# The composition of an algebraic function and a transcendental function

I say that a function $$f : \mathbb{R} \to \mathbb{R}$$ is algebraic if it is a solution of a polynomial equation, that is there exists a polynomial $$F(x,y)$$ such that $$F(x,f(x)) =0$$ for every $$x$$. I say that a function is transcendental if it is not algebraic. This should be consistent with the standard terminology.

Now, is it true that if I precompose a transcendental function with a non-constant algebraic function, the result is transcendental? If not true, is something similar true?

In symbols, given $$g : \mathbb{R} \to \mathbb{R}$$ transcendental, $$f : \mathbb{R} \to \mathbb{R}$$ algebraic and non-constant, is $$g \circ f$$ transcendental?

If needed/helpful, we can put some hypothesis on $$g$$; for instance, it can be taken to be $$C^\infty$$, even analytic everywhere except a finite set of points. Also, maybe it helps if the domains are not the whole $$\mathbb{R}$$: I would expect everything works in the same way if instead of $$\mathbb{R}$$ one takes some connected interval (or maybe some connected open set in $$\mathbb{C}$$?).

(I am not sure what kind of tags to put on this questions -- suggestions are welcome)

• For $C^\infty$ the answer is false: take $f(x) = x^2$, then you can take $g$ algebraic on $[0, \infty)$ and whatever you want on $(-\infty, -1]$ (and something smooth to connect). For analytic $g$, everything can be extended to almost $\mathbb C$, and then, as $(g \circ f)\circ f^{-1} = g$ is transecdental, $g \circ f$ can't be analytic (not sure if we can hand-wave around multi-value functions here, but probably can). Jun 7, 2022 at 16:33
• Of course! I am happy to consider a local inverse of $f$, which is indeed algebraic where it is defined. Jun 7, 2022 at 20:12
• Maybe you are asking whether "algebraic" can be a "local" property? Jun 10, 2022 at 23:09

Is it true that the result is transcendental if I precompose a transcendental function with a non-constant algebraic function?

Yes, it is true.

Statement:
Let
$$X,Y\subseteq\mathbb{R}$$ (or $$\mathbb{C}$$),
$$A\colon X\to Y$$ a bijective algebraic function,
$$T\colon Y\to\mathbb{C}$$ a transcendental function.
Then the function $$F=T\circ A$$ is transcendental.

Proof:
A function in $$\mathbb{R}$$ is a function $$D\subseteq\mathbb{R}\to\mathbb{R}$$.
A function in $$\mathbb{C}$$ is a function $$D\subseteq\mathbb{C}\to\mathbb{C}$$.
Let $$^{-1}$$ denote the inverse function.
Assume $$T\circ A$$ is algebraic. Then there is an algebraic function $$A_1$$ of one variable so that $$\forall x\in X\colon\ T(A(x))=A_1(x)$$.
$$\text{for all}\ y\in Y:$$ $$T(A(A^{-1}(y)))=A_1(A^{-1}(y))$$ $$T(y)=A_1(A^{-1}(y))\tag{1}$$ Because the bijective algebraic functions in $$\mathbb{R}$$ (or $$\mathbb{C}$$) are closed under inversion and under composition, $$A_1\circ A^{-1}$$ is an algebraic function. But $$T$$ is a transcendental function. So (1) is a contradiction, and therefore the assumption is wrong. This proves the statement.
q.e.d.

If $$A$$ is not injective, you can decompose it into different pieces of injective functions (e.g. the inverses of the partial inverses of $$A$$) and apply the statement.