# Noetherian prime localizations does not imply that whole ring is Noetherian

We know the following fact: If $$A$$ is Noetherian ring and $$S$$ is multiplicatively closed subset of $$A$$ then $$S^{-1}A$$ is also Noetherian ring.

Corollary: If $$A$$ is Noetherian ring then for any prime ideal $$P$$ the ring $$A_P$$ is also Noetherian.

However, the converse of this corollary is not true.

Let's demonstrate it in the following example: Take $$R=\prod \limits_{i=1}^{\infty}\mathbb{Z}_2$$ and we see that $$R$$ is not Noetherian because it is not finitely generated.

We see that for any $$x\in R$$ we have $$x^2=x$$, i.e. $$R$$ is a Boolean ring. Let $$P$$ be a prime ideal in $$R$$ then we can show that $$R/P\cong \mathbb{Z}_2$$ which means that $$R/P$$ is a field. Hence $$P$$ is maximal ideal.

Thus maximal and prime ideals in $$R$$ are the same

Now we want to consider the localizations $$R_P$$ for any prime ideal $$P$$.

I have the following two questions (I have spent couple of hours but I was not able to answer these questions).

1) How to show that $$R$$ is local ring? i.e. there is only one maximal ideal (i.e. one prime ideal).

2) If we can show this then $$P=0$$ because intersection of all prime ideals is the set of nilpotent elements in $$R$$ but $$R$$ being boolean has no nonzero nilpotent elements. Hence $$P=0$$. How it follows that $$R_P$$ is field?

I know that this question is already exists in forum. But they are not detailed. Could anyone help me to asnwer these questions?

Would be very grateful for help! Also please do not duplicate this question!

## 1 Answer

How to show that $$R$$ is local ring? i.e. there is only one maximal ideal (i.e. one prime ideal).

I guess you mean how to show $$R_P$$ is a local ring? This is the first thing you learn about localizations: localizing at the complement of a prime ideal results in a local ring. Just suppose that $$M$$ is a maximal ideal of $$R_P$$ different from $$PR_P=\{\frac{p}{s}\mid p\in P, s\notin P\}$$. Then there is some $$\frac{m}{s}\in M\setminus PR_P$$, and that entails that $$m\notin P$$. But $$\frac{m}{1}$$ is a unit in $$R_P$$, so $$\frac{m}{s}$$ is a unit and $$M=R_P$$, a contradiction.

If we can show this then $$P=\{0\}$$ because intersection of all prime ideals is the set of nilpotent elements in $$R$$ but $$R$$ being boolean has no nonzero nilpotent elements. Hence $$P=\{0\}$$.

What? There are infinitely many nonzero prime ideals in $$R$$. I don't know how you expect this to work.

How it follows that $$R_P$$ is field?

On one hand, $$R_P$$ is a local ring. On the other hand, everything in it is idempotent: for $$\frac{a}{b}\frac{a}{b}=\frac{aa}{bb}=\frac{a}{b}$$ for any $$\frac{a}{b}\in R_P$$. But in a local ring, idempotents must be $$0$$ or $$1$$. Therefore everything in $$R_P$$ is the additive or multiplicative identity: therefore $$R_P=F_2$$.

• Your answer indeed very nice! After a long battle I guess that I've finally got it. +1 – ZFR Jan 29 '19 at 17:31