Equivalence of Definitions of Integral Element of Algebra

Theorem
Let $A$ be a commutative ring with unity.

Let $f : A \to B$ be a commutative $A$-algebra.

Let $b\in B$.

Definition 1 implies Definition 2
Assume $b$ is a root of a monic polynomial in $A \sqbrk x$.

That is, there are $n \in \N_{>0}$ and $a_1, \ldots, a_{n-1} \in A$ be such that:
 * $b^n + a_{n - 1} b^{n-1} + \cdots + a_1 b + a_0 = 0$

That is:
 * $b^n = - a_0 - a_1 b - \cdots - a_{n-1} b^{n-1}$

Therefore $\set {1, b, \ldots, b^{n-1} }$ generates $A \sqbrk b$.

Definition 2 implies Definition 3
Choose $C = A \sqbrk b$.

Definition 3 implies Definition 4
Let $C$ be a finitely generated $A$-module such that:
 * $A \sqbrk b \subseteq C$

Then $C$ is also $A \sqbrk b$-module.

Clearly, $C$ is faithful over $A \sqbrk b$, since:
 * $1 \in A \sqbrk b \subseteq C$.

Definition 4 implies Definition 1
Let $M$ be a faithful $A \sqbrk b$-module whose restriction of scalars to $A$ is finitely generated.

Define an $A$-module endomorphism $\phi : M \to M$ by:
 * $ \map \phi m := b m$

Let $\mathfrak a := A$.

Then:
 * $\phi \sqbrk M \subseteq M = {\mathfrak a} M$

By Cayley-Hamilton Theorem for Finitely Generated Modules:
 * $\phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$

for some $a_i \in {\mathfrak a} = A$.

That is, for all $m \in M$:

as $\map {\phi^i} m = b^i m$.

As $M$ is faithful as an $A \sqbrk b$-module, we have:
 * $b^n + a_{n - 1} b^{n-1} + \cdots + a_1 b + a_0 = 0$