# Solving dimension of an affine algebraic variety.

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!

• 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. Oct 18, 2020 at 0:04
• Would there be a more 'down to earth' approach? I am reading chapter 1 of Gathmann's notes for algebraic geometry. Oct 18, 2020 at 0:06
• Are you more comfortable showing e.g. that a given containment between prime ideals has no other primes that can go "in between"? Oct 18, 2020 at 1:08
• Yes, I am comfortable with that method, which I couldn't succeed. Oct 18, 2020 at 1:09