# When is integral closure generated by one element

## Main Question or Discussion Point

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.

Related Linear and Abstract Algebra News on Phys.org
mathwonk
Homework Helper
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))?

Last edited:
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