Let $\mathbb k$ be a field of characteristic $p$ and let $\mathbb k_n$ be a 1dimensional representation of $\mathbb k^\times$, where the action is given by $t\circ v= t^n v$. Is it known what are the Extgroups $\mathrm{Ext}^1(\mathbb k_n,\mathbb k_m)$ in the category of $\mathbb F_p[\mathbb k^\times]$ modules? I'm mostly interested in the case when $\mathbb k$ is algebraically closed and $n=0$ if this makes things simplier

1$\begingroup$ For $n=0$, $\mathbb{k}_0$ is a direct sum of (infinitely if $\mathbb{k}$ is infinite) many copies of the trivial 1dimensional module $\mathbb{F}_p$, so this boils down to describing $\mathrm{Ext}^1_{\mathbb{F}_p[\mathbb{k}^\times]}(\mathbb{F}_p,\mathbb{k}_m)$. $\endgroup$– YCorMay 30, 2016 at 6:22
2 Answers
By coincidence I needed to know something about this recently, and one thing I know is that that Ext group vanishes for $0 < \vert m  n \vert < p  1$.
There is a proof in Lemma 6.1 of my paper "Cohomology of automorphism groups of free groups with twisted coefficients". The proof is written for plocal integers instead of $\mathbb{F}_p$, but it works in that case too.

$\begingroup$ Now that I think about it, the argument I had in mind might be specific to F_p. $\endgroup$ May 30, 2016 at 10:14

$\begingroup$ Thanks ! So do you mean the case k = F_p? $\endgroup$ May 31, 2016 at 1:57
So returning to the question I think I can prove that $\mathrm{Ext}^1_{\mathbb F_p[\mathbb k^\times]}(\mathbb F_p, \mathbb k_n)=0$ for $n\neq 0$ and $\mathbb k=\mathbb F_q$ or $\mathbb k=\overline{\mathbb F_p}$. First general remark: $\mathrm{Ext}^1_{\mathbb F_p[\mathbb k^\times]}(\mathbb F_p, \mathbb k_n)$ is naturally a subgroup of $\mathrm{Ext}^1_{\mathbb Z[\mathbb k^\times]}(\mathbb F_p, \mathbb k_n)$, which is a subgroup of $H^1(\mathbb k^\times, \mathbb k_n):=\mathrm{Ext}^1_{\mathbb Z[\mathbb k^\times]}(\mathbb Z, \mathbb k_n)$ (considering the short exact sequence $0\rightarrow \mathbb Z \rightarrow \mathbb Z \rightarrow \mathbb F_p \rightarrow 0$ and using the fact that $\mathbb k^{\mathbb k^\times}_n=0)$. So it is enough to prove that $H^1(\mathbb k^\times, \mathbb k_n)=0$. If $\mathbb k=\mathbb F_q$, we have $\mathbb F_q^\times\simeq \mathbb Z/(q1)\mathbb Z$ and $H^1(\mathbb k^\times, \mathbb k_n)=H^1(\mathbb Z/(q1)\mathbb Z, \mathbb k_n)=0$ as it should be killed by $p$ and $q1$ simultaniously. If $\mathbb k =\overline{\mathbb F_p}$, then any element is a root of unity of degree not divisible by $p$ and it is easy to see that $\mathbb k^\times=\mathrm{colim}_{p\nmid m} \mathbb Z/m \mathbb Z = \mathbb Z_{(p)}/\mathbb Z$. For each $n$, we have $ \mathbb k_n^{\mathbb Z/m \mathbb Z}=0$, from this it follows that $H^1(\mathbb k^\times,\mathbb k_n)=H^1(\mathrm{colim}_{p\nmid m} \mathbb Z/m \mathbb Z ,\mathbb k_n)=\lim_{p\nmid m}H^1(\mathbb Z/m \mathbb Z ,\mathbb k_n)=0$ by the same argument, as $(m,p)=1$ for all $m$ in the limit.
Now in general situation if $\mathbb k$ is algebraically closed we have $(\mathbb k^\times)_{\mathrm{tors}}=\overline{\mathbb F_p}^\times=\mathbb Z_{(p)}/\mathbb Z$, moreover the quotient $\mathbb k^\times/(\mathbb k^\times)_{\mathrm{tors}}$ is divisible and torsion free abelian group, so is isomorphic to some vector space $V_{\mathbb Q}$ over $\mathbb Q$ (as an abelian group). Moreover $\mathrm{Ext}_{\mathbb Z}^1(V_{\mathbb Q},\mathbb Z_{(p)}/\mathbb Z)=0$, so $\mathbb k^\times=\mathbb Z_{(p)}/\mathbb Z\times V_{\mathbb Q}$. So from Kunneth it would be enough to know that $H^1(V_{\mathbb Q}, \mathbb k_n)=0$. Seems like this should be true for any $p$torsion module in general. So does anybody know by any chance if it is true that $H^1(\mathbb Q, M)$ for $\mathbb F_p[\mathbb Q]$module $M$?