Injective map of schemes that is not a monomorphism

Question

I know that it is not true that for a map $f \colon X \rightarrow Y$ of schemes, injectivity (on underlying sets) of $f$ gives a monomorphism in the category of schemes.
Stronger assumptions are required, see e.g. Tag 01L6 in the Stacks project.

But I have not come up with a counterxample yet. So can someone provide me with some concrete counterexample, preferrably some well-known class of such morphisms?

Answer 1

Let $L/K$ be any proper extension of fields. The inclusion $K\to L$ gives a map of affine schemes $\mathrm{Spec}(L)\to\mathrm{Spec}(K)$. Since both schemes have just one point, this map is a bijection on underlying sets. But it is not a monomorphism, since $K\to L$ is not an epimorphism of rings. For example (using Galois theory if $L/K$ is algebraic or transcendence bases if $L/K$ is transcendental), $L$ admits multiple distinct embeddings into its algebraic closure $\overline{L}$ which all restrict to the identity on $K$.

Answer 2

The inclusion $\mathbb{Q}\to\mathbb{Q}[\sqrt{2}]$ is perhaps an easier example than the one given by Alex Kruckman. The Galois conjugation on the latter gives a non-trivial morphism that is identity on $\mathbb{Q}$. So, in general, algebraic field extensions provide some relatively easy examples.

If you want an example over algebraically closed fields, the inclusion $R=\mathbb{C}[x]\to S=\mathbb{C}[x,y]/\langle y^2\rangle$ may be the easiest. This has the automorphisms $a:S\to S$ given by $(x,y)\mapsto (x,ay)$. Note that $a$ acts as identity on $R$. In general, given a split nilpotent thickening $Y$ of $X$ (i.e. such that there is a splitting $Y\to X$) there would be non-trivial automorphisms of $Y$ that are identity on $X$.