A property $\mathcal{P}$ of a morphism $f:Y\rightarrow X$ is said to be stable under base change if given any morphism of schemes $g:X'\rightarrow X$ the induced morphism $p:X'\times_X Y\rightarrow X'$ has the property $\mathcal{P}$. $$\require{AMScd} \begin{CD} X' \times_X Y @>p>> X'\\ @VqVV @VVgV \\ Y @>f>> X \end{CD}$$ Supposing that $f$ is a closed immersion, I am trying to prove that $p$ is a closed immersion.

As $f$ is a closed immersion, we have $f(Y)$ homeomorphic to some closed subset $Z$ of $X$. I have to prove that $p$ induces a homeomorphism from $X' \times_X Y$ onto a closed subset of $X'$ and one of the clear choice for closed subset of $X'$ is the inverse image $g^{-1}(Z)$.

I am trying to prove that $p(X'\times_X Y)\cong g^{-1}(Z)$ but could not succeed. After spending half an hour, I realized this does not work.

Any suggestion on possible choice of closed set is welcome.

I want to know if this is a bad way of proving some thing is a closed immersion. Any suggestion for surjectivity of sheaves $\mathcal{O}_{X'}\rightarrow p_*\mathcal{O}_{X'\times_X Y}$ is also welcome.

I have seen that there is another thread with same question. But here, I am trying to work in a different way.

  • $\begingroup$ The clearest way to see this is perhaps to work affine locally. The statement should be clear when you translate it to commutative algebra. $\endgroup$ – Asvin Apr 1 '17 at 14:58
  • $\begingroup$ @Asvin I am sure that is one way of doing this problem. I want to know If the way that I have started with lead to somewhere. $\endgroup$ – user312648 Apr 1 '17 at 15:00
  • $\begingroup$ Homeomorphism is too weak to do what you want to do. Closed immersion is more that homeomorphism to its image. $\endgroup$ – Mohan Apr 1 '17 at 15:08
  • $\begingroup$ @Mohan : It is atleast a homeomorphism onto its image? In that case I should prove atleast a homeomorphism. Right? $\endgroup$ – user312648 Apr 1 '17 at 15:22
  • $\begingroup$ Not quite. Let $k\subset L$ be a finite extension of fields. Then $\mathrm{Spec} L\to\mathrm{Spec} k$ is a homeomorphism (juts one point). But, the base change map for $\mathrm{Spec}\overline{k}\to\mathrm{Spec} k$, ($\overline{k}$ is the algebraic closure) is not a homeomorphism. $\endgroup$ – Mohan Apr 1 '17 at 15:45

As for the surjectivity of sheaves, it follows from the fact that tensor is right exact. More precisely, to build a fiber product you are glueing spectra of rings of the form $B \otimes_A C$. More precisely, the diagram $$\require{AMScd} \begin{CD} X' \times_X Y @>p>> X'\\ @VqVV @VVgV \\ Y @>f>> X \end{CD}$$ locally corresponds to $$\require{AMScd} \begin{CD} B \otimes_A C @<p^\#<< C\\ @Aq^\#AA @AAg^\#A \\ B @<f^\#<< A \end{CD}$$

In this case, you have the surjection $A \rightarrow B$, where $\mathrm{Spec}(A)$ is an affine chart of $X$, and $\mathrm{Spec}(B)$ is an affine chart of $Y$. Then, you tensor over $A$ this surjection by $C$, where $\mathrm{Spec}(C)$ is an affine chart of $X'$. This provides you with a surjection $C \rightarrow B \otimes_A C$. These surjections patch to give you what you want. I know you were looking for a global argument, but this approach is somehow forced by the local nature of the fibre product.

Analyzing the above argument leads to the embedding part as well. Since you are looking for a more direct approach, maybe the following one is a good idea. You know that $g^{-1}(Z)$ is a closed subset of $X'$. You can endow it with a scheme structure in order to exploit the universal property of the fibre product. Since $Z$ is identified with $Y$, the map $g^{-1}(Z) \rightarrow Z$ gives you a map $g^{-1}(Z) \rightarrow Y$, and the composition of this latter one with $f$ gives back $g^{-1}(Z) \rightarrow Z$. Therefore, the universal property tells you that $g^{-1}(Z)$ maps into the fibre product. Some nonsense about universal objects should tell you these two agree.

Now, I cheated a bit in the above argument, since the universal property talks abot morphisms of shcemes, and $g^{-1}(Z)$ is just a set. The right scheme structure on it for the argument to work should be given by $\mathcal{I}_Z \cdot \mathcal{O}_{X'}$, where $\mathcal{I}_Z$ is the ideal sheaf of $Z$, and this operation is the inverse image of ideals. This construction is briefly mentioned in Hartshorne at page 163.

  • $\begingroup$ Thanks for the answer. I need to check details. I will ask if I have any doubts $\endgroup$ – user312648 Apr 2 '17 at 10:22

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