Nilradical of ring R, $\text{nil}(R)$ is intersection of prime ideals, is radical of $(0)$. So I feel the name 'radical' is natural.

But Jacobson radical, that is an intersection of maximal ideals, seem not to be radical of a certain ideal. Why Jacobson radical is nevertheless called 'radical'? Thank you.


2 Answers 2


The term "radical" here doesn't refer to the radical of some ideal, but to the general concept of a "radical of a ring" which is in some way "a set of bad elements". For nilradical, "bad" means nilpotent; for Jacobson radical, "bad" means annihilating all simple left modules.

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    $\begingroup$ I like this answer. Just to point out: for Jacobson radical there is also a stronger connection to nilpotency than both being 'bad'. An element $x$ is strongly nilpotent if $xy$ (and hence $yx$) is nilpotent for every $y$. At least for some class of rings (Artinian and/or Noetherian? I forgot), the Jacobson radical is equal to the set of all strongly nilpotent elements. $\endgroup$
    – Vincent
    Dec 1, 2020 at 16:22
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    $\begingroup$ @Vincent yep The lower nilradical (aka Baer nilradical) which is the intersection of all prime ideals is the set of strongly nilpotent elements. Artinian is definitely sufficient to get them to coincide with Jacobson radical because then the radical itself is nilpotent. $\endgroup$
    – rschwieb
    Dec 1, 2020 at 22:51

To add another viewpoint: the "bad element" description of radicals is somewhat useful, but I also feel like it might be someone's atttempt at a retrospective description that just caught on.

I tried tracking down the original usage in ring-theory prehistory just now, but I didn't get very far. A biographical article on Wedderburn hints that it may lie in the work of Cartan, Molien and Frobenius. (It would be interesting to know where to find that but I reached the end of my timebox on the subject. )

Really you should probably interpret "radical" as a cousin of "$\sqrt{\cdot}$" descended from the same latin root meaning... "root."

A little caveat: The following explanation is speculative. I don't know if this was the original intention or if I am just the next person with a clever retrospective explanation.

The following description applies pretty well:

A root is something important that supports a structure but is itself largely hidden. When viewing the elements through a certain lens, the "radical" for that lens is the set of elements which you can't see through the lens.

When your lens is representation theory, the elements that are invisible to the semisimple representations are the Jacobson radical elements. They don't act at all on any semisimple representations, so they are "hidden."

When your lens is classical algebraic geometry, the nilpotent elements are invisible to the Zariski topology.

I think this is essentially what is captured by the definition of a radical class and that the corresponding ideal is essentially what is invisible to the respective torsion theory. I am not an expert in this subject, though, so take it with a grain of salt.

The elements themselves are not always obviously "bad" in the sense of "badly behaved." Are the elements of the maximal ideal of a local domain really 'bad'? Not with respect to the ring operation, I don't think. But in terms of shedding light on semisimple modules for the ring, they don't contribute.

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    $\begingroup$ I always thought the first occurrence of radicals would have been the radical of an ideal in some polynomial ring, as per Nullstellensatz, where $\sqrt I = \{x \in R: \exists n: x^n \in I \}$ is literally the set of all "roots" of elements of the ideal, as in roots of numbers which had been around since the middle ages. Then by a strange shift of nomenclature, the radical of the zero ideal (i.e. all nilpotent elements) started to be called "the" radical, and then when ring theory started with non-noetherian and noncommutative rings and all, people realised there's various generalisations. $\endgroup$ Sep 15, 2021 at 19:07

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