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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?

user26857
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Mary Star
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2 Answers2

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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

Community
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kccu
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  • 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 08 '16 at 20:30
  • About the second part/link: Let $I_k$ be the ideals of $R$. Since $R$ is Noetherian we have that $I_1\subseteq I_2 \subseteq \dots \subseteq I_k$ with $I_k=I_{k+1}$. Then the ideals of $R/I$ are of the form $R/I_k$, right? Then do we have to show that there is a bijection between $I_k$ and $R/I_k$ ? Would we conclude then that $R/I_1\subseteq R/I_2 \subseteq \dots \subseteq R/I_k$ with $R/I_k=R/I_{k+1}$ ? – Mary Star Jun 08 '16 at 20:41
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    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 09 '16 at 00:20
  • When $I_k$ is an ideal of $R$, how exactly do we conclude that $I_k/I$ is an ideal of $R/I$ ? I got stuck right now... – Mary Star Jun 11 '16 at 15:43
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    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
  • So, we use the definition of an ideal, right? We have that since $R$ is Noetherian then $I_1\subseteq I_2\subseteq \dots I_k\subseteq I_{k+1}$ then $I_k=I_{k+1}$ for some $k$. When we consider the sequence $I_1/I\subseteq I_2/I\subseteq \dots I_k/I\subseteq I_{k+1}/I$ we have that, since $I_k=I_{k+1}$ for some $k$, $I_k/I=I_{k+1}/I$. Therefore, $R/I$ is also Noetherian, right? – Mary Star Jun 11 '16 at 21:16
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    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
  • Why do $I_i$ have to contain $I$ ? – Mary Star Jun 21 '16 at 17:33
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    You cannot form the quotient of $I_i$ by $I$ if $I_i$ does not contain $I$. – kccu Jun 21 '16 at 17:37
  • Ah ok... Thank you!! :-) – Mary Star Jul 04 '16 at 21:49
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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 $1$.)

Bernard
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