So without giving excessive background, I was working on baby Spivak & got to the stuff about exterior algebras, & now I'm going through the section on tensor algebras in my copy of Dummit/Foote to learn more. They define it (as usual?) as the R-algebra $\bigwedge(M) = T(M)/A(M)$ obtained by taking the quotient of the tensor algebra T(M) by the ideal A(M) generated by all elements $m\in M$ with the property that $m\otimes m = 0$. So that got me thinking about radicals, since Hungerford described them as a way to "mod out" unwanted elements of a ring. This is how he described the unwanted radical property P

i) the homomorphic image of a P-ring (ring with property P) is itself a P-ring
ii) every ring R contains a P-ideal P(R) that contains all other P-ideals
iii) P(R/P(R)) = 0
iv) P(P(R)) = P(R)
(& in this case P(R) is A(M) and T(M) is R of course)

So you can probably guess what I'm thinking. I can believe that i) & ii) hold at least when M is a vector space, and I can show iv), it's just iii) that I'm having trouble with, although that one doesn't seem hard to believe either. Also, if A(M) is a radical of T(M), when is T(M) itself radical and when is it semisimple? Are there any books or articles out there where someone has done it like this? Is there any advantage to doing it this way?

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No non-zero element $m$ of the tensor algebra $T(M)$ of a vector space is such that $m\otimes m=0$. – Mariano Suárez-Alvarez Apr 28 '11 at 5:16
You've misstated the definition: the ideal is generated by the elements $m \otimes m$ for all $m$. I'm not sure I know what a radical or a P-ring is; could you give definitions of these? – Qiaochu Yuan Apr 28 '11 at 6:02
@Qiaochu, an example should be explicative enough: a radical is a rule $P$ which assigns to a ring one of its ideals, and variations of that (the nilpotent radical, in the commutative case; the Jacobson radical; the torsion subgroup of an abelian groups; &c.) Usually one demands that $P$ be idempotent and that $P(X/P(X))=0$, for appropriate values of "quotient" and "zero". IIRC Anderson+Fuller discuss this for modules, but it tends to be scattered all over the place in various forms. – Mariano Suárez-Alvarez Apr 28 '11 at 9:16
@ Qiaochu ok then m such that $m\otimes m \equiv 0\ mod(A(M))$ There are other examples on wiki en.wikipedia.org/wiki/Radical_of_a_ring – sauron Apr 28 '11 at 17:48