Suppose $X$ is a smooth variety over $\mathbb{C}$, why do we have $K_X^*/O_X^*$ is a flasque sheaf? (Beauville "Complex Algebraic Surface" p.28)

(To show the surjection $K_X^*/O_X^*(X)\to K_X^*/O_X^*(U)$, given an element of the latter group, a collection of $(U_I,f_i)$ with $f_i/f_j\in O^*_{U_i\cap U_j}$. How to find the corresponding data for a global section? )

  • $\begingroup$ @GeorgesElencwajg Sorry for the confusion, but in the context he seemed to claim the sheaf $K_X^*/O_X^*$ be flasque for any smooth variety (p.28 Line -9), is there an example that $K_X^*/O_X^*$ not flasque? $\endgroup$ – user206517 Jan 8 '15 at 23:26
  • $\begingroup$ You are right: my apologies. I have upvoted you and deleted my first comment . If you delete yours, I will delete this one too. And, no, there is no example with $K_X^*/O_X^*$ not flasque: Beauville is perfectly right. $\endgroup$ – Georges Elencwajg Jan 9 '15 at 0:29

a) We have-completely tautologically- an exact sequence of sheaves in the Zariski topology $$ 1\to O_X^* \to K_X^* \to K_X^*/O_X^* \to 0 \quad (\bigstar) $$
It is clear that $K_X$ is flasque by definition of "rational function" (= regular function defined on some open subset of $X$ ), and thus that $K_X^*$ is flasque too.
But why is $K_X^*/O_X^*$ flasque?
The secret is to reinterpret that sheaf as the sheaf of Cartier divisors on $X$: $$(K_X^*/O_X^*)(U)=CaDiv(U)$$
And then to use that on a smooth variety Cartier divisors coincide with Weil divisors: $$CaDiv(U)=WDiv(U)$$
So finally we are reduced to show that $WDiv$ is a flasque sheaf on $X$.
But nothing is easier: given a prime divisor $Y\subset U$ we obtain a prime divisor $\bar Y\subset X$ on $X$ just by taking the Zariski closure $\bar Y$ of $Y$ in $X $.

b) Since flasque sheaves are acyclic, the exact sequence $(\bigstar)$ (extended by zeros) provides us with an acyclic resolution of $O_X^*$, which permits us to compute the cohomology of that sheaf.
Since all sheaves of index $\geq 2$ are zero in that resolution we conclude: $$ H^i(X, O_X^*)=0 \quad \operatorname {for all} i\geq 2 . $$

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    $\begingroup$ This is all nicely exposited on in the Tohoku paper, if anyone is curious. $\endgroup$ – Alex Youcis Jan 9 '15 at 22:46
  • $\begingroup$ Ah, interesting @Alex: I am curious. $\endgroup$ – Georges Elencwajg Jan 9 '15 at 23:22
  • $\begingroup$ I just checked Grothendieck's article: the argument Beauville uses and which I extended above is indeed in the Tohoku paper, and was probably to be found there for the first time. It is quite amusing to see Grothendieck express himself in a language that announces schemes but doesn't mention them yet : that would come about two years later ( think). $\endgroup$ – Georges Elencwajg Jan 9 '15 at 23:35
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    $\begingroup$ Yeah, I am moderately sure that is the first appearance. Indeed, in a letter to Serre (found in the famous Correspondence) Grothendieck discusses the subject with much excitement (dare I say pride). This is for the above theorems applciation to show that the canonical map $H^1(X,\text{GL}_n)\to H^1(X,\text{PGL}_{n-1})$ is surjective (since it's cokernel is $H^2(X,\mathcal{O}_X^\times)$!), and thus that every projective bundle comes from a vector bundle (at least on a factorial scheme). Beware though, if you go look for this letter: Grothendieck has it couched in non-modern language $\endgroup$ – Alex Youcis Jan 10 '15 at 0:50
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    $\begingroup$ he speaks of showing $H^2(X,k^\times)=0$ and discusses the application to showing $H^1(X,\text{GL}_n(k))\to H^1(X,\text{PGL}_{n-1}(k))$ is surjective. What he denotes by $k^\times$ is short hand for the sheaf of regular maps to $k^\times$, which is just $\mathcal{O}_X^\times$ (the same goes for the other two sheaves)! I think this notation is also explained in the Tohoku paper. Anywhoo, nice answer Georges! +1 $\endgroup$ – Alex Youcis Jan 10 '15 at 0:51

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