Real numbers with addition, multiplication, and a transcendental constant

Question

Consider the structure $(\mathbb R, +, *, r)$, where $+$ is addition, $*$ is multiplication, and $r$ is a transcendental real number. Are the associative and commutative properties for addition and multiplication, as well as the distributive property connecting the two, already sufficient to derive all the universally valid equations in that structure?

Answer 1

Hint:

Because $+$ and $\times$ are continuous and $\Bbb{Q}$ is dense in $\Bbb{R}$, an equation holds in $(\Bbb{R}, +, \times, r)$ iff it holds in the substructure $(\Bbb{Q}(r), +, \times, r)$ (where $\Bbb{Q}[r]$ is the subfield of $\Bbb{R}$ generated by $r$). Because $r$ is transcendental, $\Bbb{Q}[r]$ iff it holds in the polynomial ring $\Bbb{Q}[X]$. Consequently, an equation is universally valid from some specific transcendental $r$ (like $\pi$ or $e$) iff it is universally valid for arbitrary $r$. Hence you can drop the constant $r$ from the signature: an equation involving $r$ is universally valid iff the same equation with $r$ replaced by a fresh variable $X$ is universally valid (to prove this use continuity and the density of the transcendentals).

Using associativity, commutativity and distributivity, you can express any equation in the form $p(X_1, \ldots, X_n) = q(X_1, \ldots X_n)$ where $p$ and $q$ are polynomials with natural number coefficients. Now show that such an equation is universally valid iff $p$ and $q$ are identical. To see this, allow yourself to use negation and write $f = p - q$, so that $f$ is a polynomial in $X_1, \ldots X_n$ with integer coefficients and $p$ and $q$ have identical coefficients iff the coefficients of $f$ all vanish. Now prove by induction on $n$ that the coefficients of $f$ all vanish iff $f(x_1, \ldots, x_n) = 0$ for every tuple $(x_1, \ldots, x_n) \in \Bbb{R}^n$.

This gives you a decision procedure: take your equation, multiply out and cancel like terms. The equation is valid iff all the coefficients can be cancelled.