TITLE: Does every Lawvere theory arise in this way?
QUESTION [10 upvotes]: By a Lawvere theory, I mean a finite-product category $\mathsf{T}$ equipped with a distinguished object, such that every object of $\mathsf{T}$ can be expressed as a finite product of the distinguished object.
Now let $X$ denote an object of a finite-product category $\mathbf{C}$. Then $X$ gives rise to a Lawvere theory $\mathrm{Lawv}(X)$ with distinguished object $X$, defined as the full subcategory of $\mathbf{C}$ induced by the objects $\{X^n \mid n:\mathbb{N}\}.$
Examples.

Viewing $2$ as an object of $\mathbf{Set}$, we see that the arrows of $\mathbf{Lawv}(2)$ are "truth tables." Hence $\mathbf{Lawv}(2)$ is the Lawvere theory of Boolean algebras.
Let $\mathbb{K}$ denote a field. Write $U_{\mathrm{Mod}}(\mathbb{K})$ for the underlying $\mathbb{K}$-module. Then $\mathbf{Lawv}(U_{\mathrm{Mod}}(\mathbb{K}))$ is the Lawvere theory of $\mathbb{K}$-modules.


Question. Is it true that for all Lawvere theories $\mathsf{T}$, there exists a Lawvere theory $\mathsf{S}$ and an $\mathsf{S}$-algebra $X$ such that $\mathrm{Lawv}(X) \cong \mathsf{T}$?

REPLY [3 votes]: I can elaborate on the second example a bit more:
Let $T$ be a Lawvere theory. Then $T$ is commutative iff $T \cong Lawv(\text{hom}_T(x,-))$, where $x$ is the generic object in $T$.
A Lawvere theory is commutative if all its $n$-ary operations commute with each each other. A morphism of $T$-algebras is a natural transformation between functors; the naturality condition is precisely the requirement to commute with all $n$-ary operations. Your example of modules over a field (more generally a commutative semi-ring) is the prime example of a commutative theory.
Now commutative Lawvere theories are extremely useful, but also very restrictive. I will think about the general case some more.