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I am trying to prove that morphisms of finite type are stable under base change, but I am having some trouble moving from the case where everything is affine to the general case. Suppose $f:X \rightarrow Y$ is a morphism of finite type and $Y'$ is a $Y$-scheme. I want to show that the morphism $g: X \times_Y Y' \rightarrow Y'$ is of finite type. In the case that $X$, $Y$, and $Y'$ are affine, I understand why this is true. For the general case, by a lemma in Liu's book, it is enough to show that there is an affine open cover $\{V_i\}_i$ of $Y'$ such that for each $i$, $g^{-1}(V_i)$ is a finite union of affine open subsets $U_{ij}$ such that for each $i$ and $j$, $O_X(U_{ij})$ is a finitely generated algebra over $O_Y(V_i)$. Here is my attempt at proving this.

Choose an affine open cover $\{V_i\}_i$ of $Y'$. Is it true that $g^{-1}(V_i)=X \times_Y V_i$? I think this should follow from how we constructed the fibered product by gluing. Since $f:X \rightarrow Y$ is of finite type, we may choose an affine open cover $\{Y_j\}$ of $Y$ such that $f^{-1}(Y_j)$ is covered by a finite number of affine opens $W_{jk}$. Now, $W_{jk} \times_Y V_i$ are open subschemes that cover $X \times_Y V_i$, but since $Y$ is not necessarily affine, these schemes are not necessarily affine, right? Furthermore, if we are using the $W_{jk}$ to cover all of $X$, there could be infinitely many of them. To make the $W_{jk} \times_Y V_i$ affine, we could further cover them with $W_{jk} \times_{Y_k} V_i$, but we are not guaranteed finitely many $Y_k$ either, so while these schemes will be affine, there will not necessarily be finitely many. I have been having some trouble with these sorts of arguments where one can immediately reduce to the affine case, and some help here would be greatly appreciated.

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I would also appreciate tips on when one can reduce to either the domain or target of a morphism being affine in general. – Vitaly Lorman Feb 10 '11 at 3:30
The first claim of your second paragraph is true: if $f:X\to Y$ is a morphism of schemes and $V\subset Y$ then $f^{-1}(V)\cong X\times_Y V$; you can easily see that $f^{-1}(V)$ satisfies the universal property of the fibered product. Your statement follows then from the properties of the fibered product. – user1728 Feb 10 '11 at 10:27
up vote 7 down vote accepted

You can do a slight change to make your argument work and give you an affine cover of $X\times_Y Y'$ as follows. Call $h$ the base change morphism $Y'\to Y$. Consider a cover of $Y$ by open affines $V_i=Spec(A_i)$ and cover any $h^{-1}(V_i)$ by open affines $V_{ij}=Spec(A_{ij})$ of $Y'$. The preimage of $V_{ij}$ by $g$ is as you said $X\times_Y V_{ij}$ which by the properties of the fibered product coincides with $X_i\times_{V_i}V_{ij}$, where $X_i=f^{-1}(V_i)$ (this is crucial to end up with affine schemes as you will see in a moment). You can cover $X_i$ with open affines $X_{ik}=Spec(B_{ik})$ with all $B_{ik}$ finite type $A_i$-algebras. So $g^{-1}(V_{ij})$ is covered by the affines $X_{ik}\times_{V_i}V_{ij}=Spec(B_{ik}\otimes_{A_i}A_{ij})$, which are $A_{ij}$-algebras of finite type. This proves that the base change of $f$ is locally of finite type (actually we didn't use that $f$ is quasi compact, so we proved that locally of finite type morphisms are stable under base change).

If now $f$ is quasi compact, you just need a finite number of $X_{ik}$ to cover $X_{i}$ and so a finite number of $X_{ik}\times_{V_i}V_{ij}$ is enough to cover $g^{-1}(V_{ij})$, proving the quasi-compactness of $g$.

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The important thing that I was missing is getting the affine open cover of $Y'$ from an affine open cover of $Y$ and getting an affine open cover of $X$ from the same cover of $Y$ in a similar way such that everything works out nicely. Thanks for your help! – Vitaly Lorman Feb 10 '11 at 13:34
You're welcome! – user1728 Feb 10 '11 at 13:47

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