5
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Doesn't it imply that $(p)$ is maximal ideal in $R$? $\endgroup$ – N. S. Feb 12 '13 at 5:26
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Doesn't $p$ irreducible imply $p$ prime only in unique factorization domains? $\endgroup$ – William Stagner Feb 12 '13 at 5:19
  • $\begingroup$ No, it is more generally true in GCD domains (and I'm not sure if this is necessary either); however, irreducibility is not the issue here. $\endgroup$ – Zev Chonoles Feb 12 '13 at 5:20
  • $\begingroup$ Ahh, I see. The book I'm using is Artin, and he doesn't introduce prime ideals until the next chapter. Is there a similar argument that doesn't make use of them? $\endgroup$ – William Stagner Feb 12 '13 at 5:23
  • $\begingroup$ @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. $\endgroup$ – Zev Chonoles Feb 12 '13 at 5:25
  • $\begingroup$ @ZevChonoles Yea remembered it after posting... Anyhow, $(p)$ maximal implies $p$ prime element. $\endgroup$ – N. S. Feb 12 '13 at 5:28
0
$\begingroup$

Hint $\ $ If $\rm\ R/p\, \cong\, S/q\ $ then $\rm\ p\: prime\ in\,R\!\iff\! domain\: R/p \,\cong\, S/q\!\iff\! q\: prime\ in\ S$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.