\n\[ {\operatorname{Ext}}^{k}\left( {D, C}\right) \approx {\operatorname{Ext}}^{k + n - 1}\left( {Q, C}\right) \approx {\operatorname{Ext}}^{k + n}\left( {{D}^{\prime }, C}\right) . \]
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Yes
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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 \( \mathrm{F} - \dim {Q}_{j} \leq d \), for all \( j \), then \( \mathrm{F} - \dim B \leq d + n \) .
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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 \( A \in {\mathbf{M}}_{R} \) .
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Yes
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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.\n\niv) There exists a projective resolution \( \left\langle {{P}_{k},{d}_{k}}\right\rangle \) of \( B \) for which \( {P}_{k} = 0 \) when \( k > n \) .
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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) \( \Rightarrow \) (iv) by the discussion immediately preceding this proposition. Finally,(iv) \( \Rightarrow \) (i), since \( {\operatorname{Ext}}^{n + 1}\left( {B, C}\right) \) will be the homology at \( \cdots \leftarrow \operatorname{Hom}\left( {{P}_{n + 1}, C}\right) \leftarrow \cdots \), and \( {P}_{n + 1} = 0 \) .
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Yes
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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 \( B \) is flat.\n\niv) There exists a flat resolution of \( B \) whose nth kernel is flat.\n\nv) There exists a flat resolution \( \left\langle {{F}_{k},{d}_{k}}\right\rangle \) of \( B \) for which \( {F}_{k} = 0 \) when \( k > n \) .
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Proof: Pretty much like Proposition 4.4, except (ii) \( \Rightarrow \) (iii) uses Corollary 3.19.
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No
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Corollary 4.6 For all \( B \in {}_{R}\mathbf{M},\mathrm{F} \) -dim \( B \leq \mathrm{P} \) -dim \( B \) .
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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 \( B \leq n \) .
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Yes
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Corollary 4.7 LG-dim \( R \geq \mathrm{W} \) -dim \( R \) and RG-dim \( R \geq \mathrm{W} \) -dim \( R \) .
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Proof: Take the supremum; see also Proposition 3.16.
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No
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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.\n\niv) There exists an injective resolution of \( C \) whose nth cokernel is injective.\n\nv) There exists an injective resolution \( \left\langle {{E}_{k},{d}_{k}}\right\rangle \) of \( C \) for which \( {E}_{k} = 0 \) when \( k > n \) .
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Proof: Again, like Proposition 4.4, with arrows reversed. (ii) \( \Rightarrow \) (iii) uses Corollary 3.15.
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No
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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\\]
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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 \\( C > n \\) . By Proposition 4.8, there exists a left ideal \\( I \\) for which \\( {\\operatorname{Ext}}^{n + 1}\\left( {R/I, C}\\right) \\neq 0 \\) . But now by definition P-dim \\( \\left( {R/I}\\right) > n \\), contradicting the definition of \\( n \\) .
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No
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Corollary 4.10 If LG-dim \( R > 0 \), then LG-dim \( R = 1 + \sup \{ \) P-dim \( I \) : \( I \) a left ideal \( \} \) .
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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 \( n > \mathrm{P} \) -dim \( I \) . That is, \( \mathrm{P} \) -dim \( \left( {R/I}\right) = 1 + \mathrm{P} \) -dim \( I \) . (This is a special case of some general results; see the exercises and the next chapter.) On the other hand, if \( R/I \) is projective, then so is \( I \) by setting \( n = 1 \) . Hence, in all cases, \( 1 + \mathrm{P} \) -dim \( I \leq \mathrm{{LG}} \) -dim \( R \), while if \( \mathrm{P} \) -dim \( \left( {R/I}\right) > 0 \) , then P-dim \( \left( {R/I}\right) = 1 + \) P-dim \( I \) . Taking the (positive) supremum yields the result.
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No
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Corollary 4.12 If \( R \) is a PID, then LG-dim \( R \leq 1 \) .
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Proof: If \( I = {Ra} \neq 0 \), then \( I \) is isomorphic to \( R \) in \( {}_{R}\mathbf{M} \) since \( R \) is an integral domain \( \left( {I\text{is free on}\{ a\} }\right) \) .
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No
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Corollary 4.12 If \( R \) is a PID, then LG-dim \( R \leq 1 \) .
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Proof: If \( I = {Ra} \neq 0 \), then \( I \) is isomorphic to \( R \) in \( {}_{R}\mathbf{M} \) since \( R \) is an integral domain \( \left( {I\text{is free on}\{ a\} }\right) \) .
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No
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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 for all \( x \\in P,\\left\\{ {i \\in \\mathcal{I} : {\\varphi }_{i}\\left( x\\right) \\neq 0}\\right\\} \) is finite, and \( x = \\sum {\\varphi }_{i}\\left( x\\right) {s}_{i} \).\n\niii) There exists a generating set \( \\left\\{ {{s}_{i} : i \\in \\mathcal{I}}\\right\\} \) of \( P \) for which there exist \( {\\varphi }_{i} \\in {P}^{ * }, i \\in \\mathcal{I} \) such that for all \( x \\in P,\\left\\{ {i \\in \\mathcal{I} : {\\varphi }_{i}\\left( x\\right) \\neq 0}\\right\\} \) is finite, and \( x = \\sum {\\varphi }_{i}\\left( x\\right) {s}_{i} \).
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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 \( P \) is projective. Suppose \( \\eta : P \\rightarrow F \) is a splitting, with \( \\eta \\left( x\\right) = \\left\\langle {{\\varphi }_{i}\\left( x\\right) : i \\in \\mathcal{I}}\\right\\rangle \), that is, suppose the \( i \) th coordinate of \( \\eta \) is \( {\\varphi }_{i} \) . Then these \( {\\varphi }_{i} \) have all the required properties, including \( x = \\sum {\\varphi }_{i}\\left( x\\right) {s}_{i} \), which follows from \( {i}_{P} = {\\pi \\eta } \).\n\n(ii) \( \\Rightarrow \) (iii) is trivial, so assume (iii), and again let \( F \) be the free module on \( \\mathcal{I},\\pi : F \\rightarrow P \) defined via \( i \\mapsto {s}_{i} \) . Run the previous paragraph in reverse; define \( \\eta \\left( x\\right) = \\sum {\\varphi }_{i}\\left( x\\right) \\cdot i \), that is, define \( \\eta \\left( x\\right) \) by requiring its \( i \) th coordinate to be \( {\\varphi }_{i}\\left( x\\right) \) . Then \( x = \\sum {\\varphi }_{i}\\left( x\\right) \\cdot {s}_{i} \) implies that \( {i}_{P} = {\\pi \\eta } \), so that \( F \\rightarrow P \\rightarrow 0 \) splits. Thus, \( P \) is (isomorphic to) a direct summand of \( F \), and so is projective.
