9
$\begingroup$

$\DeclareMathOperator\Pic{Pic}$If $A$ is a ring, then we know that $\Pic(A)=\Pic(A_\text{red})$, but for a scheme $X$ it is false in general.

On the other hand, we have that $\Pic(X)=H^{1}_{et}(X,\mathbb{G}_m)$ and étale cohomology doesn't see the nilpotents, so there should be no difference of the right hand side if we replace $X$ by $X_\text{red}$.

How can we accomodate the two?

$\endgroup$
3
  • 9
    $\begingroup$ While étale cohomology with finite coefficients does not "see" the nilpotents, this is definitely not true for general coefficients, in particular for $\mathbb{G}_m$. To convince your self note that this is the same as the Zariski cohomology of $\mathcal{O}_X^*$, which is easily seen to be usually different from that of $\mathcal{O}_{X_{\mathrm{red}}}^*$. $\endgroup$
    – abx
    May 30, 2018 at 4:24
  • 2
    $\begingroup$ What bothers me is that in stacks project Tag. 04DY, it says that étale cohomology of ANY abelian sheaf is the same for $X$ and $X_{red}$. $\endgroup$
    – prochet
    May 30, 2018 at 8:05
  • 15
    $\begingroup$ @prochet Yes, but the restriction of $\mathbf{G}_{m, X}$ to $X_{\rm red}$ is not $\mathbf{G}_{m, X_{\rm red}}$. $\endgroup$ May 30, 2018 at 9:12

1 Answer 1

6
$\begingroup$

Let $f : X \to Y$ be a universal homeomorphism of schemes. Then as you noted above, the pullback functor on (small) étale sites

$$\begin{eqnarray} Y_{\text{ét}} &\to& X_{\text{ét}}\\ U &\mapsto& U \times_X Y \end{eqnarray} $$ is an equivalence of categories. In this case, the natural transformations $\text{id} \to f_\ast f^\ast$ and $f^\ast f_\ast \to \text{id}$ are isomorphisms. So now let $X = Y_{\text{red}}$. If $f^\ast \mathbf{G}_{m,Y} = \mathbf{G}_{m, Y_{\text{red}}}$, then it would follow for any etale open $U_{\text{red}}$ over $Y_{\text{red}}$ that $$\mathbf{G}_{m,{\text{red}}}(U_{\text{red}}) = \mathbf{G}_m(U), $$

where $U$ is the unique scheme étale over $Y$ that pulls back to $U_{\text{red}}$ over $Y_{\text{red}}$.

The following example shows this is already false for rings: Take $U= Y = \operatorname{Spec} k[\epsilon]/(\epsilon^2)$. Then $\mathbf{G}_m(Y) = k[\epsilon]^\times$ which is evidently not equal to $k^\times$. For instance, $1+\epsilon$ is a unit because $(1+ \epsilon)(1-\epsilon) = 1 - \epsilon^2 = 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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