3
$\begingroup$

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!

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

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

$\endgroup$
  • $\begingroup$ Thank you again. Your comment 1 is not in this book though. Can you give me a reference? $\endgroup$ – KittyL Jan 10 '16 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.