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Yes
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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 \( {i}_{P} \) .
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Proof: \( x = \sum {\varphi }_{i}\left( x\right) \cdot {s}_{i} \) for all \( x \Leftrightarrow \sum {\varphi }_{i} \otimes {s}_{i} \mapsto {i}_{P} \) .
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No
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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.
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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 \( 0 \rightarrow {C}^{ * } \rightarrow {F}_{0}^{ * } \rightarrow {F}_{1}^{ * } \) is exact. \( B \) is flat, so we have exactness of \( 0 \rightarrow {C}^{ * } \otimes B \rightarrow {F}_{0}^{ * } \otimes B \rightarrow {F}_{1}^{ * } \otimes B \) . But naturality of \( {A}^{ * } \otimes B \rightarrow \operatorname{Hom}\left( {A, B}\right) \) in the variable \( A \) gives commutativity of\n\n\n\nso that \( {C}^{ * } \otimes B \rightarrow \operatorname{Hom}\left( {C, B}\right) \) is an isomorphism by the 5-lemma.
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Yes
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Corollary 4.21 Suppose \( R \) is left Noetherian. Then LG-dim \( R = \) W-dim \( R \) .
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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.
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No
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Corollary 4.22 (Auslander) Suppose \( R \) is both right and left Noetherian. Then LG-dim \( R = \mathrm{{RG}} \)-dim \( R \) .
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Proof: Both equal W-dim \( R \) .
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No
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Lemma 4.27 Every submodule of a semisimple module is semisimple.
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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 \( A \) a submodule of \( C \), there exists \( B \) such that \( A \oplus B = D \) , since \( D \) is semisimple. Hence, \( A + \left( {B \cap C}\right) = \left( {A + B}\right) \cap C = C \) . Since \( A \cap \left( {B \cap C}\right) \subset A \cap B = \{ 0\}, A + \left( {B \cap C}\right) \) is a direct sum.
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No
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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 not containing \( x \), is maximal as a submodule. Such submodules exist.
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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 \( x \in {B}^{\prime \prime } \) and \( S \subset {B}^{\prime \prime } \Rightarrow {B}^{\prime \prime } = B \) since \( S \) and \( x \) generate \( B \) . Hence, \( {B}^{\prime } \) is maximal. It remains to show that such submodules exist. This uses Zorn’s lemma: Look at the set \( \mathcal{B} \) of all submodules of \( B \) that contain \( S \) but do not contain \( x \) ; the submodule generated by \( S \) is such a submodule. Partially order by set inclusion. The union of a nonempty chain of submodules not containing \( x \) will be a submodule not containing \( x \), and will constitute an upper bound for the chain. Since every nonempty chain in \( \mathcal{B} \) is bounded, Zorn’s lemma applies, and \( \mathcal{B} \) has a maximal element.
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Yes
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Lemma 4.29 Every nonzero semisimple module contains a simple submodule.
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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 \( {B}^{\prime }/{B}^{\prime \prime } \) will be simple. But \( {B}^{\prime } \) is semisimple by Lemma 4.27, so there exists \( {B}^{\prime \prime \prime } \) such that \( {B}^{\prime } = {B}^{\prime \prime } \oplus {B}^{\prime \prime \prime } \) . But \( {B}^{\prime \prime \prime } \approx {B}^{\prime }/{B}^{\prime \prime } \), so \( {B}^{\prime \prime \prime } \) is a simple submodule of \( B \) .
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Yes
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Lemma 4.30 Every semisimple module is the sum of its simple submodules.
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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 semisimple by Lemma 4.27, so \( {B}^{\prime \prime } \) contains a simple submodule by Lemma 4.29. This contradicts \( {B}^{\prime \prime } \cap {B}^{\prime } = 0 \) , since \( {B}^{\prime } \) is the sum of all simple submodules of \( B \) .
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Yes
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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 \( {B}^{\prime } \) of \( B \) there exists a subset \( \mathcal{J} \) of \( \mathcal{I} \) such that \( B = {B}^{\prime } \oplus \left( {{ \oplus }_{\mathcal{J}}{B}_{i}}\right) \) (direct sum).
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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 \( \mathcal{I} \) .) Zorn’s lemma applies: Given any chain \( \left\{ {\mathcal{J}}_{\alpha }\right\} \) of such subsets, their union \( \mathcal{J} = \cup {\mathcal{J}}_{\alpha } \) is such a subset, that is, \( {B}^{\prime } + \left( {{\sum }_{\mathcal{J}}{B}_{i}}\right) \) is a direct sum, \( {B}^{\prime } \oplus \left( {{ \oplus }_{\mathcal{J}}{B}_{i}}\right) \) . To see this, observe that any \( {b}^{\prime } + \left( {{\sum }_{\mathcal{J}}{b}_{i}}\right) \) (being a finite sum) has only nonzero \( {b}_{i} \) for (finitely many) \( i \) in one of the \( {\mathcal{J}}_{\alpha } \) (since \( \left\{ {\mathcal{J}}_{\alpha }\right\} \) is a chain), so \( {b}^{\prime } + \left( {{\sum }_{\mathcal{J}}{b}_{i}}\right) = 0 \Rightarrow {b}^{\prime } = 0 = {b}_{i} \), all \( i \in \mathcal{J} \) . Thus, there is a \( \mathcal{J} \subset \mathcal{I} \) which is maximal with the property that \( {B}^{\prime } + \left( {{\sum }_{\mathcal{J}}{B}_{i}}\right) \) is a direct sum, \( {B}^{\prime } \oplus \left( {{ \oplus }_{\mathcal{J}}{B}_{i}}\right) \) . Suppose \( j \in \mathcal{I} - \mathcal{J} \) . Then \( \mathcal{J} \subset \mathcal{J} \cup \{ j\} \), so \( {B}^{\prime } + \left( {{\sum }_{\mathcal{J}\cup \{ j\} }{B}_{i}}\right) \) is not a direct sum while \( {B}^{\prime } \oplus \left( {{ \oplus }_{\mathcal{J}}{B}_{i}}\right) \) is, so \( \left( {{B}^{\prime } \oplus \left( {{ \oplus }_{\mathcal{J}}{B}_{i}}\right) }\right) \cap {B}_{j} \neq 0 \) . As a submodule of the simple module \( {B}_{j} \) , \( \left( {{B}^{\prime } \oplus \left( {{ \oplus }_{\mathcal{J}}{B}_{i}}\right) }\right) \cap {B}_{j} = {B}_{j} \) . That is, \( {B}_{j} \subset {B}^{\prime } \oplus \left( {{ \oplus }_{\mathcal{J}}{B}_{i}}\right) \) . But that means that \( {B}^{\prime } \oplus \left( {{ \oplus }_{\mathcal{J}}{B}_{i}}\right) \) contains every \( {B}_{j}, j \in \mathcal{I} \), since it trivially contains every \( {B}_{j}, j \in \mathcal{J} \) . Hence, \( {B}^{\prime } \oplus \left( {{ \oplus }_{\mathcal{J}}{B}_{i}}\right) \supset {\sum }_{\mathcal{I}}{B}_{i} = B \) .
