Let $H = Z(xy-z^2) \subset \Bbb A ^ {3}$.
And let $L = Z(x,z)$.
I need to show that $L \subset H$ , that $dim(L) = dim (H) -1$ and that the ideal of regular functions on $H$ which vanishing on $L$ is not a principal ideal.
So showing $L\subset H$ is easy, and as $xy-z^2$ is irreducible, we have $dim(L) = dim(H) -1$ by principal ideal theorem.
Im not sure how to show the last part.
The regular functions on $H$ are $k[x,y,z]/(xy-z^2)$ and the ideal of regular functions vanishing on $L$ is $(x,z)$ so I need to show $k[x,y,z]/(xy-z^2) \cap (x,z)$ is not principal.
Any ideas how?
One approach is to consider the grading.
$$\require{cancel}(x,z)/(xy - z^2) = k\{x,z\} + k\{x^2,xy,xz,yz\} + k\{x^2,x^2y,x^2z,xyz,xy^2,y^2z\} +\cdots$$
So the dimensions are $2,4,6,\dots$.
On the other hand, what does the grading look like for $(f)/(xy-z^2)$? It would be
$$ k\{f\}+ \frac{k\{xf,yf,zf\}}{z^2 \sim xy} + \cdots $$
Ok, so we can't really get an explicit basis for the higher degree terms but we can see an issue with $k\{f\}$ as this is one dimensional.