It is shown that if a bilinear map f: A x B --> C of modules over a
commutative ring k is nondegenerate (i.e., if no nonzero element of A
annihilates all of B, and vice versa), and A and B are Artinian, then A and B
are of finite length.
Some consequences are noted. Counterexamples are given to some attempts to
generalize the above statement to balanced bilinear maps of bimodules over
noncommutative rings, while the question is raised whether other such
generalizations are true.
The inner automorphisms of a group G can be characterized within the category
of groups without reference to group elements: they are precisely those
automorphisms of G that can be extended, in a functorial manner, to all groups
H given with homomorphisms G --> H. Unlike the group of inner automorphisms of
G itself, the group of such extended systems of automorphisms is always
isomorphic to G.
It is shown that any finite-dimensional homomorphic image of an inverse limit
of nilpotent not-necessarily-associative algebras over a field is nilpotent.
More generally, this is true of algebras over a general commutative ring k,
with "finite-dimensional" replaced by "of finite length as a k-module".
Let k be an infinite field, I an infinite set, V a k-vector-space, and
g:k^I\to V a k-linear map. It is shown that if dim_k(V) is not too large (under
various hypotheses on card(k) and card(I), if it is finite, respectively
countable, respectively < card(k)), then ker(g) must contain elements
(u_i)_{i\in I} with all but finitely many components u_i nonzero.
We study surjective homomorphisms f:\prod_I A_i\to B of
not-necessarily-associative algebras over a commutative ring k, for I a
generally infinite set; especially when k is a field and B is
countable-dimensional over k.
Our results have the following consequences when k is an infinite field, the
algebras are Lie algebras, and B is finite-dimensional:
If all the Lie algebras A_i are solvable, then so is B.
If all the Lie algebras A_i are nilpotent, then so is B.
If the free algebra F on one generator in a variety V of algebras (in the
sense of universal algebra) has a subalgebra free on two generators, must it
also have a subalgebra free on three generators? In general, no; but yes if F
generates the variety V.