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Let $A= \mathbb{Z}/(pq)$ where $p$ and $q$ are primes, if $M=A/(p)$, $N=A/(q)$ be two $A$-modules. How can I compute $\text{Ext}_{A}^n(M,N)$ in the cases $p=q$ and $p \not= q$.

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    Why did you delete your previous post (with my comment) and ask the same question again ? – Roland Sep 24 '17 at 21:11
  • @Roland was looking for an answer, your answer was useful but it was not something I did not know. –  Sep 24 '17 at 21:47
  • Well, then you should explained what you have tried. I suggested you to construct a projective resolution of $M$, have you tried something in that direction ? – Roland Sep 24 '17 at 21:49
  • @Roland, yes, sorry! I have indeed found a resolution before I posted the question. –  Sep 24 '17 at 21:53
  • That's good, put your resolution in your question and we will see where you get stuck (or someone else cause it's late here). – Roland Sep 24 '17 at 21:54
  • Knowing little about commutative algebra, I would have guessed that when $M$ is $p$-torsion and $N$ is $q$-torsion, the Ext group would necessarily be both $p$- and $q$-torsion, and thus zero (when $p\ne q$). – Lubin Sep 24 '17 at 23:18

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There is the usual surjection $A\to A/p$, and the kernel of this is the ideal $(p)$, that can be covered by multiplication by $p$ to get the beginning of a resolution of $A/p$ $$A\stackrel{p}\to A$$ The kernel of multiplication by $p$ in $A$ is the ideal $(q)$, and this can be covered by multiplication by $q$, to get

$$A\stackrel{q}\to A\stackrel{p}\to A$$ Since the kernel of multiplication by $q$ is $(p)$, you obtain an infinite free periodic resolution $$\cdots A\stackrel{q}\to A\stackrel{p}\to A\stackrel{q}\to A\stackrel{p}\to A$$ If you apply $\hom_A(?,A/q)$ to this we obtain a cochain complex using the natural isomorphism $\hom_A(A,-)\simeq -$, $$A/q \stackrel{p}\to A/q\stackrel{q}\to A/q\stackrel{p}\to\cdots.$$ Multiplication by $q$ is manifestly $0$ on $A/q$, but there are two cases to consider if $p=q$ or not.

Pedro
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  • Thank you for your answer, I don't understand the last part using the natural isomorphism, could you explain that? I get the actual isomorphism but why does the maps in the cochain complex become $p$ and $q$? –  Sep 25 '17 at 22:03
  • @sodan Use the natural isomorphism to se what the induced map must be. – Pedro Sep 26 '17 at 10:43