1

Let $\mathscr{P}$ be a property of morphism of schemes such that:

(a) closed immersion is $\mathscr{P}$;

(b) composition of $\mathscr{P}$-morphism is $\mathscr{P}$;

(e) if $f:X\to Y$, $g:Y\to Z$ are two morphisms with $g\circ f$ is $\mathscr{P}$ and $g$ is separated, then $f$ is $\mathscr{P}$.

I'm required to show that $$f:X\to Y \text{ is }\mathscr{P}\Rightarrow f_{\text{red}}:X_{\text{red}}\to Y_{\text{red}} \text{ is } \mathscr{P},$$ where $X_{\text{red}}$, $Y_{\text{red}}$ are the corresponding reduced schemes of $X$ and $Y$.

I saw another proof on some solutions available online, but I have a much shorter one. I wonder if it's legitimate:

There is natural map $i_X:X_\text{red}\to X$, where $i^\#_X$ should be the canonical surjection $\mathcal{O}_X(U)\to \mathcal{O}_X(U)_\text{red}$ on each open subset of $X$. Combining the facts that

  • $i^\#_X$ is surjective;
  • $i_X$ induce homeomorphism as topological space (Hartshorne Ex II.2.3),

we conclude that $i_X$ and $i_Y$ are closed immersion. Now, align a commutative diagram as follows:

where the hooked arrows denote closed immersions.

Now, $i_Y\circ f_\text{red}=f\circ i_X$ is $\mathscr{P}$ from (a) and (b). $i_Y$ is $\mathscr{P}$ since closed immersion is separated. Hence $f_\text{red}$ is $\mathscr{P}$.

If I am correct that $i_X$ is a closed immersion, then I am pretty sure my argument will be correct. It will be great if someone can point out any mistakes I have made.

Thanks in advance for answering.

Ivan So
  • 797
  • I think that $i_X$ is indeed a closed immersion, but your proof doesn't work. A morphism of sheaves is said to be surjective if it's stalkwise surjective, not if its images on open sets are surjective. – Gabriel Dec 27 '21 at 20:56
  • 1
    @Gabriel surjectivity on opens implies surjectivity on stalks. (The reverse implication is not true.) – KReiser Dec 27 '21 at 20:56
  • This is also a subtle point that may be useful: https://math.stackexchange.com/a/712835/209130 – Gabriel Dec 27 '21 at 20:57
  • @KReiser Oh... of course. Thanks :) – Gabriel Dec 27 '21 at 20:57
  • Yeah, I attempted to show $i^#_X$ is surjective through the fact that it is surjective on each open subset. So, I suppose there is no problem with my proof? – Ivan So Dec 27 '21 at 21:07
  • @IvanSo I'm pretty sure that your proof is fine then :) – Gabriel Dec 27 '21 at 21:12
  • 1
    I also think your proof is correct. Another way to see that $i_X$ is a closed immersion is that $X_{red}$ is the reduced induced scheme structure on $X$ itself. – Daniel Dec 28 '21 at 06:25

0 Answers0