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?

  • $\begingroup$ 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. $\endgroup$ – Robert Shore Sep 23 '19 at 21:24
  • $\begingroup$ I mean a function including polynomials as well as any other kind of combination under +-*/ of solutions to polynomial equations with rational coefficients. $\endgroup$ – TeXnichal Sep 23 '19 at 22:00
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    $\begingroup$ That rational functions are closed under all of those operations, so they "cover the waterfront." $\endgroup$ – Robert Shore Sep 23 '19 at 22:06
  • $\begingroup$ You should state that the algebraic function is not a constant function otherwise it may be false. $\endgroup$ – 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 $p$.

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