# ring-finite integral extensions are module-finite

Theorem If $B$ is a subring of $A$ and $u_{1},\ldots,u_{s}\in A$ are integral over $B$, then $B[u_{1},\ldots,u_{s}]$ is module-finite over $B$.

Proof. If $s=1$ then $u^{n}+b_{1}u^{n-1}+\cdots+b_{n}=0$, so $\{1,u,\ldots,u^{n-1}\}$ spans $B[u]$ over $B$.

If $s>1$, use induction on $B\subset B[u_{1}]\subset B[u_{1},u_{2}]\subset\ldots\subset B[u_{1},\ldots,u_{% s}]$ and multiply the spanning sets together.

Title ring-finite integral extensions are module-finite RingfiniteIntegralExtensionsAreModulefinite 2013-03-22 17:01:28 2013-03-22 17:01:28 rm50 (10146) rm50 (10146) 6 rm50 (10146) Theorem msc 16D10 msc 13C05 msc 13B02 ModuleFiniteExtensionsAreIntegral