Is there a standard example in commutative algebra of a ring where the jacobson radical does not equal the nilradical?


2 Answers 2


For any local (meaning "has a unique maximal ideal") domain $D$ with unique maximal ideal $M\neq 0$, the Jacobson radical is the maximal ideal $M$, but its nilradical is obviously $0$.

You can easily produce such local domains as localizations of domains at well-chosen prime ideals.

Or, if you're a big fan of valuation domains, one of those would also serve your purposes. A concrete example of this would be $\Bbb F[[x]]$ for any field $F$, whose maximal ideal is $(x)$.

If you want to keep looking for more examples, you might use the following approach. Since the Jacobson radical is the intersection of maximal ideals, and the nilradical is the intersection of prime ideals, you need to look at rings where there you understand the prime ideals and maximal ideals well.

For Artinian rings, the maximal ideals and prime ideals coincide, so it's fruitless to search there.

  • $\begingroup$ Also this related question for an example where $R/J(R)$ is more complicated than just a field. $\endgroup$
    – rschwieb
    Feb 27, 2019 at 16:51

The formal power series ring $A[[x]]$, where $A$ is any Noetherian ring, is such an example. You can show that $f = \sum_{n \geq 0}a_nx^n$ is in the Jacobson radical if and only if $a_0$ is in the Jacobson radical of $A$, thus $x$ is in the Jacobson radical, but is not nilpotent.

  • 3
    $\begingroup$ Why do you require $A$ Noetherian? Am I missing something? $\endgroup$ Aug 1, 2015 at 18:36

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