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Let $\mathbb{F}$ be a field (possibly algebraically closed) and let $V\subseteq\mathbb{F}^n$ be an affine variety. Now define the cone over $V$ as follows: $$\mathrm{Cone}(V):=\{\lambda\mathbf{p}: \mathbf{p}\in V, \lambda\in\mathbb{F}\}.$$ I want to show that $\mathrm{Cone}(V)\subseteq\mathbb{F}^n$ is also an affine variety, but I'm having trouble. Any advice would be appreciated.

Drew Armstrong
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    Are you sure about your definition of $\operatorname{Cone}(V)$ ? If $V={(x,y)\in\mathbb{F}^2~|~ x=1$ then (with your definition) $\operatorname{Cone}(V)={(\lambda,\lambda y)}=(\mathbb{A}^1\setminus{0})\times\mathbb{A}^1 \cup{(0,0)}$ which is not a variety. – Roland Nov 16 '20 at 17:08
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    Roland is asking, but I'm telling: your definition of the cone is wrong. You need to take the closure of the set you've mentioned in your definition. – KReiser Nov 16 '20 at 18:22
  • Aha! Thank you. Now I know why I was having a hard time proving this; it is false. – Drew Armstrong Nov 16 '20 at 18:37

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So an affine variety is the shared zero set among some collection of polynomials $\{f_1, \dots, f_m\}\subset \mathbb{F}[x_1, \dots, x_n]$. We can show that $Cone(V)$ is an affine variety in $\mathbb{F}^{n+1}$ by taking the $f_i$ and transforming them into homogenous polynomials, i.e. polynomials $p$ satisfying $$p(\lambda x_1, \dots, \lambda x_n)=\lambda^{deg(p)}p(x_1, \dots, x_n)$$ We can homogenize any non-homogenous polynomial by appending a coordinate and doing a sort of "renormalization" on inputs for the original. Specifically, take $\lambda$ to be a new coordinate and define $$\widetilde{f_i}(\lambda,x_1, \dots, x_n)=\lambda^{deg(f_i)}f_i(\tfrac{x_1}{\lambda}, \dots, \tfrac{x_n}{\lambda})$$ Given some $(a_1, \dots, a_n) \in V$, we have $$\widetilde{f_i}(\lambda, \lambda a_1, \dots, \lambda a_n)=\lambda^{deg(f_i)}f_i(\tfrac{\lambda a_1}{\lambda}, \dots, \tfrac{\lambda a_n}{\lambda})=\lambda^{deg(f_i)}f_i(a_1, \dots, a_n)=0$$

Which means $Cone(V)$ is an affine variety in $\mathbb{F}^{n+1}$ (or, can be thought of as a projective variety in $\mathbb{P}^n$).

gdd
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  • This shows that the set ${(\lambda,a_1,\ldots,a_n)}\subseteq\mathbb{F}\times\mathbb{F}^n$ with $(\lambda a_1,\ldots,\lambda a_n)\in \mathrm{Cone}(V)$ is a variety. It seems like more is needed to conclude that $\mathrm{Cone}(V)$ itself is a variety. – Drew Armstrong Nov 16 '20 at 17:04
  • Per Roland's comment on the original question, I think your definition of $Cone(V)$ is not quite right. https://math.stackexchange.com/questions/2305750/what-is-an-affine-cone – gdd Nov 16 '20 at 17:37