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.

  • 2
    $\begingroup$ 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. $\endgroup$ – KReiser Jul 2 '18 at 1:11
  • $\begingroup$ 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. $\endgroup$ – 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 $.)

  • $\begingroup$ 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. $\endgroup$ – Suzet Jul 2 '18 at 1:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.