7
$\begingroup$

Is there a morphism of schemes, say $X\to Y$, such that the underlying topological map is injective but the morphism $X\to Y$ is not separated (i.e., the diagonal embedding $\Delta\colon X\to X\times_Y X$ is not closed)? Clearly, a monomorphism of schemes is separated, but I do not know if the same holds if we only demand injectivity.

$\endgroup$
  • $\begingroup$ @George: sorry, you're right. I have deleted it to avoid confusion. $\endgroup$ – Cla Apr 24 '17 at 14:42
3
$\begingroup$

No, there is not exist!

Proof: Let $f:X\to Y$ be a morphism of schemes such that the underlying map of topological spaces is injective. By construction of fibre product of schemes, one can assume that $Y$ is an affine scheme and that $X\times_YX$ admits a covering by affine open subschemes $U\times_YV$, where $U$ and $V$ are affine open in $X$. Let $z\in X\times_YX$ with projections $x\in U,y\in V$, then these points lie over the same point $w\in Y$ (see Stacks Project); by hypothesis $f(x)=w=f(y)\Rightarrow x=y$. Without loss of generality, let $U=V$. Restricting $\Delta:X\to X\times_YX$ to closed emebedding $\Delta_U:U\to U\times_YU$ ($U$ is affine, therefore it is separated); one has that $\Delta$ is the gluing morphism of $\Delta_U$'s for $U$ runs in the set of affine open subsets of $X$, that is $\Delta$ is a closed morphism of schemes, equivalently, $X$ is a separated scheme. (Q.E.D. $\Box$)

For more details, see Bosch - Algebraic Geometry and Commutative Algebra, proposition 7.4.9 and corollary 7.4.10.

$\endgroup$
  • $\begingroup$ Thanks for the reference! Some of the things that you state are automatic (so we don't need to assume it). $\endgroup$ – AYK Apr 25 '17 at 16:59
  • $\begingroup$ What are things authomatic? ?_? $\endgroup$ – Armando j18eos Apr 25 '17 at 17:52
  • $\begingroup$ Once you reduce the problem to the case where $Y$ is affine, it is simply true that $X\times_Y X$ has a covering as you describe (no need for a reduction here). Also your second use of the word 'assume' causes slight confusion. I guess the point you want to make is that the affines of the form $U\times_Y U$ already cover the product $X\times_Y X$. Then it follows indeed that the diagonal is closed... $\endgroup$ – AYK Apr 25 '17 at 20:23
  • $\begingroup$ No: "it is not simply true"! In general, if for any $z\in X\times_YX$ (with $Y$ affine scheme) there exists an affine open subset $U$ of $X$ such that $p_1(z)=x,p_2(z)=y\in U$, then the $U\times_YU$'s form an affine open covering of $X\times_YX$. Outside of this hypothesis, I do not know if one can find an affine open covering of $X\times_YX$ as described. $\endgroup$ – Armando j18eos Apr 26 '17 at 15:02
  • $\begingroup$ That is not what I meant: you first mention a covering with opens of the form $U\times_Y V$. (I agree with your remark, without injectivity you can't find a covering with opens of the form $U\times_Y U$ ...) $\endgroup$ – AYK Apr 26 '17 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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