TITLE: Why $y=e^x$ is not an algebraic curve?
QUESTION [5 upvotes]: Why $y=e^x$ is not an algebraic curve over $\mathbb R$? I can say that is not a algebraic curve over $\mathbb C$ because $e^x$ is a periodic function, but what about $\mathbb R$?
EDIT:
I don't want to use trascendence of $e$. Or, I can ask this question for $y=2^x$.
UPDATE:
Can we just say that algebraic curve over $K$ is also alegraic over any extension of $K$?

REPLY [5 votes]: Suppose $x$ and $e^x$ satisfy a polynomial equation $f(x,e^x)=0$ where $f(x,y)$ has minimal degree in $y$.
Write $f(x,y)=p(x)y^n+g(x,y)$, where $p(x)y^n$ is the leading term in $y$.
Differentiate $p(x)e^{nx}+g(x,e^x)=0$ and get $np(x)e^{nx}+p'(x)e^{nx}+h(x,e^x)=0$, for some $h$.
Subtract $n$ times the first equation from the second and get $p'(x)e^{nx}+h(x,e^x)-ng(x,e^x)=0$.
This equation has a leading term $p'(x)e^{nx}$ of smaller degree in $x$.
We can repeat this process until we remove the term in $y^n$, which is a contradiction.
The same proof works for $b^x$ with some $\log b$ factors that can be absorbed.