ring-finite integral extensions are module-finite

Theorem If B is a subring of A and u1,,usA are integral over B, then B[u1,,us] is module-finite over B.

Proof. If s=1 then un+b1un-1++bn=0, so {1,u,,un-1} spans B[u] over B.

If s>1, use induction on BB[u1]B[u1,u2]B[u1,,us] 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