What does Lindemann-Weierstrass-Theorem imply?

Question

I try to understand the Theorem by Lindemann and Weierstrass: If $x,y$ are variables and I look at the field $\mathbf{Q}(x,y,exp(x),exp(y))$ what does L-W theorem imply on the transcendec degree of this field? Is it 4?

Answer 1

Yes, $\mathbb{Q}(x,y,\exp(x),\exp(y))$ has transcendence degree $4$ since $x$, $y$, $\exp(x)$, and $\exp(y)$ are algebraically independent over $\mathbb{Q}$. Indeed, suppose $f\in\mathbb{Q}[a,b,c,d]$ is such that $f(x,y,\exp(x),\exp(y))=0$. Then for any $\alpha,\beta\in\mathbb{C}$, $f(\alpha,\beta,e^\alpha,e^\beta)=0$.

In particular, if $\alpha$ and $\beta$ are algebraic and linearly independent over $\mathbb{Q}$, this gives an algebraic relation between $e^\alpha$ and $e^\beta$. By Lindemann-Weierstrass, this relation must be trivial. This means that if you consider $f$ as a polynomial in $c$ and $d$ with coefficients in $\mathbb{Q}[a,b]$, the coefficients all vanish when you plug in $\alpha$ and $\beta$ for $a$ and $b$. It is easy to deduce from this that the coefficients must be $0$ as polynomials in $a$ and $b$ (fix a value one of the variables, and then use the fact that a polynomial in one variable with infinitely many roots vanishes identically). Thus actually $f=0$.