Is a finite ring map of finite presentation finitely presented? Let $\phi: A \rightarrow B$ be a ring map (of commutative unitary rings).
Assume that $\phi$ is finite, i. e. $B$ is finitely generated as an $A$-module, and $\phi$ is of finite presentation as in https://stacks.math.columbia.edu/tag/00F3.
Do we then get that $B$ is finitely presented as an $A$-module? I know the converse is true from https://stacks.math.columbia.edu/tag/0D46.
 A: Yes, this holds.
Consider the following Lemmas:

Lemma 1: Let $Q$ be a finitely presented $R$-module, and consider a surjective $R$-linear map $Q \to S$ with a finitely generated kernel. Then $S$ is also a finitely presented $R$-module.

Proof: a straightforward exercise. $\square$

Lemma 2: Consider an $R$-algebra $W$ which is finitely generated as an $R$-module. There is some $R$-algebra $Q$ which is finitely presented as an $R$-module and some surjective $R$-algebra homomorphism $Q \to W$.

Proof: we begin with a surjective $R$-linear map $f : R^n \to W$, and we call the standard generators of $R_n$ $e_1, e_2, \ldots, e_n$.
Now, we select, for each $1 \leq i, j \leq n$, some $w_{ij} \in R^n$ such that $f(w_{ij}) = f(e_i) f(e_j)$. We also select $q$ such that $f(q) = 1$.
For convenience, we define the bilinear operator $\star : (R^n)^2 \to R^n$ by $e_i \star e_j = w_{ij}$.
Now let $I$ be the ideal generated by $\{(e_i \star e_j) - (e_j \star e_i) \mid 1 \leq i, j \leq n\} \cup \{((e_i \star e_j) \star e_k) - ((e_i \star e_j) \star e_k) \mid 1 \leq i, j, k \leq n\} \cup \{(e_i \star q) - e_i \mid 1 \leq i \leq n\}$. It’s easy to see from the definition that $I$ is finitely generated.
Now consider the module $Q = R^n / I$, with projection map $\pi : R^n \to Q$. We see that $Q$ is finitely presented.
We define the operator $\cdot : Q^2 \to Q$ by $\pi(a) \cdot \pi(b) = \pi(a \star b)$. It’s straightforward, if a bit tedious, to confirm this operator is well-defined. Also straightforward but tedious are verifying that $\cdot$ is bilinear and satisfies the associative and commutative properties. We can also show that $\pi(q)$ is the unit of this operation, making $Q$ a commutative ring. Verify that for all $r \in R$ and $s \in Q$, we have $r \cdot q = (r \cdot 1) \cdot q$; this confirms we have an $R$-algebra structure on $Q$ compatible with its module structure.
Finally, we note that $I \subseteq \ker f$. Therefore, we have an $R$-module homomorphism $g : Q \to W$ defined by $g \circ \pi = f$; since $f$ is surjective, so is $g$. A careful analysis of $\star$ shows us that $f(a \star b) = f(a) \cdot f(b)$ for all $a, b \in R^n$. Therefore, we see that for all $a, b \in Q$, we have $g(a \cdot b) = g(a) \cdot g(b)$. So $g$ is a ring homomorphism and an $R$-linear map, hence an $R$-algebra homomorphism. $\square$

Lemma 3: Suppose $A$ is a finitely-presented $R$-algebra, and consider some surjective $R$-algebra map $f : R[X_1, \ldots, X_n] \to A$. Then $\ker f$ is a finitely generated ideal.

Proof: another fairly straightforward one, which I’ll leave as an exercise. Hint: if we have two surjections $f : R[X_1, \ldots, X_n] \to A$ and $g : R[Y_1, \ldots, Y_m] \to A$, we can consider a surjection $h : R[X_1, \ldots, X_n, Y_1, \ldots, Y_n]$. Show that $f$ has a finitely generated kernel iff $h$ does, iff $g$ does. $\square$

Lemma 4: suppose $A$ is an $R$-algebra which is finitely generated as an $R$-module, and suppose $I \subseteq A$ is a finitely generated ideal. Then $I$ is a finitely generated submodule of $A$.

Proof: straightforward. Combine the generators of $I$ as an ideal with the generators of $A$ as a module to get generators for $I$ as a module. $\square$
Now that the tedium is out of the way, let’s continue with our proof. We know that $W$ is an $R$-algebra which is finitely generated as a module, so we invoke Lemma 2. We have some $R$-algebra $Q$ which is finitely presented as a module, together with some surjective $R$-algebra map $f : Q \to W$.
Now since $Q$ is finitely generated as a module, it is certainly finitely generated as an algebra. So take some $R$-algebra surjection $g : R[X_1, \ldots, X_n] \to Q$. Then $f \circ g$ is an $R$-algebra surjection. By Lemma 3, $\ker (f \circ g)$ is a finitely generated ideal. That is, $g^{-1}(\ker f)$ is a finitely generated ideal.
Since $g$ is surjective, we see that $\ker f$ is a finitely generated ideal of $Q$. By Lemma 4, $\ker f$ is also finitely generated as a submodule of $Q$. By Lemma 1, $W$ is a finitely presented $R$-module. $\square$
A: The answer is yes, I also found it in the Stacks project now: https://stacks.math.columbia.edu/tag/0564
