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
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 $A$.