The fact that $H^1_{\mathrm{dR}}(\mathbb{A}^1_k) = \bigoplus_{i=0}^{\infty}k$ for $k$ of positive characteristic exemplifies what can go wrong with de Rham cohomology when the characteristic is not zero. I'm trying to do some computations with the next step, and determine the cohomology of the plane, but I'm unsure how to proceed.
Since $X = \mathbb{A}^2_k = \mathrm{Spec}(k[x,y])$ is affine, one can compute $H^{\bullet}_{\mathrm{dR}}$ without recourse to hypercohomology, and instead just take the cohomology of the (global sections of the) de Rham complex: $$k[x,y] \xrightarrow{d_0} \Omega^1_{X/k} \xrightarrow{d_1} \Omega^2_{X/k}$$
It seems that $\mathrm{im}(d_1)$ is generated by all forms $u\mathrm{d}x \wedge v\mathrm{d}y$, where $u,v$ are monomials whose degrees are not both congruent to $-1 \text{ mod } p$, and this would come close to computing $H^2_{\mathrm{dR}}$. But my argument is not rigorous, and I feel like I'm missing something.
The part I'm even more unsure about is how to calculate $H^1_{\mathrm{dR}}$. I know in characteristic $0$ it should be equal to $0$, but I was even having a hard time proving that.
If someone could help me with these calculations, or just provide a good reference, I would appreciate it.