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The other day I was reading through some slides I found online about Ext and Tor. One of the examples gave a cursory derivation for a general formula $$ \operatorname{Ext}^i_\mathbb{Z}(\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/n\mathbb{Z})= \begin{cases} \mathbb{Z}/d\mathbb{Z}, &i=0,1\\ \{0\}, &i\geq 2 \end{cases} $$ where $d=\gcd(m,n)$.

So I notice that it's very easy to calculate $\operatorname{Ext}^i_\mathbb{Z}(\mathbb{Z}/(p),\mathbb{Z}/(p))$ for instance. What happens if we change the ring from $\mathbb{Z}$ to $\mathbb{Z}/(p^n)$ where $p$ is a prime? Is there still nice formula for $\operatorname{Ext}^i_{\mathbb{Z}/(p^n)}(\mathbb{Z}/(p),\mathbb{Z}/(p))$ for $i\geq 0$? Since the derivation was very terse, I'm not quite sure how to adapt the method for $\mathbb{Z}/(p^n)$. Thanks.

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Could you give a reference to this slides? –  Norbert Apr 30 '12 at 10:04
    
@Norbert Sure! I found them here, at Algebra Seminar, UWaterloo, Spring 2010. The formula in question is a few slides from the end. –  Yong Pan Apr 30 '12 at 20:40
    
big thanks!${}{}$ –  Norbert Apr 30 '12 at 22:28
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up vote 3 down vote accepted

You can compute Ext via projective resolutions of the first argument. In this case, we have the periodic projective resolution

$\dotsc \xrightarrow{p^{n-1}} \mathbb{Z}/p^n \xrightarrow{\cdot p} \mathbb{Z}/p^n \xrightarrow{\cdot p^{n-1} } \mathbb{Z}/p^n \xrightarrow{\cdot p}\mathbb{Z}/p^n \xrightarrow{\text{pr}} \mathbb{Z}/p \to 0.$

Let us apply $\hom(-,\mathbb{Z}/p)$ to the resolution, this gives (since $\hom(\mathbb{Z}/n,\mathbb{Z}/m) \cong \mathbb{Z}/gcd(n,m))$ the periodic sequence

$\dotsc \xrightarrow{p^{n-1}} \mathbb{Z}/p \xrightarrow{\cdot p} \mathbb{Z}/p \xrightarrow{\cdot p^{n-1} } \mathbb{Z}/p \xrightarrow{\cdot p}\mathbb{Z}/p \to 0$

The maps $\cdot p$ are zero, and the same is true for the other ones - unless $n=1$ but this case is rather boring (since then you can compute Ext of something which is projective, so that Ext vanishes in all positive degrees). If $n$ is larger than $1$, we optain that every Ext group is isomorphic to $\mathbb{Z}/p$.

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Thank you Martin. So to summarize, and please correct me if I misunderstood you, $$\operatorname{Ext}_{\mathbb{Z}/(p^n)}^i(\mathbb{Z}/(p^n),\mathbb{Z}/(p^n))\con‌​g \{0\}$$ for all $i\geq 0$ if $n=1$, but $$\operatorname{Ext}_{\mathbb{Z}/(p^n)}^i(\mathbb{Z}/(p^n),\mathbb{Z}/(p^n))\con‌​g \mathbb{Z}/(p)$$ for all $i\geq 0$ if $n>1$? –  Yong Pan Apr 30 '12 at 20:48
    
Sorry, I don't know why my comment isn't rendering properly. –  Yong Pan Apr 30 '12 at 20:50
    
Yes, except for the case $i=0$, $n=1$, where Ext is just Hom and therefore also Z/p. –  Martin Brandenburg May 1 '12 at 7:08
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