Proof Verification: Morphism of Schemes is Locally of Finite Type

Question

I've been working on problem 3.1 of chapter II of Hartshorne, and have been stuck for a while. I found a proof that seems satisfactory, but every solution I have seen online uses a more involved argument, so I'm guessing that it is wrong.

Hartshorne defines a morphism $f \colon X \to Y$ of schemes to be locally of finite type if there exists an open cover of $Y$ by affine schemes $V_i = \text{Spec} \,B_i$, such that, for each $i$, $f^{- 1}(V_i)$ can be covered by rep affine subsets $U_{ij} = \text{Spec} \, A_{ij}$, such that the induced homomorphism of rings $f^\sharp \colon B_i \to A_{ij}$ gives $A_{ij}$ the structure of a finitely generated $B_i$-algebra.

The problem I am working on is proving that this is a local property, that is, for any open affine subset $V = \text{Spec} \, B$ of $Y$, $f^{-1}(V)$ may be covered by open affine subsets $U_i = \text{Spec} \, A_i$, where each $A_i$ is a finitely generated $B$-algebra. This is my tentative "proof:"

Since $f$ is locally of finite type, we have open cover $\{V_i\} = \{\text{Spec} \, B_i\}$ for $Y$ such that, for each $i$,

$$f^{-1}(\text{Spec} \, B_i) \subset \bigcup_j \text{Spec} \, A_{ij},$$ where $A_{ij}$ is a finitely generated $B_i$-algebra, for each $j$. $V \cap V_i$ may be covered by simultaneously distinguished open subsets $D_B(g) = D_{B_i}(h_i)$. Since $D_{B_i}(h_i) \cong \text{Spec} \, (B_i)_{h_i}$ , and

$$f^{-1}(\text{Spec} \, \big(B_i)_{h_i}\big) \subset \bigcup_j \text{Spec} \, (A_{ij})_{h_i},$$

and $(A_{ij})_{h_i}$ is finitely generated as a $(B_i)_{h_i}$-algebra, we see that $D_B(g) = D_{B_i}(g^i)$ satisfies the property we are trying to prove. Since $\{V_i\}$ covers $Y$, $\{V \cap V_i\}$ covers $V$, meaning we have open cover $\{D_{B_i}(h_i)\}$ for $V$, thus open cover $\{D_B(g)\}$ for $V$. As an affine scheme, $V$ is quasi-compact, meaning there exists a finite subcover. Thus we have finitely many elements $g_1, \dots , g_n$ of $B$ such that $D_B(g_1), \dots , D_B(g_n)$ cover $V = \text{Spec} \, B$, meaning $\langle g_1, \dots , g_n\rangle = B$.

Using standard inverse image properties, we have that

$$f^{- 1}(V) = f^{-1}\left(\bigcup_{k = 1}^n D_B(g_k)\right) = \bigcup_{k = 1}^n f^{-1}\big(D_B(g_k)\big),$$

Each $D_B(g_k)$ has an inverse image covered by elements of the form $\text{Spec} (A_{ij})_{h_i}$, with $(A_{ij})_{h_i}$ finitely generated as a $(B_i)_{h_i}$-algebra, thus as a $B_{g_k}$-algebra. Therefore $f^{-1}(V)$ can be covered by elements of the form $\text{Spec} (A_{ij})_{h_i}$. As $(A_{ij})_{h_i}$ is finitely generated as a $B_{g_k}$-algebra, we have that

$$ (A_{ij})_{h_i} \cong B_{g_k}[x_1, \dots , x_n]/I,$$ yet this means that $$(A_{ij})_{h_i} \cong B[x_1, \dots , x_n, 1/g_k]/I,$$ which would imply that $(A_{ij})_{h_i}$ is finitely generated as a $B$-algebra.This holds for each $A_{ij}$, meaning that $V = \text{Spec} \, B$ satisfies the desired property, and we are done.

Yet all of the other proofs of this that I have seen use the fact that if $\langle f_1, \dots , f_n \rangle = A$, and each $A_{f_i}$ is a finitely generated $B$-algebra, then so is $A$. Is my proof alright? I feel like I must have missed something obvious, but I can't find an error. These other proofs that I have seen for this property have been thoroughly unsatisfying, so if anyone has a better, cleaner proof that they know of, I'd love to see it. Thanks in advance for any help or hints or anything!

Answer 1

I think that you’re mixing indices a bit, but the proof is correct. Let’s see if we can’t rephrase it a little bit to make it clearer.

I’m using “locally of finite type” as per your definition (I believe that there are other definitions that fortunately turn out to be equivalent), and “AF type” for the stronger property of a map $\operatorname{Spec}{B} \rightarrow \operatorname{Spec}{A}$ corresponding to a ring homomorphism $A \rightarrow B$ making $B$ a finitely-generated $A$-algebra.

Let’s first digress.

The definition of “locally of finite type” can thus be written as follows: let $f: X \rightarrow Y$ be a morphism of schemes. It is of finite type if there exists an affine open cover $Y=\bigcup_i{V_i}$ such that every $f^{-1}(V_i)$ has an affine open $f^{-1}(V_i)=\bigcup_j{U_{i,j}}$ such that every $f: U_{i,j} \rightarrow V_i$ is AF type.

Also consider the following statement: if $A$ is a $B$-algebra and $f_1,\ldots,f_r \in A$ are such that $(f_1,\ldots,f_r)=A$ and every $A_{f_i}$ is finitely generated over $B$, then $A$ is a finitely generated $B$-algebra. In the scheme-theoretic language above, this is equivalent to: a morphism of affine schemes $X \rightarrow Y$ which is locally of finite type for the open cover $Y$ [“locally of finite type” is enough, given your proof] is AF type. So it’s not the same thing (and it’s an important fact as well)!

Anyway, now for the proof. Your argument can be rephrased as follows:

Fix affine open covers $Y=\bigcup_i{V_i}$ and $f^{-1}(V_i)=\bigcup_j{U_{i,j}}$. We can also find affine open covers $V_i \cap V=\bigcup_k{W_{i,k}}$ where the $W_{i,k}$ that are principal in $V_i$ and in $V$ (this deserves elaboration if you haven’t done it yourself, in my opinion).

In particular, it is formal that the $f: U_{i,j} \cap f^{-1}(W_{i,k}) \rightarrow W_{i,k}$ are AF type (as “principal base change” of $f: U_{i,j} \rightarrow V_i$). But $W_{i,k} \subset V$ is AF type (as $W_{i,k}$ is principal in $V$), so $f: U_{i,j} \cap f^{-1}(W_{i,k}) \rightarrow V$ is AF type.

Now, every $U_{i,j}\cap f^{-1}(W_{i,k})$ is affine, and $$\bigcup_{i,j,k}{U_{i,j}\cap f^{-1}(W_{i,k})} = \bigcup_{i,k}{f^{-1}(V_i) \cap f^{-1}(W_{i,k})}= f^{-1}\left(\bigcup_{i,k}{V_i \cap W_{i,k}}\right)=f^{-1}\left(\bigcup_i{V\cap V_i}\right)=f^{-1}(V). $$