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 $K$?
On
An algebraic curve over $\mathbb{Q}$ is, by definition :
a set of points on the Euclidean plane whose coordinates are zeros of some polynomial in two variables with coefficients in $\mathbb{Q}$, i.e are algebraic numbers.
but $y=e^1=e$ is not an algebraic number (this can be proved as a consequence of the Lindemann–Weierstrass theorem). so the coordinates of the point $(1,e)$ are not solutions of a polynomial equation with rational coefficients.
Added.
The above definition is valid for an algebraic curve over $\mathbb{Q}$, but can be extended to any field $K$:
An algebraic (plane) curve is the graph of an equation $f(x,y)=0$ where $f(x,y)$ is a polynomial with coefficients in $K$.
The equation $y-2^x=0$ cannot be write as a polynomial.
On
As I read in your post you accept that $y=e^x$ is not algebraic curve over $\mathbb{C}$. So assume that $y=e^x$ is algebraic over $\mathbb{R}$. That is to say that there is a non constant polynomial $p \in \mathbb{R}[X,Y]$ such that $$p(x,e^x) = 0$$ for all $x \in \mathbb{R}$. Set $f(z) := p(z,e^z)$. Then $f : \mathbb{C} \to \mathbb{C}$ is a holomorphic function i.e. an entire function. The restriction of $f|_{\mathbb{R}}$ is zero. So by the identity principle $f$ is zero for all complex number. Then $$p(z,e^{z}) = 0$$ for all complex numbers hence $y=e^x$ is algebraic over the complex numbers.
On
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.
Let $b > 1$ be a real number. If the function $y = b^{x}$ were algebraic, there would exist a positive integer $N$ and one-variable polynomial functions $p_{k}$, with $0 \leq k \leq N$ and $p_{N}$ is not identically zero, such that the polynomial $$ f(x, y) = \sum_{k=0}^{N} p_{k}(x) y^{k} $$ has zero set equal to the graph $y = b^{x}$. But this would imply $$ 0 = \sum_{k = 0}^{N} p_{k}(x) b^{kx} = b^{Nx} \sum_{k = 0}^{N} p_{k}(x) b^{(k - N) x} \quad \text{for all real $x$.} \tag{1} $$ Each term in the rightmost sum is a polynomial in $x$ multiplied by an exponential function with non-positive exponent. As $x \to \infty$, each term with $k < N$ approaches $0$. Since the term $p_{N}(x)$ is not identically $0$, equation (1) is false no matter how the $p_{k}$ are chosen.