# Dimension of variety $V(I)$ where $I=\langle xy,xz \rangle\in k[x,y,z]$ and $k$ is an infinite field.

I have some doubts about this question although I know the answer is $2$. The dimension of an affine variety is defined in many different ways. One of the definitions defines it as the maximum dimension of the coordinate subspaces if it is a union of a finite number of linear subspaces of $k^n$. It is very easy to see from this definition that the dimension of $V(I)$ is $2$.

Now the problem I was doing in the book Ideals, Varieties, and Algorithms by Cox et al. asks the following:

(a) Show that $I\cap k[x]=0$, but that $I\cap k[x,y]$ and $I\cap k[x,z]$ are not $0$.

(b) Show that $I\cap k[y,z]=0$, but that $I\cap k[x,y,z]\ne 0$

(c) What do you conclude about the dimension of $V(I)$?

Using Groebner basis, it is easy to show (a) and (b). I am confused about (c) though. It looks as if the author wants to use the following definition to conclude that $\dim (V(I))=2$:

The dimension of $V$ is the largest integer $r$ for which there exist $r$ variables $x_{i_1},\dots, x_{i_r}$ such that $I(V)\cap k[x_{i_1}, \dots,x_{i_r}]=\{0\}$.

I don't see how to use this definition in this case, since our $I$ is an arbitrary ideal defining $V$ instead of $I(V)$. Also $k$ is not necessarily algebraically closed. So what does the question (c) mean?

Thank you for any help!

1. The dimension doesn't change when replace $k$ by its algebraic closure.
2. $I$ is a radical ideal, so $I(V(I))=I$.