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Let $R$ be a noetherian domain, $Q$ its field of fraction and $u\in Q$. Could you help me to prove that

$u$ is integral over $R$ if and only if there exists $r\in R$ $r\neq0$ and $ru^n\in R$ for every $n\geq0$?

(for the reverse implication there is the hint to consider the $R$-submodule of $Q$ generated by $\frac{1}{r}$).

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An element satisfying the equivalent condition is called almost integral. –  user26857 May 26 at 19:56
@user26857 That's already mentioned in my answer (two years ago). –  Bill Dubuque May 31 at 17:39

2 Answers 2

up vote 5 down vote accepted

Some ideas that should help.

$\Rightarrow$. If $u$ is integral over $R$, then $u$ satisfies some monic polynomial, say of degree $n$, with coefficients in $R$. Show that any nonnegative power of $u$ can be written as an $R$-linear combination of $1, u, \ldots, u^{n-1}$. It follows that you really just need to find an $r$ such that $ru^i \in R$ for $i < n$.

$\Leftarrow$. The hint is good. $R\frac1r$ is certainly finitely generated as an $R$-module, so it's Noetherian. Show that $R[u] \subset R\frac1r$ using the given condition. What do you now know about $R[u]$ as an $R$-module?

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The nontrivial direction is a special case of the Lemma below, with $\rm\:D =\:$ integral closure of $\rm\:R.\:$

Lemma $\ $ Suppose that $\rm\:D = \mathbb Z\:$ (or any Noetherian integrally closed domain, e.g. any PID), and suppose that $\rm\:w\:$ is a fraction over $\rm\:D\:$ such that some unbounded sequence of powers of $\rm\:w\:$ has a common denominator $\rm\:0 \ne d\in D,\:$ i.e. $\rm\:d\!\:w^{n_i}\in D\:$ for all $\rm\:n_i.\:$ Then $\rm\:w\in D.$

Proof $\ $ By ACC the sequence of ideals $\rm (d, dw^{n_1}, dw^{n_2},\ldots)$ eventually stabilizes, which implies that for some $\rm\:k\:$ we have $\rm\: dw^{n_k}\in (dw^{n_{k-1}},\ldots, dw^{n_1}, d),\:$ which implies

$$\rm d\: w^{n_k} + c_{n_{k-1}} d\: w^{n_{k-1}} +\:\! \cdots +\: c_{n_1} d\: w^{n_1} + d\: =\: 0$$

Cancelling $\rm\:d\:$ yields $\rm\:w\:$ is integral over $\rm\:D,\:$ hence $\rm\:w\in D,\:$ since $\rm\:D\:$ is integrally closed. $\ $ QED

Elements whose powers have such a common denominator are called almost integral. It is clear that integral elements are almost integral. The above shows that the converse holds true in Noetherian domains. This is employed implicitly in Dedekind's work on ideal theory.

Here is a typical application from my answer to this prior question.

Show $\rm\:a|b^2,\: b^3|a^4,\: a^5|b^6,\: b^7|a^8 \:\cdots\Rightarrow\: a = b\:$ for $\rm\:a,b\in\mathbb Z_+$

Hint $\rm\ \ \forall\: n\in\mathbb N:\ \ a\:\!\left(\dfrac{a}b\right)^{4n+3}\!\in\mathbb Z,\:\ b\:\!\left(\dfrac{b}a\right)^{4n+1}\!\in \mathbb Z\ \ \Rightarrow\ \dfrac{a}b,\:\dfrac{b}a\in\mathbb Z\ \ \Rightarrow\ \ a = \pm b\ \ \ $ QED

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This answer was unfairly overlooked at first, I think. It's a very cool proof. –  Dylan Moreland Jun 5 '12 at 2:37

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