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Suppose $R$ is a domain. We say an element $x\in R$ is "irreducible" if $x=yz$ implies that $y$ or $z$ is a unit or both are units. I want to know if an irreducible element is still an irreducible element when taking localization. If not, what is the "most simplest counterexample"? :)

Let $R$ be a commutative domain with 1. Let $x$ be an irreducible element of $R$. Could we find some prime ideal $\mathfrak{p}$, or a multiplicative set $S$, such that $x$ is not irreducible in $R_S$?

I allow $x$ to be a unit because it may become a unit in $R_x$.

Note: the definition of irreducible here is not standard.

Thanks.

As ones comment this it is an easy exercise, and should do it by myself. So I spend some times to think about it.

The $A=\mathbb{Z}[\sqrt{-5}]$ works. The element $2$ is irreducible in $A$, but it is reducible in $A_3$. Since $2\times 3=(1-\sqrt{-5})(1+\sqrt{-5})$. And $(1-\sqrt{-5})$ is not a unit in $A_3$, suppose it is a unit, then $(1-\sqrt{-5})*a=3^n$ for some $a\in A$ and $n$. Then taking norm will give a contradiction. Similar argument for $(1+\sqrt{-5})$.

Zev Chonoles
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wxu
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    I think your definition for irreducible is wrong, the equality $x=yz$ implies $y$ or $z$ have an inverse, not neccasarily that they are units. – Belgi Apr 09 '12 at 15:57
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    @Belgi In a commutative ring, what is your distinction between having an inverse and being a unit? (In other words, I think wxu's definition is fine.) – David E Speyer Apr 09 '12 at 16:01
  • Not answering wxu's question because it is a pretty basic homework question. Hint: You should try out a few of the basic examples of rings where irreducibles behave in an interesting way. – David E Speyer Apr 09 '12 at 16:04
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    Isn't wxu's definition incorrect because it allows $x$ to be a unit? (wikipedia page) – Zev Chonoles Apr 09 '12 at 16:04
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    @wxu: One does not usually consider units to be irredudible. You are really asking whether an irreducible will either stay irreducible or become a unit when passing to a localization. – Arturo Magidin Apr 09 '12 at 16:11
  • Well , consider for example $\mathbb{R}[x]$, all non-zero elements of $\mathbb{R}$ have an inverse – Belgi Apr 09 '12 at 16:19
  • @Belgi, UFD is still UFD under localization. – wxu Apr 09 '12 at 16:21
  • Indeed, $2x$ is irreducible in $\mathbb{R}[x]$ , but in $\mathbb{Z}[x]$ it decompose to $2*x$, note that $2$ does not have an inverse in $\mathbb{Z}$ – Belgi Apr 09 '12 at 16:21
  • I just said the definition for irreducible is wrong, I didn't talk about this case... – Belgi Apr 09 '12 at 16:22
  • Please also note that $\mathbb{Z}$ is a UFD hence this is still worng for UFD's... – Belgi Apr 09 '12 at 16:24

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