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.

  • 2
    $\begingroup$ You can start by noticing $f$ and $g$ do satisfy $f^b=g^a$. $\endgroup$
    – orion
    Nov 16, 2016 at 18:12
  • $\begingroup$ This question already has an answer here: math.stackexchange.com/questions/2017106/… $\endgroup$
    – IV_
    Apr 1, 2018 at 15:46
  • $\begingroup$ Algebraic independence over $\mathbb{C}$ does not follow from linear independence over $\mathbb{C}$. $\endgroup$
    – IV_
    Jul 11, 2020 at 17:18

1 Answer 1


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.


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