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Yes
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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.
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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 being \( \oplus J{B}_{i} \) .
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Yes
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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.
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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 \( \operatorname{im}\varphi = {B}^{\prime } \) since \( {B}^{\prime } \) is simple. Hence, \( \varphi \neq 0 \Rightarrow \varphi \) is bijective \( \Rightarrow {\varphi }^{-1} \) exists in \( \operatorname{Hom}\left( {{B}^{\prime }, B}\right) \) .
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Yes
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Lemma 4.35 \( \operatorname{End}\left( {B}^{n}\right) \approx {M}_{n}\left( {\operatorname{End}\left( B\right) }\right) \) .
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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.
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No
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Lemma 4.36 Suppose \( {B}_{1},\ldots ,{B}_{N} \) are pairwise nonisomorphic simple \( R \) - modules. Then\n\n\[ \operatorname{End}\left( {{B}_{1}^{{n}_{1}} \oplus \cdots \oplus {B}_{N}^{{n}_{N}}}\right) \approx {M}_{{n}_{1}}\left( {\operatorname{End}\left( {B}_{1}\right) }\right) \oplus \cdots \oplus {M}_{{n}_{N}}\left( {\operatorname{End}\left( {B}_{N}\right) }\right) ,\] \n\na finite sum of matrix rings over division rings.
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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\[ \approx {\bigoplus }_{i = 1}^{N}\operatorname{Hom}\left( {{B}_{i}^{{n}_{i}},{B}_{i}^{{n}_{i}}}\right) \]\n\n\[ = {\bigoplus }_{i = 1}^{N}\operatorname{End}\left( {B}_{i}^{{n}_{i}}\right) \]\n\n\[ \approx {\bigoplus }_{i = 1}^{N}{M}_{{n}_{i}}\left( {\operatorname{End}\left( {B}_{i}\right) }\right) \]
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Yes
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Lemma 4.37 Suppose \( B \) is a finitely generated semisimple \( R \) -module. Then \( B \) is a finite direct sum of simple modules.
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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 \( \mathcal{F} = \cup {\mathcal{F}}_{j} \) . Then each \( {x}_{j} \in {\sum }_{\mathcal{F}}{B}_{i} \), so \( B = {\sum }_{\mathcal{F}}{B}_{i} \) . By Lemma 4.31, \( \mathcal{F} \) has a (necessarily finite) subset \( \mathcal{J} \) for which \( B = { \oplus }_{\mathcal{J}}{B}_{i} \) .
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Yes
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Lemma 4.39 \( \operatorname{End}\left( {{}_{R}R}\right) \approx {R}^{\mathrm{{op}}} \) .
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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 \( \Phi \left( {rs}\right) = \Phi \left( s\right) \Phi \left( r\right) \) ; this is left to the reader. Finally, observe that \( \ker \Phi = 0 \), since \( \Phi \left( r\right) = 0 \Rightarrow {\varphi }_{r} = 0 \Rightarrow {\varphi }_{r}\left( 1\right) = 0 \Rightarrow r = 1 \cdot r = 0 \), while \( \Phi \) is onto, since \( \psi \in \operatorname{End}\left( {{}_{R}R}\right) \) implies that \( \psi \left( x\right) = \psi \left( {x \cdot 1}\right) = {x\psi }\left( 1\right) = {\varphi }_{r}\left( x\right) \) , where \( r = \psi \left( 1\right) \), so that \( \psi = \Phi \left( r\right) \) .
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No
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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 sequence of left \( R \) -modules splits.\n\nvi) Every left ideal is injective.\n\nvii) Every maximal left ideal is injective.\n\nviii) Every maximal left ideal is a direct summand of \( R \).\n\nix) For every left ideal \( I, R/I \) is projective.\n\nx) Every simple left \( R \) -module is projective.\n\nxi) \( R \) is semisimple as a left \( R \) -module.\n\nxii) \( R \) is a finite direct sum of matrix rings over division rings.
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Proof: It's hard to follow the logic without a scorecard. The implications are proved as follows:\n\n\n\nThe implications marked with a check are trivial as statements. Most of the rest are quick. If \( B \\in {}_{R}\\mathbf{M} \), then P-dim \( B = 0 \\Leftrightarrow B \) is projective, so (i) \\(\\Rightarrow\\) (ii) by definition of left global dimension. Similarly,(i) \\(\\Rightarrow\\) (iii), using Proposition 4.1(a). (ix) \\(\\Rightarrow\\) (i) by the global dimension theorem. (iii) \\(\\Rightarrow\\) (v), since if \( 0 \\rightarrow A \\rightarrow B \\rightarrow C \\rightarrow 0 \) is short exact, then it must split when \( A \) is injective. (v) \\(\\Rightarrow\\) (iv) since if \( A \) is a submodule of \( B \), then the splitting of \( 0 \\rightarrow A \\rightarrow B \\rightarrow B/A \\rightarrow 0 \) will guarantee that \( A \) is a direct summand of \( B \). (vii) \\(\\Rightarrow\\) (viii) since injectives are absolute direct summands. (xi) \\(\\Rightarrow\\) (ix), since if a left ideal \( I \) is a direct summand of \( R \), then the other summand is \( R/I \), which will then be projective, since \( R \) is projective. We now have all the implications except (ix) \\(\\Rightarrow\\) (x) \\(\\Rightarrow\\) (viii) \\(\\Rightarrow\\) (xi) \\(\\Leftrightarrow\\) (xii).\n\nSuppose \( B \) is a simple left \( R \) -module, and \( 0 \\neq x \\in B \). Then \( B = {Rx} \\approx \) \( R/\\operatorname{ann}\\left( x\\right) \). Hence,(ix) \\(\\Rightarrow\\) (x). But if \( I \) is a maximal left ideal, then \( R/I \) is simple, so \\( \\left( \\mathrm{x}\\right) \\Rightarrow 0 \\rightarrow I \\rightarrow R \\rightarrow R/I \\rightarrow 0 \) splits when \( I \) is a maximal left ideal, and this implies (viii). We are now left with (viii) \\(\\Rightarrow\\) (xi) \\(\\Leftrightarrow\\) (xii).\n\n(viii) \\(\\Rightarrow\\) (xi): Suppose (viii). Let \( I \) be the (left ideal) sum of all simple submodules (that is, simple left ideals) in \( R \). Suppose \( I \\neq R \). Let \( J \) be a maximal left ideal with \( J \\supset I \). Then \( J \) is a direct summand of \( R \), say \( R = J \\oplus {J}^{\\prime } \). But \( {J}^{\\prime } \\approx R/J \) is simple, so \( {J}^{\\prime } \\subset I \) by definition. But now \( {J}^{\\prime } \\subset I \\subset J \), while \( J \\oplus {J}^{\\prime } \) is a direct sum. Hence, \( {J}^{\\prime } = 0 \), contradicting the definition of \
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Yes
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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.
