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Let $k$ be a perfect field of characteristic $p>0$. For any positive integers $n$, let $W_n(k)$ be the truncated Witt vectors of length $n$ with coefficients in $k$. For any positive integers $a,b$, is $\textrm{Ext}(W_a(k), W_b(k))$ as an abelian group well understood? Are there any references? Thank you in advance!

Here $\textrm{Ext}(W_a(k), W_b(k))$ is the abelian group that contains all the $W(k)$-modules $M$ (up to isomorphisms) such that $0 \to W_b(k) \to M \to W_a(k) \to 0$ is a short exact sequence of $W(k)$-modules.

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  • $\begingroup$ Are you sure you don't mean to understand the extensions of commutative group schemes over $k$, and not the extensions of $W(k)$-modules (after taking $k$-points)? The former is described in Oort's book, "Commutative Group Schemes", I think, as well as Ch VII of Serre's "Algebraic Groups and Class Fields. The latter seems to depend heavily on the choice of field. $\endgroup$
    – Marty
    Feb 27, 2012 at 19:20
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    $\begingroup$ The short exact sequence $0\to W\to W\to W_a\to 0$, where the map $W\to W$ is multiplication by $p^a$, is a projective resolution of $W_a$ over $W$. Applying $\mathrm{Hom}(-, W_b)$ to this resolution, we get that $\mathrm{Ext}_W(W_a, W_b)$ is just $W_b/p^a W_b$. $\endgroup$
    – Piotr Achinger
    Feb 27, 2012 at 19:24
  • $\begingroup$ I do mean the extensions of $W(k)$-modules. $\endgroup$
    – Xiao
    Feb 27, 2012 at 19:26
  • $\begingroup$ @Piotr Achinger: Thanks! That is the answer I am looking for! $\endgroup$
    – Xiao
    Feb 27, 2012 at 19:44

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Answer reposted from comment:

The short exact sequence $0\to W\to W\to W_a\to 0$, where the map $W\to W$ is multiplication by $p^a$, is a projective resolution of $W_a$ over $W$. Applying $\mathrm{Hom}_W(-, W_b)$ to this resolution, we get that $\mathrm{Ext}_W(W_a, W_b)$ is just $W_b/p^a W_b = W_{min(a, b)}$.

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