## 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 $\bigwedge(M)$.