## The tensor algebra of a direct factor is a subalgebra

Let be a commutative ring with 1 and let and be -bialgebras such that . Suppose further that is a direct factor of . Prove that is (isomorphic to) a subalgebra of . Prove the analogous result for the symmetric and exterior algebras.

Say for convenience that .

Note that the usual injection induces a module homomorphism ; we claim that this map is injective. It suffices to prove injectivity on simple tensors. To that end, suppose and . Then we have in . Now recall that tensor products distribute over direct sums, so that is module isomorphic to the direct sum over all possible -factor tensor products of and via the map which selects the – or -coordinate as appropriate. (For example, .) Note that under this map, is zero. In particular, in . Thus is injective.

Now note that is naturally a submodule of , and so by the universal property of tensor algebras we have an algebra homomorphism . Note that the restriction of to is precisely as described above; in particular, is injective. So is naturally isomorphic to a subalgebra of .

Analogously, let be the natural homomorphism; we claim that this map is injective. To see this, let be a simple tensor and suppose . Then we have . Using the decomposition of as a direct sum from above, we see that . In particular, .

Now the induced map is injective on each summand, and so is injective.

An analogous argument holds for .

### Like this:

Like Loading...

*Related*

or leave a trackback:

Trackback URL.