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I am trying to prove that the following is a short exact sequence $$ 0 \rightarrow H^0(X,\Omega_X) \rightarrow H^1_{\text {dR}}(X/k) \rightarrow H^1(X,\mathcal O_X) \rightarrow 0, $$ where $k$ is an algebraically closed field and $X$ is a smooth projective curve.

Using Cech cohomology I have derived the long exact sequence $$ 0 \rightarrow H^0_{\text{dR}}(X/k) \rightarrow H^0(X,\mathcal O_X) \rightarrow H^0(X,\Omega_x) \rightarrow H^1_{\text{dR}}(X/k) \rightarrow H^1(X,\mathcal O_X) \rightarrow H^1(X,\Omega_X) \rightarrow H^2_{\text{dR}}(X/k) \rightarrow 0 $$ and if I can show the first and last (non-trivial) maps are isomorphisms I am done.

I can show the injectivity of the short exact sequence anyway, but the surjectivity is eluding me.

Any help would be useful, but in particular I thought there may be some duality theorem. Since $H^1(X,\Omega_X) = k$ then all I need to do is show $H^2_{\text {dR}} (X/k) \neq 0$ and I will be done.

Since I know that $H^0_{\text {dR}}(X/k)$ is non-zero I thought a duality theorem may complete this, but I can only find analytic duality theorems, and certainly none that clearly hold in positive characteristic.

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"Since $H^1(X,O_X)=k \ldots$" Not true; did you mean to have an $\Omega$ there instead? – user64687 Feb 7 '14 at 14:18
Yep, thank you for pointing that out – Joe Tait Feb 7 '14 at 14:21
up vote 1 down vote accepted

Since $X$ is projective, a global function is a constant, so it's killed by $d$, which gives the injectivity of the first map.

For the surjectivity of the second, note that a projective curve can always be covered by two open affines (since a point is ample by Riemann-Roch). Write down the Cech complex with respect to a double cover (i.e. resolve $\mathcal{O}$ and $\Omega$ separately, then take the total complex), and you will get surjectivity directly.

As far as duality is concerned, see Tate's (still very readable) paper on residues for an algebraic computation. The one-line summary is: you can define residues formally as coefficients of $a_{-1}$ in Laurent expansions without actually doing any complex analysis, and this recovers classical duality theory for algebraic curves. Note however that this doesn't work in characteristic $p$, and in general algebraic de Rham cohomology is broken in characteristic $p$.

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Thank you. Can I check, with reference to the paper by Tate, are you referring to "Residues of differential on curves"? If so, does this not pertain to Serre duality, rather than a duality of algebraic de Rham cohomology? I was hoping I could use this as a shortcut to avoid the computation you described, but if doing it directly is the only way then I guess that will have to do :) – Joe Tait Feb 7 '14 at 14:52
that's right -- he doesn't directly mention adrC in the paper, instead, he interprets serre duality in terms of residues, using the Tate complex of sheaves (which I will descirbe in a second) -- but thepoint is that you can use this complex, rather than the Cech complex, to compute adRC and recover duality for curves – hunter Feb 7 '14 at 14:57
in general adRC does satisfy a poincare duality statement (i.e. not just for curves) and it doesn't use tate's paper, so maybe that reference was misleading. The Tate complex goes 0 ---> sheaf of regular functions ---> constant sheaf of meromorphic functions ---> the sheaf that eats U and gives the direct sum over P in U of formal Laurent expansions of meromorphic functions at P mod formal Taylor expansions of holomorphic functions at P ---> 0 – hunter Feb 7 '14 at 14:59
this complex is obviously a flasque resolution. you can do the same thing with Omega (replace meromorphic functions with meromorphic 1-forms and sum K_p/O_p with sum Omega_{K_p}/Omega_{O_P}.) these resolutions are compatible, and that will gives you a double complex to compute adRC with. – hunter Feb 7 '14 at 14:59
note that capitalizations of adRC in the preceding comments are motivated more by when I happen to be holding the shift key than by logic :) – hunter Feb 7 '14 at 15:00

I assume that you are working over $\mathbb{C}$, otherwise it does not make sense to talk about DeRham cohomology. By hodge decomposition theory, $H^2(X,\mathbb{C} )=H^{0,2}\oplus H^{1,1}\oplus H^{2,0}$. By GAGA principal, $H^{0,2}$ and $H^{2,0}$ are isomorphic to cohomology of sheaf of algebraic differentials. $H^2(X,\mathcal{O}_X)=0$ by Grothendieck Vanishing. $H^0(X,\Omega_{X/\mathbb{C}}^2)=0$, since $dimX=1$. thus $H^2(X,\mathbb{C} )=H^{1,1}$, we are done.

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