# Thursday, April 21 ## Artin Rees :::{.exercise title="?"} Show that if $A\in\CRing$ and $m\in \mspec A$ then $A\complete{m}$ is local. ::: :::{.solution} Show that $x\in A\complete{m}\sm \ker(A\complete{m} \mapsvia{\pi} A/m)$ is invertible. If $\pi(x) = 1$ then $1-x\in \ker\pi$ and $y = {1\over 1-(1-x)} = \sum_k (1-x)^k$ is an inverse. ::: :::{.remark} Note that $\alpha\dash$adic completion is not generally exact, but *is* exact in most cases of interest, e.g. for $M\in\amod^\fg$. There is a derived functor used in e.g. Bhatt-Scholze. ::: :::{.theorem title="Artin-Rees"} For $M\in \amod^\fg$ for $A\in \Noeth\CRing$ with $M'\leq M$ and $\alpha\in \Id(A)$, the $\alpha\dash$adic topology on $M'$ coincides with the topology on $M'$ induced by the $\alpha\dash$adic topology on $M$. ::: :::{.corollary title="?"} Taking the $\alpha\dash$adic completion for finitely-generated modules on Noetherian rings is exact. ::: :::{.proof title="?"} This coincides with the induced topology, and we showed that taking the *induced* completions is exact. ::: :::{.definition title="?"} For $M\in \amod$ and $\Fil M$ a descending filtration, - $\Fil M$ is **compatible** with $\alpha$ if $\alpha \Fil_i M \subseteq \Fil_{i+1} M$. - Equivalently, $\gr_i M\in \mods{\gr A}$. - $\Fil M$ is **$\alpha\dash$good** if for $i\gg 0$ this is an equality. - Equivalently, $\gr_i M\in \mods{\gr A}^{\fg}$. ::: :::{.observation} Recall that for Hilbert dimension, we took a good filtration and used the eventual degree. If $\Fil M, \Fil' M$ are two $\alpha\dash$good filtrations, \[ \Fil'_{i-k} M \subseteq \Fil_i M \subseteq \Fil'_{i-k} M .\] As a corollary, any two $\alpha\dash$good filtrations induce the same topology, since this yields a containment of basis elements. ::: :::{.theorem title="Artin-Rees reprised"} If $M, M', A, \alpha$ as in the original statement above, $M' \intersect \alpha^i M$ is $\alpha\dash$good. ::: :::{.remark} Why this implies the previous version: the induced filtration is $\alpha\dash$good and induces the same topology as the $\alpha\dash$adic filtration. ::: :::{.definition title="Rees algebra/blowup algebra"} For $\alpha\in \Id(A)$, define the **Rees algebra** \[ \Rees A \da A \oplus \alpha[1] \oplus \alpha^2[2] \oplus \cdots .\] ::: :::{.exercise title="?"} Show that if $A$ is Noetherian then $\Rees A$ is Noetherian. ::: :::{.solution} Show that if $\ts{a_i}$ generate $\alpha$, there is a surjection $A[x_1,\cdots, x_n] \surjects (\Rees A)[1]$ where $x_i\mapsto a_i$ and apply the Hilbert basis theorem. ::: :::{.definition title="Rees modules"} Let $(M, \Fil)$ be an $\alpha\dash$compatible filtered $A\dash$module. Define \[ \Rees M \da M \oplus (\Fil_1 M)[1] \oplus (\Fil_2 M)[2] \oplus \cdots \qquad \in \mods{\Rees A} .\] ::: :::{.proposition title="?"} Let $A\in\Noeth\CRing$, then TFAE: - $M\in \mods{A}^\fg$ with an $\alpha\dash$good filtration, - $\Rees M \in \mods{\Rees A}^\fg$. ::: :::{.proof title="?"} $2\implies 1$: restrict generators to degree zero, then use that finitely generated implies that generators are in bounded degree to get $\alpha\dash$goodness. $1\implies 1$: find $i_0$ such that $\alpha M_i = M_{i+1}$ for $i\geq i_0$, so $\Rees M$ is generated by $\tau_{\leq i_0}\Rees M$. Each $M_i$ is finitely generated since it's a submodule of a finitely-generated module over a Noetherian ring. ::: :::{.proof title="of Artin-Rees reprised"} Let $\Rees M'$ be the induced filtration, then $\Rees M$ is $\alpha\dash$good and finitely-generated over $A^*$ and we want to show $\Rees M' \subseteq \Rees M$. Conclude using that $\Rees A$ is Noetherian. ::: ## Injections into completions :::{.corollary title="?"