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Let $K$ be a commutative ring. Is it true that $T(V \oplus W) \cong T(V) \otimes_K T(W)$ as $K$-algebras for any $K$-modules $V$ and $W$?

The reason I ask is that I have heard it mentioned (I don't remember where) that the canonical comultiplication on the tensor algebra (which gives it a Hopf algebra structure) is the image of the diagonal $\Delta: V \to V \oplus V$ by the tensor algebra functor. In order for this to be true, we would need $T(\Delta): T(V) \to T(V \oplus V) \cong T(V) \otimes_K T(V)$. Also, for Lie $K$-algebras I know that $U(\mathfrak{g} \times \mathfrak{h}) \cong U(\mathfrak{g}) \otimes_K U(\mathfrak{h})$ so I would hope that this isomorphism is inherited from a similar isomorphism of Tensor algebras.

But I have some doubts that this is true. The tensor algebra functor $T: K \text{Mod} \to K\text{Alg}$ is left adjoint to the forgetful functor, and thus preserves coproducts, so $T(V \oplus W) \cong T(V) + T(W)$ where the right hand side is the coproduct of algebras. But then we would need $T(V) \otimes_K T(W) \cong T(V) + T(W)$ as $K$-algebras in order for this to work. But the coproduct of $K$-algebras is the tensor product if we are working in the category of commutative $K$-algebras; otherwise the coproduct is more complicated.

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This is indeed not true. As you observe, $T(V\oplus W)$ is the coproduct of $T(V)$ and $T(W)$ in the category of $K$-algebras, which is typically larger than $T(V)\otimes_K T(W)$.

For a very simple example, take $V=W=K$. Then $T(V)$ and $T(W)$ are both polynomial rings in one variable over $K$; write $T(V)=K[x]$ and $T(W)=K[y]$. Then $T(V)\otimes_K T(W)\cong K[x,y]$, but $T(V\oplus W)\cong K\langle x,y\rangle$, the ring of noncommutative polynomials in two variables.

However, there still is a canonical ring-homomorphism $T(V\oplus W)\to T(V)\otimes_K T(W)$ given by the universal property of the coproduct. So the diagonal map $\Delta:V\to V\oplus V$ induces a map $T(V)\to T(V\oplus V)$ which can then be composed with this map to give a map $T(V)\to T(V)\otimes T(V)$. This is the comultiplication of the tensor algebra.

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    $\begingroup$ The "obvious" homomorphism comes rather from the universal property of the free algebra, no? $\endgroup$ – Mariano Suárez-Álvarez Jul 16 '16 at 4:47
  • $\begingroup$ Sure, that's a more direct way to get it. $\endgroup$ – Eric Wofsey Jul 16 '16 at 4:59

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