3 added 13 characters in body

I'm writing this as an "answer" because (a) there are a number of comments, and (b) I don't know if it would fit in a comment.

Let $F$ a finite field, and let $V$ a finite dimensional $F$-vector space, and view $V$ as an $F^\times$-module via multiplication. Then as pointed out in Andy Putman's answer, $H^i(F^\times,V) = 0$ for all $i \ge 0$ provided $|F| > 2$.

Well, it is clear enough under the assumption $|F|>2$ that $H^0(F^\times,V) = V^{F^\times} = 0$. For the higher cohomology vanishing, there is no need to use the description of "cohomology of cyclic groups" to obtain this vanishing; the point is just that $|F^\times|$ is invertible in $F$. Use the following generality:

Let $H$ be a subgroup of finite index $n$ in a group $G$. If $M$ is a $\mathbf{Z}G$-module, then $\operatorname{Cor} \circ \operatorname{Res}$ is multiplication by $n$ on $M$, H^\bullet(G,M)$, where$\operatorname{Cor}:H^\bullet(H,M) \to H^\bullet(G,M)$denotes the corestriction and$\operatorname{Res}:H^\bullet(G,M) \to H^\bullet(H,M)$the restriction; see e.g. Serre's Local Fields VII.7, VIII.2. Let now$k$be a commutative ring (with 1), suppose that$H=1$and that$n = [G:1]= |G|$is invertible in$k$. If$M$is a$kG$-module (i.e. a$k$-module with$k$-linear$G$action), then all$H^i(G,M)$are$k$-modules and$H^i(H,M) = H^i(1,M) = 0$for$i>0$. For$i>0$, the preceding result shows these$k$-modules to be annihilated by the unit$n$of$k$; thus$H^i(G,M) = 0$for$i>0$. To apply this result in the original setting, take$k=F$,$M=V$and$G=F^\times$; we find that$H^i(F^\times,V) = 0$for$i>0$. 2 edited body I'm writing this as an "answer" because (a) there are a number of comments, and (b) I don't know if it would fit in a comment. Let$F$a finite field, and let$V$a finite dimensional$F$-vector space, and view$V$as an$F^\times$-module via multiplication. Then as pointed out in Andy Putnam's Putman's answer,$H^i(F^\times,V) = 0$for all$i \ge 0$provided$|F| > 2$. Well, it is clear enough under the assumption$|F|>2$that$H^0(F^\times,V) = V^{F^\times} = 0$. For the higher cohomology vanishing, there is no need to use the description of "cohomology of cyclic groups" to obtain this vanishing; the point is just that$|F^\times|$is invertible in$F$. Use the following generality: Let$H$be a subgroup of finite index$n$in a group$G$. If$M$is a$\mathbf{Z}G$-module, then$\operatorname{Cor} \circ \operatorname{Res}$is multiplication by$n$on$M$, where$\operatorname{Cor}:H^\bullet(H,M) \to H^\bullet(G,M)$denotes the corestriction and$\operatorname{Res}:H^\bullet(G,M) \to H^\bullet(H,M)$the restriction; see e.g. Serre's Local Fields VII.7, VIII.2. Let now$k$be a commutative ring (with 1), suppose that$H=1$and that$n = [G:1]= |G|$is invertible in$k$. If$M$is a$kG$-module (i.e. a$k$-module with$k$-linear$G$action), then all$H^i(G,M)$are$k$-modules and$H^i(H,M) = H^i(1,M) = 0$for$i>0$. For$i>0$, the preceding result shows these$k$-modules to be annihilated by the unit$n$of$k$; thus$H^i(G,M) = 0$for$i>0$. To apply this result in the original setting, take$k=F$,$M=V$and$G=F^\times$; we find that$H^i(F^\times,V) = 0$for$i>0$. 1 I'm writing this as an "answer" because (a) there are a number of comments, and (b) I don't know if it would fit in a comment. Let$F$a finite field, and let$V$a finite dimensional$F$-vector space, and view$V$as an$F^\times$-module via multiplication. Then as pointed out in Andy Putnam's answer,$H^i(F^\times,V) = 0$for all$i \ge 0$provided$|F| > 2$. Well, it is clear enough under the assumption$|F|>2$that$H^0(F^\times,V) = V^{F^\times} = 0$. For the higher cohomology vanishing, there is no need to use the description of "cohomology of cyclic groups" to obtain this vanishing; the point is just that$|F^\times|$is invertible in$F$. Use the following generality: Let$H$be a subgroup of finite index$n$in a group$G$. If$M$is a$\mathbf{Z}G$-module, then$\operatorname{Cor} \circ \operatorname{Res}$is multiplication by$n$on$M$, where$\operatorname{Cor}:H^\bullet(H,M) \to H^\bullet(G,M)$denotes the corestriction and$\operatorname{Res}:H^\bullet(G,M) \to H^\bullet(H,M)$the restriction; see e.g. Serre's Local Fields VII.7, VIII.2. Let now$k$be a commutative ring (with 1), suppose that$H=1$and that$n = [G:1]= |G|$is invertible in$k$. If$M$is a$kG$-module (i.e. a$k$-module with$k$-linear$G$action), then all$H^i(G,M)$are$k$-modules and$H^i(H,M) = H^i(1,M) = 0$for$i>0$. For$i>0$, the preceding result shows these$k$-modules to be annihilated by the unit$n$of$k$; thus$H^i(G,M) = 0$for$i>0$. To apply this result in the original setting, take$k=F$,$M=V$and$G=F^\times$; we find that$H^i(F^\times,V) = 0$for$i>0\$.