Let $R$ be a non-reduced commutative ring(not necessarily Noetherian) with unit. Let the nilradical $\mathcal{N}$ of $R$ be a prime ideal with the property that $\mathcal{N}^2=0$. Do we know about the structure of such rings or any kind of characterization of such rings. Is anyone aware of the structure of such rings with the additional hypothesis that $R$ is a non-reduced local ring such that the nilradical does not coincide with the maximal ideal. An example of such a ring is: Let $V$ be a valuation ring and let $\mathcal{N}$ be its nilradical. Then $R=V/{\mathcal{N}^2}$ satisfies the above properties.
Nilradical that is a prime ideal
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commutative-algebra
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0What more do you want to know about the structure of a ring which is local, has Krull dimension >1 and has a nilradical which squares to zero? That is already very specific, and local rings are very "basic" already. – 2015-05-26
1 Answers
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Take any ring R' and an arbitrary prime ideal P' in it, then R=R'/(P')^2 is such a ring with nilradical P=P'/(P')^2.
Edit: the question asks for nonreduced ring. For that to hold, in our example we need that P'\neq (P')^2. In general it will be tricky, for example, take R' to be a product of two field, then any prime ideal of it satisfies (P')^2=P'. In the Noetherian case,if you assume that \mathrm{Spec} R' is connected, in other words, if R' contains no idempotents other than $0$ and $1$, then (P')^2\neq P' for any prime ideal P'.
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0But is there any characterization of such rings? $P'$ should not be a maximal ideal because i dont want the nilradical to coincide with the maximal ideal. – 2011-05-01