2
$\begingroup$

Let $S$ be a scheme and let $f:X\to Y$ be a morphism of separated $S$-schemes. This MSE answer shows that, when $X$ is reduced, the obvious morphism $g_f:\Gamma_f \to X\times_S Y\to X$ is an isomorphism, where $\Gamma_f$ is the scheme-theoretic image of $\Delta_f:X\to X\times_s Y$.

Can we show that $g_f$ without assuming that $X$ is reduced?

I see how the reduced induced structure is being used there, but I do not see if $X$ being reduced affects the statament, $X$ cannot have double/triple/... points if $\Gamma_f$ does not! So, I wonder if there is a probably more elaborate proof that shows $g_f$ to be an isomorphism.

The only assumption I have is that $S$ is Noetherian.

A hint or a reference would be appreciated.

$\endgroup$
  • $\begingroup$ I think one problem is that typically you give the graph the image scheme structure, which is the reduced induced structure... and this is an obstruction to the projection being an isomorphism. Note that if $Y$ is reduced, then $f$ induces a map from the reduction of $X$ to $Y$, and for this you can study the graph and get an isomorphism as before. How are you defining the graph? $\endgroup$ – Lorenzo Najt Feb 19 '17 at 23:03
  • $\begingroup$ @AreaMan: $\Gamma_f$ is not necessarily reduced, I take it to be the scheme-theoretic image of $\Delta_f$, that is the smallest subscheme in $X\times_S Y$ that factorises $\Delta_f$, as in Hartshorne's II.Ex.3.11(d). I think that $X\to \Gamma_f$ is a closed immersion, using a similar argument to stacks-project Tag 024T, and their underlying topological spaces are isomorphic as in the other MSE answer, referred to in the question. I just do not see how $\Gamma_f$ can be bigger. A counterexample would be appreciated! $\endgroup$ – user24453 Feb 19 '17 at 23:25
  • 1
    $\begingroup$ Oh, you're right, I got my arrows backwards -- the reduction is a closed subscheme, which makes sense because it is smaller... I remembering thinking about this example: $Spec k[[x]] \to Spec k[x]$, or $\bigcup_{n \geq 0} k[x]/x^n \to Spec k[x]$. The scheme theoretic image in both cases is the entire line, which is certainly initially counter intuitive (though makes sense from the Taylor seris pov). Maybe you can use this example somehow? $\endgroup$ – Lorenzo Najt Feb 19 '17 at 23:32
1
$\begingroup$

For any morphism of $S$-schemes $f:X\to Y$ the graph morphism $\Gamma_f:X\to X\times _SY$ is an immersion (but not necessarily a closed immersion).
Its associated image subscheme $G_f\subset X\times _SY$ is indeed isomorphic to $X$ under the restriction $g:G_f\to X$ of the first projection $p_X:X\times _SY \to X$.
This is explained in Corollaire (5.3.11) page 133 of $EGA_I$ and the seven following lines .

$\endgroup$
  • $\begingroup$ As a note for future readers, the Corollaire referenced in this answer is numbered 4.1.10 in the version hosted on NUMDAM's website. This has caused me some confusion. $\endgroup$ – KReiser Mar 19 '18 at 5:34
  • $\begingroup$ Dear @KReiser, I have added the correct page to my reference and a link to NUMDAM, so that just clicking will show the user what result I mean. I think my reference number *Corollaire (5.3.11) is correct, so could you please add in a comment below a link to the version where the result has the number you claim? $\endgroup$ – Georges Elencwajg Mar 19 '18 at 8:09
  • $\begingroup$ Your updated reference is correct - I got very confused when I was trying to find 5.3.11 and ended up going with the page number instead of the lemma number. $\endgroup$ – KReiser Mar 19 '18 at 15:54
  • $\begingroup$ Dear @ KReiser, I'm happy to see all is well now :-) $\endgroup$ – Georges Elencwajg Mar 19 '18 at 17:27

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.