Left adjoint to forgetful functor between varieties of algebras Given algebraic theories $S$ and $T$ for which there is a forgetful functor $U : S_{mod} \to T_{mod}$ (e.g. $U : \textbf{Rng} \to \textbf{Ab}$), it is known that $U$ is monadic, and hence has a left adjoint $F : T_{mod} \to S_{mod}$. Is it known how to compute this left adjoint in general? I.e. given a $T$-model M, is it known how to compute the $S$-model $F$(M) in general?
 A: Only just saw this question. I'll give two answers: the first is more abstract but more economizing on conceptual effort, the second is somewhat more concretely tied to Lawvere theories. 
The question means more precisely that the forgetful functor $U_S: S_{mod} \to Set$ lifts through the forgetful functor $U_T: T_{mod} \to Set$ via a $T$-algebra structure $T U_S \to U_S$; equivalently, via a morphism of monads $\phi: T \to U_S F_S = S$. There is a right action of $T$ on $S$ via the composite 
$$S T \stackrel{S \phi}{\to} S S \stackrel{\mu_S}{\to} S$$ 
and for any $T$-algebra $X$, we have by definition a left $T$-action $\alpha: T X \to X$. Thus we have an evident parallel pair of maps 
$$STX \rightrightarrows SX,$$ 
one involving the right action and the other involving the left action, and notice this is actually a reflexive pair with common right inverse $SX \to STX$ formed using the unit $u: 1_{Set} \to T$. Assuming that $S_{mod}$ has reflexive coequalizers (which it does), the coequalizer $S \circ_T X$ in the category $S_{mod}$ gives the value of the left adjoint to $S_{mod} \to T_{mod}$ at the $T$-algebra $X$. This is exactly parallel to the situation in module theory where for ring maps $A \to B$, the left adjoint to the forgetful functor $Mod_B \to Mod_A$ is given by a tensor product construction $B \otimes_A M$. The proof this works is also exactly like the one in module theory; details can be found at the nLab here. 
For the second answer, in terms of a morphism of Lawvere theories, where we have a product-preserving functor $i: Th(T) \to Th(S)$ (which I'll abbreviate to $i: T \to S$, slightly abusing language), the left adjoint can also be expressed "more concretely" as taking a $T$-algebra $X$ to a (reflexive) coequalizer of a pair of maps 
$$\sum_{m,n} S(n, 1) \cdot T(m, n) \cdot X^m \rightrightarrows \sum_n S(n, 1) \cdot X^n$$ 
where $m, n$ are natural numbers and $T(m,n)$ refers to an appropriate hom-set. These maps again involve a "left action" $T(m, n) \cdot X^m \to X^n$ and a "right action"  
$$S(n, 1) \cdot T(m, n) \to S(n, 1) \cdot S(m, n) \to S(m,1)$$ 
whose descriptions should be self-evident. We underline the fact that the reflexive coequalizer is computed setwise. More precisely, viewing a $T$-algebra $X$ as a (product-preserving) functor $T \to Set$, this is a coend 
$$\int^{n \in T} S(n, 1) \cdot X^n$$ 
or equivalently, the value at $1$ of a tensor product 
$$S(i-, -) \otimes_T X$$ 
as computed in the presheaf category $[S, Set]$ (i.e., computed objectwise in $Set$). To see this, notice the assignment $X \mapsto S(i-, -) \otimes_T X$ is the formula for the left Kan extension that is left adjoint to the forgetful functor $[i, Set]: [S, Set] \to [T, Set]$ between presheaf categories. Since the inclusion $S_{mod} = \text{Prod}(S, Set)$ into $[S, Set]$ is full and faithful, the same tensor product formula will indeed give the correct left adjoint $\text{Prod}(T, Set) \to \text{Prod}(S, Set)$ -- so long as $S(i-,-) \otimes_T X$ is a product-preserving functor $S \to Set$. I think one simple way of seeing this is to recognize that a functor $X: T \to Set$ is product-preserving iff it is a sifted colimit of representables, say 
$$X \cong \text{colim}_{(t, x) \in El(X)^{op}} T(t, -)$$ 
and then $S(i-, -) \otimes_T -$ will preserve this sifted colimit, hence it takes $X$ to 
$$\text{colim}_{(t, x) \in El(X)^{op}} S(i-, -) \otimes_T T(t, -) \cong \text{colim}_{(t, x) \in El(X)^{op}} S(it, -)$$ 
which is a sifted colimit of representables $S \to Set$, hence product-preserving.  
