# Cohomology group of the affine plane without origin: need help with algebraic computation

Let $k$ be a field and $X=\mathbb{A}^2_k\setminus \{(0,0)\}$ be the affine plane without origin. I would like to compute the first cohomology group $H^1(X,\mathcal{O})$ of this scheme. For this, I proceeded exactly the same way as in the answer of this related question. Everything is fine until the purely algebraic statement to which the problem reduces. It is the following:

Consider the morphism $d:k[X,Y,X^{-1}]\oplus k[X,Y,Y^{-1}]\rightarrow k[X,Y,X^{-1},Y^{-1}]$ defined by sending the pair $(\frac{P}{X^i},\frac{Q}{Y^j})$ to $\frac{P}{X^i}-\frac{Q}{Y^j}$, where $P,Q\in k[X,Y]$. Then, we have $$k[X,Y,X^{-1},Y^{-1}]/\operatorname{Im}(d)=\operatorname{Span}_k\left(\frac{1}{X^iY^j}\right)_{i>0,j>0}$$

Surprisingly (for me), I fail to show this statement which shouldn't be that much of a problem normally. I understand why we have a surjective map $$\operatorname{Span}_k\left(\frac{1}{X^iY^j}\right)_{i>0,j>0}\rightarrow k[X,Y,X^{-1},Y^{-1}]/\operatorname{Im}(d)$$ however I can't see why injectivity holds. I couldn't show it by direct computation, and I don't see any kind of short elegant argument for it. Any help or hint with this problem would be gladly appreciated.

• Hint: it's enough to show that the map is injective on a basis. Suppose $\frac{1}{X^iYj}$ and $\frac{1}{X^lY^m}$ have the same image. Then what must be true? Write out the equation that they must satisfy. – KReiser Jul 2 '18 at 1:11
• Okay, I managed to write it down eventually, treating all the cases. It is quite a fastidious computing though, but indeed rather straightforward. Thank you for pointing this out. – Suzet Jul 2 '18 at 1:41

It's simply that $$k[X, Y, X^{-1}, Y^{-1}] = {\rm Span}_k \left( \frac{1}{X^iY^j}\right)_{i, j \in \mathbb Z}$$ while $$k[X, Y, X^{-1}] = {\rm Span}_k \left( \frac{1}{X^iY^j}\right)_{i, j \in \mathbb Z {\rm \ with \ }j \leq 0 }$$ and $$k[X, Y, Y^{-1}] = {\rm Span}_k \left( \frac{1}{X^iY^j}\right)_{i, j \in \mathbb Z {\rm \ with \ } i \leq 0 }$$ so \begin{align} {\rm Im}(d) &= {\rm Span}_k \left( \frac{1}{X^iY^j}\right)_{i, j \in \mathbb Z {\rm \ with \ } j \leq 0 } + {\rm Span}_k \left( \frac{1}{X^iY^j}\right)_{i, j \in \mathbb Z {\rm \ with \ } i \leq 0 } \\ &= {\rm Span}_k \left( \frac{1}{X^iY^j}\right)_{i, j \in \mathbb Z {\rm \ with \ }i \leq 0 {\rm \ or \ }j \leq 0}.\end{align} Hence $$k[X,Y,X^{-1},Y^{-1}]/{\rm Im}(d) \cong {\rm Span}_k \left( \frac{1}{X^iY^j}\right)_{i, j \in \mathbb Z {\rm \ with \ }i \geq 1 {\rm \ and \ }j \geq 1}.$$ (And yes, I do mean $i,j\geq 1$, rather than $i,j > 1$.)
• Ah right. Of course, I edited my question to correct the bounds for $i$ and $j$, thank you for your attention. Indeed, written like this, the solution is both convincing and straightforward. Thank you very much. – Suzet Jul 2 '18 at 1:17