I would like to prove that $$Ext^*_{\mathbb F_2C_2}(\mathbb F_2,\mathbb F_2)\cong Ext^*_{\mathbb ZC_2}(\mathbb Z,\mathbb F_2)$$ I was told to use that the tensor functor $-\otimes_{\mathbb ZC_2}\mathbb F_2C_2:\mathbb Z_2C_2-Mod\rightarrow\mathbb F_2C_2-Mod$ is left adjoint of the restriction of scalars (considering a $\mathbb ZC_2$-module as a $\mathbb F_2C_2$-module by reduction mod 2). This yields $$ Hom_{\mathbb F_2C_2}(\mathbb F_2,\mathbb F_2)\cong Hom_{\mathbb ZC_2}(\mathbb Z,\mathbb F_2)$$ as $\mathbb Z\otimes_{\mathbb ZC_2}\mathbb F_2C_2\cong \mathbb F_2$. I therefore would like to prove that $Ext^i_{\mathbb F_2C_2}(\mathbb F_2,\mathbb F_2)\cong Ext^i_{\mathbb ZC_2}(\mathbb Z,\mathbb F_2)$ for all $i$ as it would induce the isomorphism on graded rings. However I have no idea how to do this. I tried using the definition of $Ext$ as the cohomology of the Hom of the projective resolution but didn't have the right modules to use the $Hom$ isomorphism.
Does anyone have an idea? Thanks!
Preparation. It is generally false that $\mathrm{Ext}_{\mathbb{Z}C_2}^i(M,N)\cong\mathrm{Ext}^i_{\mathbb{F}_2C_2}(M\otimes_{\mathbb{Z}C_2}\mathbb{F}_2C_2,N)$ for a $\mathbb{Z}C_2$-module $M$ and an $\mathbb{F}_2C_2$-module $N$, but in a sense it's not far off. Let $\mathscr{F}$ be a free resolution of $M$. Then $$ \mathrm{Ext}_{\mathbb{Z}C_2}^i(M,N)\cong \mathrm{H}_{-i}(\mathrm{Hom}_{\mathbb{Z}C_2}(\mathscr{F},N))\cong\mathrm{H}_{-i}(\mathrm{Hom}_{\mathbb{F}_2C_2}(\mathscr{F}\otimes_{\mathbb{Z}C_2}\mathbb{F}_2C_2,N))\cong\mathrm{Ext}^i_{\mathbb{F}_2C_2}(\mathscr{F}\otimes_{\mathbb{Z}C_2}\mathbb{F}_2C_2,N). $$ (Here we generalize the definition of $\mathrm{Ext}$ to also allow its first input to be a free resolution. You can take the third expression to be a definition of the last one. Also, I don't switch between chain complexes and cochain complexes but consider everything a chain complex.) Now, $\mathscr{F}\otimes_{\mathbb{Z}C_2}\mathbb{F}_2C_2$ is the derived tensor product $M\otimes^\mathbb{L}_{\mathbb{Z}C_2}\mathbb{F}_2C_2$, so we find $\mathrm{Ext}_{\mathbb{Z}C_2}^i(M,N)\cong\mathrm{Ext}^i_{\mathbb{F}_2C_2}(M\otimes^\mathbb{L}_{\mathbb{Z}C_2}\mathbb{F}_2C_2,N))$. (I will later say a bit more about this ''derived adjunction'', because there is a more natural way to phrase it.)
Computations. Now, write $C_2=\{e,\sigma\}$, and define two maps $f,g\colon \mathbb{Z}C_2\to\mathbb{Z}C_2$ by declaring that $f(e)=e+\sigma$ and $g(e)=e-\sigma$. In our case where $M=\mathbb{Z}$ with trivial $C_2$-action, we can then take $\mathscr{F}$ to be the chain complex $$ \require{AMScd} \begin{CD} \ldots @>{f}>> \mathbb{Z}C_2 @>{g}>>\mathbb{Z}C_2 @>{f}>> \mathbb{Z}C_2 @>{g}>> \mathbb{Z}C_2@>>> 0@>>>\ldots \end{CD} $$ with the last copy of $\mathbb{Z}C_2$ in degree 0. You can check that this is indeed a free resolution of $\mathbb{Z}$. If we write $h\colon\mathbb{F}_2C_2\to\mathbb{F}_2C_2$ for the map with $h(e)=e+\sigma$, then $\mathscr{F}\otimes_{\mathbb{Z}C_2}\mathbb{F}_2C_2$ takes the form $$ \require{AMScd} \begin{CD} \ldots @>{h}>> \mathbb{F}_2C_2 @>{h}>>\mathbb{F}_2C_2 @>{h}>> \mathbb{F}_2C_2 @>{h}>> \mathbb{F}_2C_2@>>> 0@>>>\ldots \end{CD} $$ You can check that its homology is given by $\mathrm{H}_i(\mathscr{F}\otimes_{\mathbb{Z}C_2}\mathbb{F}_2C_2)\cong 0$ when $i\neq 0$, and $\mathrm{H}_0(\mathscr{F}\otimes_{\mathbb{Z}C_2}\mathbb{F}_2C_2)\cong\mathbb{F}_2$, where $C_2$ acts trivially on $\mathbb{F}_2$. Therefore, we see that $\mathscr{F}\otimes_{\mathbb{Z}C_2}\mathbb{F}_2C_2$ is a free resolution of $\mathbb{F}_2$ (in other words, $\mathbb{Z}\otimes^\mathbb{L}_{\mathbb{Z}C_2}\mathbb{F}_2C_2\simeq\mathbb{F}_2$), and therefore $$ \mathrm{Ext}_{\mathbb{Z}C_2}^i(\mathbb{Z},\mathbb{F}_2)\cong \mathrm{Ext}^i_{\mathbb{F}_2C_2}(\mathscr{F}\otimes_{\mathbb{Z}C_2}\mathbb{F}_2C_2,\mathbb{F}_2)\cong \mathrm{Ext}^i_{\mathbb{F}_2C_2}(\mathbb{F}_2,\mathbb{F}_2). $$ This solves your question.
