My opinion is that the statement, for a real number $X$:
"$X$ is transcendental."
is decidable. The sketch of the "proof" should be the following. If transcendency or algebraicity of $X$ is undecidable in a theory O, then we could derive two new different theories A and B:
- X is algebraic in A, so a polynomial with integer coefficients $P(x)$ exists, and $X$ is a zero of it;
- X is transcendental in B, so no polynomial P(X) exists with X zero;
But we know that the set of the polynomials with integer coefficients is countable, so a complete list of such polynomials exists. So, how is it possible that in theory A a polynomial $P$ is present, but $P$ is NOT present in B?
In conclusion, because of the set of all polynomials with integer coefficients is countable, :
- OR such polynomial exists also in O (as well as in A), and then it is present also in B (and it should be a contradiction!), and X is algebraic,
- EITHER P does not exist at all, so it can not be present neither in A (another contradiction!), and X is transcendental.
So transcendence can not be considered undecidable, because its undecidability should lead to contradictions.
Is the sketch correct?