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Piotr Achinger
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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_{max(0, b-a)}$$W_b/p^a W_b = W_{min(a, b)}$.

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 $W_b/p^a W_b = W_{max(0, b-a)}$.

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 $W_b/p^a W_b = W_{min(a, b)}$.

Source Link
Piotr Achinger
  • 16.1k
  • 2
  • 49
  • 90

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 $W_b/p^a W_b = W_{max(0, b-a)}$.