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?

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Oct 29, 2020 at 12:26
  • $\begingroup$ I think the answers have elaborated my comment. $\endgroup$ Commented Oct 29, 2020 at 14:51

2 Answers 2


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.


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