# Algebraic independence of functions

Do you know of a neat proof of the algebraic independence (over $$\mathbb{C}$$) of the functions $$f(t) = e^{at}$$ and $$g(t)=e^{bt}$$ when $$a$$ and $$b$$ are linearly independent over $$\mathbb{Q}$$?

IMPORTANT: $$f$$ and $$g$$ are algebraic independent if for any nonzero polynomial $$P \in \mathbb{C}[x, y]$$, the function $$P(f(t),g(t))$$ is not the zero function.

• You can start by noticing $f$ and $g$ do satisfy $f^b=g^a$. Nov 16, 2016 at 18:12
– IV_
Apr 1, 2018 at 15:46
• Algebraic independence over $\mathbb{C}$ does not follow from linear independence over $\mathbb{C}$.
– IV_
Jul 11, 2020 at 17:18

Lemma. If $r_1,r_2,\ldots,r_k$ are distinct real numbers, then the functions $f_i(t)=e^{r_it}$ are linearly independent over $\Bbb{C}$.
Proof. Induction on $k$. Clear with $k=1$. If there is a linear dependence relation between the functions $f_i$ then by differentiating that relation w.r.t. $t$ we get another one. A suitable linear combination of those two relations then involves a smaller number of functions $f_i$, and the induction hypothesis kicks in. QED.
Assume that there exists a polynomial $$P(x,y)=\sum_{i,j} a_{i,j}x^iy^j\in\Bbb{C}[x,y]$$ such that $P(e^{at},e^{bt})=0$ for all $t\in\Bbb{R}$. Then we have $$0=P(e^{at},e^{bt})=\sum_{i,j} a_{i,j}e^{[ai+bj]t}\qquad(*)$$ for all $t$. Here $i,j$ are integers, so as $a$ and $b$ are linearly independent over $\Bbb{Q}$ the numbers $$r_{i,j}=ai+bj$$ are all distinct, i.e. different pairs of coefficients $(i,j)$ give distinct numbers $r_{i,j}$.
But, unless $P$ is the constant polynomial $0$, $(*)$ shows that the functions $e^{[r_{i,j}]t}$ are linearly independent over $\Bbb{C}$. This contradicts the Lemma, and proves your claim.