Equivalence of Definitions of Integral Dependence

Theorem
Let $R \subseteq A$ be an extension of commutative rings with unity.

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


 * 1. $x$ is integral over $R$


 * 2. The $R$-module $R[x]$ is finitely generated


 * 3. There exists a subring $B$ of $A$ such that $x \in B$, $R \subseteq B$ and $B$ is a finitely generated $R$-module


 * 4. There exists a subring $B$ of $A$ such that $xB \subseteq B$ and $B$ is finitely generated over $R$


 * 5. There exists a faithful $R[x]$ module $B$ that is finitely generated as an $R$-module

Proof
1. $\Rightarrow$ 2.

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


 * $x^n + r_{n-1} x^{n-1} + \cdots + r_1x + r_0 = 0$

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


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

therefore this set generates $R[x]$.

2. $\Rightarrow$ 3.

$B = R[x]$ trivially satisfies the required conditions.

3. $\Rightarrow$ 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 $xB \subseteq B$ as required.

4. $\Rightarrow$ 5.

By 4. we have an $R[x]$-module $B$ that is finitely generated over $R$.

Moreover, $1 \in B$, so if $y$ lies in the annihilator $\operatorname{Ann}_{R[x]}(B)$ then in particular $y\cdot 1 = 0$, and $y = 0$.

Therefore, $B$ is faithful over $R[x]$.

5. $\Rightarrow$ 1.

Let $B$ be as in 5., say generated by $m_1,\dots, m_n \in B$.

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


 * $\displaystyle x \cdot m_i = \sum_{j = 1}^n r_{ij}m_j$

Therefore, if $b_{ij} = x\delta_{ij} - r_{ij}$ where $\delta_{ij}$ is the kronecker delta, then


 * $\displaystyle \sum_{j = 1}^n b_{ij}m_j = 0,\quad i=1,\ldots,n$

Therefore if $M = (b_{ij})_{1\leq i,j\leq n}$, by Cramer's Rule we have $(\det M)m_i = 0$, $i = 1,\ldots, n$.

Since $\det M \in R[x]$, also $\det M \in \operatorname{Ann}_{R[x]}(B)$, so $\det M = 0$ by hypothesis.

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