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?

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    $\begingroup$ 0 need not be prime in an Artin ring. $\endgroup$ – dc2814 Jun 10 '13 at 5:35
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    $\begingroup$ 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

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.


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