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.

  • $\begingroup$ @George: sorry, you're right. I have deleted it to avoid confusion. $\endgroup$ – User3773 Apr 24 '17 at 14:42

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.

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  • $\begingroup$ 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

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