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 $\mathrm{Lawv}(X) \cong \mathsf{T}$?

oppositeof any Lawvere theory embeds in the category of algebras in a canonical way. $\endgroup$ – Zhen Lin Jul 28 '15 at 2:32