# Equivalence of Definitions of Integral Dependence

## Theorem

Let $A$ be an extension of a commutative ring with unity $R$.

For $x \in A$, the following are equivalent:

 $\text {(1)}: \quad$ $\ds$  $\ds$ $x$ is integral over $R$  $\text {(2)}: \quad$ $\ds$  $\ds$ The $R$-module $R \sqbrk x$ is finitely generated  $\text {(3)}: \quad$ $\ds$  $\ds$ There exists a subring $B$ of $A$ such that $x \in B$, $R \subseteq B$ and $B$ is a finitely generated $R$-module  $\text {(4)}: \quad$ $\ds$  $\ds$ There exists a subring $B$ of $A$ such that $x B \subseteq B$ and $B$ is finitely generated over $R$  $\text {(5)}: \quad$ $\ds$  $\ds$ There exists a faithful $R \sqbrk x$-module $B$ that is finitely generated as an $R$-module 

## Proof

### $(1) \implies (2)$

By hypothesis, there exist $r_0, \ldots, r_{n - 1} \in R$ such that:

$x^n + r_{n - 1} x^{n - 1} + \cdots + r_1 x + r_0 = 0$

So the powers $x^k$, $k \ge n$ can be written as an $R$-linear combination of:

$\set {1, \ldots, x^{n - 1} }$

Therefore this set generates $R \sqbrk x$.

$\Box$

### $(2) \implies (3)$

$B = R \sqbrk x$ trivially satisfies the required conditions.

$\Box$

### $(3) \implies (4)$

By $(3)$ we have an $R$-module $B$ such that $R \subseteq B$, $B$ is finitely generated over $R$.

Also, $x \in B$, so $x B \subseteq B$ as required.

$\Box$

### $(4) \implies (5)$

By $(4)$ we have an $R \sqbrk x$-module $B$ that is finitely generated over $R$.

Let $y$ lie in the annihilator $\map {\operatorname {Ann}_{R \sqbrk x} } B$

We have that $1 \in B$.

Then in particular $y \cdot 1 = 0$, and $y = 0$.

Therefore, $B$ is faithful over $R \sqbrk x$.

$\Box$

### $(5) \implies (1)$

Let $B$ be as in $(5)$, say generated by $m_1, \ldots, m_n \in B$.

Then there are $r_{i j} \in R$, $i, j = 1,\ldots, n$ such that:

$\ds x \cdot m_i = \sum_{j \mathop = 1}^n r_{i j} m_j$

Let $b_{i j} = x \delta_{i j} - r_{i j}$ where $\delta_{i j}$ is the Kronecker delta.

Then:

$\ds \sum_{j \mathop = 1}^n b_{i j} m_j = 0, \quad i = 1, \ldots, n$

So, let $M = \sqbrk {b_{i j} }_{1 \le i, j \le n}$.

Then by Cramer's Rule:

$\paren {\det M} m_i = 0$, $i = 1, \ldots, n$

Since $\det M \in R \sqbrk x$, also $\det M \in \map {\operatorname {Ann}_{R \sqbrk x} } B$.

So $\det M = 0$ by hypothesis.

But $\det M = 0$ is a monic polynomial in $x$ with coefficients in $R$.

Thus $x$ is integral over $R$.

$\blacksquare$