The tensor algebra of a direct factor is a subalgebra

Let R be a commutative ring with 1 and let M and N be (R,R)-bialgebras such that rm = mr. Suppose further that M is a direct factor of N. Prove that \mathcal{T}(M) is (isomorphic to) a subalgebra of \mathcal{T}(N). Prove the analogous result for the symmetric and exterior algebras.

Say for convenience that N = M \oplus T.

Note that the usual injection \varphi_k : M^k \rightarrow (M \oplus T)^k induces a module homomorphism \mathcal{T}^k(M) \rightarrow \mathcal{T}^k(M \oplus T); we claim that this map is injective. It suffices to prove injectivity on simple tensors. To that end, suppose z = m_1 \otimes \cdots \otimes m_k and \varphi_k(z) = 0. Then we have (m_1,0) \otimes \cdots \otimes (m_k,0) = 0 in \mathcal{T}^k(M \oplus T). Now recall that tensor products distribute over direct sums, so that \mathcal{T}^k(M \oplus T) is module isomorphic to the direct sum over all possible k-factor tensor products of M and T via the map which selects the M– or T-coordinate as appropriate. (For example, (a_1,b_1) \otimes (a_2,b_2) \mapsto a_1 \otimes a_2 + a_1 \otimes b_2 + b_1 \otimes a_2 + b_1 \otimes b_2.) Note that under this map, \varphi_k(z) is zero. In particular, z = 0 in \mathcal{T}^k(M). Thus \varphi_k is injective.

Now note that M is naturally a submodule of \mathcal{T}(N), and so by the universal property of tensor algebras we have an algebra homomorphism \varphi : \mathcal{T}(M) \rightarrow \mathcal{T}(N). Note that the restriction of \varphi to \mathcal{T}^k(M) is precisely \varphi_k as described above; in particular, \varphi is injective. So \mathcal{T}(M) is naturally isomorphic to a subalgebra of \mathcal{T}(N).

Analogously, let \psi_k : \mathcal{S}^k(M) \rightarrow \mathcal{S}^k(M \oplus T) be the natural homomorphism; we claim that this map is injective. To see this, let z \in \mathcal{T}^k(M) be a simple tensor and suppose \psi_k(\overline{z}) = 0. Then we have \varphi_k(z) = \sum z_i \otimes ((m_{i,1},t_{i,1}) \otimes (m_{i,2},t_{i,2}) - (m_{2,i},t_{2,i}) \otimes (m_{i,1},t_{i,1})) \otimes w_i. Using the decomposition of \mathcal{T}^k(M \oplus T) as a direct sum from above, we see that z \in \mathcal{C}^k(M). In particular, \overline{z} = 0.

Now the induced map \psi : \mathcal{S}(M) \rightarrow \mathcal{S}(M \oplus T) is injective on each summand, and so is injective.

An analogous argument holds for \bigwedge(M).

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