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Let $f : A \to B$ be a monomorphism of commutative rings. Is it necessarily true that the canonical map $\operatorname{Spec}(f) : \operatorname{Spec} B \to \operatorname{Spec} A$ is epic? Similarly, if $f$ is epic, is it necessarily true that $\operatorname{Spec}(f)$ is monic?

It seems obvious to me that the answer to both of these questions is true in the category of affine schemes, though I cannot prove it, but I'm not sure if this holds in the full category of schemes.

Thank you in advance.

  • Since $\mathsf{AffSch} \cong \mathsf{cRing}^{\textrm{op}}$ through the Spec functor, this is true by unfolding the definitions of mono, epi and dual categories. – Trebor Oct 11 '24 at 06:03
  • @Trebor That was (roughly speaking) the intuitive reasoning I was using to my belief that the claim holds in $\mathsf{AffSch}$, but I can't see how to generalize it to the category of schemes as a whole. If I recall correctly, a monomorphism of a subcategory is not always a monomorphism of the the entire category. – Brandon Harad Oct 11 '24 at 06:11

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These questions are trivial for the category of affine schemes (since equivalences of categories preserve monomorphisms and epimorphisms), so I assume your question refers to the category of all schemes.

Since $\mathrm{Spec} : \mathbf{CRing}^{\mathrm{op}} \to \mathbf{Sch}$ is right adjoint to the global sections functor (see for example here), it preserves all limits (see here), and hence it preserves all monomorphisms (see here). This means that an epimorphism of commutative rings is mapped to a monomorphism of the associated schemes. Notice that epimorphisms of commutative rings are much more general than surjective homomorphisms, for example every localization is an epi. See MO/109.

The Spec functor does not preserve epimorphisms, though. In other words, if $A \to B$ is an injective homomorphism of commutative rings, there might be schemes $X$ with two different morphisms $\mathrm{Spec}(A) \to X$ which agree when composed with $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$. (So, the map $X(A) \to X(B)$ is not injective.) Namely, when $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$ happens to be an open immersion which is not an isomorphism, we can construct $X$ as the pushout (glueing) of that open immersion with itself. A concrete example therefore is $A =k[x]$, $B=k[x,x^{-1}]$. The scheme $X$ also known as the "affine line with doubled origin".

Notice that the scheme constructed above is not separated. In fact, we can prove that if $X$ is a separated scheme, the induced map $X(A) \to X(B)$ is injective (hint: consider the equalizer of two $A$-points and use that it is closed). This is related to the valuative criterion for separated morphisms, which proves that the converse is also true, at least when we assume that $X$ is quasi-separated. Therefore, $\mathrm{Spec}$ does preserve epimorphisms when corestricted to the category of separated schemes.

Related questions:

  • Of course, one does not have to go as far as separated, but quasiseparated is not enough. It suffices to have affine diagonal, but there seems to be no name for this condition. – Zhen Lin Oct 11 '24 at 13:39
  • Such schemes are called semi-separated. You are saying that they have the property that their functor of points preserves monos? – Martin Brandenburg Oct 11 '24 at 14:37
  • Oops, somehow I missed that we need the equaliser to be a regular monomorphism of affine schemes rather than just any monomorphism. – Zhen Lin Oct 11 '24 at 22:28