Transitivity of Integrality

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

Suppose $A \subseteq B$ is a ring extension, and $x_1,\ldots,x_n \in B$ are integral over $A$.

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

Then $A[x_1,\ldots,x_n]$ 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

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

lie in $C$, i.e. are integral over $A$.

But the set of such sums is precisely $A[x_1,\ldots,x_n]$.

$\Box$

Now let $x \in C$, so $x$ is supposed integral over $B$.

That is, we can find an expression

$x^n + b_{n-1}x^{n-1} + \cdots + b_1x + b_0 = 0,\quad b_i \in B,\ i = 0,\ldots,n-1 \qquad (1)$

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

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

Moreover, $D[x]$ is finitely generated over $D$ because of the equation $(1)$.

Therefore by Transitivity of Finite Generation, $D[x]$ is a finitely generated $A$-module.

Finally by 2. $\Rightarrow$ 1. of Equivalent Definitions of Integral Dependence, $x$ is integral over $A$.

$\blacksquare$

Note

Integrality is a real word.