Transitivity of Integrality/Lemma
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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.