# Base change of normalization map and scheme-theoretic surjectivity

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.
• Sorry, forgot to add the condition that the morphism $g$ is surjective. I have edited the question. – Jana May 27 at 13:53
• 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. – Johan May 27 at 19:46
• 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 $(f,f \mod t^5)$. – Jana May 27 at 21:11
• @Jana you consider the map $A \to B'$ instead of the map $A' \to B'$. – Joshua Mundinger Jun 27 at 1:17
• @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. – Jana Jun 27 at 6:25