# 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 principal ideals are flat. Apr 19, 2019 at 20:10

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. May 19, 2012 at 11:57

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] May 19, 2012 at 11:50

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

• Thank you George for the quick example. May 19, 2012 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. May 19, 2012 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. Oct 16, 2017 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. Oct 16, 2017 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) Oct 16, 2017 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. Oct 16, 2017 at 21:08