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.
Sources
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- 2011: J.S. Milne: Algebraic Number Theory: Chapter $2$ Ring of Integers: Integral elements: Proposition $2.15$