# Transitivity of Integrality/Lemma

## Lemma for Transitivity of Integrality

Let $A \subseteq B$ be a ring extension.

Let $x_1, \dotsc, x_n \in B$ be integral over $A$.

Let $A \sqbrk {x_1, \dotsc, x_n}$ be the subring of $B$ generated by $A \cup \set {x_1, \dotsc, x_n}$ over $A$.

Then $A \sqbrk {x_1, \dotsc, x_n}$ is integral over $A$.

## Proof

Let $C$ be the integral closure of $A$ in $B$.

Since the $x_i$ are integral over $A$, they lie in $C$.

So by Integral Closure is Subring, all sums of the form:

$\ds \sum_{\text {finite} } r x_1^{\alpha_1} \dotsm x_n^{\alpha_n}, \quad r \in A,\ \alpha_j \in \N \cup \set 0$

lie in $C$.

That is, they are integral over $A$.

But the set of such sums is precisely $A \sqbrk {x_1, \dotsc, x_n}$.

$\blacksquare$

## Linguistic Note

Integrality is a real word.