Why the right branch of an hyperbole is not an algebraic curve? 
Prove that the subset of $\mathbb{R}^2$ defines by $X^2-Y^2=1$, $X>0$ is not an algebraic curve. 

It's clear that the hyperbole is an algebraic curve because I can define the set 
$$V(F)=\{(X,Y)\in\mathbb{R}^2 : F(X, Y)= X^2-Y^2-1=0\}.$$
However, I don't know why the right branch is not an algebraic curve. Any help would be aprecciate.
 A: Suppose your half-hyperbola (that I will call $C$) is a curve defined over some finitely generated field $K$ and let $\sigma$ be a field automorphism of $\Bbb R$ fixing $K$.
Then $C$ is fixed by $\sigma$. Since $\pi_1(C) = [1 ; \infty)$, $\sigma$ must induce a permutation of $[1 ; \infty)$.
In particular, $\forall x \in \Bbb R, x\ge 1 \iff \sigma(x)\ge 1$. Since $\sigma$ is also a field morphism, it fixes $\Bbb Q$ and so for any rational $r$, by replacing $x$ with $x+1-r$ and doing a few manipulations, you get that $\forall x \in \Bbb R, x \ge r \iff \sigma(x) \ge r$.
But since every real number is entirely determined by which rational numbers are lesser (or greater) than it, you deduce from there that $\sigma$ has to be the identity on $\Bbb R$.
So, any field automorphism fixing $K$ has to fix $\Bbb R$, and therefore $K=\Bbb R$
.
This contradicts the fact that $K$ is finitely generated (you can't shove an uncountable number of coefficients in a polynomial equation !)
And so $C$ cannot be an algebraic curve.

You can turn this around and pick a point on $(x,y) \in C$ transcendant over $K$, then there is a $K$-automorphism sending $x$ to $-x$, and so $(-x,y)$ and $(-x,-y)$ also satisfy the equation and have to be in $C$ but they are on the left branch.
A: I will use the fact that the curve has a rational parametrization. Every point on the curve except $(1,0)$ can be written as 
$$(x,y) = \left( \frac{1+t^2 }{1-t^2}, \frac{2 t}{1-t^2}\right) $$ for a unique $t \in \mathbb{R} \backslash\{\pm 1\}$. The part with $x >0$ corresponds to $t \in (-1, 1)$. 
Say we have a polynomial $P$ so that $P( \frac{1+t^2 }{1-t^2}, \frac{2 t}{1-t^2}) = 0$ for all $t \in (-1,1)$. Then $P( \frac{1+t^2 }{1-t^2}, \frac{2 t}{1-t^2}) = 0$ for all $t \in \mathbb{R} \backslash \{\pm 1\}$. That is, if $P$ is $0$ on the one branch of the hyperbola, it is $0$ on the hyperbola. Therefore, the positive branch is not an algebraic curve. 
Note: We use the existence of a rational parametrization. In fact, even for other curves this is true: a piece of a curve is not algebraic. For instance, take $y^2 = x(x+1)(x+2)$. The positive branch is not an algebraic curve again. One needs to use an analytic parametrization in this case. 
