# 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.

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 $$M_1/Sb_1$$)

• Here it seems we're using the classification theorem for finitely generated modules over PIDs, so that they can be written as sums of cyclic modules, so that since $Km_1\cap M_1$ is a finitely generated $S$-module that injects into a cyclic $K$-module it can be seen that $Km_1\cap M_1$ has to be $S$-cyclic Commented Dec 10, 2021 at 16:14
• Yes. $M_1$ is finitely generated so there is some $f\in S$ such that $f(Km_1\cap M_1)\subset Sm_1$. So $I = \{ g\in S, \frac{g}{f}m_1\in M_1\}$ is an ideal of $S$, ie. $I=(h)$ and $Km_1\cap M_1=Sb_1, b_1=\frac{gh}{f} m_1$. @Blunka Commented Dec 10, 2021 at 16:40
• Just realised if you use the classification theorem for fg modules over PIDs the exercise is pretty much immediate anyway Commented Dec 10, 2021 at 21:59