3
$\begingroup$

By dimension, I used this definition: algebraic set $V$ has dimension d if maximum length of chains $V_0\subset V_1 \subset\cdots\subset V_d$ is $d$ where $V_i's$ are irreducible subvariety of $V$, and all of them are distinct.

Let $k$ be an algebraicly closed field, and consider $\mathbb{A}^3$. Let $X=Z(y-x^2,z-x^2)$. I proved that $X$ is an affine variety by showing that $(y-x^2, z-x^2)$ is a prime ideal in $k[x,y,z]$. However, I am struggling to show $X$ has dimension $1.$ Intuitively, it makes sense because $X$ is just a curve, but I don't know how to prove it. More specifically, how can I show that there is no irreducible subvariety between a point and the curve itself?

Thanks!

$\endgroup$
4
  • $\begingroup$ One approach would be to use the theorem linking the transcendence degree of the field of rational functions to dimension. It is easily shown here that 2 of the three variables are algebraic over the remaining 1. $\endgroup$ Oct 18, 2020 at 0:04
  • $\begingroup$ Would there be a more 'down to earth' approach? I am reading chapter 1 of Gathmann's notes for algebraic geometry. $\endgroup$ Oct 18, 2020 at 0:06
  • $\begingroup$ Are you more comfortable showing e.g. that a given containment between prime ideals has no other primes that can go "in between"? $\endgroup$ Oct 18, 2020 at 1:08
  • $\begingroup$ Yes, I am comfortable with that method, which I couldn't succeed. $\endgroup$ Oct 18, 2020 at 1:09

0

You must log in to answer this question.