1. Mar 4, 2014

coquelicot

I think I have proved the following theorem:

"If R is an integral domain, x is integral of degree n over R, and K is the fraction field of R, then the intersection of R[x] with K is included in 1/n R[X] (the set of elements r/n where r belongs to R). Furthermore, if R is Dedekind, then this intersection is R."

My simple proof can be found here:

My question is: Where can I found reliable references to this theorem, or at least reliable references to another theorem implying it ?

Thx.

Last edited by a moderator: May 6, 2017
2. Mar 4, 2014

micromass

The statement about Dedekind domains seems to be the same as saying that Dedekind domains are integrally closed. Chapter 9 of Atiyah-MacDonald should be relevant here.

3. Mar 4, 2014

micromass

So let's take an integral domain $R$ and take any $\theta\in K$ which is integral over $R$. Since $\theta\in K$, it follows that $n=1$. So your theorem proves that

$$R[\theta]\cap K\subseteq R$$

So your theorem implies that any integral domain is integrally closed. There are counterexamples to that, so where did I go wrong?

4. Mar 4, 2014

coquelicot

If x belongs to K and is integral over R, then obviously x belongs to R. so, the intersection of R[X] with R is R. In your statement, you infer from a particular case to the general case, and this is what is wrong.

5. Mar 4, 2014

micromass

Consider $R= \mathbb{C}[t^2,t^3]$, then $K = \mathbb{C}(t)$. But then $t\in K$ is integral over $R$ since it's the root of

$$X^2 - t^2 = 0$$

but $t$ is not in $R$.

6. Mar 4, 2014

coquelicot

Sorry, you are right about the fact that an element of K integral over R needs not belong to R. In fact, the problem in your counterexample above is that you assert that n=1, which is, of course wrong unless x belong to R.

7. Mar 4, 2014

micromass

I just take an arbitrary element $\theta$ in $K\setminus R$ that is integral over $R$. My previous post shows that such situation exist.
Then since $\theta\in K$, it follows that $n=1$ (since your definition of $n$ was the degree of $\theta$ over $K$). Why is this wrong?

8. Mar 4, 2014

coquelicot

You are, once more right. I have understood my mistake, and that my theorem is false. Thank you so many for your help.

9. Mar 4, 2014

micromass

You did prove something. So I think that you should carefully check the proof and see where your error lies. Maybe with some additional assumptions, it might still work?

10. Mar 4, 2014

coquelicot

Of course, if it is supposed that the minimal polynomial of theta is also monic with coefficients in R, this works. But this makes the theorem much more uninteresting.