Computing $R^1f_*\mathcal{O}_\hat{X}(\pm E)$ for blow-up of $\mathbb{A}^2_k$ at the origin

Question

Consider the blowup of the affine plane at the origin: $f:\hat{X}\to X=\mathbb{A}^2_k$. I want to show that $\mathcal{L}:=R^1f_*\mathcal{O}_\hat{X}(\pm E)=0$ in the most elementary way possible. It's easily seen that $\mathcal{L}|_{X-0}=0$, since the morphism is affine when restricted to this open set, so the problem is to show that its stalk at the origin is zero as well.

I've thought about two possible strategies, both which end up having to deal with some cumbersome commutative algebra I'm not comfortable with:

1) Either showing that the sheaf $\mathcal{L}$ is S2 (meaning that $\operatorname{depth}_x(\mathcal{L}_x)=\min(\dim \mathcal{L}_x,2)$ for all $x\in X$, and the depth of a module is the supremum of the lenghts of regular sequences) so that, by this answer in MO, we can conclude that the sheaf $\mathcal{L}|_{X-0}=0$ extends uniquely to the zero sheaf over the whole space.

2) Or on the other hand showing that $\mathcal{O}_\hat{X}(\pm E)$ is flat over $X$. Then all the hypothesis are met so that the stalk of the sheaf at the origin is: $$ \mathcal{L}_0\cong H^1(E,\mathcal{O}_\hat{X}(\pm E)|_E)\cong H^1(\mathbb{P}^1,\mathcal{O}(\mp 1))=0. $$

I guess my question is what is the best approach? Is it one of these two or am I missing something more straighforward? If it's either of these some additional pointers would be appreciated.

Answer 1

I'm not sure what the "best" approach would be - but I think the following argument works:

Consider the short exact sequence $$0\rightarrow \mathcal{O}_{\hat{X}} (-E) \rightarrow \mathcal{O}_{\hat{X}} \rightarrow \mathcal{O}_E \rightarrow 0$$

and apply $f_*$, we get the following long exact sequence

$$0 \rightarrow f_*\mathcal{O}_{\hat{X}}(-E) \rightarrow f_*\mathcal{O}_{\hat{X}} \rightarrow f_*\mathcal{O}_E \rightarrow R^1f_*\mathcal{O}_{\hat{X}} (-E) \rightarrow R^1 f_*\mathcal{O}_{\hat{X}}$$

Note that $f_*\mathcal{O}_{\hat{X}} = \mathcal{O}_{X}$ and $f_*\mathcal{O}_E$ is $\mathcal{O}_p$, so we deduce that

$$R^1f_*\mathcal{O}_{\hat{X}} (-E) \rightarrow R^1 f_*\mathcal{O}_{\hat{X}}$$

is injective. It suffices to show that $R^1 f_*\mathcal{O}_{\hat{X}} = 0$. But this is the sheaf associated to the module $\widetilde{H^1(\hat{X},\mathcal{O}_{\hat{X}}})$, which I think can be computed to be equal to $0$ using Cech cohomology on $\hat{X}$.

Similarly by twisting the above short exact sequence by $\mathcal{O}(E)$ and looking at the associated long exact sequence, we deduce that $R^1f_*\mathcal{O}_X(E) \cong R^1 f_*\mathcal{O}_E(E)$, and you can compute $R^1f_*\mathcal{O}_E(E) = \widetilde{H^1(\hat{X}, \mathcal{O}_{E}(E))} = \widetilde{H^1(E, \mathcal{O}_E(E))}$ rather easily.

Edit: Also I don't think $\mathcal{O}_{\hat{X}}(\pm E)$ is flat over $X$..

Answer 2

One can realize $\hat{X}$ as a hypersurface in $\mathbb{A}^2 \times \mathbb{P}^1$ given by the equation $$ g = xu - yv = 0, $$ where $(x,y)$ are coordinates on $\mathbb{A}^2$ and $(u:v)$ are homogeneous coordinates on $\mathbb{P}^1$. Let $p \colon \hat{X} \to \mathbb{P}^1$ be the projection. Then $$ O_{\hat{X}}(\pm E) \cong p^*O_{\mathbb{P}^1}(\mp 1) $$ (since $Pic(\hat{X})$ is generated on one hand by $O(E)$, on the other by $p^*O(1)$, and the fact that the sign should be the opposite follows from self-intersection of $E$ being $-1$). Thus, you need to compute $R^1f_*p^*O(\mp 1)$.

For this use the Koszul resolution on $\mathbb{A}^2 \times \mathbb{P}^1$: $$ 0 \to O(-1) \stackrel{g}\to O \to O_{\hat{X}} \to 0. $$ Tensoring it by $p^*O(\mp 1)$ and writing the long exact sequence of pushforwards, one gets $$ 0 \to 0 \to 0 \to f_*p^*O(-1) \to O \to 0 \to R^1f_*p^*O(-1) \to 0 $$ and $$ 0 \to O \stackrel{g}\to O \oplus O \to f_*p^*O(1) \to 0 \to 0 \to R^1f_*p^*O(1) \to 0 $$ respectively. In both cases the vanishing of the $R^1f_*$ is clear.