If $S$ is a Noetherian ring and $R\subset S$ is a sub-ring, prove that $R$ is Noetherian or a counterexample

Question

If $S$ is a Noetherian ring and $R\subset S$ is a sub-ring, prove that $R$ is Noetherian or a counterexample if this assertion is not always certain.

I have thought for a long time about this and I come to the following conclusion that I do not know if it is correct:

I think this is not true in general and I think the following can work: we have that $\mathbb{Z}\subset\mathbb{R}$, is there an ideal that is not finitely generated in $\mathbb{Z}$ and that is finitely generated in $\mathbb{R}$?

Answer 1

Every field is Noetherian, and every integral domain embeds as a sub-ring of its field of fractions. So one source of counterexamples is non-Noetherian integral domains.

For example, for any field $k$, the field $k(x_1,x_2,\dots)$ of rational functions in infinitely many variables is Noetherian, but its subring $k[x_1,x_2,\dots]$ is not Noetherian.

Answer 2

Since I run into this very same question for didactical reasons, I believe it could be useful to have other examples at hand.

Personally, I would contribute with the ring $\mathbb{Z}\left[\sqrt{p}\mid p \text{ is a positive prime number}\right]$ obtained by adding to $\mathbb{Z}$ all the real roots of positive prime numbers, which is a subring of $\mathbb{R}$ but it is not Noetherian. Indeed, the ideal $$\langle \sqrt{p} \mid p \text{ is a positive prime number} \rangle$$ is not finitely generated.