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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 \( {re} = s - {se} \Rightarrow s = {se} + {re} \Rightarrow \) \( {se} = \left( {{se} + {re}}\right) e = s{e}^{2} + r{e}^{2} = {se} + {re} \Rightarrow {re} = 0 \) . Hence, \( I \cap {Rf} = 0 \) and \( I \oplus {Rf} = R \) .\n\nIf \( R = I \oplus J \) for some left ideal \( J \), then \( 1 = e + f \) for some \( e \in I \) and \( f \in J \) . Thus, \( {Re} \subset I \) and \( {Rf} \subset J \) ; also, if \( x \in R \), then \( x = x \cdot 1 = \) \( x \cdot \left( {e + f}\right) = {xe} + {xf} \) . Hence,
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Yes
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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 \( {Re} + {Rf} = R\left( {e + f}\right) \) .
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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 \( f = f\left( {e + f}\right) \in R\left( {e + f}\right) \), so that \( {Re} + {Rf} \subset R\left( {e + f}\right) \) .
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Yes
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Lemma 4.44 Suppose \( R \) is a ring. Then \( {Ra} + {Rb} = {Ra} + {Rb}\left( {1 - a}\right) \) .
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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, \( b = {ba} + b\left( {1 - a}\right) \in {Ra} + {Rb}\left( {1 - a}\right) \), so \( {Rb} \subset {Ra} + {Rb}\left( {1 - a}\right) \) . Again \( {Ra} \subset {Ra} + {Rb}\left( {1 - a}\right) \), so \( {Ra} + {Rb} \subset {Ra} + {Rb}\left( {1 - a}\right) \) .
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Yes
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Lemma 4.45 Suppose \( R \) is regular. Then every finitely generated left ideal is principal (and generated by an idempotent).
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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 that if \( R \) is regular, then every principal left ideal is generated by an idempotent.) For this purpose, we successively modify \( e \) and \( f \) without changing the sum \( {Re} + {Rf} \) until we are in the situation of Lemma 4.43.\n\nFirst of all, \( {Re} + {Rf} = {Re} + {Rf}\left( {1 - e}\right) \) by Lemma 4.44. Write \( f\left( {1 - e}\right) = \) \( f\left( {1 - e}\right) {rf}\left( {1 - e}\right) \), and set \( {f}^{\prime } = {rf}\left( {1 - e}\right) \) . Then \( {Rf}\left( {1 - e}\right) = R{f}^{\prime } \), so \( {Re} + {Rf} = {Re} + R{f}^{\prime } \) . Also, \( {f}^{\prime }e = {rf}\left( {1 - e}\right) e = {rf}\left( {e - {e}^{2}}\right) = 0 \) . Next, do the same to \( e : R{f}^{\prime } + {Re} = R{f}^{\prime } + {Re}\left( {1 - {f}^{\prime }}\right) \) . Set \( {e}^{\prime } = e\left( {1 - {f}^{\prime }}\right) \) . This \( {e}^{\prime } \) is already idempotent: \( {\left( {e}^{\prime }\right) }^{2} = e\left( {1 - {f}^{\prime }}\right) e\left( {1 - {f}^{\prime }}\right) = e\left( {e - {f}^{\prime }e}\right) \left( {1 - {f}^{\prime }}\right) = {e}^{2}\left( {1 - {f}^{\prime }}\right) = \) \( e\left( {1 - {f}^{\prime }}\right) = {e}^{\prime } \) . Furthermore, \( {e}^{\prime }{f}^{\prime } = e\left( {1 - {f}^{\prime }}\right) {f}^{\prime } = e\left( {{f}^{\prime } - {\left( {f}^{\prime }\right) }^{2}}\right) = 0 \), while \( {f}^{\prime }{e}^{\prime } = {f}^{\prime }e\left( {1 - {f}^{\prime }}\right) = 0 \) . Hence, \( {Re} + {Rf} = {Re} + R{f}^{\prime } = R{e}^{\prime } + R{f}^{\prime } = R\left( {{e}^{\prime } + {f}^{\prime }}\right) \) , the last equality by Lemma 4.43.
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Yes
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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) \( {\\operatorname{Tor}}_{1}\\left( {R/J, R/I}\\right) = 0 \) for every finitely generated right ideal \( J \) and every finitely generated left ideal I.\n\n\[ \n\\text{v)}{\\operatorname{Tor}}_{1}\\left( {R/{aR}, R/{Ra}}\\right) = 0\\text{for every}a \\in R\\text{}.\n\]\n\nvi) \( R \) is regular.
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\( \\textbf{Proof: We prove that (i)} \\Rightarrow \) (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) \( \\Rightarrow \) (vi) by Exercise 9(c) of Chapter 3, which states, \( {}^{\\omega }{\\operatorname{Tor}}_{1}\\left( {R/J, R/I}\\right) \\approx \\left( {J \\cap I}\\right) /{JI} \) \
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No
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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\[ \text{b)}\mathrm{F} - {\dim }_{R}F\left( B\right) \leq \mathrm{P} - {\dim }_{S}B + \mathrm{F} - {\dim }_{R}F\left( S\right) \text{.} \]
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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- \( {\dim }_{R}F\left( S\right) \) (Chapter 4, Exercise 11), or zero if \( \mathcal{I} \) is empty. Flat dimension also works this way.\n\nSecond case: \( B \) is projective. Then \( B \oplus C \) is free for some \( C \), and \( \mathrm{P} \) - \( {\mathrm{{dim}}}_{R}F\left( B\right) \leq \mathrm{P} \) - \( {\mathrm{{dim}}}_{R}F\left( B\right) \oplus F\left( C\right) = \mathrm{P} \) - \( {\mathrm{{dim}}}_{R}F\left( {B \oplus C}\right) \leq \mathrm{P} \) - \( {\mathrm{{dim}}}_{R}F\left( S\right) \) by the first case. Flat dimension also works this way.\n\nFinally, arbitrary \( B \) . If \( \mathrm{P} - {\dim }_{S}B = \infty \), there is nothing to prove, so suppose \( n = \mathrm{P} - {\dim }_{S}B < \infty \) . Let \( 0 \rightarrow {P}_{n} \rightarrow {P}_{n - 1} \rightarrow \cdots \rightarrow {P}_{0} \rightarrow B \rightarrow 0 \) be a projective resolution of \( B \) , possible by the projective dimension theorem. Then \( 0 \rightarrow F\left( {P}_{n}\right) \rightarrow F\left( {P}_{n - 1}\right) \rightarrow \cdots \rightarrow F\left( {P}_{0}\right) \rightarrow F\left( B\right) \rightarrow 0 \) is exact since \( F \) is exact, so \( \mathrm{P} - {\dim }_{R}F\left( B\right) \leq n + d \), where \( d = \mathrm{P} - {\dim }_{R}F\left( R\right) \) by Corollary 4.3 (and the second case). Flat dimensions follow in a similar way.
