Give an example of a ring $A$ such that $\dim(A) =0$ but $A$ is not a field.
One thing is sure, $A$ cannot be an integral domain.
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2Hint: Consider a very small finite ring. – Tobias Kildetoft Oct 30 '15 at 09:57
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let $k$ be a field. consider $k \times k$ (pointwise multiplication).
hunter
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But $k\times k$ will be an integral domain. But for the conditions to be satisfied, we need an example that should not be an integral domain. – gamma Oct 30 '15 at 16:28
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@chi $k\times k$ is certainly not an integral domain; $(1, 0) \cdot (0, 1) = (0, 0)$. – hunter Oct 30 '15 at 16:28
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But $<(k,0)>$ will be a prime ideal in $k \times k$ . Therefore, dimension $\neq 0$ – gamma Oct 30 '15 at 16:42
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1@chi the dimension is the number of primes in a maximal prime chain minus one. this ring contains exactly two primes, and they're not nested, so the dimension is zero. The spectrum of this ring is two closed points. – hunter Oct 30 '15 at 17:04
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@chi < (0, 0) > is not prime, because, as we just discussed, $k \times k$ is not an integral domain – hunter Oct 30 '15 at 17:22
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"Steps in Commutative Algebra" by "R. Y. Sharp" has:
8.39 LEMMA. Let $R$ be a commutative Artinian ring. Then every prime ideal of $R$ is maximal.
So every Artinian ring which is not a field, is an answer to your question.
For examples of Artinian rings see here and here. Also if you are familiar with localization, you can easily see that if $p$ is a minimal prime ideal of $R$, then $R_p$ is artinian.
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Let $R$ be a Boolean ring other than $\mathbb{Z}/2\mathbb{Z}$.
Lemma The only integral domain that is Boolean is $\mathbb{Z}/2\mathbb{Z}$, for otherwise, if $R$ an integral domain but $R \neq Z/2Z$, we may take $a \neq 0, 1$. But $a \cdot (a-1) = a^2 - a = a - a = 0$ a contradiction.
Hence, Every prime ideal $P$ of $R$ is maximal, since $R/P$ is boolean and integral implies $R/P \simeq \mathbb{Z}/2\mathbb{Z}$ a field. Therefore, $dim R = 0$.
Weaam
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