13B21-IntegralityIsTransitive.tex

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\pmtitle{integrality is transitive}
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\pmtype{Theorem}
\pmcomment{trigger rebuild}
\pmclassification{msc}{13B21}

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\begin{document}
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 $C$.

\textbf{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 $C$.

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