# Is an injective proper morphism of schemes a closed immersion

I have a proper morphism $f:X\to Y$ between proper $S$-schemes, where $S$ is a Noetherian base. I can show that for every field $k$ the induced map $$f_\ast:S\text{-}\mathbf{Sch}(\mathop{\mathrm{Spec}} k, X)\to S\text{-}\mathbf{Sch}(\mathop{\mathrm{Spec}} k, Y)$$ is bijective. Does that imply that $f$ is a closed immersion?

A reference or a counterexample would be appreciated.

• I think the map induced from the inclusion $\mathbb{C}[t^2,t^3]\subset\mathbb{C}[t]$ as $\mathbb{C}$ schemes satisfy all your conditions, but it is not a closed immersion. Mar 16, 2017 at 18:01
• @Mohan: The affine line $\mathop{\mathrm{Spec}} \mathbb{C}[t]$ is not proper over $\mathbb{C}$! Mar 16, 2017 at 18:07
• This can be easily rectified by taking the map $\mathbb{P}^1\to C$, where $C$ is the plane cuspidal cubic. The issues in your situation are local on the base, so it would have made no difference. Mar 16, 2017 at 20:10
• @Mohan's example is perfect and is the one I, as a geometer, would prefer. However, just for fun, I have given a completely different answer. Mar 16, 2017 at 20:40

Let $S=\mathbb Z,\: Y=\operatorname {Spec} \mathbb Q,\: X=\operatorname {Spec} \mathbb Q[\epsilon ]$ and let $f:X\to Y$ be the unique scheme morphism, dual to the unique ring morphism $\mathbb Q\hookrightarrow\mathbb Q[\epsilon ]$ (as usual $\mathbb Q[\epsilon ]=\mathbb Q[T]/\langle T^2\rangle$).
a) The map $f:X\to Y$ is proper because it is finite.
b) The induced maps $f_*:X(k)\to Y(k)$ are all bijective: they are maps from the empty set to the empty set if $char(k)\neq 0$ and between two singleton sets if $char(k)=0$.
c) Nevertheless the morphism $$f:X=\operatorname {Spec} \mathbb Q[\epsilon ]\to Y=\operatorname {Spec} \mathbb Q$$ is not a closed immersion.