GIT-Quotient of $\mathbb{C}^*$ acting on $\mathbb{C}^{m+1}$ I'm new to GIT and I came across the easy example of $G=\mathbb{C}^*$ acting on $V=\mathbb{C}^{m+1}$ by scalar multiplication, given a character $\chi(t) = t^k$ of $\mathbb{C}^*$ for some integer $k$. It was defined that for a group $G$ acting on a vector space $V$ we have $\mathbb{C^*}[V]^{G,\chi^n}$ the set of polynomials $f$ satisfying $f(g\cdot v) = \chi(g)^nf(v)$ for all $g,v$. Then the GIT-quotient was defined as
$$ V/_{\chi}G = \text{Proj}\left( \bigoplus_{n\geq 0} \mathbb{C^*}[V]^{G,\chi^n} \right), $$
which is projective over the ring of $G$-invariant functions. 
Now to me it would seem that for $k=0$, we have $ \mathbb{C^*}[V]^{G,\chi^n} 
 =\mathbb{C}$ for any $n$. Namely, only the constants are invariant. This gives us
$$ V/_{\chi}G = \text{Proj}\left( \bigoplus_{n\geq 0} \mathbb{C} \right) = \text{Proj}(\mathbb{C}). $$ 
For $k<0$, I would say that again $ \mathbb{C^*}[V]^{G,\chi^0} 
 =\mathbb{C}$ but now $ \mathbb{C^*}[V]^{G,\chi^n} 
 =\{0\}$ for $n > 0$, since polynomials have non-negative degree. This again gives us
$$ V/_{\chi}G = \text{Proj}\left( \mathbb{C} \oplus \left( \bigoplus_{n\geq 1} \{0\} \right) \right) = \text{Proj}(\mathbb{C}). $$
Lastly, for $k>0$ we have that $\mathbb{C^*}[V]^{G,\chi^n}$ equals the space of homogeneous polynomials of degree $kn$, so that
$$ V/_{\chi}G = \text{Proj}\left( \{ \text{polynomials in $m+1$ variables with all terms degree divisible by } k \} \right).$$
In the text it was stated that
$$V/_{\chi}G = \begin{cases} \mathbb{C}P^m \ \text{ for } k>0; \\ \{*\} \quad \text{ for } k = 0; \\ \varnothing \quad \ \ \ \text{ for } k<0.\end{cases}$$
Therefore I have two questions:


*

*How can it be that for $k=0$ and for $k<0$ we have different GIT-quotients?

*How can one prove that the GIT-quotient is $\mathbb{C}P^m$ for all $k>0$?
 A: For $k>0$, if $k=1$ this is the definition of the projective space. If $k>1$ this is just the Veronese embedding, more precisely if $S$ is a graded ring and $k>0$, we can form a graded ring $R$ by keeping elements of degree divisible by $k$. We have $Proj(S) \cong Proj(R)$, for a proof of this see "Graded rings and weighted projective spaces" by Miles Reid.
For $k = 0$, we can write $S = \Bbb C[u]$ seen as a graded ring in one variable with $\deg(u) = 1$, and it follows that $Proj(S) = Spec(\Bbb C) $ is a point. Indeed a point in Proj should corresponds to a homogeneous ideal which do not contain $\mathfrak m = (u)$, the ideal composed of elements of positive degree. We have only one such ideal (the zero ideal).
For $k<0$ we obtain the following : points in Proj should corresponds to prime homogenous ideals which do not contains the irrelevant ideal $\mathfrak m := \bigoplus_{n > 0}R^{(n)}$. But here $\mathfrak m = 0$ and also it is the only ideal of $\mathbb C$ so it follows that the corresponding Proj is empty.
Also the book by Mukai "Introduction to Invariant and Moduli" is excellent.
