Here's a classic definition: let $R\subseteq S$ be commutative rings. An element $s\in S$ is called **integral** over $R$ if $f(s)=0$ for some monic polynomial $f\in R[x]$. It's classic because appending the solutions of polynomials to base rings goes way back to the ancient pasttime of finding solutions to polynomial equations.

For example, consider $\Z\subseteq \Z[\sqrt{2}]$. Every element of $\Z[\sqrt{2}]$ is integral over $\Z$, which essentially comes down to the fact that $\sqrt{2}$ satisfies $x^2 – 2$. On the other hand, the only elements of $\Q$ integral over $\Z$ are the integers themselves.

The situation is much different for finite commutative rings. If $R\subseteq S$ are finite rings, then every element of $S$ is integral over $R$. **Proof**: suppose $s\in S$ and set $T = \{ f(s): f\in R[x]\}$. For each $t\in T$ fix a polynomial $f$ such that $f(s) = t$. The set of all such polynomials is finite so we can define $m$ as the maximum degree of all these polynomials. Then $s^{m+1}\in T$ and so there is an $f$ of degree at most $m$ such that $s^{m+1} – f(s) = 0$. Thus $s$ satisfies the monic polynomial $x^{m+1} – f(x)$. QED.

Cool right? However, this is just a more general case of the following theorem: *let $R\subseteq S$ be commutative rings. Then $S$ is finitely generated as an $R$-module if and only if $S$ is finitely generated as an $R$-algebra and every element of $S$ is integral over $R$.*