TITLE: Cone over a variety is a variety?
QUESTION [2 upvotes]: 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.

REPLY [2 votes]: 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 $\mathbb{P}^n$).
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