} If $A\in \Loc\Noeth\CRing$ with $m\in \mspec A$ and $M\in \amod^\fg$, \[ M \injects M\complete{m} .\] Note that this is not true for arbitrary modules! ::: :::{.proof title="?"} Let $N$ be the kernel, use: \begin{tikzcd} N && M \\ \\ {\hat N} && {\hat M} \\ \\ {N/mN} \arrow["0", from=1-1, to=3-1] \arrow["0", from=1-1, to=3-3] \arrow[from=1-3, to=3-3] \arrow[from=3-1, to=3-3] \arrow[from=1-1, to=1-3] \arrow["0"', curve={height=30pt}, dashed, from=1-1, to=5-1] \arrow[from=3-1, to=5-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCJOIl0sWzAsMiwiXFxoYXQgTiJdLFsyLDAsIk0iXSxbMiwyLCJcXGhhdCBNIl0sWzAsNCwiTi9tTiJdLFswLDEsIjAiXSxbMCwzLCIwIl0sWzIsM10sWzEsM10sWzAsMl0sWzAsNCwiMCIsMix7ImN1cnZlIjo1LCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMSw0XV0=) So $N/mn = 0 \implies N=0$ by Nakayama. ::: :::{.example title="of when injectivity fails"} Let $A = k[t]$ and $M = k[t]/\gens{t-1}$ with $m = \gens{t}$. Then $M/t^i M = 0$ for all $i$, so $M\complete{m} = 0$. ::: :::{.exercise title="?"} Show that for $A\in \Noeth\CRing, M\in \amod^\fg, \alpha\in \Id(A)$, \[ \hat{A} \tensor_A M \iso \hat{M} .\] ::: :::{.solution} Hint: $(a_i)_{i\in I}\tensor m \mapsto (a_i m)_{i\in I}$. Now carry out a diagram chase and use the 5 lemma (or snake lemma) on the following diagram: \begin{tikzcd} 0 & {M'} & {A^n} & M & 0 \\ \\ {} & {M'\tensor_A \hat{A}} & {\hat{A}^n} & {M\tensor\hat{A}^n} & 0 \\ \\ 0 & {\hat{M'}} & {\hat{A}^n} & {\hat{M}} & 0 \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=5-1, to=5-2] \arrow[from=5-2, to=5-3] \arrow[from=5-3, to=5-4] \arrow[from=5-4, to=5-5] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow["\therefore", two heads, from=3-2, to=5-2] \arrow[from=3-3, to=5-3] \arrow["\therefore", two heads, from=3-4, to=5-4] \arrow["{(\wait)\tensor_A \hat{A}}"{description}, squiggly, from=1-3, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) ::: :::{.exercise title="?"} Show that if $A\in \Noeth\CRing$ and $\alpha\in \Id(A)$ then $\hat{A}\in \amod^\flat$. ::: :::{.solution} Hint: take $0\to M_1\to M_2\to M_3\to 0$, tensor and use the proposition, then it STS $0\to \hat M_1\to \hat M_2\to \hat M_3\to 0$ is exact. It suffices to check flatness for $M_i\in \amod^\fg$ by a previous HW exercise. ::: ## Motivations :::{.theorem title="Cohen structure theorem"} If $k = \bar{k} \in \Field$ and $A\in \Noeth\kalg$ is *regular* which is local and complete with respect to its maximal ideal, then $A \iso k\fps{x_1,\cdots, x_n}$. ::: :::{.example title="?"} A non-regular example: take a complete local ring like $k\fps{x, y}/\gens{xy}$ and localize to zoom in on $\vector 0\in \AA^2\slice k$. An important example in algebraic curves: if $A\slice k$ for $A$ Dedekind and $m\in \mspec A$, the completion is $\hat{A} \cong k\fps{x}$. ::: :::{.slogan} Here *regular* means smooth, so $\T_m\dual = \dim A$. ::: :::{.remark} A general pattern for studying rings: - Start with a ring $A$. - Localize to achieve smoothness. - Complete to reduce to questions about power series. ::: :::{.remark} What would show up in a 2nd course on commutative algebra: singularity theory. See Grothendieck duality, Cohen–Macaulay rings. :::