# If the localization of a ring $R$ at every prime ideal is an integral domain, must $R$ be an integral domain?

Let $R$ be a commutative ring. Suppose that for every prime ideal $\mathfrak p$ of $R$, the localization $R_{\mathfrak p}$ is an integral domain. Must $R$ be a integral domain?

I was trying to think of counter-examples, but kept getting $R_{(0)}$ = the zero ring, which is not a domain.

Any guidance would be much appreciated. Thanks.

• A ring $R$ is locally a domain iff it is reduced and its principle ideals are flat. – Badam Baplan Apr 19 at 20:10

## 6 Answers

One simple counterexample is $R=\mathbb{Z}/6\mathbb{Z}$. The prime ideals of $R$ are $P=2\mathbb{Z}/6\mathbb{Z}$ and $Q=3\mathbb{Z}/6\mathbb{Z}$ and $R_P$ is an integral domain (the argument easily carries over for $R_Q$), while $R$ clearly is not.

• Thank you Zev. This was the example that I came up with myself, but for some reason, I kept thinking that 0Z/6Z was prime in R, which it is clearly not. [Hence we have this counter-example to start with] – Conan Wong May 19 '12 at 11:50

Zev's and Georges' answers are complete. I would like to give you a deeper sight. This is a guide line through easy claims that you should be able to prove on your own. Let $A$, $B$ be two rings.

1. Every ideal of $A \times B$ is of the form $I \times J$, for unique ideals $I \subseteq A$ and $J \subseteq B$. If this is the case, $(A \times B) / (I \times J) \simeq (A / I) \times (B / J)$.

2. $A \times B$ is a domain iff ($A$ is a domain and $B = 0$) or ($B$ is a domain and $A = 0$).

3. Every prime ideal of $A \times B$ is of the form $\mathfrak{p} \times B$, for some prime ideal $\mathfrak{p}$ of $A$, or $A \times \mathfrak{q}$, for some prime ideal $\mathfrak{q}$ of $B$.

4. The localization $(A \times B)_{\mathfrak{p} \times B}$ is isomorphic to $A_\mathfrak{p}$, for every prime ideal $\mathfrak{p}$ of $A$.

This proves that if a ring $R$ is the product of a finite number ($\geq 2$) of integral domains, then $R$ is not a domain but every its localization at primes is a domain.

This has also a geometric interpretation, if you know Zariski topology on the prime spectrum of a ring. From (3) you have a homeomorphism $\mathrm{Spec} (A \times B) \simeq \mathrm{Spec} A \coprod \mathrm{Spec} B$, then $\mathrm{Spec} (A \times B)$ is disconnected and is locally the spectrum of a domain, if $A$ and $B$ are domains.

• Andrea - this is fantastic. I had to prove Claims (1) & (3) [following your numbering above] in another assignment. But I didn't see the link to my main question until you wrote this. Thank you so much. – Conan Wong May 19 '12 at 11:57

No: $R=\mathbb Q\times \mathbb Q$ .

• Thank you George for the quick example. – Conan Wong May 19 '12 at 11:51

While as the other answers have pointed out that such a ring $R$ need not be an integral domain in general, if $R$ has only one minimal prime ideal then the claim holds. To see this, let $I$ be the minimal prime. Since $R_P$ is an integral domain for every prime ideal $P$ and $I_P$ is a minimal prime in $R_P$, we must have that $I_P=(0)_P$ for all prime ideals $P$. Thus by the local-global principle, $I=(0)$ so $R$ is an integral domain.

• Thanks Alex - your additional proof is helpful to me too. – Conan Wong May 19 '12 at 11:47

Those examples here are with disconnected spectrum.

So we might hope that if $R$ has no non-trivial idempotent elements and $R_{\mathfrak{p}}$ is a domain for every prime ideal $\mathfrak{p}$, does this conditions imply that $R$ is a domain?

We know that adding one more condition that assumes $R$ is Noetherian or has only finitely many minimal prime ideals the implication holds.

Suppose $\mathfrak{p}_i,i=1,2,\ldots,n$ be all the minimal prime ideals of $R$. Then the condition $R_{\mathfrak{p}}$ is integral for every prime ideal $\mathfrak{p}$ is exactly saying that $R$ is reduced and $R=\mathfrak{p}_i+\mathfrak{p}_j$ when $i\neq j$. So by Chinese remainder theorem, $R=R/\mathfrak{p}_1\times\cdots\times R/\mathfrak{p}_n$. And if we assume $R$ has no non-trivial idempotent elements, then $R$ has only one minimal prime ideal which is zero, namely, $R$ is a domain.

However, I do not know if whether it is still true without the assumption that $R$ has finitely many minimal prime ideals.

I am not clear whether the continuous functions ring on $[0,1]$, i.e., $A=C[0,1]$, is a counter-example.

It is clear that $A$ is not a domain, $A$ is reduced and $A$ is connected! So the only one problem is that whether for any two minimal primes $\mathfrak{p}$ and $\mathfrak{q}$ are comaximal. (I remember this is true, but I do not remember from where I know this result.)

• In fact, you want to prove that $A_{\mathfrak p}$ is an integral domain for every prime ideal $\mathfrak p$, and I think this is wrong. – user26857 Oct 16 '17 at 9:27
• @GeorgesElencwajg Dear Georges, I know you were interested in the ring of continuous functions. Could you show me a short proof for $A_{{\mathfrak m}_x}$ is not an integral domain? (Here ${\mathfrak m}_x$ is the maximal ideal of continuous functions vanishing at $x\in[0,1]$. The localization is, of course, the ring of germs of continuous functions at $x$.) Thank you. – user26857 Oct 16 '17 at 9:50
• Dear @user26857: If, say, $x=1/2$ take the functions $f,g$ defined by [$f(t)=t-1/2$ for $t\gt1/2$, $f(t)=0$ else] and [$g(t)=1/2-t$ for $t\lt 1/2$, $g(t)=0$ else] to prove that the ring of germs of continuous functions at $1/2$ is not a domain, since the germs of $f$ and $g$ are not zero while $fg=0$. The same idea works for all $x\in (0,1)$ (To be continued) – Georges Elencwajg Oct 16 '17 at 21:08
• @user26857 (Continuation) The ring of germs of continuous functions is not a domain at $x=0$ either , but the construction is a bit more complicated. You might take $f(x)=\max \;[\; x\sin (1/x),0\;]$ and $g(x)=\min\; [\; x\sin (1/x),0\;]$. Then the germs of $f,g$ at zero are non-zero but $fg=0$. The same idea works for $x=1$ too, of course. – Georges Elencwajg Oct 16 '17 at 21:08

A broad class of answers that encompasses some of the specific examples already mentioned:

It is known that a commutative ring is von Neumann regular iff its localizations at prime ideals are all fields.

But a von Neumann regular ring is only a domain if it is a field. Von Neumann regular rings are a huge class with lots of examples which aren't bound by chain conditions. The main reason they make good candidates is because their prime ideals are maximal (so the result of the localization is rigged.)