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.

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.