# Tuesday, February 15 :::{.exercise title="?"} Check that $\Ext_R^i(M, \wait)$ is independent of injective resolutions, and $\Ext_R^i(\wait, N)$ is independent of projective resolutions. ::: :::{.exercise} Check that $\cocomplex{\Ext}$ is determined by - $\Ext^0 = \Hom$ - $\Ext^{i>0}(P, I) = 0$ if $P$ is projective or $I$ is injective. - It extends SESs to LESs ::: :::{.exercise title="?"} Show \[ \cocomplex{\Ext}_{k[t]/\gens{t^2} } (k, k) = \bigoplus_{i\geq 0} k[i] .\] Use the projective resolution with entries $k[t]/\gens{t^2}$ with differential $\del = \cdot t$ ::: :::{.exercise title="?"} Compute $\cocomplex{\Ext}_{\kxn}(k, k)$ using the Koszul resolution $\cocomplex{\Extalgebra} \kxn \covers k$. ::: :::{.remark} Defining Noetherian rings and modules: - $A\in \CRing$ is Noetherian iff every $\Id(A) \subseteq \rmod^\fg$. - $M\in \amod$ is Noetherian iff every $N\leq M$ satisfies $N\in \amod^\fg$ > Thank you Emmy Noether!! ::: :::{.exercise title="?"} Show that TFAE: - $R$ is Noetherian - Every $M\in \rmod^\fg$ is Noetherian. ::: :::{.solution} For $1\implies 2$, it STS that $R^n$ is Noetherian. To reduce, use the diagram \begin{tikzcd} && {R^n} && M && 0 \\ \\ {R^n} && {\pi\inv(N)} && N \arrow["\exists", two heads, from=3-1, to=3-3] \arrow[hook, from=3-3, to=1-3] \arrow["\pi", two heads, from=1-3, to=1-5] \arrow[from=3-3, to=3-5] \arrow[hook, from=3-5, to=1-5] \arrow[from=1-5, to=1-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwwLCJSXm4iXSxbNCwwLCJNIl0sWzYsMCwiMCJdLFs0LDIsIk4iXSxbMiwyLCJcXHBpXFxpbnYoTikiXSxbMCwyLCJSXm4iXSxbNSw0LCJcXGV4aXN0cyIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFs0LDAsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMSwiXFxwaSIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFs0LDNdLFszLDEsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMl1d) To show $R^n$ is Noetherian, use induction since we know $R^1$ is Noetherian. Use the following diagram, using the snake lemma on $s_1, s_3$ to show $s_2$ is surjective: \begin{tikzcd} && 0 && {R^{n-1}} && {R^n} && R && 0 \\ \\ && 0 && {N \intersect R^{n-1}} && N && {N \over N \intersect R^{n-1}} && 0 \\ &&&& {\fg \text{ by IH}} && {\therefore \fg} && {\fg \text{ as a submodule of } R} \\ \\ \textcolor{rgb,255:red,92;green,92;blue,214}{0} && \textcolor{rgb,255:red,92;green,92;blue,214}{R^a} && \textcolor{rgb,255:red,92;green,92;blue,214}{R^{a+b}} && \textcolor{rgb,255:red,92;green,92;blue,214}{R^b} && \textcolor{rgb,255:red,92;green,92;blue,214}{0} \arrow[hook, from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[two heads, from=1-7, to=1-9] \arrow[two heads, from=3-7, to=3-9] \arrow[from=3-5, to=3-7] \arrow[hook, from=3-3, to=3-5] \arrow[from=1-9, to=1-11] \arrow[from=3-9, to=3-11] \arrow["{s_1}", color={rgb,255:red,92;green,92;blue,214}, curve={height=-12pt}, two heads, from=6-3, to=3-5] \arrow["{s_3}", color={rgb,255:red,92;green,92;blue,214}, curve={height=-12pt}, two heads, from=6-7, to=3-9] \arrow["{\therefore s_2}", color={rgb,255:red,92;green,92;blue,214}, curve={height=-12pt}, dashed, two heads, from=6-5, to=3-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=6-1, to=6-3] \arrow[color={rgb,255:red,92;green,92;blue,214}, hook, from=6-3, to=6-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, two heads, from=6-5, to=6-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=6-7, to=6-9] \arrow[hook, from=3-5, to=1-5] \arrow[hook, from=3-7, to=1-7] \arrow[hook, from=3-9, to=1-9] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) ::: :::{.exercise title="?"