Definition: Let $X$ be an algebraic set and let $K(X)$ denote the rational function field of $X$. If $X$ is irreducible, then the dimension of $X$ is defined to be the transcendence degree of $K(X)$ over $F$. If $X$ is reducible, then the dimension of $X$ is defined to be the maximum dimension of its irreducible components.
Question: Suppose $F$ is algebraically closed. I want to find the dimension of the algebraic set $X=V(y-x^2,y^2-y+z^2)$.
Attempt: According to the definition, the first step should be determine whether $X$ is irreducible or not. I have tried to find a polynomial isomorphism between the field $F$ and $X$ but that didn't work. So I thought maybe I should try a rational map. Say $x=\frac{1-t^2}{1+t^2}$, $y=x^2=\frac{1-t^2}{1+t^2}$ and $z=x \frac{2t}{1+t^2}=\frac{(1-t^2)(2t)}{1+t^2}$. But is this rational map an isomorphism? If it is, then I will get $\dim X=\dim F=1$.
Can anyone help me with this? Thank you !
On $X$, $y=x^2$ and so $z^2=y-y^2=x^2(1-x^2)$ and so $(z/x)^2=1-x^2$ (as long as $x\ne0$). If we set $z=xu$ then $u^2+x^2=1$ which is an irreducible conic (if we are not in characteristic $2$). When $x=0$ we have just the point $(0,0,0)$. So the highest-dimension irreducible component of $X$ is a curve; do $X$ has dimension $1$.