Domain containing polynomial ring over field, if finitely generated generated as module over it, is free as module over it

This is the first part of exercise 4.2 from Eisenbud's "Commutative Algebra" textbook. The problem is:

Let $R$ be a domain containing a polynomial ring in one variable over a field, say $R \supseteq S = k[t]$. Show that if $R$ is a finitely generated $S$-module, then $R$ is free as an $S$-module.

The relevant chapter of the book, chapter 4, deals with integral extensions, the Cayley-Hamilton theorem, Nakayama's lemma, lying over and going up, Jacobson rings and the Nullstellensatz, so it's likely Eisenbud has a solution in mind involving at least one of these.

One thing I tried was viewing $R$ as a quotient $\frac{k[x_1,\ldots,x_n]}{J}$ for some $n$, so that $J$ is prime and contains a monic in $x_n$ (by Proposition 4.1 of the book), and seeking to show that $J$ is of the form $\langle p_2,\ldots,p_n\rangle$ with $p_i\in k[x_1,\ldots,x_i]$ monic. However this has proved difficult and now I'm not sure if it's even true.

Solution 1:

Let $K=k(t)$, $W_1 = K \otimes_S R,M_1=R$. Then $W_1$ is a finite dimensional $K$-vector space and $M_1$ is a finitely generated sub-$S$-module of it, where $S=k[t]$ is a PID.

Take $m_1\ne 0\in M_1$, then $K m_1\cap M_1 = S b_1$ for some $b_1\in M_1$ (this is where we need that $S$ is a PID and that $M_1$ is finitely generated). Let $$M_2 = M_1/Sb_1, \qquad W_2 = W_1/Kb_1$$

And then repeat with $M_2,W_2$ which satisfy the same hypothesis, obtaining $b_2$ and then $M_3,W_3$, and so on.

$\dim_K(W_n)$ decreases at each step so the process terminates, and the obtained $b_1,\ldots,b_d$ are a free $S$-basis of $M_1=R$.

(take any representative in $M_1$ of $b_2$ which at first is in $M_1/Sb_1$)