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I am writing the proof that locally closed immersions are of finite type but I am stuck at minor detail. I would like that either (1) preimages of open affines by open immersions be quasicompact or (2) that given an affine $Spec\, A$ in the source of a closed immersion it would be possible to find an affine in the target whose preimage is exactly $Spec \, A$.

Here is why:

To show that the locally closed immersion $Z\overset{\rho}{\rightarrow} W\overset{\tau}{\rightarrow} X$ is of finite type, where $\rho$ is closed and $\tau$ is open, by definition, we must show that for any open affine $Spec\,C$ in the target and open affine $Spec\,A$ in the preimage the induced map of structure sheaves $(\tau\rho)^\#:C\rightarrow A$ makes $A$ into a finitely generated $C$-algebra.

Now, what I need is to get my hands on an open affine $Spec\,B \subset \tau^{-1}(Spec\,C)$ such that $\rho^{-1} (Spec\, B) = Spec\, A$. If this were true, then it would follow trivially from $A\rightarrow B$ being a surjection that $A$ is a finitely generated $B$-algebra. Since $Spec\,B$ is quasicompact, we would be able to cover it by finitely many distinguished open sets $D(f_i)$ such that $\tau (\bigcup D(f_i)) = \bigcup D(\tau^\# (f_i)) = A_{\prod f_i}\Rightarrow B=A_{\prod f_i}$, so that $A$ is clearly a finitely generated $C$ algebra.

But how do fill the gap in my argument?

Rodrigo
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1 Answers1

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Sorry I haven't read your proof (it looks a little bit clumsy). I would prove it as follows:

Locally of finite type morphisms are closed under composition. Hence, it suffices to prove that closed immersions and open immersions are of finite type. For closed immersions, this is clear: Simply use that $A/I$ is a finitely generated $A$-algebra for every ideal $I$ of a commutative ring $A$ (in fact, $\emptyset$ is a generating system!).

Now let $U \to X$ be an open immersion. By may work locally on $X$, so let us assume that $X=\mathrm{Spec}(A)$ is affine. Then $ U = \cup_i D(f_i)$ is a union of basic-open subsets. Each $D(f_i)$ is affine and of finite type over $X$, since $A_{f_i}$ is a finitely generated $A$-algebra (namely, generated by $f_i^ {-1}$). Hence, $U \to X$ is locally of finite type.

Martin Brandenburg
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  • Okay, the point that I was actually struggling with was to show that finite-type morphisms are closed under composition, but now I got it. If $Spec,A \subset (\tau\rho)^{-1} (Spec , C)$ I can cover $\rho(Spec,A)$ by open affines $V_i \subset \tau^{-1} (Spec , C)$, and cover $Spec,A$ by open affines $U_{ij}$ such that $\rho (U_{ij}) \subset U_i$. Then I get that the $\Gamma(U_{ij},\mathcal O_{U_{ij}})$ are f.g. $C$-algebras, and I use quasicompactness of $Spec,A$ to cover it by finitely many $D(f_{ijk})$ all of which are f.g. over $C$. – Rodrigo Jan 28 '14 at 23:21
  • Perhaps you should first prove that the condition of being locally of finite type can be tested with respect to any open affine cover. Because then it is trivial(!) that these morphisms are closed under composition. But in any case, this should follow from the definitions. Which one are you using? – Martin Brandenburg Jan 29 '14 at 00:12
  • I am down with the fact that being of locally finite type can be checked over any open affine cover, but I don't think you can get it more trivially than how I stated in my last comment. You have to check (at least once in your life) that it doesn't matter what form (open/closed/affine...) the image of Spec $A$ in $W$ has, you can still cover it by open affines there. That's basically what I said in the last post. --Okay, I think I see what your saying, I can skip checking that I can cover Spec A by open affines whose images lie in the open affines that form the cover of $\rho$ Spec $A$. – Rodrigo Jan 29 '14 at 23:39