# Exceptional Set and Schanuel's conjecture

I was reading an article about transcendental funtions (Algebraic values of transcendental functions at algebraic points, by Huang, J., Marques, D., Mereb, M.).

The authors gave an example that says:

"Assuming Schanuel's conjecture to be true, it is easy to prove that if $f(z) = \sin(\pi z)e^z$, $g(z) = 2^{3^z}$ and $h(z)=2^{2^{2^{z-1}}}$, then $S_f=S_g=\mathbb{Z}$ and $S_h=\mathbb{N}$."

Well, it certainly wasn't easy for me hehe

First, one definition: We define the exceptional set of a function $f: D \subseteq \mathbb{C} \to \mathbb{C}$ as $S_f=\{\alpha \in D\cap\overline{\mathbb{Q}}: f(\alpha) \in \overline{\mathbb{Q}}\}$.

Schanuel's conjecture:

If $x_1,\ldots, x_n \in \mathbb{C}$ are linearly independent over $\mathbb{Q}$, then $$grtr(\mathbb{Q}(x_1,\ldots,x_n,e^{x_1},\ldots,e^{x_n})|\mathbb{Q})\geq n,$$ where $grtr(\mathbb{Q}(x_1,\ldots,x_n,e^{x_1},\ldots,e^{x_n})|\mathbb{Q})$ means of transcendental degree of $\mathbb{Q}(x_1,\ldots,x_n,e^{x_1},\ldots,e^{x_n})|\mathbb{Q}$.

(grtr is how we denote it in portuguese, I don't know if "dgtr" would make any sense).

The fact is I don't understand the conjecture I think. What does $grtr(\mathbb{Q}(x_1,\ldots,x_n,e^{x_1},\ldots,e^{x_n})|\mathbb{Q})\geq n$ imply? I think that if I understood the conjecture I would understand the exemples.

So, I was hoping to get some help so I could understand the conjecture :)

Thanks!

In the US we use $\text{trdeg}$ for transcendence degree.

The notation means that among the $2n$ numbers in the list $x_1,\ldots x_n, e^{x_1},\ldots e^{x_n}$, there are at least $n$ of them algebraically independent of one another, i.e. there are at least $n$ of them among which there are no rational polynomials which vanish on those $n$, i.e. $\exists n$ numbers from among those $2n$ such that $\forall p(t_1,\ldots, t_n)\in \Bbb Q[t_1,\ldots, t_n]$,

$$p(x_1,\ldots, x_n)\ne 0.$$

Are you looking for more on those specific examples or just the clarification?

• I think I understood, but I still don't know how to use this to find the excptional set of the functions $f$, $g$ and $h$. Could you explain that? For just one of the functions.. The other two I try by myself..
– Anna
Commented Aug 7, 2014 at 19:21

Schanuel's conjecture, if proven, would generalize most known results in transcendental number theory.

$$2^{3^z}=2^{(3^z)}$$:
Assuming Schanuel's conjecture to be true, Baker's conjecture follows and further Gelfond–Schneider theorem follows.
Let $$z$$ be algebraic. Apply Gelfond-Schneider theorem:
If $$z$$ would be algebraic and not rational, then $$3^z$$ would be transcendental. Because $$3^z$$ would be transcendental, $$2^{(3^z)}$$ would be transcendental.
If $$z$$ is rational, then $$3^z$$ is algebraic. If $$3^z$$ would be algebraic and not rational, then $$2^{3^z}$$ would be transcendental. If $$3^z$$ is rational, then $$2^{3^z}$$ is algebraic.
If $$z$$ is rational and not integer, then $$3^z$$ is not rational. If $$z$$ is integer, then $$3^z$$ is rational.
Therefore $$z$$ must be integer.

$$2^{2^{2^{z-1}}}=2^{(2^{(2^{z-1})})}$$:
Assuming Schanuel's conjecture to be true, Baker's conjecture follows and further Gelfond–Schneider theorem follows.
Let $$z$$ be algebraic. Apply Gelfond-Schneider theorem:
If $$z$$ would be algebraic and not rational, then $$2^{z-1}$$ would be transcendental. Because $$2^{z-1}$$ would be transcendental, $$2^{(2^{z-1})}$$ would be transcendental. Because $$2^{(2^{z-1})}$$ would be transcendental, $$2^{(2^{(2^{z-1})})}$$ would be transcendental.
If $$z$$ is rational, then $$2^{z-1}$$ is algebraic. If $$2^{z-1}$$ would be algebraic and not rational, then $$2^{(2^{z-1})}$$ would be transcendental. Because $$2^{(2^{z-1})}$$ would be transcendental, $$2^{(2^{(2^{z-1})})}$$ would be transcendental.
If $$2^{z-1}$$ is rational, then $$2^{(2^{z-1})}$$ is algebraic. If $$2^{(2^{z-1})}$$ would be algebraic and not rational, then $$2^{(2^{(2^{z-1})})}$$ would be transcendental.
If $$2^{(2^{z-1})}$$ is rational, then $$2^{(2^{(2^{z-1})})}$$ is algebraic.
If $$2^{(2^{z-1})}$$ would be algebraic and not rational, then $$2^{(2^{(2^{z-1})})}$$ would be transcendental.
If $$2^{(2^{z-1})}$$ is rational and not integer, then $$2^{(2^{(2^{z-1})})}$$ is not rational. If $$2^{(2^{z-1})}$$ is integer, then $$2^{(2^{(2^{z-1})})}$$ is rational.
$$2^{(2^{z-1})}$$ is integer iff $$z$$ is a nonzero natural number. Therefore $$z$$ must be a nonzero natural number.