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Let $R$ be a commutative ring with $1$ (we take $R$ not to be a field for this post). Must $R$ contain at least one prime ideal that is not maximal?

The question is equivalent to the following: For a ring $R$ (commutative with 1) is there necessarily a integral domain $S$ which is not a field such that there is a surjective ring homomorphism $R \twoheadrightarrow S$?

I feel that it's not true. Some help would be appreciated. Thanks.

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  • $\begingroup$ How about a field $R$? Or, let $R$ be Artinian. $\endgroup$ – user714630 May 9 '13 at 14:28
  • $\begingroup$ This answers my question! Thanks. $\endgroup$ – nigel May 9 '13 at 14:30
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    $\begingroup$ Note also that the Krull dimension is a measure of the lengths of chains of prime ideals, and any zero-dimensional ring has this property. $\endgroup$ – user714630 May 9 '13 at 14:32
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    $\begingroup$ @user1 Please don't clutter the comments with answers: clutter the answers with your answers, please :) $\endgroup$ – rschwieb May 9 '13 at 14:37
  • $\begingroup$ @rschwieb I will keep that in mind, thank you. $\endgroup$ – user714630 May 9 '13 at 14:48
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As your intuition suggests, the answer is no. Any commutative ring with the property that all prime ideals are maximal necessarily has Krull dimension zero, and conversely.

In fact, any commutative Artinian ring will do, since any prime ideal $\mathfrak p$ contains a product of maximal ideals. This implies $\mathfrak p$ is itself a maximal ideal.

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    $\begingroup$ Boolean rings, products of fields ... there are lots of zero-dimensional rings. $\endgroup$ – Martin Brandenburg May 9 '13 at 15:12
  • $\begingroup$ I have not seen this notion of Krull dimension in my undergrad algebra courses. Both your comment and the answer were very informative. $\endgroup$ – nigel May 9 '13 at 15:56

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