I would like someone to verify my computation of $\operatorname{Ext}^n$.

Problem: Let $p$ be a prime, $k$ a field of characteristic $p$, $G = \langle x \mid x^p = 1 \rangle$, $B = kG$, $S = k(1 + x + \dotsb + x^{p - 1})$. What's $\operatorname{Ext}_B^n(S, S)$?

My answer: Take a projective resolution $P_\bullet$ as follows. $$B \twoheadrightarrow S,\ b \to b(1 + x + \dotsb + x^{p - 1}),$$ $$N \colon B \to B,\ b \mapsto b(1 + x + \dotsb + x^{p - 1}),$$ $$T \colon B \to B,\ b \mapsto b(1 - x),$$ $$P_\bullet \colon \dotsb \stackrel{T}{\to} B \stackrel{N}{\to} B\stackrel{T}{\to} B \stackrel{N}{\to} B\stackrel{T}{\to} B \to 0.$$

Then the induced cochain complex of $k$-vector space is $$ \operatorname{Hom}_B(P_\bullet, S) \colon 0 \to \operatorname{Hom}_B(B, S) \stackrel{\operatorname{Hom}_B(T, S)}{\to} \operatorname{Hom}_B(B, S) \stackrel{\operatorname{Hom}_B(N, S)}{\to}\operatorname{Hom}_B(B, S) \stackrel{\operatorname{Hom}_B(T, S)}{\to} \dotsb. $$ As $(1 + x + \dotsb + x^{p - 1})^2 = p(1 + x + \dotsb + x^{p - 1}) = 0$, $$ \ker\operatorname{Hom}_B(N, S) = \operatorname{Hom}_B(B, S), \qquad \operatorname{im}\operatorname{Hom}_B(N, S) = 0.$$ And as $(1 + x + \dotsb + x^{p - 1})(1 - x) = 1 - x^p = 0$, $$ \ker\operatorname{Hom}_B(T, S) = \operatorname{Hom}_B(B, S), \qquad \operatorname{im}\operatorname{Hom}_B(T, S) = 0.$$

Therefore, $$\operatorname{Ext}_B^0(S, S) = H^0(\operatorname{Hom}_B(P_\bullet, S)) = \ker\operatorname{Hom}_B(T, S)/\operatorname{im}\operatorname{Hom}(0, S) \cong \operatorname{Hom}_B(B, S),$$ $$\operatorname{Ext}_B^{2n + 1}(S, S) = H^{2n + 1}(\operatorname{Hom}_B(P_\bullet, S)) = \ker\operatorname{Hom}_B(N, S)/\operatorname{im}\operatorname{Hom}_B(T, S) \cong \operatorname{Hom}_B(B, S),$$ $$\operatorname{Ext}_B^{2n}(S, S) = H^{2n}(\operatorname{Hom}_B(P_\bullet, S)) = \ker\operatorname{Hom}_B(T, S)/\operatorname{im}\operatorname{Hom}_B(N, S) \cong \operatorname{Hom}_B(B, S).$$

Since $\operatorname{Hom}_B(B, S) \to S,\ f \mapsto f(1)$ gives an isomorphism and $B$ acts $S$ trivially, $\operatorname{Hom}_B(B, S) \cong S \cong k_B$.

In summary, $\operatorname{Ext}_B^{n}(S, S)\cong k_B$ for all $n \geq 0$.

Added: Is there any trick like "casting out nines" to check this kind of computation subsequently? Or is there some interpretation of this outcomes? I cannot see this intuitively. So I cannot check/confirm the result by myself other than recalculation.

  • $\begingroup$ @PedroTamaroff What I can see is an isomorphism $\operatorname{Hom}_B(B, S) \stackrel{\cong}{\to} S, f \mapsto f(1)$. Is this what you suggesting? $\endgroup$ – Orat Jun 12 '15 at 4:18
  • 1
    $\begingroup$ Looks OK. Yes, indeed. $\endgroup$ – Pedro Tamaroff Jun 12 '15 at 4:19
  • 1
    $\begingroup$ You can check the computation (in this special case) by comparing the outcome with the well-known cohomology of the cyclic group $G$. For, as a $kG$-Module, $S$ is just $k$ with trivial $G$-action. Hence $Ext^i_B(S,S)=Ext^i_{kG}(k,k)=H^i(G,k)=k$ for $i \ge 0$. $\endgroup$ – Todd Leason Jun 12 '15 at 8:26
  • $\begingroup$ @ToddLeason I'm not familiar with group cohomology too. But a wikipedia article says $\operatorname{Ext}_{\mathbb{Z}G}^n(\mathbb{Z}, M) = H^n(G, M)$. Does your equation still hold other than integral group ring? $\endgroup$ – Orat Jun 12 '15 at 12:38
  • $\begingroup$ See also Carlson (1996, Proposition 7.1) for a (minimal) projective resolution. $\endgroup$ – Orat Dec 11 '17 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.