# Equivalence of Definitions of Integral Element of Algebra

## Theorem

The following definitions of the concept of Integral Element of Algebra are equivalent:

Let $A$ be a commutative ring with unity.

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

Let $b\in B$.

### Definition 1

The element $b$ is integral over $A$ if and only if it is a root of a monic polynomial in $A \sqbrk x$.

### Definition 2

The element $b$ is integral over $A$ if and only if the generated subalgebra $A \sqbrk b$ is a finitely generated module over $A$.

### Definition 3

The element $b$ is integral over $A$ if and only if the generated subalgebra $A \sqbrk b$ is contained in a subalgebra $C \le B$ which is a finitely generated module over $A$.

### Definition 4

The element $b$ is integral over $A$ if and only if there exists a faithful $A \sqbrk b$-module whose restriction of scalars to $A$ is finitely generated.

## Proof

### 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 $A \sqbrk b$.

$\Box$

### Definition 2 implies Definition 3

Choose $C = A \sqbrk b$.

$\Box$

### 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:

$1 \in A \sqbrk b \subseteq C$.

$\Box$

### 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:

$\phi \sqbrk M \subseteq M = {\mathfrak a} M$
$\phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$

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

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

 $\ds \paren {b^n + a_{n - 1} b^{n-1} + \cdots + a_1 b + a_0} m$ $=$ $\ds \map {\phi^n} m + a_{n - 1} \map {\phi^{n-1} } m + \cdots + a_1 \map \phi m + a_0$ $\ds$ $=$ $\ds 0$

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

As $M$ is faithful as an $A \sqbrk b$-module, we have:

$b^n + a_{n - 1} b^{n-1} + \cdots + a_1 b + a_0 = 0$

$\blacksquare$