The Stacks project

Lemma 50.20.3. Let $k$ be a field. Let $X$ be a smooth proper scheme over $k$ equidimensional of dimension $d$. The map

\[ \text{d} : H^ d(X, \Omega ^{d - 1}_{X/k}) \to H^ d(X, \Omega ^ d_{X/k}) \]

is zero.

Proof. It is tempting to think this follows from a combination of Lemmas 50.20.2 and 50.20.1. However this doesn't work because the maps $\mathcal{O}_ X \to \Omega ^1_{X/k}$ and $\Omega ^{d - 1}_{X/k} \to \Omega ^ d_{X/k}$ are not $\mathcal{O}_ X$-linear and hence we cannot use the functoriality discussed in Duality for Schemes, Remark 48.27.3 to conclude the map in Lemma 50.20.2 is dual to the one in this lemma.

We may replace $X$ by a connected component of $X$. Hence we may assume $X$ is irreducible. By Varieties, Lemmas 33.25.4 and 33.9.3 we see that $k' = H^0(X, \mathcal{O}_ X)$ is a finite separable extension $k'/k$. Since $\Omega _{k'/k} = 0$ (see for example Algebra, Lemma 10.158.1) we see that $\Omega _{X/k} = \Omega _{X/k'}$ (see Morphisms, Lemma 29.32.9). Thus we may replace $k$ by $k'$ and assume that $H^0(X, \mathcal{O}_ X) = k$.

Assume $H^0(X, \mathcal{O}_ X) = k$. We conclude that $\dim H^ d(X, \Omega ^ d_{X/k}) = 1$ by Lemma 50.20.1. Assume first that the characteristic of $k$ is a prime number $p$. Denote $F_{X/k} : X \to X^{(p)}$ the relative Frobenius of $X$ over $k$; please keep in mind the facts proved about this morphism in Lemma 50.19.5. Consider the commutative diagram

\[ \xymatrix{ H^ d(X, \Omega ^{d - 1}_{X/k}) \ar[d] \ar[r] & H^ d(X^{(p)}, F_{X/k, *}\Omega ^{d - 1}_{X/k}) \ar[d] \ar[r]_{\Theta ^{d - 1}} & H^ d(X^{(p)}, \Omega ^{d - 1}_{X^{(p)}/k}) \ar[d] \\ H^ d(X, \Omega ^ d_{X/k}) \ar[r] & H^ d(X^{(p)}, F_{X/k, *}\Omega ^ d_{X/k}) \ar[r]^{\Theta ^ d} & H^ d(X^{(p)}, \Omega ^ d_{X^{(p)}/k}) } \]

The left two horizontal arrows are isomorphisms as $F_{X/k}$ is finite, see Cohomology of Schemes, Lemma 30.2.4. The right square commutes as $\Theta _{X^{(p)}/X}$ is a morphism of complexes and $\Theta ^{d - 1}$ is zero. Thus it suffices to show that $\Theta ^ d$ is nonzero (because the dimension of the source of the map $\Theta ^ d$ is $1$ by the discussion above). However, we know that

\[ \Theta ^ d : F_{X/k, *}\Omega ^ d_{X/k} \to \Omega ^ d_{X^{(p)}/k} \]

is surjective and hence surjective after applying the right exact functor $H^ d(X^{(p)}, -)$ (right exactness by the vanishing of cohomology beyond $d$ as follows from Cohomology, Proposition 20.20.7). Finally, $H^ d(X^{(d)}, \Omega ^ d_{X^{(d)}/k})$ is nonzero for example because it is dual to $H^0(X^{(d)}, \mathcal{O}_{X^{(p)}})$ by Lemma 50.20.1 applied to $X^{(p)}$ over $k$. This finishes the proof in this case.

Finally, assume the characteristic of $k$ is $0$. We can write $k$ as the filtered colimit of its finite type $\mathbf{Z}$-subalgebras $R$. For one of these we can find a cartesian diagram of schemes

\[ \xymatrix{ X \ar[d] \ar[r] & Y \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & \mathop{\mathrm{Spec}}(R) } \]

such that $Y \to \mathop{\mathrm{Spec}}(R)$ is smooth of relative dimension $d$ and proper. See Limits, Lemmas 32.10.1, 32.8.9, 32.18.4, and 32.13.1. The modules $M^{i, j} = H^ j(Y, \Omega ^ i_{Y/R})$ are finite $R$-modules, see Cohomology of Schemes, Lemma 30.19.2. Thus after replacing $R$ by a localization we may assume all of these modules are finite free. We have $M^{i, j} \otimes _ R k = H^ j(X, \Omega ^ i_{X/k})$ by flat base change (Cohomology of Schemes, Lemma 30.5.2). Thus it suffices to show that $M^{d - 1, d} \to M^{d, d}$ is zero. This is a map of finite free modules over a domain, hence it suffices to find a dense set of primes $\mathfrak p \subset R$ such that after tensoring with $\kappa (\mathfrak p)$ we get zero. Since $R$ is of finite type over $\mathbf{Z}$, we can take the collection of primes $\mathfrak p$ whose residue field has positive characteristic (details omitted). Observe that

\[ M^{d - 1, d} \otimes _ R \kappa (\mathfrak p) = H^ d(Y_{\kappa (\mathfrak p)}, \Omega ^{d - 1}_{Y_{\kappa (\mathfrak p)}/\kappa (\mathfrak p)}) \]

for example by Limits, Lemma 32.19.2. Similarly for $M^{d, d}$. Thus we see that $M^{d - 1, d} \otimes _ R \kappa (\mathfrak p) \to M^{d, d} \otimes _ R \kappa (\mathfrak p)$ is zero by the case of positive characteristic handled above. $\square$

Comments (2)

Comment #4774 by Noah Olander on

Typo: In sentence 2 of paragraph 3 you write instead of .

There are also:

  • 2 comment(s) on Section 50.20: PoincarĂ© duality

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FW6. Beware of the difference between the letter 'O' and the digit '0'.