} Show TFAE: - $M \in \amod$ is Noetherian. - The ACC for submodules. ::: :::{.solution} Hint: for $1\implies 2$, set $M_\infty = \Union_i M_i \leq M \in \amod$. Write $M_i = \gens{x_1,\cdots, x_n}$ and use that each $x_k \in M_{i_k}$ to choose $N\gg 1$ with $M_\infty = M_N$. For $2\implies 1$, for $M$ non-Noetherian find $S\leq M$ infinitely generated as $S = \gens{x_1,\cdots}$ and take the chain $\ts{S_k}_{k\geq 0}$ where $S_k = \gens{x_1,\cdots, x_k}$. ::: :::{.exercise title="?"} Show that the following are Noetherian: - Fields - PIDs - $R/I$ for $R$ Noetherian and $I$ arbitrary - For $A$ Noetherian, $A[x]$ (Hilbert's basis theorem).[^invariant_theory] - Similarly $A\fps{x}$. - Localizations of Noetherian rings [^invariant_theory]: A very important result! Marks the end of invariant theory historically. ::: :::{.theorem title="?"} If $A$ a Noetherian local ring and $M \in \amod^{\fg, \proj}$, then $M$ is free. ::: :::{.exercise title="?"} Prove this! ::: :::{.solution} Hints: - Surjectivity: - Choose a basis $M/\mfm M = \gens{\ts{ \bar x}_{k\leq n} }$ and lifts $x_k$ to $M$. - Take a surjective map $A^n \to M$ where $e_i \mapsto x_i$. - Take the SES $A^n \mapsvia{f} M \surjects \coker f\to 0$ and apply $(\wait)\tensor_A A/\mfm$; use that $(A/\mfm)^n \iso M/\mfm M$ and apply Nakayama. - Injectivity: - Write $0\to K = \ker f \to A^n \mapsvia{f} M \to 0$ and apply $(\wait)\tensor_A A/\mfm$ to get \[ \cdots \to \Tor^0(M, A/\mfm) \to K/\mfm K \to (A/\mfm)^n \iso M/\mfm M\to 0 ,\] then $K/\mfm K = 0$ and apply Nakayama to conclude $K=0$. ::: :::{.remark} Try an example: $k[x]\in \mods{k}$, which is free after reduction mod $\mfm = \gens{x}$ but not free before reduction. ::: :::{.remark} Note that this works with projective replaced by flat. ::: :::{.remark} Why care about Noetherian rings? Hilbert studied group actions $G$ on modules over (say) polynomial rings and wanted to find the submodules of $G\dash$invariants. There was an industry of writing down generating sets in order to show existence and finiteness, and the basis theorem (which is partially effective) showed that this is no longer necessary -- finite generating sets always exist. ::: :::{.theorem title="Hilbert's basis theorem"} If $A$ is Noetherian then $A[x]$ is Noetherian. ::: :::{.proof title="?"} Fix $U \subseteq A[x]$ an ideal, so $I = \ts{f\in I\st f = \sum a_{f, i} x^i}$. Write $J = \gens{a_{f, \deg f}}$ be the ideal generated by all leading coefficients of elements in $I$. Since $A$ is Noetherian, $J$ is finitely-generated, so write $J = \gens{a_1,\cdots, a_n}$ and choose $\ts{f_1,\cdots, f_n}$ so that $a_i$ is the leading coefficient of $f_i$. Note that these exist since $J = A\ts{a_1,\cdots, a_n}$ (i.e. these already form an ideal). Consider $L\da I \intersect A[x]^{\deg \leq d}$: since $A$ is Noetherian, $A[x]^{\deg \leq d} \in \amod^\fg$, and $L$ also forms a finitely generated $A\dash$module. Write the generators as $L = \ts{g_1,\cdots, g_m}$. :::{.claim} \[ I = I_{fg} \da \gens{f_1,\cdots, f_n, g_1,\cdots, g_m} .\] ::: :::{.proof title="?"} If not, pick $f\in I\sm I_{fg}$ of minimal degree. Then $\deg f > d$ by construction of $L$. Write $f = \sum b_i x^i$ where $b_{\deg f} = \sum c_i a_i$, and check $f - \sum c_i f_i x^{\deg f - \deg f_i}\in I$. ::: ::: :::{.remark} Idea: the $f_k$ take care of low degree elements, the $g_k$ knock things down in degree. ::: :::{.corollary title="?"} Any quotient $\kxn/I$ is Noetherian, as is $\ZZ[x_1,\cdots, x_n]/I$. ::: :::{.example title="Non-Noetherian rings"} Some examples: - $k[x_1,\cdots, ]$ is not Noetherian: take $I = \gens{x_1,\cdots}$. - $k[t. t^{1\over 2}, t^{1\over 3}, t^{1\over 4},\cdots]$. - $\ZZ\adjoin{ \ts{ 2^{1\over n}}_{n\geq 0} }$. Note this is countable! ::: :::{.remark} Next time: other finiteness conditions, integrality, the Nullstellensatz. :::