In this short post, I will give a proof of the following result:
Theorem. If \(f : X \to Y\) is a monomorphism of schemes, then \(f\) is injective.
In the above, injective just means injectivity on the level of functions. To prove this, it is worthwhile to study the (set-theoretic) points of \(X\) from a scheme-theoretic perspective. We’ll need the following construction: For a point \(x \in X\), let \(\kappa(x)\) denote its residue field.
Lemma. For each \(x \in X\), there is a unique morphism \(\DeclareMathOperator{\Spec}{Spec}\Spec \kappa(x) \to X\) sending the unique point of \(\Spec \kappa(x)\) to \(x\), and inducing the identity on residue fields.
Proof. Pick an affine open say \(\Spec A\) containing \(x\). Then \(x\) corresponds to some prime \(\DeclareMathOperator{\fp}{\mathfrak{p}}\fp \in \Spec A\). The morphism is just \(\Spec (A_\fp / \fp A_\fp) \to \Spec A \to X\). \(\blacksquare\)
Lemma. Let \(f : X \to Y\) be a morphism and let \(x \in X, y \in Y\). Then \(f(x) = y\) iff there is a morphism \(\tilde{f}\) making the diagram \[\begin{CD} \Spec \kappa(x) @>{\tilde{f}}>> \Spec \kappa(y) \\ @VVV @VVV \\ X @>{f}>> Y \end{CD}\]commute.
Proof. The if part is trivial. For the only if part, we can assume \(X\) and \(Y\) are affine. \(\blacksquare\)
Now we prove the theorem.
Proof of Theorem. Suppose \(x, x' \in X\) both map to \(y \in Y\). Thus we have morphisms \(\Spec \kappa(x) \to \Spec \kappa(y)\) and \(\Spec \kappa(x') \to \Spec \kappa(y)\). These correspond to field extentions \(\kappa(x)/ \kappa(y)\) and \(\kappa(x')/ \kappa(y)\). The trick (see also Stacks tag 01J5): there is a field extention \(K/\kappa(y)\) containing both \(\kappa(x)\) and \(\kappa(x')\). Thus we have a commutative diagram \[\begin{CD} \Spec K @>>> \Spec \kappa(x) \\ @VVV @VVV \\ \Spec \kappa(x') @>>> \Spec \kappa(y) \end{CD}\] which forms part of the commutative diagram \[\begin{CD} \Spec K @>>> \Spec \kappa(x) @>>> X \\ @VVV @VVV @VV{f}V \\ \Spec \kappa(x') @>>> \Spec \kappa(y) @>>> Y \\ @VVV @VVV @| \\ X @>>{f}> Y @= Y \\ \end{CD}\]Since \(f\) is monic, the two morphisms \(\Spec K \to X\), given by composition in the top row and composition in the left column, are equal. The first has image \(x\), while the second has image \(x'\), so \(x = x'\). \(\blacksquare\)
Since monomorphisms are preserved under base change, we immediately get:
Corollary. Monomorphisms are universally injective.
See Stacks tag 01S2 for more precise results on universal injectivity. In particular, in tag 01S4, we can use (4) implies (2) to give an alternate proof of the theorem, since a morphism is monic iff its diagonal is an isomorphism.