Extension of the fields of fractions of integral domains

Question

Let $A$ and $B$ be integral domains, and let $\varphi:A\to B$ a ring homomorphism. We can give $B$ a structure of $A$-module by saying $ab = \varphi(a)b$. Suppose that $B$ is a finitely generated $A$-module. Let $F_1 = \text{Frac } A$ and $F_2 = \text{Frac }B$.

Is it true that $F_2$ is (isomorphic to) a finite extension of $F_1$?

Answer 1

First notice that $\phi$ must be supposed injective, else it is impossible to extend $\phi$ to the fraction fields.
If this is the case then, yes, the extended morphism $\text{Frac }\phi:\text{Frac } A\to \text{Frac }B$ is a finite extension.
The trick is to consider the multiplicative set $S=A\setminus \{0\}$ and the morphism $S^{-1}\phi: S^{-1}A=\text{Frac } A=F_1\to S^{-1}B$, for which $S^{-1}B$ is finitely generated over $S^{-1}A.$ (generators of $B$ over $A$ remain generators for $S^{-1}B$ over $S^{-1}A$ !).
Since $S^{-1}B$ is a finite dimensional domain over the field $F_1$, it is a field (see here) and thus $S^{-1}B=\text{Frac }B=F_2$, so that $F_2$ is finite dimensional over $F_1$ as requested.

Answer 2

This is false in general. For example if $B=\mathbb Z/p\mathbb Z$ and $A=\mathbb Z$, then certainly $B$ is a finitely generated $A$-module, but $\mathbb F_p$ is not even a field extension of $\mathbb Q$. Alternatively, if $A=\mathbb Z[X]$, $B=\mathbb Z$ and $\varphi: f(X)\mapsto f(0)$, then $F_1$ is isomorphic to an infinite extension of $F_2$.

However, this is true if $\varphi$ is injective, in which case we can view $A$ as a subring of $B$.

Indeed, pick a set of generators $x_1,\ldots, x_n\in B$ of $B$ as an $A$-module. It follows that $F_2 = F_1(x_1,\ldots, x_n)$.

Indeed:

• Clearly $F_1(x_1,\ldots, x_n)\subset F_2$, since $F_1\subset F_2$ and $x_1,\ldots, x_n\in B\subset F_2$.
• Conversely, if $y\in F_2$ is non-zero, then there exists a non-zero $b\in B$ such that $by\in B$. Since $b$ and $by$ can both be expressed as $A$-linear combinations of the $x_i$, it follows that $b,by\in F_1(x_1,\ldots,x_n)$. Since $F_1(x_1,\ldots,x_n)$ is a field, it follows that $y\in F_1(x_1,\ldots,x_n)$.

It remains to check that each $x_i$ is algebraic over $F_1$. This is clear, since if $1,x_i,x_i^2,\ldots$ were linearly independent over $F_1$, they would also be linearly independent over $A$, which would contradict the fact that $B$ is finitely generated.