Hartshorne exercise II.3.3(c) asks the reader to prove that if $f:X\to Y$ is finite type, then for any $\mathrm{Spec}A\subset Y$ and $\mathrm{Spec}B\subset X$ with $\mathrm{Spec}B\subset f^{-1} \mathrm{Spec}A$, we have $A\to B$ is of finite type.

However, the proof I produced seems to work in the case that $f$ is only locally of finite type, but exercise 2 of that section of Hartshorne asks to prove a strictly weaker condition in that case. (That is, for any affine of $Y$, the preimage can be covered by affines giving us finite-type homomorphisms.)

Is the property of local finite-typeness affine-local on the source? My proof seemed very straightforward:

Take any affine open $\mathrm{Spec}\ B$ of $X$, and also choose a cover $\mathrm{Spec}\ B_i$ of $X$ by affine opens so that each $A\to B_i$ is finite type. (This cover is guaranteed to exist by part II.3.2.(b).) Then by the "Affine Communication Lemma" (5.3.2 of Vakil's FOAG - page 155 in the Feb 25, 2013 version) it suffices to show that if $A\to B$ is finite type and $f\in B$, so is $A\to B_f$, and if $f_1,\dots,f_n\in B$ generate the unit and each $A\to B_{f_i}$ is finite type, then so is $A\to B$.

The first claim is trivial. As for the second, take $b\in B$, and choose $b_{i1},\dots,b_{ik}\in B$ which generate $B_{f_i}$ over $A$. Then we can choose $n$ and $g_{ij}\in A$ for each $i$ we have $f_i^nb=\sum_j g_{ij}b_{ij}$. Some linear combination of the $f_i^n$ is equal to one, and so we can write $b$ as an $A$-linear combination of all the $g_{ij}$.

Does this reasoning make sense?

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    $\begingroup$ "Is “locally of finite type” affine-local on the source?" Yes, see any good introduction to algebraic geometry (and Hartshorne is not good, although standard). $\endgroup$ Jun 7 '13 at 10:05
  • $\begingroup$ Okay thanks! I've been using Hartshorne, Qing Lui, and Vakil together. I don't usually even read the text of Hartshorne other than to get definitions, and I otherwise just use it for the exercises, but I couldn't seem to find anything about it in Lui or Vakil. Vakil doesn't seem to do it, but I did find it in Lui, page 87, Definition 2.1 and Proposition 2.2, for those interested. $\endgroup$ Jun 7 '13 at 15:13
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    $\begingroup$ Hartshorne isn't a good idea even for definitions (some are simply "wrong"). $\endgroup$
    – user314
    Jun 8 '13 at 21:07
  • $\begingroup$ I've noticed that; I only get definitions from Hartshorne for purposes of Hartshorne's exercises (and unfortunately they are "right" in that context). $\endgroup$ Jun 11 '13 at 1:44

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