Transitivity of Integrality

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Theorem

Let $A \subseteq B \subseteq C$ be extensions of commutative rings with unity.

Suppose that $C$ is integral over $B$, and $B$ is integral over $A$.

Then $C$ is integral over $A$.


Proof

First, a lemma:


Lemma

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

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

Let $A \left[{x_1, \dotsc, x_n}\right]$ be the subring of $B$ generated by $A \cup \left\{ {x_1, \dotsc, x_n}\right\}$ over $A$.

Then $A \left[{x_1, \dotsc, x_n}\right]$ is integral over $A$.


Proof of Lemma

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:

$\displaystyle \sum_{\operatorname{finite}} r x_1^{\alpha_1} \dotsm x_n^{\alpha_n}, \quad r \in A,\ \alpha_j \in \N \cup \left\{ {0}\right\}$

lie in $C$.

That is, they are integral over $A$.

But the set of such sums is precisely $A \left[{x_1, \dotsc, x_n}\right]$.

$\Box$


Now let $x \in C$.

Thus $x$ is supposed integral over $B$.

That is, we can find an expression:

$(1): \quad x^n + b_{n-1} x^{n-1} + \dotsb + b_1 x + b_0 = 0, \quad b_i \in B, \ i = 0, \dotsc, n-1$

Let $D$ be the subring of $C$ generated by $A \cup \left\{ {b_0, \dotsc, b_{n-1} }\right\}$.

By the lemma, $D$ is finitely generated over $A$.

Moreover, $D \left[{x}\right]$ is finitely generated over $D$ because of the equation $(1)$.

Therefore by Transitivity of Finite Generation:

$D \left[{x}\right]$ is a finitely generated $A$-module.

Finally by Equivalent Definitions of Integral Dependence, $x$ is integral over $A$.

$\blacksquare$


Linguistic Note

Integrality is a real word.