Terminology: Algebras where long strings of products are 0? - MathOverflow

Terminology: Algebras where long strings of products are 0?

2010-04-02T18:08:29Z

I'd like a name for an augmented algebra $A = \langle 1\rangle \oplus A_+$ for which there is an $N$ so that any product of more than $N$ elements in the augmentation ideal is $0$, i.e., $(A_+)^N = 0$. Is there a name? It seems related to nilpotence, and it implies that all elements in $A_+$ are nilpotent, but is stronger than that. There is a uniform bound on the degree of nilpotence, but that's not enough either, as the example of the exterior algebra in infinitely many variables over $\mathbb{Z}/2$ shows.

MathWorld defines a nilpotent algebra or nilalgebra to be one where every element is nilpotent. (They are therefore not considering unital algebras, contrary to an earlier discussion here.) Is this standard? Is there a better term?

Answer by David Jordan for Terminology: Algebras where long strings of products are 0?

2010-04-02T18:44:07Z

I believe the terminology you want is that "the augmentation ideal is locally nilpotent." See, e.g. http://planetmath.org/encyclopedia/NilAndNilpotentIdeals.html, although there are surely places in actual literature where this is used. I think an ideal being nilpotent means that the powers of the ideal are eventually zero (and is stronger than just requiring it to be a nil-ideal, meaning it's comprised of individually nilpotent elements). So a nilpotent ideal is one where there is a uniform bound, i.e. I^N=0 for some sufficiently large N. Locally nilpotent means that for any finitely generated subalgebra there exists such an N.

Is that the notation you want? I don't know a name for an augmented algebra whose ideal is locally nilpotent other than just that.

Answer by Vladimir Dotsenko for Terminology: Algebras where long strings of products are 0?

2010-04-03T08:10:35Z

Just to confirm the accepted answer, I link here relevant definitions from Springer Online Encyclopaedia of Mathematics (definitely more reliable source than MathWorld and even PlanetMath...):

Nilpotent algebra

Locally nilpotent algebra

Nil algebra

So the most appropriate thing to say would be "the augmentation ideal is nilpotent". This terminology is very standard.

Answer by Greg Marks for Terminology: Algebras where long strings of products are 0?

2010-06-30T05:23:32Z

It seems to me that Jordan and Dotsenko are giving different answers from one another, and I agree with Dotsenko’s. The condition Thurston has stated is the definition of $A_{+}$ being nilpotent. “Locally nilpotent” is a weaker condition. There are many examples of nonunital rings $A_{+}$ that are locally nilpotent (meaning for any finite set $a_1, \ldots, a_k \in A_{+}$ there exists $n \in \mathbb{N}$ such that $a_{i_1} \cdots a_{i_n} = 0$ provided every $i_j \in \lbrace 1, \ldots, k\rbrace$) but not nilpotent (meaning there exists $n \in \mathbb{N}$ such that $a_1 \cdots a_n = 0$ provided every $a_i \in A_{+}$, which is the condition Thurston stated). Even better: two nice (and quite different) examples of a locally nilpotent prime nonunital ring can be found in E. I. Zelmanov, “An example of a finitely generated primitive ring,” Sibirsk. Mat. Zh. 20 (1979), no. 2, 423, 461, and J. Ram, “On the semisimplicity of skew polynomial rings,” Proc. Amer. Math. Soc. 90 (1984), no. 3, 347–351. (Of course, if one merely wants an example where $A_{+}$ is locally nilpotent but not nilpotent—and so does not satisfy Thurston’s condition—one could take something like $A_{+} = \bigoplus_{i=2}^{\infty} 2\mathbb{Z}/2^i\mathbb{Z}$.)

N.B. Mathematical Reviews incorrectly lists the title of Zelmanov’s paper as “An example of a finitely generated primary ring.” It’s listed correctly in Zentralblatt. Possibly the problem lies in the translation from the original Russian; the condition primitive in the English translation of the paper (Siberian Math. J. 20 (1979), no. 2, 303–304) is what we would today call prime.