# 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

 $\displaystyle u\omega_{1}$ $\displaystyle=b_{11}\omega_{1}+b_{12}\omega_{2}$ $\displaystyle u\omega_{2}$ $\displaystyle=b_{21}\omega_{1}+b_{22}\omega_{2}$

Consider

 $C=\left(\begin{array}[]{cc}u-b_{11}&-b_{12}\\ -b_{21}&u-b_{22}\end{array}\right)$

and let $C^{\mathrm{adj}}$ be the adjugate of $C$. Then $C\left(\begin{array}[]{c}\omega_{1}\\ \omega_{2}\end{array}\right)=0$, so $C^{\mathrm{adj}}C\left(\begin{array}[]{c}\omega_{1}\\ \omega_{2}\end{array}\right)=0$.

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

 $\left(\begin{array}[]{cc}f(u)&0\\ 0&f(u)\end{array}\right)\left(\begin{array}[]{c}\omega_{1}\\ \omega_{2}\end{array}\right)=\left(\begin{array}[]{c}0\\ 0\end{array}\right)$

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 $\mathbb{Z}$.

Title module-finite extensions are integral ModulefiniteExtensionsAreIntegral 2013-03-22 17:01:23 2013-03-22 17:01:23 rm50 (10146) rm50 (10146) 9 rm50 (10146) Theorem msc 16D10 msc 13C05 msc 13B02 RingFiniteIntegralExtensionsAreModuleFinite