In a commutative algebraic theory, do all constant symbols necessarily represent the same value?

Question

Let $T$ denote a commutative algebraic theory with two nullary function symbols $a$ and $b$ (i.e. constants). Is it an automatic consequence of the definitions that $a=b$ is a theorem of $T$?

My thoughts. I think the answer is yes, simply by "following the pattern." In particular, let $a()$ represent a nullary function. Then we have the following.

• Suppose $f(*,*)$ represents a binary function. Then I think it is a consequence of the definitions that $f(a(),a()) = a()$ in every interpertation.

• Similarly, if $f(*)$ represents a unary function, then I think that $f(a()) = a()$ in every interpretation.

• Taking the pattern to its logical conclusion, I think that if $f()$ is a nullary function, then $f()=a()$ in every interpertation.

Answer 1

By definition of "$a$ and $b$ commute", you start with the $0\times 0$ matrix (in effect "nothing"), apply $a$ to "each" row (as there are no rows, you do nothing), then apply $b$ to the ($0$ size) row of results and obtain $b()$. In the other order you obtain $a()$ and commutativity thus states that $b()=a()$.

Answer 2

That two constants i.e. 0-ary operations commute with each other says exactly that they are equal, by definition. You don't need any "logic" or higher-ary operations in order to see this. Details:

Recall that that two maps $t : A^n \to A$ and $t' : A^m \to A$ commute if for all $x=(x_{ij})_{1 \leq i \leq n,~ 1 \leq j \leq m} \in A^{n \times m}$ we have $$t(t'(x_{11},\dotsc,x_{1m}),\dotsc,t'(x_{n1},\dotsc,x_{nm})) = t'(t(x_{11},\dotsc,x_{n1}),\dotsc,t(x_{1m},\dotsc,x_{nm})).$$ This has a nice visualization through matrices, see the nlab article.

For $n=0$ we have $A^n=\{\star\}$, the set with only one element (which we denote always by $\star$ for all $n$). Hence $t,t'$ correspond to constants $t(\star),t'(\star)$, and $\star$ is the only possible input for $t$ and $t'$. Hence, commutativity reduces to $$t = t'.$$