Well one of the key thing seemed was (on a cursory look) to take the grading of $k[x,y,z]$ as $p,q,r$ respectively and then have the ideal generated by just homogenous elements, and then you start looking at various $\Bbb{N}$ linear combinations which come up as homogenous elements
But these intersection problems are weird problems. I think it is an open problem to show that every space curve in $\Bbb{P}^3$ is the intersection of two surfaces
@kunalCh. Yes. For example, cycling can happen, in which one gets stuck at a vertex (bfs) in the polyhedron but the entering and exiting variables keep permuting around.
In any case, even without cycling, I don't think reappearance of the same basic variable is an issue.
In this video —> (youtu.be/aMLl6jUlpqA), around 5:22, the author says that y hat spans a 2D vector space by the vector equation $\hat{Y}=a(1,2,4)+b$ for a scalar $a$ and a constant vector $b$. Does this equation really span a plane?
let $B \le A$ be finite abelian groups. how do I show that every irreducible character $\phi: B \to \mathbb{C}$ lifts to precisely $[A:B]$ irreducible characters $\psi: A \to \mathbb{C}$?
I started a math blog once but I have since not posted anything :L
I always want to sit there and polish all my expository articles, and I do this in LaTeX since it's a lot more flexible than mathjax or katex in html I find.
@schn $\operatorname{sp} \{a(1,2,4)+b\}_{a \in \mathbb{R}}$ spans at most a 2 dimensional subspace. (It is not hard to show that $\operatorname{sp} \{a(1,2,4)+b\}_{a \in \mathbb{R}} = \operatorname{sp} \{(1,2,4),b\}$.)
linear span in the legalese :-)
but really, you would have to figure out what the youtube person meant by span