# Thursday, February 10 :::{.exercise title="?"} Let $R= k[x,y]$ and $M = k \cong k[x,y]/\gens{x, y}$, and compute $\complex{\Tor}^{\rmod}(k, k)$. ::: :::{.solution} Hint: use the following resolution \begin{tikzcd} && 1 && {(y,-x)} && 1 && 1 \\ 0 && {k[x,y]} && {k[x,y]\tensorpower{R}{2}} && {k[x,y]} && k && 0 \\ && {} && {e_1} && x \\ &&&& {e_2} && y \arrow[from=2-1, to=2-3] \arrow[from=2-3, to=2-5] \arrow[from=2-5, to=2-7] \arrow[from=2-7, to=2-9] \arrow[from=2-9, to=2-11] \arrow[maps to, from=1-3, to=1-5] \arrow[maps to, from=3-5, to=3-7] \arrow[maps to, from=4-5, to=4-7] \arrow[maps to, from=1-7, to=1-9] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) ::: :::{.exercise title="?"} Compute $\complex{\Tor}^{\rmod}(k, k)$ for $R=\kxn$ and $M = k = \kxn/\gens{x_1,\cdots, x_n}$. ::: :::{.remark} $\Tor$ measures failure of injectivity of tensoring against a module $M$, $\cocomplex{\Ext}$ measures failure of surjectivity when mapping against $M$. ::: :::{.exercise title="?"} Show that for $R\in\CRing$, every $M\in\rmod$ admits an injective resolution. Hint: it suffices to show any $M$ injects into an injective object. Use the following diagram: \begin{tikzcd} 0 && M && {I^0} && {I^0/M} && 0 \\ \\ &&&& {I'} \\ \\ &&&& {\coker f} && {I^2} & \cdots \\ \\ &&&& 0 \arrow[from=1-1, to=1-3] \arrow["\exists", color={rgb,255:red,92;green,92;blue,214}, hook, from=1-3, to=1-5] \arrow[two heads, from=1-5, to=1-7] \arrow["\exists", color={rgb,255:red,92;green,92;blue,214}, hook, from=1-5, to=3-5] \arrow[two heads, from=3-5, to=5-5] \arrow["\exists", color={rgb,255:red,92;green,92;blue,214}, hook, from=5-5, to=5-7] \arrow[from=5-5, to=7-5] \arrow["{\therefore \exists}", dashed, from=1-3, to=3-5] \arrow[from=1-7, to=1-9] \arrow["{\therefore \exists}", dashed, from=3-5, to=5-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) First show this for $R=\ZZ$ using that $\QQ\sumpower{J}/K$ is divisible and thus injective: \begin{tikzcd} && 0 & 0 && 0 \\ \\ && 0 & K && K & 0 \\ \\ && 0 & {\ZZ\sumpower{J}} && {\QQ\sumpower{J}} & {\coker g} \\ \\ 0 && \textcolor{rgb,255:red,214;green,92;blue,92}{\ker f} & M && {\QQ\sumpower{J}/K} & {\coker f} \\ \\ &&& 0 && 0 & 0 \arrow[from=1-4, to=3-4] \arrow[from=1-6, to=3-6] \arrow[hook, from=3-4, to=5-4] \arrow[hook, from=3-6, to=5-6] \arrow[two heads, from=5-4, to=7-4] \arrow[two heads, from=5-6, to=7-6] \arrow[from=7-4, to=9-4] \arrow[from=7-6, to=9-6] \arrow[hook, two heads, from=3-4, to=3-6] \arrow["g", hook, from=5-4, to=5-6] \arrow[from=7-1, to=7-3] \arrow[from=7-3, to=7-4] \arrow["f", from=7-4, to=7-6] \arrow[from=3-3, to=5-3] \arrow[from=5-3, to=7-3] \arrow[from=3-7, to=5-7] \arrow[from=5-7, to=7-7] \arrow[dotted, from=7-3, to=3-7] \arrow[from=5-3, to=5-4] \arrow[from=5-6, to=5-7] \arrow[from=7-6, to=7-7] \arrow[from=3-3, to=3-4] \arrow[from=3-6, to=3-7] \arrow[from=7-7, to=9-7] \arrow[from=1-3, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Now reduce to $R=\ZZ$ using that $M\injects D$ for $D$ a divisible abelian group, and show $M\injects I \da \Hom_{\zmod}(R, D)\in\rmod$. Form the map as the composition: \begin{tikzcd} && 0 && M && D \\ \\ 0 && M && {\Hom_\ZZ(R, M)} && {\Hom_\ZZ(R, D)} \\ && m && {\mult_m: r\mapsto rm} \\ &&&& f && {i\circ f} \arrow[""{name=0, anchor=center, inner sep=0}, "i", hook, from=1-5, to=1-7] \arrow["\exists", dashed, from=3-3, to=3-5] \arrow[""{name=1, anchor=center, inner sep=0}, "{i^*}", hook, from=3-5, to=3-7] \arrow[maps to, from=4-3, to=4-5] \arrow[maps to, from=5-5, to=5-7] \arrow[from=1-3, to=1-5] \arrow[dashed, from=3-1, to=3-3] \arrow["\iota", curve={height=-24pt}, dotted, from=3-3, to=3-7] \arrow["{\Hom_\ZZ(R, \wait)}", shorten <=9pt, shorten >=9pt, Rightarrow, from=0, to=1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Then show that $I\da \Hom_{\zmod}(R, D)$ is injective using the universal property: \begin{tikzcd} 0 & X & Y \\ \\ {\Hom_{\zmod}(R, D)} \\ \\ {\Hom_{\rmod}(Y, \Hom_{\zmod}(R, D))} && {\Hom_\ZZ(Y, D)} \\ f && {y\mapsto f(y)(1)} \\ {y\mapsto(r\mapsto g(ry))} && g \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-1, to=3-1] \arrow["{?}"', dashed, from=1-2, to=3-1] \arrow[tail reversed, from=5-1, to=5-3] \arrow[maps to, from=6-1, to=6-3] \arrow[maps to, from=7-3, to=7-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) ::: :::{.exercise title="?"} Show $\ZZ\in \zmod$ has an injective resolution. ::: :::{.solution} \[ 0 \to \ZZ \injects \QQ \surjects \QQ/\ZZ \to 0 .\] ::: :::{.exercise title="?"} Show $\cocomplex{\Ext}_{\zmod}(C_3, \ZZ) \cong C_3[1]$. > Hint: apply $\Hom_{\zmod}(C_3, \wait)$ to the above resolution and use $\Hom_{\zmod}(C_3, \QQ/\ZZ) \cong C_3$ ::: :::{.exercise title="?"} Show that if $f\in \rmod(M, N)$ then there is an induced chain map $\tilde f\in \Ch\rmod(\cocomplex{I}_M, \cocomplex{I}_N)$ which is unique up to homotopy. Conclude that $\cocomplex{\Ext}_{\rmod}(M, N)$ is independent of injective resolution. > Hint: take $f=\id_M$. ::: :::{.exercise title="?"} Show that if $0\to A\to B\to C\to 0$ is a SES in $\rmod$ with $\cocomplex{I}_A \cocovers A, \cocomplex{I}_C\cocovers B$, then there is a complex $\cocomplex{I}_B\cocovers B$ making $0\to \cocomplex{I}_A \to \cocomplex{I}_B \to \cocomplex{I}_C \to 0$ a SES in $\Ch\rmod$. ::: :::{.exercise title="?"} Show that a SES in $\rmod$ induces a LES in $\cocomplex{\Ext}_{\rmod}$. Do this for both homs: start with $0\to A\to B\to C\to 0$, and produce a LES for $\cocomplex{\Ext}_{\rmod} (M, \wait)$ and $\cocomplex{ \Ext}_{\rmod}(\wait, M)$. > Hint: for the first case, apply $\Hom_{\rmod}(M, \wait)$ to the SES of chain complexes of injective resolutions, and use that if $I_1$ is injective then the SES splits. > For the second, use that $\Hom(\wait, I)$ is exact iff $I$ is injective and take an injective resolution of $M$. ::: :::{.exercise title="?"} Show that $\cocomplex{ \Ext}_{\rmod}(M, N)$ is uniquely characterized by 1. $\Ext^0_{\rmod}(M, N) \cong \Hom_{\rmod}(M, N)$ 2. $\tau_{\geq 1} \cocomplex{\Ext}_{\rmod}(M, N) = 0$ if $M$ is projective or $N$ is injective. 3. The two LESs above exist. ::: :::{.solution} Hints: - Resolve $\cocomplex{I}_N \cocovers N$, apply $\Hom_{\rmod}(M,\wait)$, and identify $\Ext^0_{\rmod}$ as a kernel. - $N$ is its own injective resolution when $N$ is injective. - $\Hom_{\rmod}(M,\wait)$ is exact when $M$ is projective. - For uniqueness, use that if $0\to N\to I \to C\to 0$ with $I$ injective, then the middle terms in the LES vanish to get isomorphisms. :::