module-finite extensions are integral

Theorem Suppose $B\subset A$ is module-finite. Then $A$ is integral over $B$.

Proof.Choose $u\in A$.

For clarity, assume $A$ is spanned by two elements ${\omega }_1,{\omega }_2$. The proof given clearly generalizes to the case where a spanning set for $A$ has more than two elements.

Write

$u{\omega }_1$$=b_{11}{\omega }_1+b_{12}{\omega }_2$

$u{\omega }_2$$=b_{21}{\omega }_1+b_{22}{\omega }_2$

Consider

and let $C^{\mathrm{adj}}$ be the adjugate of $C$.Then $C\left(\begin{array}{c}\hfill {\omega }_1\hfill \\ \hfill {\omega }_2\hfill \end{array}\right)=0$, so$C^{\mathrm{adj}}C\left(\begin{array}{c}\hfill {\omega }_1\hfill \\ \hfill {\omega }_2\hfill \end{array}\right)=0$.

Now, $C^{\mathrm{adj}}C$ is a diagonal matrix with $detC$ on the diagonal, so

where $f\in B[x]$ is monic.

But neither ${\omega }_1$ nor ${\omega }_2$ is zero, so $f(u)$ must be.

Note that, as with the field case, the converse is not true. For example, the algebraic integers are integral but not finite over $\mathbb{Z}$.