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Yes
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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\[ \mathrm{{SP}} - \dim \left( {{B}^{\prime \prime }, B}\right) = \max \left\{ {\mathrm{{SP}} - \dim \left( {{B}^{\prime \prime },{B}^{\prime }}\right) ,\mathrm{{SP}} - \dim \left( {{B}^{\prime }, B}\right) }\right\} .\n\]
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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 \( \left( {B}^{\prime \prime }\right. \) , \( B) \geq \max \left\{ {\mathrm{{SP}}\text{-dim }\left( {{B}^{\prime \prime },{B}^{\prime }}\right) ,\mathrm{{SP}}\text{-dim }\left( {{B}^{\prime }, B}\right) }\right\} \) . Suppose this inequality is strict. Then \( \exists C \) with P-dim \( C > \max \left\{ {\operatorname{SP-dim}\left( {{B}^{\prime \prime },{B}^{\prime }}\right) ,\operatorname{SP-dim}\left( {{B}^{\prime }, B}\right) }\right\} \) and \( {B}^{\prime \prime } \subset C \subset B \) . Since \( {B}^{\prime \prime } \subset C \cap {B}^{\prime } \subset {B}^{\prime } \) and \( {B}^{\prime } \subset C + {B}^{\prime } \subset B \), it follows that \( \mathrm{P} - \dim C > \max \left\{ {\mathrm{P} - \dim \left( {C \cap {B}^{\prime }}\right) ,\mathrm{P} - \dim \left( {C + {B}^{\prime }}\right) }\right\} \) . This cannot happen, though: Choose \( n \) finite, P-dim \( C \geq n > \max \{ \) P-dim \( (C \cap \) \( \left. {{B}^{\prime }),\mathrm{P}\text{-dim}\left( {C + {B}^{\prime }}\right) }\right\} \) ; and choose \( D \) so that \( {\operatorname{Ext}}^{n}\left( {C, D}\right) \neq 0 \) . By the algebraic Mayer-Vietoris sequence (Chapter 3, Exercise 12), we have an exact sequence \( 0 = {\operatorname{Ext}}^{n}\left( {C \cap {B}^{\prime }, D}\right) \leftarrow {\operatorname{Ext}}^{n}\left( {C, D}\right) \oplus {\operatorname{Ext}}^{n}\left( {{B}^{\prime }, D}\right) \leftarrow \) \( {\operatorname{Ext}}^{n}\left( {C + {B}^{\prime }, D}\right) = 0 \), a contradiction.
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Yes
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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 \( R = 1 + \max \left\{ {\mathrm{{SP}}\text{-dim }\left( {{I}_{j - 1},{I}_{j}}\right) }\right. \) : \( j = 1,\ldots, n\} \) .
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Proof: See comments preceding Proposition 5.2.
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No
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Proposition 5.4 Suppose \( B, C \in {}_{R}\mathbf{M} \) . Then\n\n\[ \n\text{ SP-dim }\left( {B \oplus C}\right) = \max \{ \text{ SP-dim }B,\text{ SP-dim }C\} .\n\]
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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 isomorphism theorems, so any module between \( B \oplus 0 \) and \( B \oplus C \) has the form \( B \oplus {C}^{\prime } \) for a submodule \( {C}^{\prime } \) of \( C \) . Also P-dim \( \left( {B \oplus {C}^{\prime }}\right) = \max \left\{ {\text{P-dim}B,\text{P-dim}{C}^{\prime }}\right\} \) by Exercise 11, Chapter 4. Taking suprema over \( {C}^{\prime } \), SP-dim \( (B \oplus 0, B \oplus C\} = \) \( \max \{ \mathrm{P} - \dim B,\mathrm{{SP}} - \dim C\} \), so that\n\n\[ \n\text{ SP-dim }\left( {B \oplus C}\right) = \max \{ \text{ SP-dim }B,\text{ SP-dim }C\}\n\]\n\n(since SP-dim \( B \geq \) P-dim \( B \) by definition).
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Yes
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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 \( R = 1 + \max \left\{ {\mathrm{{SP}}\text{-dim }{I}_{j} : j = 1,\ldots, n}\right\} \) .
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Proof: Again, see the comments preceeding Proposition 5.2.
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No
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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.
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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 surjection \( \widehat{\pi } : \widehat{P} \rightarrow \widehat{B} \) . There is a splitting over \( R \) , that is, an \( \eta : \widehat{B} \rightarrow \widehat{P} \) such that \( \widehat{\pi }\eta = {i}_{\widehat{B}} \), since \( \widehat{B} \) is \( R \) -projective. But \( \phi \) surjective \( \Rightarrow \eta \in {\mathrm{{Hom}}}_{\widehat{R}}\left( {\widehat{B},\widehat{P}}\right) \) : \( \eta \left( {\phi \left( r\right) \widehat{b}}\right) = \eta \left( {r \cdot \widehat{b}}\right) = r \cdot \eta \left( \widehat{b}\right) = \phi \left( r\right) \eta \left( \widehat{b}\right) \) for all \( r \in R \) and \( \widehat{b} \in \widehat{B} \) ,( \( \eta \) is an \( R \) -module homomorphism), so \( \eta \left( \widehat{rb}\right) = \widehat{r}\eta \left( \widehat{b}\right) \) , since \( \phi \) is onto. Thus, \( \widehat{B} \) is (isomorphic to) a direct summand of \( \widehat{P} \), and so is \( \widehat{R} \) -projective. That is, \( \widehat{B} \) is \( R \) -projective \( \Leftrightarrow \widehat{B} \) is \( \widehat{R} \) -projective.
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Yes
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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 \( R \) \( \geq 1 + \operatorname{LG} - \dim \widehat{R} \) .
