# Prove that $p$ is prime in $\mathbb{Z}[\sqrt{-3}]$ if and only if $x^2+3$ is irreducible in $\mathbb{F}_p[x]$.

I'm having trouble with this particular homework problem. I think I have one direction:

Let $R=\mathbb{Z}[\sqrt{-3}]$. If $p$ is prime, then $p$ is irreducible in $R$, since $R$ is an integral domain. Thus, $(p)$ is maximal, so $$R/(p) \cong \mathbb{F}_p[x]/(x^2+3)$$ is a field. Hence, $(x^2+3)$ is maximal in $\mathbb{F}_p[x]$, so $x^2+3$ is irreducible.

However, if we assume $x^2+3$ is irreducible in $\mathbb{F}_p[x]$, then that only implies that $p$ is irreducible in $R$ by the same reasoning. Since $R$ isn't a UFD, this doesn't imply $p$ is prime, so I'm stuck.

Thanks so much!

• Doesn't it imply that $(p)$ is maximal ideal in $R$? – N. S. Feb 12 '13 at 5:26

If $R/(p)$ is a domain, then that does imply that $p$ is prime in $R$ (see the Wikipedia article). In other words, an element is prime precisely when the ideal it generates is prime, which is the case precisely when quotienting by that ideal produces an integral domain. If you assume that $x^2+3$ is irreducible, then you have that $$\mathbb{Z}[\sqrt{-3}]/(p)\cong \mathbb{Z}[x]/(p,x^2+3)\cong \mathbb{F}_p[x]/(x^2+3)$$ is an integral domain (indeed, a field in this case).
• Doesn't $p$ irreducible imply $p$ prime only in unique factorization domains? – William Stagner Feb 12 '13 at 5:19
• @N.S. No, for example $2$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$ but it is not prime because $(1+\sqrt{-5})(1-\sqrt{-5})$ is divisible by $2$ even though neither factor is. – Zev Chonoles Feb 12 '13 at 5:25
• @ZevChonoles Yea remembered it after posting... Anyhow, $(p)$ maximal implies $p$ prime element. – N. S. Feb 12 '13 at 5:28
Hint $\$ If $\rm\ R/p\, \cong\, S/q\$ then $\rm\ p\: prime\ in\,R\!\iff\! domain\: R/p \,\cong\, S/q\!\iff\! q\: prime\ in\ S$