What is a basis for the common equational identities of $(\mathbb{R};*,0)$, $(\mathbb{R};*,1)$, and $(\mathbb{R};*,-1)$?

Question

Consider the three algebraic structures $(\mathbb{R};*,0)$, $(\mathbb{R};*,1)$, and $(\mathbb{R};*,-1)$, where $*$ denotes multiplication. What is a basis for the equational identities those structures have in common? I conjecture that these identities are enough: $x*y=y*x$, $x*(y*z)=(x*y)*z$, and $r*(r*r)=r$, where $r$ is the constant symbol. Is this true, or do we need more?

Answer 1

Yes, this is correct.

Via the associative and commutative identities, any term can be thought of as having the form $$\left(\prod_{i\in I}x_i^{a_i}\right)\cdot r^b$$ for some finite set of variable indices $I$, some choices of positive integer powers $a_i$ ($i\in I$), and some nonnegative integer power $b$.

Now suppose I have two terms of the above form, $$t=\left(\prod_{i\in I}x_i^{a_i}\right)\cdot r^b\quad\mbox{and}\quad t'=\left(\prod_{i\in I'}x_i^{a_i'}\right)\cdot r^{b'}.$$ It's easy to see that $t=t'$ holds in $(\mathbb{R};*,-1)$, and this latter condition holds iff we have $I=I'$ and $a_i=a_i'$ for all $i\in I$ and $b\equiv_{mod\, 2} b'$ (apologies for nonstandard notation, but the standard notation is quite inconvenient in my opinion). Meanwhile, if exactly one of $b,b'$ is zero, then $t=t'$ fails in $(\mathbb{R};*,0)$, and if both $b$ and $b'$ are nonzero then $t=t'$ holds in $(\mathbb{R};*,0)$ if $t=t'$ holds in $(\mathbb{R};*,-1)$.

Putting all of the above together, the common equational theory of the three structures in question consists of the commutativity and associativity equations plus all equations of the form $t=t'$ for $t,t'$ as above with either $b=b'=0$ or $b\not=0\not=b'$. Since this can be deduced from the theory you've written (exercise), your guess is correct.