Suppose that $M$ is a free, rank $n$ $R$-module and furthermore, suppose a group $G$ acts on $M$ and the $G$-action commutes with the $R$-action. Hence we have a homomorphism $\rho: G \rightarrow \text{GL}_n(R)$. This in turn gives us a $G$-action on $\text{End}_R(M)$ via $g\cdot \phi = \rho(g) \phi \rho(g)^{-1}$. Now is $$\text{Ext}^1_{\mathbb{Z}[G]}(\mathbb{Z}, \text{End}_R(M)) \cong \text{Ext}^1_{R[G]}(M,M)?$$ I know that the sets are in bijection via a rather roundabout argument, but I am wondering if there is a simple group isomorphism hiding here. Also, are there corresponding isomorphisms of higher Ext groups?
Isomorphism with Ext functor
homological-algebrahomology-cohomology
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$\DeclareMathOperator{\Aut}{Aut} \newcommand{\p}{\pi_1X} \newcommand{\Z}{\mathbb{Z}} \newcommand{\XX}{\tilde{X} }$Ok, I finally figured it out! So, the cellular chain complex of $\XX$ is given by $$ 0\longrightarrow \Z[t, t^{-1}] \overset{\delta}\longrightarrow \Z[t, t^{-1}] \longrightarrow 0, $$ where the boundary map $\delta$ is just multiplication by $t-1$, for geometrical reasons.
The new boundary map $D:\Z_3\to\Z_3$ which we obtain afte tensoring with $\Z_3$ over $\Z[t,t^{-1}]$ can be computed by looking at the sequence of maps $$ \Z_3 \overset{\cong}\longrightarrow \Z[t,t^{-1}]\underset{\Z[t,t^{-1}]}\otimes\Z_3 \; \overset{\delta\otimes id}\longrightarrow \Z[t,t^{-1}]\underset{\Z[t,t^{-1}]}\otimes\Z_3 \overset{\cong}\longrightarrow \Z_3, $$ where the first map is $a\mapsto 1\otimes a$ and last map is $1\otimes a \mapsto a $. Of course the point is that we need reduce to the form $1\otimes a$ before applying the last map. Therefore we get $$ a \mapsto 1\otimes a \mapsto (t-1)\otimes a = 1 \otimes (ta - a) \mapsto ta - a. $$ Hence by a direct computation $D$ turns out to be the identity of $\Z_3$: $$ D(0) = 0, \quad D(1)=t\cdot1 - 1 = 2-1 = 1, \quad D(2)=t\cdot 2 - 2 = 1-2 \equiv 2 \mod 3. $$
Therefore we conclude that the homology groups of $S^1$ with coefficients in $\Z_3$ twisted by the nontrivial $\rho$ are trivial: $$ H_0(S^1\Z_3)_\rho \cong H_1(S^1\Z_3)_\rho \cong 0 $$
Taking an explicit free resolution of $\mathbb{Q}$ is not impossible but difficult. Instead we can use the functoriality of $\operatorname{Ext}^i$ to determine $\operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2)$. Consider the map $$ \varphi\colon \operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2)\to \operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2);\quad x\mapsto 2x. $$ This map is induced by the isomorphism $\mathbb{Q}\to \mathbb{Q};\;x\mapsto 2x$, so $\varphi$ is an isomorphism. However, it is also induced by the zero map $\mathbb{Z}/2\to \mathbb{Z}/2;\;x\mapsto 2x$, so $\varphi$ is a zero map. Putting these together we get $$ \operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2)=0 $$ for all $i\geq 0$.
Best Answer
First let $A$ be any $\mathbb{Z}[G]$-module, and $M$ and $N$ any $R[G]$-modules.
There is a natural isomorphism of $R$-modules $$\operatorname{Hom}_{\mathbb{Z}[G]}\bigl(A,\operatorname{Hom}_R(M,N)\bigr)) \cong \operatorname{Hom}_{R[G]}\bigl(A\otimes_{\mathbb{Z}}M,N\bigr)$$ (where $G$ acts on $\operatorname{Hom}_R(M,N)$ by $(g\cdot\varphi)(m)=g\varphi(g^{-1}m)$ and on $A\otimes_{\mathbb{Z}}M$ by $g\cdot(a\otimes m)=(ga)\otimes(gm)$), given by $$\theta\mapsto[a\otimes m\mapsto \theta(a)(m)]$$ with inverse $$\psi\mapsto [a\mapsto[m\mapsto\psi(a\otimes m)]].$$
From now on assume that $A$ is a projective $\mathbb{Z}[G]$-module and $M$ is projective as an $R$-module. Then as functors from $R[G]$-modules to $R$-modules, $$\operatorname{Hom}_{R[G]}(A\otimes_{\mathbb{Z}}M, -)\cong \operatorname{Hom}_{\mathbb{Z}[G]}\bigl(A,\operatorname{Hom}_R(M,-)\bigr),$$ which is exact, so $A\otimes_{\mathbb{Z}}M$ is a projective $R[G]$-module.
Applying these functors of $A$ to a projective resolution $P_*$ of the $\mathbb{Z}[G]$-module $\mathbb{Z}$, we get isomorphic chain complexes $$\operatorname{Hom}_{\mathbb{Z}[G]}\bigl(P_*,\operatorname{Hom}_R(M,N)\bigr)) \cong \operatorname{Hom}_{R[G]}\bigl(P_*\otimes_{\mathbb{Z}}M,N\bigr).$$
The homology of the first one is $\operatorname{Ext}^*_{\mathbb{Z}[G]}\bigl(\mathbb{Z}, \operatorname{Hom}_R(M,N)\bigr)$, and (since $P_*\otimes_{\mathbb{Z}}M$ is a projective resolution of$M$ as an $R[G]$-module) the homology of the second is $\operatorname{Ext}^*_{R[G]}(M,N)$.
In particular, if $M=N$ this gives natural isomorphisms of $R$-modules $$\operatorname{Ext}^n_{\mathbb{Z}[G]}\bigl(\mathbb{Z},\operatorname{End}_R(M)\bigr)\cong \operatorname{Ext}^n_{R[G]}(M,M)$$ for every $n$.