Composition of finitely presented ring morphisms.

Question

Let $A, B$ be a commutative rings with identity and let $h:A\to B$ be a ring homomorphism. We say $h$ is finitely presented if $A[x_1,...,x_n]/(f_1,...,f_k)\simeq B$ for some $n\in\mathbb Z_+$ and $f_1,...,f_k\in A[x_1,...,x_n].$ I would like to prove the following statement:

If $A\to B$ and $B\to C$ are finitely presented, then their composition $A\to C$ is also finitely presented.

Attempts: By definition let $A[x_1,...,x_n]/I \simeq B$ and $B[y_1,...,y_m]/J\simeq C$ where $I$ and $J$ are f.g. over $A[x_1,...,x_n]$ and $B[y_1,...,y_m]$, respectively. Then we would have $$(A[x_1,...,x_n]/I)[y_1,...,y_n]\simeq A[x_1,...,x_n,y_1,...,y_n]/I[y_1,...,y_n]\simeq B[y_1,...,y_n].$$

Name $R=A[x_1,...,x_n,y_1,...,y_n]$ and $K=I[y_1,...,y_n]$ then $K$ is obviously f.g. as an ideal over the ring $R$. Name $T=B[y_1,...,y_n]$ then by the isomorphism $R/K\simeq T$, there is a finitely generated ideal $P/K\simeq J, P\supset K$ in the ring $R/K$ such that $(R/K)/(P/K)\simeq T/J\simeq C$. But by the third isomorphism theorem $R/P\simeq (R/K)/(P/K)$ so finally it suffices to show that $P$ is f.g. over R.

The last step is finished as follows: $P/K$ is f.g. over $R/K$, thus f.g. over $R$. Since both $P/K$ and $K$ are f.g. over $R$ in the exact sequence $0\to K\to P\to P/K$ it follows that $P$ is f.g. over $R$.

This is kind of lengthy so I wonder if there is a much simplier proof to this fact. Since most books/references I read mark this as trivial.

Answer 1

If $A\to B$ and $B\to C$ are finitely presented, then by definition there exists surjections $f: A[x_1,...,x_n]\to B$ and $g: B[y_1,...,y_m]\to C$ with finitely generated kernels for both(as an ideal). The ring map $A[x_1,...,x_n]\to B$ induces another ring map $f^*:A[x_1,...,x_n,y_1,...,y_m]\to B[y_1,...,y_m]$ with $\mathrm{Ker}(f^*)=\mathrm{Ker}(f)[y_1,...,y_m]$ obviously a finitely generated ideal in $A[x_1,...,x_n,y_1,...,y_m]$. It remains to show that $\mathrm{Ker}(g\circ f^*)$ is finitely generated. We prove a more general result below:

Let $A\to B\to C$ are ring maps such that $A\to B$ is surjective. If both the kernel of $f: A\to B$ and $g: B\to C$ are finitely generated as an ideal, say $\mathrm{Ker}(f)=(k_1,...,k_n)$ and $\mathrm{Ker} (g)=(b_1,...,b_n)$ then the kernel of the composition map $A\to C$ is also finitely generated.

Say $g(f(x))=0$ for $x\in A$ then we can write $f(x)=\sum b_ib'_i$ where $b_i,b'_i\in B$ and by surjectivity $f(x)=\sum f(a_ia'_i)$ for some $a_i, a'_i\in A$, or $f(x-\sum a_ia'_i)=0.$ Thus $x-\sum a_ia'_i\in \mathrm{Ker} (f)$ which means that $x-\sum a_ia'_i=\sum k_ik'_i$ for some $k'_i\in A.$ Thus $\mathrm{Ker}(g\circ f)$ is generated by the $a_i$ and $k_i$ in $A$, as desired.