I know that if $r$ is a transcendental constant, then $(\mathbb{R},+,*,r)$ has no identities beyond commutativity, associativity, and distributivity. My intuition tells me that the converse is false. So, given an arbitrary algebraic constant $r$, what is an explicit identity or set of identities that generate the equational identities of that structure?


The converse is true. Basic answer: if $r$ is algebraic, then it is a solution of a polynomial equation $c_0 + c_1r + \cdots + c_nr^n = 0$ for $c_0, ... , c_n \in \mathbb{R}$, not all zero. So there's your relation.

More in depth: let $F$ be any field, including $\mathbb{R}$. Let $\Omega$ be a field containing $F$, and $r \in \Omega$. Define $F[r]$ to be the set of elements of $\Omega$ of the form $f(r)$, where $f$ is a polynomial with coefficients in $F$. There is a surjective ring homomorphism

$$F[X] \rightarrow F[r]$$

$$f(X) \mapsto f(r)$$

which is the identity on $F$.

To say that $r$ "satisfies no relations" is to say that this map is an isomorphism, which is the case if and only if $f(r)$ is never zero for any nonzero polynomial $f$, if and only if $r$ is transcendental.

Otherwise, $r$ is algebraic, and the above homomorphism has nontrivial kernel: general ring theory tells you that this kernel is generated by an irreducible polynomial $\mu \in F[X]$, and so you can view $F[r]$ as the polynomial ring $F[X]$ modulo the relations coming from the ideal $(\mu)$.

  • $\begingroup$ I am afraid this does not quite answer my question. I need an identity explicity within the signature $(+,*,r)$ For example, if $r$ was 1, we would need at least the identity $(x * 1) = x$. A natural follow-up question would be the same thing with regard to the signature $(+,-,*,r)$. $\endgroup$ – user107952 Jul 4 '17 at 17:38

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