1
$\begingroup$

One of the properties of an Artinian ring $R$ is that every prime ideal is maximal. So, if $\mathfrak{m}$ is a nonzero prime ideal, $(0)\subseteq \mathfrak{m}$ is a length-$1$ chain of prime ideals, meaning that $\dim R\geq 1$. Since every prime ideal is maximal, no longer chains exist, meaning $\dim R=1$. But Artinian rings are supposed to have dimension $0$. What am I missing?

$\endgroup$
  • 7
    $\begingroup$ 0 need not be prime in an Artin ring. $\endgroup$ – dc2814 Jun 10 '13 at 5:35
  • 2
    $\begingroup$ Therefore, an artin ring where (0) is prime (hence maximal) is a field. $\endgroup$ – dc2814 Jun 10 '13 at 5:37
  • $\begingroup$ @dc2814 Perhaps you should put those thoughts you have expressed via comments in an answer. $\endgroup$ – Karl Kronenfeld Jun 10 '13 at 5:38
8
$\begingroup$

Being an Artin ring does not require that $(0)$ is a prime ideal. In fact, if $(0)$ is prime in the Artin ring $R\neq 0$, then $(0)$ is the only maximal ideal in $R$. Therefore $R$ is a field since every nonzero element is not contained in a maximal ideal, hence they are all units.

$\endgroup$

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.