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$\DeclareMathOperator{\Spec}{Spec}$ Let $X = \Spec A$, $Y = \Spec B$ be affine complex varieties, that is reduced $\mathbb{C}$-schemes of finite type. Equivalently we can say that $A$ and $B$ are reduced finitely generated $\mathbb{C}$-algebras.

There should be a ring-theoretic condition on $\phi \in \operatorname{Hom}_\mathbb{C}(B, A) \cong \operatorname{Hom}_\mathsf{Sch}(X, Y)$ detecting whether the corresponding morphism of schemes is injective. Is there a nice way to phrase this condition? (besides the obvious thing saying that for every maximal ideal $\mathfrak{m} \subset B$ there is at most one maximal ideal $\mathfrak{n} \subset A$ such that $\phi^{-1}(\mathfrak{n}) = \mathfrak{m}$).

My thoughts: This condition has to be a generalization of both surjective morphisms and localizations and is stable under composition. But not every injective morphism of varieties is an immersion, e.g. the normalization of the cuspidal cubic $y^2 = x^3$.

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    $\begingroup$ Epimorphism in the category of rings? $\endgroup$ Commented Dec 8, 2021 at 10:17
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    $\begingroup$ I was actually unable to check that an epimorphism in the category of reduced commutative rings is also an epimorphism in the category of all commutative rings. $\endgroup$
    – YCor
    Commented Dec 8, 2021 at 12:19
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    $\begingroup$ An epimorphism of rings corresponds to a monomorphism of affine schemes (in the category of affine schemes or, equivalently, of all schemes). Even for reduced affine schemes, this is a stronger condition than injectivity, as the normalization of the cusp demonstrates. $\endgroup$ Commented Dec 8, 2021 at 13:44
  • $\begingroup$ @YCor: If $K\to L$ is a field extension adjoining a p-th root to a characteristic p field, then this is epi in reduced rings but not in rings. (The kernel of $L\otimes_K L\to L$ is nilpotent.) $\endgroup$ Commented Dec 8, 2021 at 22:35
  • $\begingroup$ @LaurentMoret-Bailly are you saying that the normalization of the cusp is not a monomorphism (in the category of reduced affine $\mathbb{C}$-schemes)? It seems like monomorphism and injectivity should be equivalent because morphisms of varieties are determined by their map on topological spaces (so injective => monomorphism) and monomorphism have to be injective (otherwise you have a contradiction by using two different maps from $\operatorname{Spec} \mathbb{C}$). $\endgroup$ Commented Dec 9, 2021 at 2:21

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