# Algebraic functions on transcendental numbers

An algebraic function of an algebraic number yields an algebraic number, but does an algebraic function of a transcendental number always yield a transcendental number?

• If by "algebraic" function you mean a non-constant rational function (quotient of polynomials with rational coefficients), then the answer to your question is yes. – Robert Shore Sep 23 '19 at 21:24
• I mean a function including polynomials as well as any other kind of combination under +-*/ of solutions to polynomial equations with rational coefficients. – TeXnichal Sep 23 '19 at 22:00
• That rational functions are closed under all of those operations, so they "cover the waterfront." – Robert Shore Sep 23 '19 at 22:06
• You should state that the algebraic function is not a constant function otherwise it may be false. – Somos Sep 23 '19 at 22:48

Yes, this is true. If $$f$$ is an algebraic function, then it satisfies an irreducible polynomial $$p(x,f(x))=0$$ with algebraic coefficients. Plugging in $$x=t$$, our transcendental number, we see that $$p(t,f(t))=0$$. But this says that a transcendental number $$t$$ satisfies a polyonmial with algebraic coefficients. The only way for this to be true is if that polynomial is zero: $$p(x,f(t))=0$$ for any $$x$$. But this means that $$p(x,y)$$ is divisible by $$(y-f(t))$$, contradicting irreducibility of $$p$$.