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A composition of integral ring maps is integral

Lemma 10.36.6. Suppose that $R \to S$ and $S \to T$ are integral ring maps. Then $R \to T$ is integral.

Proof. Let $t \in T$. Let $P(x) \in S[x]$ be a monic polynomial such that $P(t) = 0$. Apply Lemma 10.36.4 to the finite set of coefficients of $P$. Hence $t$ is integral over some subalgebra $S' \subset S$ finite over $R$. Apply Lemma 10.36.4 again to find a subalgebra $T' \subset T$ finite over $S'$ and containing $t$. Lemma 10.7.3 applied to $R \to S' \to T'$ shows that $T'$ is finite over $R$. The integrality of $t$ over $R$ now follows from Lemma 10.36.3. $\square$

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