Unneeded extra information. Allow me to finish by stating the derived adjunction from above in its proper setting. It may introduce concepts you haven't heard of, but it may be interesting to see. We let $\mathcal{D}(\mathbb{Z}C_2)$ and $\mathcal{D}(\mathbb{F}_2C_2)$ be the derived ($\infty$-)categories of $\mathbb{Z}C_2$ and $\mathbb{F}_2C_2$, respectively. These are basically categories of chain complexes over $\mathbb{Z}C_2$ and $\mathbb{F}_2C_2$, but our natural notion of isomorphism is replaced by quasi-isomorphism, i.e. we consider a map between chain complexes to be an equivalence if it induces isomorphisms on all homology groups. This for instance means that we have an equivalence between any module $M$ (considered as a chain complex in degree 0) and a free resolution of it. Since quasi-isomorphisms are a homotopical notion, the natural setting to study them is in an $\infty$-category, and that is what $\mathcal{D}(\mathbb{Z}C_2)$ is for.
For $A,B\in\mathcal{D}(\mathbb{Z}C_2)$ there is an enriched hom object $\mathrm{hom}_{\mathcal{D}(\mathbb{Z}C_2)}(A,B)\in\mathcal{D}(\mathbb{Z}C_2)$, which basically is the chain complex of maps between $A$ and $B$ (you may know this construction already). Now, if $A=M[0]$ and $B=N[0]$ are actually just $\mathbb{Z}C_2$-modules turned into chain complexes by putting them in degree 0, then $H_{-i}\mathrm{hom}_{\mathcal{D}(\mathbb{Z}C_2)}(M[0],N[0])$ is the Ext-group $\mathrm{Ext}^i_{\mathbb{Z}C_2}(M,N)$, so we can see $\mathrm{hom}_{\mathcal{D}(\mathbb{Z}C_2)}(-,-)$ as a higher object whose lower representation (on homology) recovers $\mathrm{Ext}$-groups. It would therefore be great to state the derived adjunction in terms of these higher hom objects, as it would recover our isomorphisms above by taking homology, showing that our isomoprhisms above are just shadows of a higher adjunction. To do so, as a final ingredient we need the derived tensor product $-\otimes^\mathbb{L}_\mathbb{ZC_2}-$ on $\mathcal{D}(\mathbb{Z}C_2)$. This is the derived functor of the usual tensor product of chain complexes (which namely usually does not preserve quasi-isomorphisms, so we derive it to make sure it does).
The adjunction now takes the form of an equivalence $$\mathrm{hom}_{\mathcal{D}(\mathbb{F}_2C_2)}(K\otimes^\mathbb{L}_{\mathbb{Z}C_2}\mathbb{F}_2C_2,L)\simeq\mathrm{hom}_{\mathcal{D}(\mathbb{Z}C_2)}(K,L),$$ in $\mathcal{D}(\mathbb{Z}C_2)$ for $K\in\mathcal{D}(\mathbb{Z}C_2)$ and $L\in\mathcal{D}(\mathbb{F}_2C_2)$. Upon putting $K=M[0]$ and $L=N[0]$ and taking homology, we get $\mathrm{Ext}_{\mathbb{F}_2C_2}^i(M\otimes^\mathbb{L}_\mathbb{ZC_2}\mathbb{F}_2C_2,N)\cong\mathrm{Ext}_{\mathbb{Z}_2C_2}^i(M,N)$ for a $\mathbb{Z}C_2$-module $M$ and an $\mathbb{F}_2C_2$-module $N$, recovering our previous version of the derived adjunction.