4
$\begingroup$

I have been reading Ravi Vakil notes on algebraic geometry, and one exercise asks if $phi:X\rightarrow Y$ and $\pi:X\rightarrow Y$ are morphisms of $k$-schemes and $\ell/k$ a field extension, if $\pi_\ell=\rho_\ell$, then show that $\pi=\rho$, where $\pi_\ell$ and $\rho_\ell$ mean the induced maps after base change.

I can seem to reduce it to the affine case, and since $X \times_k \ell\rightarrow X$ is surjective, we get (as Ravi Vakil suggests in the notes) that $\pi$ and $\rho$ agree on the level of sets. However, I cannot seem to complete the proof that they must be the same, as agreeing on the level of sets is not enough.

Thank you for any guidance, help, or advice.

$\endgroup$
  • $\begingroup$ Could you tell me how to reduce the problem to affine case? I am really confused about that. Thank you $\endgroup$ – Mike Sep 28 at 14:32
4
$\begingroup$

The crucial fact you need is that $X \otimes_k \ell \to X$ is an epimorphism in the sense of category theory, and for this, it suffices to know that $X \otimes_k \ell \to X$ is an effective epimorphism, such as a faithfully flat morphism. But the property of being faithfully flat is preserved by base change, so it suffices to verify that $\operatorname{Spec} \ell \to \operatorname{Spec} k$ is a faithfully flat morphism, and this is easy: it is obviously surjective on points, and any morphism with codomain $\operatorname{Spec} k$ is flat.

That said, the above is overkill. As you say, the problem is local on $X$ and $Y$, so we may assume that both $X$ and $Y$ are affine. The problem then reduces to the following statement in commutative algebra:

If $f, g : A \to B$ are $k$-algebra homomorphisms such that $f \otimes_k \ell = g \otimes_k \ell$ as $\ell$-algebra homomorphisms $A \otimes_k \ell \to B \otimes_k \ell$, then $f = g$.

Indeed, for each $k$-algebra homomorphism we have a commutative diagram $$\begin{array}{ccc} A & \to & A \otimes_k \ell \\ \downarrow & & \downarrow \\ B & \to & B \otimes_k \ell \end{array}$$ and it is easy to verify that the horizontal arrows are injective homomorphisms, so the claim follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.