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Let $R$ be a Noetherian domain with quotient field $K$,
$I \subseteq R$ a finitely generated ideal, $I \neq (0)$ and
$x \in K$ such that $x \cdot I \subseteq I$.
I want to show that $x$ is integral over $R$.

Supposedly, this is also true when dropping the requirement of $R$ being Noetherian.

My ideas so far:

  • Let $x = \frac a b$ with $a,b \in R$. If we could show that $\frac 1 b$ is integral over $R$, then also $\frac a b$ is integral over $R$.
  • Using something similar as https://math.stackexchange.com/a/371657/514331 :
    We have $\frac 1 b \cdot I \subseteq \frac 1 {b^2} \cdot I \subseteq \dots$
    If we could show that $\frac 1 b \cdot I \subseteq I$, then by $R$ being Noetherian, we would get $\frac 1 {b^n} \cdot I = \frac 1 {b^{n+1}} \cdot I$ for some $n$. But even with this, I fail to see that $\frac 1 b$ is integral over $R$.
Ben
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2 Answers2

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Here is a proof which uses Noetherianness (not that it is much simpler). You have inclusions of $R$-algebras, $R\subset R[x]\subset \mathrm{End}_R(I)$. Noetherian property implies the last is a finite type $R$-module and hence again by Noetherian property, so is $R[x]$. This immediately implies $x$ is integral over $R$.

Mohan
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  • Thanks! Just to clarify, the last implication follows by looking at the chain of ideals $(\frac a b) \subseteq (\frac a b, \frac {a^2} {b^2}) \subseteq \dots$ – Ben Jan 07 '18 at 11:37
  • @Ben If $R[x]$ is finitely generated as an $R$-module, let $f_1,\ldots,f_k$ be generators and thinking of them as polynomials in $x$, there is an $n$ such that $f_i=\sum_{l=0}^n a_{il}x^l$ for all $i$ with $a_{il}\in R$. Then, it is clear that $1,x,\ldots, x^n$ also generates $R[x]$ and thus we can write $x^{n+1}=\sum_{l=0}^n b_lx^l$, which is an integral equation for $x$. – Mohan Jan 07 '18 at 14:33
  • Ah, I wanted to exploit that $R[x]$ as a finitely generated module of a Noetherian ring is again Noetherian. But looking at your comment, I see that this is not necessary. Could you also explain why $\mathrm{End}_R(I)$ is a finitely generated $R$-module? – Ben Jan 07 '18 at 16:11
  • @Ben If $M$ is any finitely generated module over $R$, write a surjection $R^n\to M$. Then for any $R$-module $N$, you have an inclusion $\mathrm{Hom}_R(M,N)\subset \mathrm{Hom}_R(R^n,N)=N^n$. Letting $M=N=I$, we see that $\mathrm{End}_R(I)\subset I^n$ and by Noetherian property, we get $\mathrm{End}_R(I)$ to be finitely generated. – Mohan Jan 07 '18 at 17:08
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Your statement is not quite correct, you need to assume that $I$ is also nonzero. In that case however, there is a very elementary way to see this.

So you have $I=(f_1,\ldots,f_n)$ and $x\cdot I\subseteq I$. In particular, $$ x\cdot f_i = \sum_{j=1}^n x_{ij} f_j $$ for a matrix $X:=(x_{ij})\in R^{n\times n}$. Denote by $f=(f_1,\ldots,f_n)\in R^n$ the vector containing the $f_i$, then we can write this as $X\cdot f = x\cdot f$. Let $\chi=\sum_{k=0}^n a_k t^k \in R[t]$ be the characteristic polynomial of $X$, then we have $$ \chi(x)\cdot f= \sum_{k=0}^n a_k x^k\cdot f = \sum_{k=0}^n a_k X^k \cdot f = \chi(X)\cdot f = 0 $$ Since $I$ is not the zero ideal, we may assume that $f_1\ne 0$ and the above implies $\chi(x)\cdot f_1 = 0$, which as an equation inside the field $K$ implies $\chi(x)=0$. Now $\chi$ is the integral relation for $x$ which you seek.

PS: Do you have any interesting examples of such a situation?

Jesko Hüttenhain
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  • Thanks! Do you know of a simpler proof in the situation where $R$ is Noetherian? The result of this theorem is supposedly useful for exercise 4 on http://www.math.uni-bonn.de/people/franke/A10.pdf – Ben Jan 06 '18 at 22:21
  • @Ben: I am afraid I don't immediately see how to exploit the Noetherian condition to simplify this a whole lot, but I will give it some thought and edit my answer in the event of any epiphany. – Jesko Hüttenhain Jan 06 '18 at 22:25