Let $C$ be an affine, integral curve and $f: \widetilde{C} \to C$ be its normalization. Let $g:D \to C$ be a finite, affine, surjective morphism (note $D$ need not be reduced, but can assume generically reduced). Denote by $D'$ the base change of $\widetilde{C}$ by the morphism $g$ and $g': D' \to D$ the resulting morphism. Is the morphism $g'$ scheme-theoretically surjective i.e., the induced ring homomorphism $\mathcal{O}_D \to \mathcal{O}_{D'}$ is injective?


No. Let $A \to B$ be a ring map. Let $M$ be a finite $A$-module such that $M \to M \otimes_A B$ is not injective. Then with $A' = A \oplus M$ where $M$ is an ideal of square zero and $B' = A' \otimes_A B$ the base change, we see that $A' \to B'$ is not injective.

Apply this with $A = k[t^2, t^3]$ where $k$ is a field, $B = k[t]$ is the normalization of $A$ and $M = k[t^2, t^3]/I$ where $I$ is the ideal generated over $A$ by $t^5$. Then $t^6 \not \in I$ but $M \otimes_A B = B/J$ where $J$ is the ideal in $B$ generated by $t^5$ and so $t^6 \in J$. Hence $M \to M \otimes_A B$ is not injective.

| cite | improve this answer | |
  • $\begingroup$ Sorry, forgot to add the condition that the morphism $g$ is surjective. I have edited the question. $\endgroup$ – Jana May 27 at 13:53
  • $\begingroup$ The morphism $\text{Spec}(A \oplus M) \to \text{Spec}(A)$ is surjective as it has a section. This is my last comment on this question. $\endgroup$ – Johan May 27 at 19:46
  • $\begingroup$ I am sorry, I do not think your example works. Ofcourse $M \to M \otimes_A B$ is not injective. But, this does not imply $A' \to B'$ is not injective. In particular, in your example you claim that the morphism $k[t] \to (k[t^2, t^3] \oplus k[t^2, t^3]/(t^5)) \otimes k[t]$ is not injective, where the tensor product is over $k[t^2, t^3]$. But, note that, $t^a$ for any $a$ maps to $(1 \oplus 1) \otimes t^a=(t^a \oplus t^a) \otimes 1$ which is never zero. I am assuming that the ring homomorphism from $k[t^2, t^3] \to k[t^2, t^3] \otimes k[t^2, t^3]/(t^5)$ sends $f$ to $(f,f \mod t^5)$. $\endgroup$ – Jana May 27 at 21:11
  • $\begingroup$ @Jana you consider the map $A \to B'$ instead of the map $A' \to B'$. $\endgroup$ – Joshua Mundinger Jun 27 at 1:17
  • $\begingroup$ @JoshuaMundinger Sorry, I wrote it in a confusing manner, but you can use the arguments to check that the map from $A'$ to $B'$ is injective. The important point to note is that as $g$ is assumed to be surjective, the most natural map from $A$ to $A \oplus M$ is the one that takes $f$ to $(f,f \mod t^5)$ (I think this is the map that Johan considers). But, for such a map, the induced map from $A'$ to $B'$ is injective. Probably, you can come up with a different injective map from $A$ to $A \oplus M$, under which $A' \to B'$ is not injective, but I could not think of such a map. $\endgroup$ – Jana Jun 27 at 6:25

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.