# When is integral closure generated by one element

Hello,
This is not a homework problem, nor a textbook question. Please do not remove.
Is there a concrete example of the following setup :
$R$ is an integrally closed domain,
$a$ is an integral element over $R$,
$S$ is the integral closure of $R[a]$ in its fraction field,
$S$ is not of the form $R{[}b{]}$ for any element $b$ in $S$.

For example, the integral closure of ${\mathbb Z}(\sqrt 5)$ is the set of elements of the form $(m + \sqrt 5 n)/2$, where $m^2 - 5n^2$ is a multiple of 4. So, $S$ is generated by $\sqrt 5/2$ over $\mathbb Z$: This does not fulfill the desired conditions.

## Answers and Replies

mathwonk
Science Advisor
Homework Helper
2020 Award
i don't understand. in your example apparently R is not integrally closed. in fact under your conditions, that R is integrally closed and a is integral over R, this implies that R = R[a], doesn't it?

Oh sorry, you meant integral over it but in some finite field extension of its fraction field. so your R was Z and your a was sqrt(5). got it. and maybe you meant Z[sqrt(5)] instead of Z(sqrt(5))?

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Yes, I meant ${\mathbb Z}{[}\sqrt 5{]}.$
In my example, $R = {\mathbb Z},\ a = \sqrt 5$ and $S$ is the integral closure of ${\mathbb Z}[\sqrt 5]$ in ${\mathbb Z}(\sqrt 5)$, which has the form given above.

mathwonk
Science Advisor
Homework Helper
2020 Award
the answer is there does exist such a counter example Z[a] to the integral closure equaling Z[ b]. all your examples are numbers fields, and all number fields fall into your examples. You are asking for the structure of the ring of integers, which is known to have a finite basis as an abelian group. You want that basis to be the powers, up to some finite power, of a single integral element.
It is well known since Richard Dedekind that not all rings of integers in number fields have so called "power bases". here is a reference: (or just search on "non monogenic fields".)

http://wstein.org/129-05/final_papers/Yan_Zhang.pdf

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• coquelicot
Very useful. I was completely unaware of this topic. Thank you so many mathwonk.