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There is something I don't get about Krull's dimension. If $K$ is a field, then the only ideals are $\{ 0 \}$ and $K$ so the only prime ideal is $\{0\}$. So the Krull's dimension is $0$. But if we have an Artinian ring, we have that every prime ideal is maximal. So if we take P a maximal ideal, $\{0 \} \subsetneq P$ is a chain of ideal which seems to be of length one. So why is the Krull's dimension of an Artinian ring $0$?

roi_saumon
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    An Artinian ring in which $(0)$ is a prime ideal is a field. So if $P$ is a nonzero prime ideal, then $(0)$ isn't a prime ideal. – Qiaochu Yuan Nov 19 '18 at 10:33
  • Oh, for some reason I thought that $(0)$ was prime in any ring but clearly it is only true when R is integral. – roi_saumon Nov 19 '18 at 11:10

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So if we take P a maximal ideal, $\{0 \} \subsetneq P$ is a chain of ideal which seems to be of length one.

It is a chain of length one. But it isn't a chain of prime ideals. And for Krull dimension we look only at chains of prime ideals.

For a (commutative) ring $R$ the ideal $\{0\}$ is prime if and only if $R$ is an integral domain. And an Artinian ring is an integral domain if and only if it is a field. So if $R$ is Artinian but not field then $\{0\}$ is not prime. On the other hand if $R$ is a field than it has no proper ideal. So the chain $\{0\}\subsetneq P$ is never a chain of prime ideals when $R$ is artinian.

Also note that fields are special Artinian rings. And indeed they all have Krull dimension $0$.

freakish
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