integrality is transitive

Let $C\subset B\subset A$ be rings. If $B$ is integral over $C$ and $A$ is integral over $B$, then $A$ is integral over $C$.

Proof. Choose $u\in A$. Then $u^n+b_1{u}^{n-1}+\mathrm{\cdots }+b_n=0,b_i\in B$. Thus $C[b_1,\mathrm{\dots },b_n,u]$ is integral and thus module-finite over $C[b_1,\mathrm{\dots },b_n]$. Each $b_i$ is integral over $C$, so $C[b_1,\mathrm{\dots },b_n]$ is integral hence module-finite over $C$. Thus $C[b_1,\mathrm{\dots },b_n,u]$ is module-finite, hence integral, over $C$, so $u$ is integral over $C$.

Title	integrality is transitive
Canonical name	IntegralityIsTransitive
Date of creation	2013-03-22 17:01:25
Last modified on	2013-03-22 17:01:25
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	6
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 13B21