Q:

For integral domains $R$ and $S$, if $S \simeq R^n$ as $R$-modules, does $\mathrm{Quot}(S) / \mathrm{Quot}(R)$ have degree $n$?

Let $\phi : R \rightarrow S$ be an injective morphism of integral domains. Then, $\phi$ extends to a morphism $\mathrm{Quot}(R) \rightarrow \mathrm{Quot}(S)$, so corresponds to a field extension $\mathrm{Quot}(S) / \mathrm{Quot}(R)$.

Now suppose that as $R$-modules, $S \simeq R^n$ for some positive integer $n$. Does the field extension then have degree $n$?

My idea is that there are elements $e_1, ... , e_n \in S$ that correspond to the standard basis vectors of $R^n$. Then, I expect that $e_1, ... , e_n$ will be a basis of $\mathrm{Quot}(S)$ over $\mathrm{Quot}(R)$.
Consider $S \otimes_R \mathrm{Quot}(R) = R^n \otimes_R \mathrm{Quot}(R) = \mathrm{Quot}(R)^n$ as $R$-modules. Multiplication gives a morphism $S \otimes_R \mathrm{Quot}(R) \rightarrow \mathrm{Quot}(S)$. However, I can't figure out if this is an isomorphism, and I'm suspecting that it's not, so maybe the answer is no.

A:

Surprisingly, the answer is yes. Set $U = R \setminus \{0\}$. Then there is a natural isomorphism of $R$ modules $S \otimes_{R} \mathrm{Frac}(R) 
 \to U^{-1}S$ given by multiplication, and up to this identification, the multiplication map $S \otimes_{R} \mathrm{Frac}(R) \to \mathrm{Frac}(S)$ is just the inclusion map $\varphi \colon U^{-1}S \to \mathrm{Frac}(S)$. This map is clearly $R$-linear, hence $\mathrm{Frac}(R)$-linear, and so it remains to show that $\varphi$ is an isomorphism, as you say. It is injective,  so we focus on showing that $\varphi$ is surjective.
This is tantamount to showing that for any $s/t \in \mathrm{Frac}(S)$, there exists some $x \in S, r \in R \setminus \{0\}$ such that $s/t = x/r$, i.e. $sr = tx$. Since the map $R \to S$ is finite, it is integral, so let $p(X) \in R[X]$ be a monic polynomial such that $p(t) = 0$. Note that we may assume that $p(X)$ has nonzero constant term; if it didn't, then we would have $p(X) = X^{k}q(X)$ for some $k \in \mathbb{N}, q(X) \in R[X]$ monic, and since $S$ is a domain, $p(t) = t^{k}q(t) = 0$ forces $q(t) = 0$.
Write $p(X) = Xq(X) - r$ for some nonzero $r \in R$. Then $tq(t) - r = 0$, so $tq(t) = r$. But then $tq(t)s = sr$, so take $x = q(t)s$, which completes the proof.

