No simplifying identities for any single nonzero number under addition.

Question

Consider the structure $(\mathbb{R}, +, r)$, where r is a nonzero real number. Are the commutative and associative identities already sufficient to derive all universally valid equations in that structure? Basically, is 0 the only number that behaves in a special manner under addition?

Answer 1

Yes, this is true. Suppose $s(x_0,x_1,\dots,x_n)$ and $t(x_0,x_1,\dots,x_n)$ are terms in the language of addition such that $s(r,x_1,\dots,x_n)=t(r,x_1,\dots,x_n)$ for all $x_1,\dots,x_n\in\mathbb{R}$. We can choose $a_1,\dots,a_n\in\mathbb{R}$ such that $r,a_1,\dots,a_n$ are linearly independent over $\mathbb{Q}$. Let $F\subset\mathbb{R}$ be the subsemigroup generated by $r,a_1,\dots,a_n$. The linear independence of $r,a_1,\dots,a_n$ implies that $F$ is freely generated by $r,a_1,\dots,a_n$ as a commutative semigroup (here we use the fact that the free commutative semigroup on a set $\{x_0,x_1,\dots,x_n\}$ is the set of formal expressions $\sum m_i x_i$ where $m_i\in\mathbb{N}$ and at least one $m_i$ is nonzero). Thus the identity $s(r,a_1,\dots,a_n)=t(r,a_1,\dots,a_n)$ implies that actually $s(x_0,x_1,\dots,x_n)=t(x_0,x_1,\dots,x_n)$ whenever $x_0,\dots,x_n$ are elements of any commutative semigroup.