1

An algebraic function in one variable $x$, say $y=f(x)$, is, by definition, a solution of the polynomial equation $$\sum_{k=0}^n a_k(x)y^k=0$$ where $a_k$'s are polynomial functions with integer coefficients.

A rooty function can, by definition, be written as an expression using a finite number of terms using only addition, subtraction, mulplication, division and raising to a rational power. Transcendental numbers are not allowed in rooty functions.

For example, $$f(x)=\frac{\sqrt{1+x^3}}{x^{3/7}-\sqrt{7}x^{1/3}}$$ is a rooty function.

Question: Is every rooty function an algebraic function? If so, then it is not at all obvious to me (because of the fractional powers). The converse of the question is known to be false (by the Abel—Ruffini theorem).

Nomas2
  • 483
  • 2
    It may be possible to adapt the proofs that if $a$ and $b$ are algebraic numbers, then $a+b$, $ab$, and $a/b$ ($b \neq 0$) are algebraic. – aschepler Oct 06 '23 at 16:45
  • @aschepler But how would you "get rid of" the fractional powers and construct a polynomial equation with a given rooty function as a solution? – Nomas2 Oct 06 '23 at 16:46
  • That $a$ algebraic implies $a^{1/n}$ is algebraic is an easier one: if $P(a)=0$ then $a^{1/n}$ is a root of $P(x^n)=0$. – aschepler Oct 06 '23 at 16:48
  • 1
    These are also known by the term "explicit algebraic function", although for these the coefficients can be any real numbers (not just rational numbers). Aside from domain issues that probably need to be addressed or shoved under the rug (see here or here), I believe the same method one uses to show that any explicit algebraic number is an algebraic number works. – Dave L. Renfro Oct 06 '23 at 16:56
  • @DaveL.Renfro Richman focuses only on the case when the coefficients are in $\mathbb{R}[x]$. Does his proof (of Theorem 3 and Corollary 4) work also for the case when the coefficients are in $\mathbb{C}[x]$? – Nomas2 Oct 06 '23 at 17:48
  • I haven't looked at Richman's paper in many years and don't have time now to dig into it, so I don't know to what extent his proofs might continue to hold for (or are easily modified to allow for) complex coefficients, but I would guess that the results themselves are still valid. – Dave L. Renfro Oct 06 '23 at 18:10

1 Answers1

2

Let's do this inductively, on induction by the maximal "depth" of nested radicals. So for example $\frac{1}{2+X}$ has depth $0$, $\sqrt{42x^2-5}$ has depth $1$ etc. (I really don't want to define this formally, I think the formal way to define this uses structural induction.) If it's unclear what I mean by depth, please leave a comment.

So in depth $0$, there is nothing to prove.

Assume that for all depth $n$ functions for a fixed $n$, we have proved the statement. Let $f$ be a rooty function of depth $n+1$. Then, choosing a common denominator for the outermost roots, we may write $f$ as $f(X)=g(h(X)^{1/r})$, where $g$ is a rational function and $h(X)$ has depth at $n$. Then because $g$ is a rational function, to show that $f$ is algebraic, it suffices to show that $h(X)^{1/r}$ is algebraic (because sums, quotients, products... of algebraic functions are again algebraic). But by the inductive hypothesis, $h$ is algebraic, so we only need to show that the $r$-th root of an algebraic function is algebraic. This is easy, as was remarked in the comments.

Lukas Heger
  • 20,801
  • I'm not sure that your 'depth' property is (yet) well-defined since one number may have expressions with different degrees of nesting, eg, $\sqrt[3]{7+5\sqrt{2}}=1+\sqrt{2}$ is nested twice on the LHS and once on the RHS. So your 'depth' presupposes some least-possible degree of nesting. Is that definitely valid? – Jam Oct 06 '23 at 17:22
  • 1
    @Jam yes, the minimum over all equivalent expressions for the function, should work. Maybe one needs to modify the inductive step slightly, so that we assume that all rooty functions of depth $\leq n$ are proven to be algebraic – Lukas Heger Oct 06 '23 at 17:26