I have an argument about splitting the trivial rank 2 vector bundle. Something is clearly wrong with it but I can't figure it out. Let $\mathbb{P}^2$ be the two-dimensional projective space over $\overline{\mathbb{F}_p}$. We have the following short exact sequence of vector bundles:
$$0\rightarrow \mathcal{O}(-1)\rightarrow \mathcal{O}^2 \rightarrow \mathcal{O}(1)\rightarrow 0$$
$\mathcal{O}^2$ lies in the extension of kernel and co-kernel, so it is an element of $H^1(\mathbb{P}^2,\mathcal{O}(-1)\otimes \mathcal{O}(1)^*)=H^1(\mathbb{P}^2,\mathcal{O}(-2))$. By Serre duality this is equal to $H^1(\mathbb{P}^2,\mathcal{O}(2)\otimes \omega_{\mathbb{P}^2})^*$. So now let the Frobenius act on this. By $n$ times pulling back the short exact sequence along the Frobenius the cohomology group $H^1(\mathbb{P}^2,\mathcal{O}(2)\otimes \omega_{\mathbb{P}^2})$ turns into $H^1(\mathbb{P}^2,\mathcal{O}(2p^n)\otimes \omega_{\mathbb{P}^2})$. If $n$ is large by Serre vanishing theorem this cohomology group vanishes. This implies that short exact sequence splits. So we get a split short exact sequence in the following form:
$$0\rightarrow \mathcal{O}(-p^n)\rightarrow \mathcal{O}^2\rightarrow \mathcal{O}(p^n)\rightarrow 0$$ But such a sequence cannot be split. Because $\mathcal{O}(p^n)$ will have large global sections which cannot be injected into $\mathcal{O}^2$. So we get a contradiction. I'd really appreciate if someone could point me to my silly mistake.