Is the induced map $V(I) \longrightarrow V(J)$ is surjective for $I=(xy-1,y^2-z)\subset\Bbb C[x,y,z]$ and $J=I\cap \Bbb C[y,z]$?

Question

Let $\mathbb{C}[x, y, z]$ be a polynomial ring, and set $I = (xy - 1, y^2 - z)$. Compute $J = I \cap \mathbb{C}[y, z]$. Is the induced map $V(I) \longrightarrow V(J)$ surjective?

I have trouble deciding and showing the second part, i.e. if $V(I) \longrightarrow V(J)$ is surjective or not. I don't know how to begin even.

For the first part, we first compute the Groebner basis of $I$ with regards to the lexicographic ordering $x > y > z$. We have $G := \{xy - 1, y^2 - z, xz - y\}$. Then, $J$ is generated by the intersection of $G$ and $\mathbb{C}[y, z]$ so $J = (y^2 - z)$. (Hopefully this is correct.)

Could we maybe say since $V(I) = \{\mathfrak{p} \in \operatorname{Spec}(R) \mid I \subset \mathfrak{p}\}$ and $J \subset I$ that $V(J) \subset V(I)$ and hence the map must be surjective (but this seems silly)? Or do we calculate the dimension of the ideals? Thank you in advance.

Answer 1

You have a few options for checking surjectivity. The most basic one is to literally just describe a point in $V(J)$ and try and find a point in $V(I)$ which maps to it under the projection $\Bbb A^3\to\Bbb A^2$. For instance, suppose we take the point $y=1$, $z=1$ in $V(J)$ - then $(1,1,1)$ in $V(I)$ maps to it. Can you see where to go from here? Next step under the spoiler.

All points in $V(J)$ are of the form $(a,a^2)$. If $a\neq 0$ and we look at $(a,a^2)\in V(J)$, then $(a^{-1},a,a^2)\in V(I)$ is a preimage. What happens if $a=0$?