ring-finite integral extensions are module-finite

Theorem If $B$ is a subring of $A$ and $u_1,\mathrm{\dots },u_s\in A$ are integral over $B$, then $B[u_1,\mathrm{\dots },u_s]$ is module-finite over $B$.

Proof. If $s=1$ then $u^n+b_1{u}^{n-1}+\mathrm{\cdots }+b_n=0$, so $\{1,u,\mathrm{\dots },u^{n-1}\}$ spans $B[u]$ over $B$.

If $s>1$, use induction on $B\subset B[u_1]\subset B[u_1,u_2]\subset \mathrm{\dots }\subset B[u_1,\mathrm{\dots },u_s]$ and multiply the spanning sets together.

Title	ring-finite integral extensions are module-finite
Canonical name	RingfiniteIntegralExtensionsAreModulefinite
Date of creation	2013-03-22 17:01:28
Last modified on	2013-03-22 17:01:28
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	6
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 16D10
Classification	msc 13C05
Classification	msc 13B02
Related topic	ModuleFiniteExtensionsAreIntegral