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!