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Proof: Take the supremum over \( {}_{\widehat{R}}\mathbf{M} \) in Proposition 5.8.
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No
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Corollary 5.10 If LG-dim \( R < \infty \) , then LG-dim \( R\left\lbrack x\right\rbrack \geq 1 + \) LG-dim \( R \) .
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Proof: Take \( a = x \in \widehat{R} = R\left\lbrack x\right\rbrack : \widehat{R}/\widehat{R}x \approx R \) .
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No
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Proposition 5.11 For all \( B \in {}_{R}\mathbf{M},\mathrm{P} - {\dim }_{R\left\lbrack x\right\rbrack }B\left\lbrack x\right\rbrack = \mathrm{P} - {\dim }_{R}B \) .
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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 \( \widehat{B} \in \underset{\widehat{R}}{}\mathbf{M} \) . Setting \( \widehat{B} = B\left\lbrack x\right\rbrack = F\left( B\right) \), where \( F \) is the polynomial functor,\n\n\[ \mathrm{P} - {\mathrm{{dim}}}_{R}B\left\lbrack x\right\rbrack \leq \mathrm{P} - {\mathrm{{dim}}}_{\widehat{R}}B\left\lbrack x\right\rbrack = \mathrm{P} - {\mathrm{{dim}}}_{R\left\lbrack x\right\rbrack }B\left\lbrack x\right\rbrack \leq \mathrm{P} - {\mathrm{{dim}}}_{R}B. \]\n\nBut \( \;P - {\dim }_{R}B\left\lbrack x\right\rbrack = P - {\dim }_{R}B\; \) by \( \; \) Exercise \( \;{11},\; \) Chapter \( \;4.\; \) Hence \( \mathrm{P} - {\dim }_{R}B = \mathrm{P} - {\dim }_{\widehat{R}}B\left\lbrack x\right\rbrack . \)
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No
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Theorem 5.13 If \( R \) is any ring, then LG-dim \( R\left\lbrack x\right\rbrack = 1 + \) LG-dim \( R \) .
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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 \( R\left\lbrack x\right\rbrack \geq \) \( 1 + \mathrm{{LG}} \) -dim \( R \) by Corollary 5.10. It therefore suffices to show that \( 1 + \) LG-dim \( R \geq \mathrm{P} \) -dim \( {}_{\widehat{R}}\widehat{B} \) for all \( \widehat{B} \in {}_{\widehat{R}}\mathbf{M} \) (with \( \widehat{R} = R\left\lbrack x\right\rbrack \) ).\n\nSuppose \( \widehat{B} \in \widehat{R} \) M. By Lemma 5.12, we have a short exact sequence of \( \widehat{R} \) - modules \( 0 \rightarrow \widehat{B}\left\lbrack x\right\rbrack \rightarrow \widehat{B}\left\lbrack x\right\rbrack \rightarrow \widehat{B} \rightarrow 0 \) . By Proposition 5.11, P-dim \( {}_{\widehat{R}}\widehat{B}\left\lbrack x\right\rbrack = \) P- \( {\dim }_{R}\widehat{B} \leq \) LG-dim \( R \) . By Corollary 4.3 (with \( n = 1 \) and \( d = \) LG-dim \( R \) ), \( \mathrm{P} - {\dim }_{\widehat{R}}\widehat{B} \leq 1 + \mathrm{{LG}} - \dim R. \)
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Yes
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Corollary 5.14 (Hilbert’s Syzygy Theorem) If \( K \) is a field, then LG-dim \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack = n \) .
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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.
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No
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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.
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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.
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Yes
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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 homomorphism, then \( b \in \ker \psi \Leftrightarrow S \cap \operatorname{ann}\left( b\right) \neq \varnothing \) . If \( \psi \) is one-to-one, then \( b/s \in \psi \left( B\right) \Leftrightarrow b \in {sB} \) .
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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 \( {s}^{ * } \in \operatorname{ann}\left( b\right) .b/s = \psi \left( c\right) \Leftrightarrow b/s = c/1 \Leftrightarrow \exists {s}^{ * } \in S \) for which \( {s}^{ * }\left( {b - {sc}}\right) = 0 \Leftrightarrow b = {sc} \) if \( \psi \) is one-to-one: By the first sentence, \( \ker \psi = \) \( 0 \Rightarrow \left( {{s}^{ * }\left( {b - {sc}}\right) = 0 \Rightarrow b - {sc} = 0}\right) .
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Yes
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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\[ \n\text{b)}\mathrm{F} - {\dim }_{R}\widehat{B} \leq \mathrm{P} - {\dim }_{{S}^{-1}R}\widehat{B}\text{for any}\widehat{B} \in {}_{{S}^{-1}R}\mathbf{M}\text{.}\n\]\n\n\[ \n\text{c)}\operatorname{LG} - \dim \left( {{S}^{-1}R}\right) \leq \operatorname{LG} - \dim R\text{.}\n\]
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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- \( {\dim }_{R}{S}^{-1}R = 0 \), yielding the inequality.\n\n(c) If \( \widehat{C}{ \in }_{{S}^{-1}R}\mathbf{M} \), then\n\n\[ \n\mathrm{P} - {\dim }_{{S}^{-1}R}\widehat{C} = \mathrm{P} - {\dim }_{{S}^{-1}R}{S}^{-1}\widehat{C}\n\]\n(Prop. 5.17(b))\n\n\[ \n\leq \mathrm{P} - {\dim }_{R}\widehat{C}\n\]\n(part (a) above)\n\n\[ \n\leq \text{LG-dim}R\text{.}\n\]\n\nTake the supremum over \( {}_{{S}^{-1}R}\mathbf{M} \) .
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Yes
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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.
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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 \( x \in \ker \left( {A \otimes j}\right), x \neq 0 \) . \( A \otimes B \) is an \( R \) -module, and \( \operatorname{ann}\left( x\right) \) is a proper ideal; choose a maximal ideal \( M \supset \operatorname{ann}\left( x\right) \) . We shall show that \( M \) works.\n\nFirst of all, \( B \mapsto {B}_{M} \) is an exact functor, so \( 0 \rightarrow {B}_{M} \rightarrow {C}_{M} \) is exact, that is, \( {j}_{M} : {B}_{M} \rightarrow {C}_{M} \) is one-to-one. The diagram\n\n\n\ncommutes by naturality of Proposition 5.17(e) (see Exercise 6). But \( {\psi }^{\prime }(A \otimes \) \( j)\left( x\right) = 0 \Rightarrow {\left( A \otimes j\right) }_{M}\psi \left( x\right) = 0 \), and \( \psi \left( x\right) \neq 0 \) by Proposition \( {5.17}\left( \mathrm{a}\right) \) . Hence, \( {A}_{M} \otimes {j}_{M} \) is not one-to-one. Since \( {A}_{M} \otimes \) sends the exact sequence\n\n\( 0 \rightarrow {B}_{M} \rightarrow {C}_{M} \) to the nonexact sequence \( 0 \rightarrow {A}_{M} \otimes {B}_{M} \rightarrow {A}_{M} \otimes {C}_{M} \) , \( {A}_{M} \) is not flat.
