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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 $\mathbb{R}$?

user424241
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    https://math.stackexchange.com/questions/274590/how-to-prove-that-this-subring-is-not-noetherian – Tsemo Aristide Sep 02 '18 at 01:01
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    You're correct that the assertion is not true, in general, but your counterexample does not work, since $\mathbb{Z}$ is a noetherian ring. In fact, $\mathbb{Z}$ is a PID, so all of its ideals are principal. – Zilliput Sep 02 '18 at 01:12

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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.

Alex Kruckman
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  • @AlexKruckman Why the field $k(x_1,x_2,…)$ of rational functions in infinitely many variables is Noetherian? – user424241 Sep 03 '18 at 12:37
  • @user424241 Because its a field! Fields have only two ideas: the $0$ ideal and the unit ideal. Why? As soon as an ideal $I$ contains a nonzero element $a$, it contains $a^{-1}a=1$, and then it contains $b1=b$ for every element $b$. – Alex Kruckman Sep 03 '18 at 13:54
  • @AlexKruckman What is the inverse element of an element in $k(x_1,x_2,...)$ and why is it in $k(x_1,x_2,...)$? – user424241 Sep 04 '18 at 19:28
  • I assume you mean multiplicative inverse? It is the reciprocal... $k(x_1,x_2,\dots)$ is the field of rational functions in the variables $x_1,x_2,\dots$. That is, it's the field of fractions of the polynomial ring $k[x_1,x_2,\dots]$. So an element of $k(x_1,x_2,\dots)$ is a fraction of the form $p/q$, where $p$ and $q$ are polynomials in the variables $x_1,x_2,\dots$, and the inverse of this element is $q/p$ (unless $p = 0$, in which case $p/q = 0$). – Alex Kruckman Sep 04 '18 at 22:53
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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.

Ender Wiggins
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