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I have a question about the resolution of this problem: Let $R=\mathbb{F}[x,y]$ be the polynomial ring in two variables over an algebraically closed field $\mathbb{F}$. If $M$ is a module such that $\operatorname{Ann}(M) \supseteq (x-a,y-b)$ , prove that $ \operatorname{Ann} \left( \operatorname{Ext}^i(M,N) \right) \supseteq (x-a,y-b) $ for every $ R$-module $N$. Hint: Consder the maps $ M \ni m \mapsto (x-a) m \in M $ and $ m \mapsto (y-b)m $. Apply then $ \operatorname{Ext}^i(-,N) $

I tried to solve it in the following way but i'm pretty sure that is wrong. Since i think that the way I did it is wrong, then if somebody could give me more hint to understand how to procede in this problem. Thank you very much. And maybe help to understand where are errors in my "solution".

Let consider the following chain $$ F_{\bullet} : M \xrightarrow{\iota} M \xrightarrow{f} M $$ where $ \iota $ is the inclusion map and $ m \mapsto f(m)=(x-a)m $. We saw that is in fact a free resolution since it is exact seaquence since $ \operatorname{im} \iota = M = \ker f $ since $ (x-a) \subseteq \operatorname{Ann}(M)= \{ r \in R : rm=0\} $. Thus applying the contravariant functor to this chain we have $$ \operatorname{Hom}(F_{\bullet},N) : \operatorname{Hom}(M,N) \xleftarrow{\operatorname{Hom}(\iota,N)} \operatorname{Hom}(M,N) \xleftarrow{\operatorname{Hom}(f,N)} \operatorname{Hom}(M,N) $$ And we have that $$ \operatorname{im} \operatorname{Hom}(f,N) = \{ 0 \} $$ since $$ \operatorname{Hom}(M,N) \ni \phi \mapsto \operatorname{Hom}(f,N)(\phi)= \phi \circ f $$ and thus for every $m \in M$ we obtain that $$ \operatorname{Hom}(f,N)(\phi)(m)=(\phi \circ f)(m)=\phi(f(m))=\phi(0) = 0 $$ Moreover we have that $ \ker \operatorname{Hom}(\iota,N) = \{ 0 \} $ since for every $m \in M$ and for every $ \phi \in \operatorname{Hom}(M,N) $ we have that $$ \operatorname{Hom}(\iota,N)(m)=(\phi \circ \iota) (m) = \phi(m) $$ thus $ 0= \phi \circ \iota = \phi $ Then we have that the cohomology group of the cochain complex $ \operatorname{Hom}(F_{\bullet},N) $ above, is by definition the $\operatorname{Ext}(M,N)$ $R$-module thus $$ \operatorname{Ext}(M,N) = H(\operatorname{Hom}(F_{\bullet},N))= \ker\operatorname{Hom}(\iota,N) / \operatorname{im} \operatorname{Hom}(f,N) = 0 $$

Thus $$ \operatorname{Ann} \left( \operatorname{Ext}(M,N) \right) = \{ r \in \mathbb{F}[x,y] : r \phi = 0 , \forall \phi \in \operatorname{Ext}(M,N) \} = \mathbb{F}[x,y] $$

This prove that $ \operatorname{Ann}\left( \operatorname{Ext}(M,N) \right) \supset (x-a) $. Similarly arguments with $ G_{\bullet} : M \xrightarrow{\iota} M \xrightarrow{g} M $, where $ m \mapsto g(m) = (y-b)m $ should prove that $ \operatorname{Ann}\left( \operatorname{Ext}(M,N) \right) \supset (y-b) $.

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  • $\begingroup$ I think you should ask this on mathoverflow. $\endgroup$
    – Tedebbur
    Jul 24, 2021 at 10:36

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