# Is the irreducibility of a ring preserved by localization at a prime ideal?

Let $R$ be a commutative ring and $\mathfrak p$ a prime ideal of $R$. Suppose that $R$ satisfies the following property: the intersection of two nonzero ideals is always nonzero. Is the property also true for the ring $R_{\mathfrak p}$, the ring $R$ localized at $\mathfrak p$ ?

Can someone give a proof this fact or show a counterexample ? Thanks a lot in advance!

• I think that property is preserved if for example $\;R\;$ is principal since then $\;R_p\;$ , being a local ring with maximal ideal $\;pR_p\;$ principal, has all its ideals being generated by a power of the generator of $\;p\;$ . Feb 13, 2016 at 23:48
• Doesn't every non-zero ideal of $R_p$ contain a corresponding non-zero ideal in $R$? As such this should be trivial no? Feb 14, 2016 at 0:25
• @ joanpemo. If R is principal of course is obviously but I want to prove or disprove it in ageneral case. Feb 14, 2016 at 0:47
• @GregoryGrant. The problem is solved if, given two nonzero ideals $I_1, I_2$ in $R$ can take a nonzero element $a\in I_1 \cap I_2 \cap p$ SUCH THAT $\frac{a}{1}$ is nonzero in $R_p$. Feb 14, 2016 at 1:02
• The case in which the zero ideal is primary is obvious of course. So I try to suppose (0) not primary and/or A not noetherian... Feb 14, 2016 at 1:09

First I want to introduce a (standard) terminology: A ring $A$ for which $I \cap J = 0$ implies $I=0$ or $J=0$ for two ideals $I,J$ of $A$ is called "irreducible". So the question is, if for $A$ irreducible every localization $A_p$ is irreducible. $\newcommand{\Ann}{\mathrm{Ann}}$ $\newcommand{\Ass}{\mathrm{Ass}}$ $\newcommand{\ideal}[1]{{\mathfrak{#1}}}$

It can be shown, that for every noetherian irreducible ring $A$, also $A_p$ is irreducible.

First let $A$ be irreducible (but not necessarily noetherian). Then the set of associated primes $\Ass(A)$ is either empty or consists of exactly one prime $\mathfrak{p} \subseteq A$. Namely, let $\mathfrak{p} = \Ann (x)$ and $\mathfrak{p}' = \Ann( y)$ with $x, y \in A$ and $\ideal{p}, \ideal{p}'$ prime. Then, as $A$ is irreducible, we have $a, b \in A$ with $ax = b y \neq 0$. Now $\Ann(x) = \Ann(a x) = \ideal{p}$ and $\Ann(y) = \Ann(by) = \ideal{p}'$, so $\ideal{p} = \ideal{p}'$.

Now let $A$ be irreducible and $\Ass(A) = \{\ideal{p}\}$. Furthermore let $S \subseteq A$ multiplicatively closed and $S \cap \ideal{p} = \emptyset$. Then $S^{-1}A$ is irreducible too.

The proof goes as follows: Let $I = (f/1)$ and $J = (g/1)$ be two non-zero principal ideals in $S^{-1}A$ and let $x \in A$ with $\Ann(x) = \ideal{p}$. Then, as $A$ is irreducible, we have $a f = q x \neq 0$ and $b g = q' x \neq 0$ with $q, q' \notin \ideal{p}$. So $a' f = a q' f = q'' x \neq 0$ and $b' g = b q g = q'' x \neq 0$ with $q'' = q q' \notin \ideal{p}$. So $a' f = b' g \neq 0$ in $A$. But it is even $a'f/1 = b' g/1 \neq 0$ in $S^{-1}A$. Otherwise $s a' f = 0$ with $s \in S$. So we had $s a' f = s q'' x = 0$ and $s q'' \in \ideal{p}$. But it is $s, q'' \notin \ideal{p}$.

