Finding an affine cover of a certain surface in $\mathbb{A}_k^4$ (11.2.D vakil) In 11.2.D in http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf, a surface $S\subset\mathbb{A}_k^4$ is defined as cut out by the equations $wz-xy=0$, $wy-x^2=0$, and $xz-y^2=0$. In computing the dimension of the surface, I feel like we need to find an affine open cover and then compute the dimension of each piece of the cover (so that 11.1.B can be applied). One can easily choose pieces $D(w)$, $D(x)$, $D(z)$, $D(y)$, but these seem not to cover the point $(x,y,z,w)$. For this remaining point, what affine open should be chosen? I feel like something like $D(x-1)$ could be used, but this feels a bit contrived.
Is there an easier affine to use, and is this even the right approach to the problem?

Your proposed cover is fine: for varieties over fields, if $V$ a closed subset and $U$ an open subset of some ambient space, we have that $\dim V \leq \max(\dim V\cap U, \dim U^c)$. Why? Write $V$ as a finite union of irreducible components $V_i$. Then $\dim V = \max_i \dim V_i$, and either:

*

*$V_i\cap U$ is a dense open subset of $V_i$ which has the same dimension

*$V_i\subset U^c$, so $\dim V_i\leq \dim U^c$.

Combining the two, we have the stated result.
You could also prove directly that your variety is the image of $\Bbb A^2\to\Bbb A^4$ under the map $(s,t)\mapsto (s^3,s^2t,st^2,t^3)$. (Really, what's going on here is you're computing the dimension of the cone on the twisted cubic.)
