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

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

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

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

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