Duality in algebraic de Rham cohomology

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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 $X$ 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 $p$.