I just finished my first time learning étale cohomology, and now would like to compute some simple examples, here is one :
Let $k=\overline{\mathbb{F}_p}$, and let $X$ be the curve $\text{Spec}\,k[X,Y]/(Y^2-X^2(X-1))\subseteq \mathbb{A}_k^2$, what is the first étale cohomology group $H^1_{ét}(X,\underline{\mathbb{Z}}_X)$ of the constant sheaf $\underline{\mathbb{Z}}_X$ on $X$ ?
Here is what I try to do. Let $f:\mathbb{A}_k^1\rightarrow X$ be the normalization morphism, then as $f$ is finite, we have $R^qf_*\mathcal{F}=0$ for every étale sheaf $\mathcal{F}$ on $\mathbb{A}_k^1$ and $q\geq 1$, hence $H^1_{ét}(\mathbb{A}_k^1,\mathcal{F})=H^1_{ét}(X,f_*\mathcal{F})$. So if there's a $\mathcal{F}$ with $f_*\mathcal{F}=\underline{\mathbb{Z}}_X$ we will be happy. My first thought was to take $\mathcal{F}=\underline{\mathbb{Z}}_{\mathbb{A}_k^1}$, but I realized that $f$ corresponds to the map
$$f:\mathbb{A}_k^1\rightarrow X,\,\,\,\,\,\,\,\,t\mapsto (t^2+1,t(t^2+1))$$
who is one-to-one except that the two points $t=\pm 1$ maps to $(0,0)$, so
$$(f_*\underline{\mathbb{Z}}_{\mathbb{A}_k^1})_{\overline{(0,0)}}=\mathbb{Z}\oplus\mathbb{Z},$$
that is, this choice doesn't work.
Now I don't know how to continue. I would be grateful to anyone answering this question, or, giving any hint.
Best Answer
Are you aware of the fact that $H^1_{et}(X,\underline{\mathbb{Z}})$ is not a 'good group' to consider? For example, if $X$ is normal then it's zero since it's $\mathrm{Hom}_\text{cont}(\pi_1^{et}(X),\mathbb{Z}))$, and since $\pi_1^{et}(X)$ is profinite.
For your example you can see the proof of Angelo Vistoli here--it's essentially the same technique you were trying to apply. Perhaps the only non-obvious points in his explanation are why $H^1_{et}(X,\pi_\ast \underline{\mathbb{Z}}))$ and $H^1_{et}(X,i_\ast\underline{\mathbb{Z}})$ are trivial. But, since $\pi$ and $i$ are finite morphisms, they're acyclic for the etale topology, and so agree with $H^1_{et}(\mathbb{A}^1,\underline{\mathbb{Z}})$ and $H^1_{et}(\mathrm{pt},\underline{\mathbb{Z}})$ respectively. But, since $\mathbb{A}^1$ and $\mathrm{pt}$ are normal, I already explained why these are zero.