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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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.