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 $(1+\sqrt{-5})$.