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Yes
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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 \( R = \) \( \sup \left\{ {\mathrm{{LG}} - \dim {R}_{M} : M\text{ a maximal ideal }}\right\} \) .
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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 maximal ideal \( \} \) . Then \( n < \infty \), and it suffices to show that, in fact, LG-dim \( R \leq n \) .\n\nLet \( I \) be any (left) ideal. Let \( \cdots \rightarrow {F}_{n} \rightarrow {F}_{n - 1} \rightarrow \cdots \rightarrow {F}_{1} \rightarrow {F}_{0} \rightarrow \) \( R/I \rightarrow 0 \) be a free resolution of \( R/I \) consisting of finitely generated \( R \) - modules. If \( M \) is any maximal ideal, and \( {F}_{j} = { \oplus }_{\mathcal{J}}R \), then \( {\left( {F}_{j}\right) }_{M} = { \oplus }_{\mathcal{J}}{R}_{M} \) , a free \( {R}_{M} \) -module. Since \( A \mapsto {A}_{M} \) is an exact functor,\n\n\[ \cdots \rightarrow {\left( {F}_{n}\right) }_{M} \rightarrow {\left( {F}_{n - 1}\right) }_{M} \rightarrow \cdots \rightarrow {\left( {F}_{1}\right) }_{M} \rightarrow {\left( {F}_{0}\right) }_{M} \rightarrow {\left( R/I\right) }_{M} \rightarrow 0 \]\n\nis an \( {R}_{M} \) -free, hence \( {R}_{M} \) -projective, resolution of \( {\left( R/I\right) }_{M} \) . Let \( {K}_{n} \) denote the \( n \) th kernel of \( \cdots \rightarrow {F}_{n} \rightarrow {F}_{n - 1} \rightarrow \cdots \rightarrow {F}_{1} \rightarrow {F}_{0} \rightarrow R/I \rightarrow 0.{K}_{n} \) is finitely presented since \( R \) is Noetherian. Also,\n\n\[ 0 \rightarrow {\left( {K}_{n}\right) }_{M} \rightarrow {\left( {F}_{n - 1}\right) }_{M} \rightarrow \cdots \rightarrow {\left( {F}_{1}\right) }_{M} \rightarrow {\left( {F}_{0}\right) }_{M} \rightarrow {\left( R/I\right) }_{M} \rightarrow 0 \]\n\nis exact, so \( {\left( {K}_{n}\right) }_{M} \) is isomorphic to the \( n \) th kernel of\n\n\[ \cdots \rightarrow {\left( {F}_{n}\right) }_{M} \rightarrow {\left( {F}_{n - 1}\right) }_{M} \rightarrow \cdots \rightarrow {\left( {F}_{1}\right) }_{M} \rightarrow {\left( {F}_{0}\right) }_{M} \rightarrow {\left( R/I\right) }_{M} \rightarrow 0. \]\n\nSince \( n \geq \operatorname{LG} \) -dim \( {R}_{M} \geq \mathrm{P} - {\dim }_{{R}_{M}}{\left( R/I\right) }_{M},{\left( {K}_{n}\right) }_{M} \) is projective by the projective dimension theorem.\n\nNow let \( M \) vary over all maximal ideals. \( {\left( {K}_{n}\right) }_{M} \) is always projective, hence flat. Thus \( {K}_{n} \) is flat by Proposition 5.19. But now \( {K}_{n} \) is projective by Theorem 4.19. Thus, \( \mathrm{P} - \dim \left( {R/I}\right) \leq n \) by the projective dimension theorem.\n\nNow let \( I \) vary: \( P \) -dim \( \left( {R/I}\right) \leq n \) for all (left) ideals \( I \), so LG-dim \( R \leq n \) by the global dimension theorem.
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Yes
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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 \( {B}_{1},{B}_{2} \in {}_{R}\mathbf{M} \) .\n\niii) \( F\left( {B \oplus B}\right) \approx F\left( B\right) \oplus F\left( B\right) \) for all \( B \in {}_{R}\mathbf{M} \) .
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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\[ {B}_{1}\overset{{\pi }_{1}}{ \leftarrow }{B}_{1} \oplus {B}_{2}\overset{{\pi }_{2}}{ \rightarrow }{B}_{2}\;{B}_{1}\overset{{\varphi }_{1}}{ \rightarrow }{B}_{1} \oplus {B}_{2}\overset{{\varphi }_{2}}{ \leftarrow }{B}_{2}. \]\n\nSince \( F \) is covariant, we get arrows\n\n\[ F\left( {B}_{1}\right) < \frac{F\left( {\pi }_{1}\right) }{}F\left( {{B}_{1} \oplus {B}_{2}}\right) \xrightarrow[]{F\left( {\pi }_{2}\right) }F\left( {B}_{2}\right) \]\n\n\[ F\left( {B}_{1}\right) \xrightarrow[]{F\left( {\varphi }_{1}\right) }F\left( {{B}_{1} \oplus {B}_{2}}\right) \xleftarrow[]{F\left( {\varphi }_{2}\right) }F\left( {B}_{2}\right) . \]\n\nFurthermore, \( F\left( {\pi }_{j}\right) F\left( {\varphi }_{j}\right) = F\left( {{\pi }_{j}{\varphi }_{j}}\right) = F\left( {i}_{{B}_{j}}\right) = {i}_{F\left( {B}_{j}\right) } \) for \( j = 1,2 \) ; also \( F\left( {\varphi }_{1}\right) F\left( {\pi }_{1}\right) + F\left( {\varphi }_{2}\right) F\left( {\pi }_{2}\right) = F\left( {{\varphi }_{1}{\pi }_{1}}\right) + F\left( {{\varphi }_{2}{\pi }_{2}}\right) = F\left( {{\varphi }_{1}{\pi }_{1} + {\varphi }_{2}{\pi }_{2}}\right) \) since \( F \) is additive. But \( F\left( {{\varphi }_{1}{\pi }_{1} + {\varphi }_{2}{\pi }_{2}}\right) = F\left( {i}_{{B}_{1} \oplus {B}_{2}}\right) = {i}_{F\left( {{B}_{1} \oplus {B}_{2}}\right) } \) .\n\n(ii) \( \Rightarrow \) (iii) is trivial, so suppose (iii). Suppose \( f, g \in \operatorname{Hom}\left( {B, C}\right) \) . We have diagrams\n\n\n\n\[ B\xleftarrow[]{{\pi }_{1}}B \oplus B\xrightarrow[]{{\pi }_{2}}B \]\n\n\[ f \oplus g = f{\pi }_{1} + g{\pi }_{2} \]\n\n(The formula for \( f \oplus g \) appears in the proof of Proposition 2.1.) Applying \( F \) yields diagrams \n\nBut \( F\left( {B \oplus B}\right) \) is assumed to be a direct sum, hence must be the biproduct in \( {}_{S}\mathbf{M} \) . (Note: There is a technical point here that needs to be addressed. Exercise 1 does just that.)\n\nSince \( F\left( {f \oplus g}\right) \) is the (one and only) filler, \( F\left( {f \oplus g}\right) = F\left( f\right) F\left( {\pi }_{1}\right) + \) \( F\left( g\right) F\left( {\pi }_{2}\right) \) (again by the formula in the proof of Proposition 2.1). Thus, we have that \( F\left( {f{\pi }_{1} + g{\pi }_{2}}\right) = F\left( f\right)
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Yes
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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.
