# What is the tensor product for the Eilenberg-Moore category of a commutative monad?

In Linear logic, monads and the lambda calculus (DRAFT), Proposition 3.0.2 says that the Eilenberg-Moore category for a commutative monad has the structure of a symmetric monoidal closed category. My question is: How do you construct the tensor product for the EM-category? When I followed the reference to Kiegher's paper "Symmetric monoidal closed categories generated by commutative adjoint monads", that paper only seems to give the construction for monads of the form A ⊗ -. I don't see how to generalize the construction to arbitrary commutative monads.

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One reference that says this explicitly is Gavin J. Seal, "Tensors, monads and actions", arxiv.org/abs/1205.0101, see beginning of section 2.2 and then theorem 2.5.5. On the nLab see here: ncatlab.org/nlab/show/… –  Urs Schreiber Feb 12 '14 at 22:42

Let $T \colon C \to C$ be your monad. Being commutative, it comes with maps $\mathrm{dst} \colon T(A) \otimes T(B) \to T(A \otimes B)$. Let $\phi \colon TA \to A$ and $\psi \colon TB \to B$ be algebras. Then $\phi \otimes \psi$ is the coequalizer in $\mathrm{Alg}(T)$ of $T(\phi \otimes \psi)$ and $\mu \circ T(\mathrm{dst})$ (which is a reflexive pair of morphisms from the free algebra on $T(A) \otimes T(B)$ to the free algebra on $A \otimes B$). The unit $I$ in $\mathrm{Alg}(T)$ is the free algebra $\mu \colon T^2(I) \to T(I)$. Moreover, the free functor $C \to \mathrm{Alg}(T)$ preserves monoidal structure.
A good example to keep in mind is where $T$ is the free vector space monad on the category of sets. The coequalizers then is pretty much directly the usual tensor product construction with bilinear maps.