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

Question

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?

Answer 1

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

Answer 2

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 $1$.)