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.

  • $\begingroup$ What 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. $\endgroup$
    – rschwieb
    May 26, 2015 at 13:40

1 Answer 1


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 $P'$.

  • $\begingroup$ But 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. $\endgroup$
    – Zac
    May 1, 2011 at 10:56

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