# 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 $u^{n}+b_{1}u^{n-1}+\cdots+b_{n}=0,b_{i}\in B$. Thus $C[b_{1},\ldots,b_{n},u]$ is integral and thus module-finite over $C[b_{1},\ldots,b_{n}]$. Each $b_{i}$ is integral over $C$, so $C[b_{1},\ldots,b_{n}]$ is integral hence module-finite over $C$. Thus $C[b_{1},\ldots,b_{n},u]$ is module-finite, hence integral, over $C$, so $u$ is integral over $C$.

Title integrality is transitive IntegralityIsTransitive 2013-03-22 17:01:25 2013-03-22 17:01:25 rm50 (10146) rm50 (10146) 6 rm50 (10146) Theorem msc 13B21