# 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 Answer

1. The dimension doesn't change when replace $k$ by its algebraic closure.

2. $I$ is a radical ideal, so $I(V(I))=I$.

• Thank you again. Your comment 1 is not in this book though. Can you give me a reference? Jan 10, 2016 at 22:20