Terminology: Algebras where long strings of products are 0? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:32:07Z http://mathoverflow.net/feeds/question/20174 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20174/terminology-algebras-where-long-strings-of-products-are-0 Terminology: Algebras where long strings of products are 0? Dylan Thurston 2010-04-02T18:08:29Z 2010-06-30T05:23:32Z <p>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.</p> <p>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 <a href="http://mathoverflow.net/questions/15107/algebra-unital-associative-algebra-better-terminology" rel="nofollow">earlier discussion</a> here.) Is this standard? Is there a better term?</p> http://mathoverflow.net/questions/20174/terminology-algebras-where-long-strings-of-products-are-0/20178#20178 Answer by David Jordan for Terminology: Algebras where long strings of products are 0? David Jordan 2010-04-02T18:44:07Z 2010-04-02T18:44:07Z <p>I believe the terminology you want is that "the augmentation ideal is locally nilpotent." See, e.g. <a href="http://planetmath.org/encyclopedia/NilAndNilpotentIdeals.html" rel="nofollow">http://planetmath.org/encyclopedia/NilAndNilpotentIdeals.html</a>, 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.</p> <p>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.</p> http://mathoverflow.net/questions/20174/terminology-algebras-where-long-strings-of-products-are-0/20217#20217 Answer by Vladimir Dotsenko for Terminology: Algebras where long strings of products are 0? Vladimir Dotsenko 2010-04-03T08:10:35Z 2010-04-03T08:10:35Z <p>Just to confirm the accepted answer, I link here relevant definitions from <a href="http://eom.springer.de/" rel="nofollow">Springer Online Encyclopaedia of Mathematics</a> (definitely more reliable source than MathWorld and even PlanetMath...):</p> <p><a href="http://eom.springer.de/n/n066700.htm" rel="nofollow">Nilpotent algebra</a></p> <p><a href="http://eom.springer.de/l/l060470.htm" rel="nofollow">Locally nilpotent algebra</a></p> <p><a href="http://eom.springer.de/n/n066640.htm" rel="nofollow">Nil algebra</a></p> <p>So the most appropriate thing to say would be "the augmentation ideal is nilpotent". This terminology is very standard.</p> http://mathoverflow.net/questions/20174/terminology-algebras-where-long-strings-of-products-are-0/30006#30006 Answer by Greg Marks for Terminology: Algebras where long strings of products are 0? Greg Marks 2010-06-30T05:23:32Z 2010-06-30T05:23:32Z <p>It seems to me that Jordan and Dotsenko are giving different answers from one another, and I agree with Dotsenko&#8217;s.&#160; The condition Thurston has stated is the definition of $A_{+}$ being nilpotent.&#160; &#8220;Locally nilpotent&#8221; is a weaker condition.&#160; 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).&#160; Even better: two nice (and quite different) examples of a locally nilpotent <i>prime</i> nonunital ring can be found in E. I. Zelmanov, &#8220;An example of a finitely generated primitive ring,&#8221; <i>Sibirsk. Mat. Zh.</i> <b>20</b> (1979), no. 2, 423, 461, and J. Ram, &#8220;On the semisimplicity of skew polynomial rings,&#8221; <i>Proc. Amer. Math. Soc.</i> <b>90</b> (1984), no. 3, 347&#8211;351.&#160; (Of course, if one merely wants an example where $A_{+}$ is locally nilpotent but not nilpotent&#8212;and so does not satisfy Thurston&#8217;s condition&#8212;one could take something like $A_{+} = \bigoplus_{i=2}^{\infty} 2\mathbb{Z}/2^i\mathbb{Z}$.)</p> <p><i>N.B.</i> <i>Mathematical Reviews</i> incorrectly lists the title of Zelmanov&#8217;s paper as &#8220;An example of a finitely generated primary ring.&#8221;&#160; It&#8217;s listed correctly in <i>Zentralblatt</i>.&#160; Possibly the problem lies in the translation from the original Russian; the condition <i>primitive</i> in the English translation of the paper (<i>Siberian Math. J.</i> <b>20</b> (1979), no. 2, 303&#8211;304) is what we would today call <i>prime</i>. </p>