So I am given that
Let $G = Z/dZ$ where d ≥ 1. Let w be a generator for G and let
G act on $A^
{n+1}$ via
$w(x_{0}, . . . , x_{n})$ = $(wx_{0}, . . . , wx_{n})$.
How can I Show that the categorical quotient of $A^{n+1}$ by G is the affine cone over the projective variety $ν_{d}(P^ {n}) ⊂ P^ {\begin{pmatrix}n+d\\d\end{pmatrix}}$ .
First off: what is the affine cone here? $v_d(\mathbb P^n)$ is the image of all monomials of degree $d$ in $n+1$ variables. So the affine cone is $Y = \mathrm{Spec}k[x^I]$, where $x^I$ ranges through all monomials of degree $d$.
According to Wikipedia, we have to check two things. First, that $Y$ is invariant under $G$, and secondly that any invariant morphism from $\mathbb A^n$ to another variety $Z$ factors through $Y$.
The first condition is clear, since the action is given by multiplying the coordinates by a $d$'th rooth of unity.
Now we have to show that given an invariant morphism, it must factor through $Y$. We can assume that $Z$ is affine, and hence that $Z=\mathbb A^m$ for some $m$. This further reduces to $ Z = \mathbb A^1$. Hence we are reduced to showing that any invariant polynomial belongs to the coordinate ring of $Y$, that is $$ k[x_1,\ldots,x_n]^G = k[X^I]. $$
The inclusion $\supseteq$ is clear. On the other hand, suppose $f(x)$ is homogeneous and invariant under $G$. We only need to show that the degree of $f$ is a multiple of $d$. Note that a polynomial is homogeneous of degree $d$ if and only if $f(\lambda x)=\lambda^d f(x)$ for all $\lambda \in k^\ast$. But since $f$ is invariant under $G$, we have $$ f(x)=f(\omega x) = \omega^{deg f} f(x). $$ This implies that $d \mid deg f$, as wanted.