Let $f:Y\to X$ and $g:Z\to Y$ be morphisms of schemes and take the product $Z\times_X Y$ with respect to the composition $g\circ f:Z\to X$ and $f$.

What property of a scheme morphism does $f$ have to fulfill such that the projection $Z\times_X Y\to Z$ is an isomorphism of schemes?

I think ''open immersion'' is a possibility. Are there other?


1 Answer 1


It is enough to assume that $f$ be a monomorphism : this is a purely categorical result which has nothing to do with schemes.
If that is the case, the projection $p:Z\times_X Y\to Z$ has as inverse the morphism $$s=(id_Z,g):Z\to Z\times_X Y$$

Yes, but what are the monomorphisms in the category of schemes?
As usual the Master has the answer( in EGA $IV_4$, 17.2.6 ) :

A morphism $f : Y \to X$ locally of finite type is a monomorphism if and only if for every $x \in X$, the fiber $f^{-1}(x)$ is either empty or isomorphic to $Spec (\kappa(x))$.

So a monomorphism is injective but the converse is completely false: given a non trivial extension of fields $k\subsetneq K$, the corresponding (trivially bijective!) morphism of schemes $Spec(K)\to Spec(k)$ is never a monomorphism.

However to finish on an optimistic note, let me remark that closed or open immersions ( you were right about those) are monomorphisms.

  • 4
    $\begingroup$ Thanks. As usual the Master has the answer. $\endgroup$ Sep 3, 2012 at 19:58
  • $\begingroup$ it should be noted that the Master appears to have an answer only for morphisms locally of finite type (which is a serious restriction, by the Master's standards). I do not know if monomorphisms of schemes are always injective. $\endgroup$
    – user691994
    Jul 30, 2019 at 10:38

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