Now if $A$ is noetherian and irreducible then $\Ass(A) =\ideal{p}$ where $\ideal{p}$ is the unique minimal prime of $A$. So for every other prime $\ideal{q} \subseteq A$ we have $S \cap \ideal{p} = (A - \ideal{q}) \cap \ideal{p} = \emptyset$ and $S^{-1}A = A_\ideal{q}$ is irreducible too.

So a counterexample must consist of a non-noetherian ring $A$ which has either $\Ass(A) = \emptyset$ or has $\Ass(A) = \{\ideal{p}\}$ and $\ideal{p} \cap (A-\ideal{q}) \neq \emptyset$ for a certain prime $\ideal{q} \subseteq A$. Furthermore it must not be integral (trivial) and not reduced: Let $A$ be reduced and irreducible: Let $I=(f)$ and $J=(g)$. If $f g = 0$ then $\sqrt{I \cap J} = \sqrt{I J} = \sqrt{0} = N_A$ with $N_A = 0$ the nilradical. So $I \cap J = 0$ and $I=0$ or $J = 0$, that is $f = 0$ or $g = 0$. So a reduced irreducible ring is an integral domain and has the permanence property sought for.

A counterexample can be constructed as follows: Let $(A,\ideal{m})$ be a noetherian local ring with minimal prime ideals $\ideal{p}, \ideal{q}$ and prime ideal $\ideal{r} \subseteq \ideal{m}$, such that

$$\ideal{p}, \ideal{q} \subsetneq \ideal{r} \subsetneq \ideal{m}$$

and $\ideal{p} \cap \ideal{q} = (0)$.

Furthermore let $E=E(A/\ideal{m})$ be the injective hull of $A/\ideal{m}$, an $A$--module.

Construct the ring $A + E = B$ with componentwise addition and

$$(a,e) \cdot (a', e') = (a a', a e' + a' e)$$

It is called in [1, Example 2.4] "idealization" and in modern terminology (Hartshorne, Algebraic Geometry, II, Ex. 8.7) would be called "trivial infinitesimal extension of $A$ by $E$".

In [1, Example 2.4] it is contended without proof that $B$ is an irreducible ring - I omit the proof here too, as it is easy to find.

From the exact sequence of $B$--Modules $0 \to E \to B \to A \to 0$ we see, that $\ideal{P} = (\ideal{p}, E)$, $\ideal{Q} = (\ideal{q}, E)$, $\ideal{R} = (\ideal{r},E)$ are prime ideals of $B$ with $\ideal{P}, \ideal{Q} \subseteq \ideal{R}$. Furthermore, we have $\ideal{P} \cap \ideal{Q} = (\ideal{p} \cap \ideal{q}, E) = (0, E) \subseteq B$.

I contend, that the ring $B_\ideal{R}$ is not irreducible, albeit it is the localization of the irreducible ring $B$: We have $\ideal{P}_\ideal{R}, \ideal{Q}_\ideal{R} \neq (0)$ as ideals of $B_\ideal{R}$. Now it is

$$\ideal{P}_\ideal{R} \cap \ideal{Q}_\ideal{R} = (\ideal{P} \cap \ideal{Q})_\ideal{R} = (0, E)_\ideal{R}$$

Now from [2, Theorem 2.4], I draw the fact, that (in $A$ and for $A$--modules) for every $e \in E$ there is a power $\ideal{m}^t$ that annihilates $e$. As $\ideal{m}^t \not\subseteq \ideal{r}$, we can find a $s \notin \ideal{r}$, $s \in \ideal{m}^t$, $s \in A$, such that $s e = 0$. So $(s, 0) \in B - \ideal{R}$ and $(0,e) = 0$ in $B_\ideal{R}$. So $(0,E)_\ideal{R} = 0$ and we have found in $\ideal{P}_\ideal{R}$ and $\ideal{Q}_\ideal{R}$ two ideals of $B_\ideal{R}$ which are nonzero, but have intersection zero. Q.E.D.

• Are you sure that the statement "B is an irreducible ring" is so easy to prove? I tried it, without success. Jun 1, 2016 at 20:38