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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, \[ F\left( {P}_{1}\right) \overset{F\left( {d}_{1}\right) }{ \rightarrow }F\left( {P}_{0}\right) \overset{F\left( \pi \right) }{ \rightarrow }F\left( B\right) \rightarrow 0 \] is exact. Hence \( F\left( B\right) \approx F\left( {P}_{0}\right) /\operatorname{im}F\left( {d}_{1}\right) = {\mathcal{L}}_{0}F\left( B\right) \) . That was the first part of the proof of Proposition 3.2, with \( F \) replacing \( A \otimes \), and \( {\mathcal{L}}_{0}F \) replacing \( {\operatorname{Tor}}_{0}\left( {A, \bullet }\right) \) . Contravariant functors work the same way, with some arrows reversed. \( ▱ \)
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Yes
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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 \( {\pi }^{\prime } : {P}_{0} \oplus {P}_{0}^{\prime \prime } \rightarrow {B}^{\prime } \) and \( {d}_{n}^{\prime } : {P}_{n} \oplus {P}_{n}^{\prime \prime } \rightarrow {P}_{n - 1} \oplus {P}_{n - 1}^{\prime \prime } \) such that the diagram is commutative with exact rows and columns. (The vertical maps are the obvious ones.)
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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 \( {\pi }^{\prime }\left( {{x}_{0},{x}_{0}^{\prime \prime }}\right) = {j\pi }\left( {x}_{0}\right) + f\left( {x}_{0}^{\prime \prime }\right) \). In the diagram \( {\pi }^{\prime }\left( {{x}_{0},0}\right) = {j\pi }\left( {x}_{0}\right) \), while \( p{\pi }^{\prime }\left( {{x}_{0},{x}_{0}^{\prime \prime }}\right) = p\left( {{j\pi }\left( {x}_{0}\right) + f\left( {x}_{0}^{\prime \prime }\right) }\right) = {pj\pi }\left( {x}_{0}\right) + {pf}\left( {x}_{0}^{\prime \prime }\right) = 0 + {\pi }^{\prime \prime }\left( {x}_{0}^{\prime \prime }\right) = {\pi }^{\prime \prime }\left( {x}_{0}^{\prime \prime }\right) \). That is, the diagram commutes. Hence \( {\pi }^{\prime } \) is onto by the short 5-lemma (Chapter 2, Exercise 14(b)). To construct \( {d}_{n}^{\prime } \) recursively, given \( {\pi }^{\prime } \) and \( {d}_{0}^{\prime },\ldots ,{d}_{n - 1}^{\prime } \), let \( {K}_{n},{K}_{n}^{\prime },{K}_{n}^{\prime \prime } \) be the \( n \) th kernels: The left hand column is exact by Proposition 6.4 (condition(iv) is satisfied). We now have By the earlier argument constructing \( {\pi }^{\prime } \), the filler \( {d}_{n}^{\prime } \) defined by \( {d}_{n}^{\prime }\left( {x,{x}^{\prime \prime }}\right) = \left( {{d}_{n}\left( x\right) ,0}\right) + f\left( {x}^{\prime \prime }\right) \) has the required properties. (See Exercise 6.)
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No
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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 \( {}_{R}\mathbf{M} \).
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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 \( F\left( B\right), F\left( {B}^{\prime }\right) \), and \( F\left( {B}^{\prime \prime }\right) \) :\n\n\n\nColumns are (split) exact, since, for all \( n,0 \rightarrow {E}_{n} \rightarrow {E}_{n} \oplus {E}_{n}^{\prime \prime } \rightarrow {E}_{n}^{\prime \prime } \rightarrow 0 \) is split exact. Applying Theorem 3.1 yields the long exact sequence of part (d).
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No
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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 \[ \text{-}{\mathcal{R}}^{n + 1}F\left( B\right) \leftarrow \cdots \] \[ {\mathcal{R}}^{n}F\left( B\right) \leftarrow {\mathcal{R}}^{n}F\left( {B}^{\prime }\right) \leftarrow {\mathcal{R}}^{n}F\left( {B}^{\prime \prime }\right) . \] \[ \cdots \leftarrow {\mathcal{R}}^{n - 1}F\left( {B}^{\prime \prime }\right) \] \[ {\mathcal{R}}^{1}F\left( B\right) \leftarrow \cdots \] \[ 0 \leftarrow {\mathcal{R}}^{0}F\left( B\right) \leftarrow {\mathcal{R}}^{0}F\left( {B}^{\prime }\right) \leftarrow {\mathcal{R}}^{0}F\left( {B}^{\prime \prime }\right) . \)
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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 \( F\left( B\right), F\left( {B}^{\prime }\right) \), and \( F\left( {B}^{\prime \prime }\right) \) :\n\n\n\nColumns are (split) exact, since, for all \( n,0 \rightarrow {E}_{n} \rightarrow {E}_{n} \oplus {E}_{n}^{\prime \prime } \rightarrow {E}_{n}^{\prime \prime } \rightarrow 0 \) is split exact. Applying Theorem 3.1 yields the long exact sequence of part (d).
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Yes
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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 \( {\mathcal{R}}_{n}F \) is half exact for all \( n \) .\n\nc) If \( F \) is contravariant, then \( {\mathcal{L}}^{0}F \) is left exact, and \( {\mathcal{L}}^{n}F \) is half exact for all \( n \) .\n\nd) If \( F \) is contravariant, then \( {\mathcal{R}}^{0}F \) is right exact, and \( {\mathcal{R}}^{n}F \) is half exact for all \( n \) .
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Proof: Again they all work out the same way; the first clause after \
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No
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