Equivalence of Definitions of Integral Dependence
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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 $R \left[{x}\right]$ 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 $R \left[{x}\right]$-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:
- $\left\{ {1, \ldots, x^{n-1} }\right\}$
Therefore this set generates $R \left[{x}\right]$.
$\Box$
$(2) \implies (3)$
$B = R \left[{x}\right]$ 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 $R \left[{x}\right]$-module $B$ that is finitely generated over $R$.
Let $y$ lie in the annihilator $\operatorname{Ann}_{R \left[{x}\right]} \left({B}\right)$
We have that $1 \in B$.
Then in particular $y \cdot 1 = 0$, and $y = 0$.
Therefore, $B$ is faithful over $R \left[{x}\right]$.
$\Box$
$(5) \implies (1)$
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:
- $\displaystyle x \cdot m_i = \sum_{j \mathop = 1}^n r_{i j} m_j$
Let $b_{i j} = x \delta_{i j} - r_{i j}$ where $\delta_{i j}$ is the Kronecker delta.
Then:
- $\displaystyle \sum_{j \mathop = 1}^n b_{i j} m_j = 0, \quad i = 1, \ldots, n$
So, let $M = \left({b_{i j} }\right)_{1 \le i, j \le n}$.
Then by Cramer's Rule:
- $\left({\det M}\right) m_i = 0$, $i = 1, \ldots, n$
Since $\det M \in R \left[{x}\right]$, also $\det M \in \operatorname{Ann}_{R \left[{x}\right]} \left({B}\right)$.
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 $R$.
$\blacksquare$