# 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

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 $A$.

$\Box$

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 $A \cup \set {b_0, \dotsc, b_{n - 1} }$.

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

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

Therefore by Transitivity of Finite Generation:

- $D \sqbrk x$ 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.