I have a question about a step in the following proof:
To show that $y-e^x +1 =0$ cannot be written as the solution to a system of polynomial equations $F_1=F_2=...=F_n=0$, first note that any such polynomial must vanish at the points $2 \pi i n$ for integer n (I guess this means the points ( $2 \pi i n,0$) ) And so by Bezout, each $F_i$ must be divisible by x, contradiction. Now this last step I don't see. By Bezout, since there is an infinite number of intersections between the polynomials and the curve, there is an infinite number of intersections between each of the polynomials, so they have a common factor. Now why is it x? And then where is the contradiction? Does this imply that x also divides $y-e^x +1 =0$?
Here are two approaches. Assume the $F_i$ are distinct.
An easy approach is to note that if $$y-e^x+1=0$$
were defined by the vanishing of $F_i(x,y)=0$, then $F_i(x,0)$ is a polynomial is $x$ with infinitely many zeros, and hence must be identically 0. Thus each $F_i$ must vanish at $y=0$, i.e. $F_i(x,y)=yG_i(x,y)$ for some polynomials $G_i(x,y)$. But then the set $\{y=0\}$ is in the vanishing set of the curve, which it is not as $1-e^x\neq 0$ for all $x$.
Alternatively, suppose $F_i(x,y)$ has degree $d_i$. $F_i(x,y)=0$ defines a plane curve. But by Bezouts Theorem two plane curves of degree $d,e$ intersect in at most $de$ points, or are divisible by a common factor .
You have infinitely many common zeroes, and hence the $F_i$ must all have a common factor $F_0$. But then the vanishing set of $F_0$ must be a subset the vanishing set of $y-e^x+1$, but it's not hard to see that this is irreducible (since $y$ is uniquely determined by $x$)