# How can a polynomial ring be finitely generated?

I am struggling to see this:

The following are equivalent: (1) $$x \in B$$ is integral over $$A$$; (2) $$A[x]$$ is finite over $$A$$.

Can a polynomial ring be a module and how can a polynomial rings be finitely generated modules? They have powers so how can it be written as an A-linear combination?

• If $x$ is integral over $A$ it is a root of a monic polynomial, which enables the higher powers of $x$ to be written in terms of lower powers. Commented Oct 29, 2020 at 12:26
• I think the answers have elaborated my comment. Commented Oct 29, 2020 at 14:51

Your sentence in the comment does not quite make sense: "It is obvious I can write $$x$$ equals to a polynomial with degree $$n$$".
But on the other hand, suppose you know that $$x$$ is a root of a monic polynomial over $$A$$, which means that $$x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 = 0$$ where the lower term coefficients $$a_0,...,a_{n-1}$$ are all in $$A$$.
Then it is obvious that you can write $$x^n$$ equal to a polynomial of degree $$\le n-1$$, namely $$x^n = - a_{n-1} x^{n-1} - ... - a_1 x - a_0$$ And then an easy induction shows for each $$k \ge n$$ that $$x^k$$ is also equal to a polynomial of degree $$\le n-1$$. For instance, in the next step of the induction multiply both sides of that equation by $$x$$, so the new equation has $$x^{n+1}$$ on the left hand side, and the leading term on the right hand side will be $$-a_{n-1} x^n$$, and then use the above equation to substitute for $$x^n$$ on the right hand side.
It follows that $$A[x]$$ is finitely generated over $$A$$, the generating set being $$1,x,....,x^{n-1}$$.
Consider $$A=\mathbb Z$$ and $$B=\mathbb Z[\sqrt[3]{2}]$$. Then $$B$$ a finitely generated $$A$$-module: $$\{1, \sqrt[3]{2}, \sqrt[3]{4}\}$$ is a generating set.
Note that $$B$$ is not a polynomial ring; it is the ring of polynomial expressions in $$\sqrt[3]{2}$$, which is not the same thing because all powers greater than $$2$$ can be reduced.