# Let $R$ be a ring, $S$ a subring and $I$ an ideal. If $R$ is Noetherian, are then $S$ and $R/I$ also Noetherian?

Let $R$ be a ring, $S$ a subring and $I$ an ideal. If $R$ is Noetherian are then $S$ and $R/I$ also Noetherian?

I have done the following:

$R$ is Noetherian iff each increasing sequence of ideal $I_1\subseteq I_2 \subseteq I_3 \subseteq \dots \subseteq I_k\subseteq \dots$ stops, i.e., $\exists k$ such that $I_k=I_{k+1}$, right?

Then since $S$ is a subring of $R$, not all $I_i$ are contained in $S$.

Therefore, the above condition isn't necessarily satisfied.

So, $S$ is not necessarily Noetherian.

Is this correct?

What can we say in that case of $R/I$? Does the increasing sequence stop?

A subring of a Noetherian ring need not be Noetherian: Subring of a finitely generated Noetherian ring need not be Noetherian?

Now for $R/I$ start with an increasing sequence of ideals in $R/I$. Can you use these to get an increasing sequence of ideals in $R$? What can you conclude about the original sequence? Hint: Bijection between sets of ideals

• About the first part/link: Could you explain to me why the subring generated by the set $\{xy^i : i>0\}$ is not finitely generated, i.e., non-Noetherian? Is it maybe because the ideals are $I_i=(xy^i)$ and so $I_1 \subseteq I_2 \subseteq \dots \subseteq I_i\subseteq I_{i+1} \subseteq \dots$ and this sequence never stops? – Mary Star Jun 8 '16 at 20:30
• You're correct about the first link. For the second, start with a chain of ideals $J_1\subseteq J_2\subseteq \cdots$ in $R/I$. These will be of the form $J_i=I_i/I$ for ideals $I_i$ of $R$ containing $I$ such that $I_1\subseteq I_2\subseteq \cdots$ (that is what the link is about - showing that ideals of the quotient actually have this form). Now you know the chain of ideals in $R$ stabilizes, so the chain of ideals in the quotient must stabilize as well. – kccu Jun 9 '16 at 0:20
• It's certainly a subgroup. If $x \in I_k$ and $r \in R$, then $x \cdot r \in I_k$. Hence if $xI \in I_k /I$ and $rI \in R/I$, $(xI)(rI)=(xr)I \in I_k/I$. – kccu Jun 11 '16 at 15:48
• Almost. You need to start with a chain of ideals $J \subseteq J_2 \subseteq \cdots$ in $R/I$. Then show these ideals are actually of the form $J_1=I_1/I$, $J_2=I_2/I$, $\dots$ for ideals $I_1 \subseteq I_2 \subseteq \cdots$ in $R$ containing $I$. (The second link in my answer is about showing why $J_i$ must be of the form $I_i/I$.) Then the rest of the proof goes the way you've described in your last comment. – kccu Jun 11 '16 at 22:28
• You cannot form the quotient of $I_i$ by $I$ if $I_i$ does not contain $I$. – kccu Jun 21 '16 at 17:37

The ring of integer-valued polynomials (polynomials with rational coefficients which take integer values on integers) is known to be a non-noetherian subring of the P.I.D. $\mathbf Q[X]$.

(Actually it is a Prüfer domain, with Krull dimension 2, whereas $\mathbf Q[X]$ has dimension $1$.)