I am trying to understand the torus action on the Hilbert Scheme.
If we have an ideal in the polynomial ring $\mathbb{C}[x,y]$, then there is an action of $(\mathbb{C}^*)^2$ on I defined as $(t_1,t_2) \cdot f(x,y) = f(t_1 x, t_2 y)$, where $f \in I, (t_1,t_2) \in (\mathbb{C}^*)^2$(two dimensional algebraic torus group).
The the following statement I want to understand
$I$ is fixed by this action if and only if $I$ is a monomial ideal.
Then the action acting on $\mathbb{C}^2$ produces an action of $(\mathbb{C}^*)^2$ on the Hilbert scheme of $n$ points. That is, $H_n$ = Hilb${^{n}}$ $(\mathbb{C}^2)$ of $n$ points in the plane consists of ideals $I \subset \mathbb{C}[x,y]$ for which $\mathbb{C}[x,y]/I$ has dimension $n$ as a vector space over $\mathbb{C}$.
I know that the general linear group( and all its subgroups) acts on polynomial ring as follows.
If $g = (g_{ij}) \in GL_2 (\mathbb{C})$ and a polynomial $f = p(x_1,x_2) \in \mathbb{C}[x_1,x_2]$, then $g$ acts on $f$ by $g \cdot p = p(gx_1, gx_2),$ where $gx_i = \Sigma_{j = 1}^2 g_{ij}x_j$. (having trouble understanding this)
Given an ideal $I$ in the polynomial ring with two variables, we get a new ideal by $g \cdot I = {\{g \cdot p} | p \in I\}$.
So since algebraic torus is a subgroup of the general linear group, it would act this way too.
However, I am having a hard time understanding this. I know that the fixed points of the torus action on the Hilbert scheme corresponds to monomial ideal which corresponds to partitions of $n$, that can be represented as Ferrer diagrams with $n$ boxes.
Can someone please help me understand the torus action on the Hilbert scheme in most simple way, and why the fixed points corresponds to the monomial ideals. thanks for any help.
To see that fixed points correspond to monomial ideals, suppose that $I$ is a fixed point; the aim is to show that $I$ is a monomial ideal. Suppose that $f \in I$. We can group terms of $f$ to write $f$ in the form $$ f = x_1^d p_d(x_2,\dotsc,x_n) + x_1^{d-1} p_{d-1}(x_2,\dotsc,x_n) + \dotsb + p_0(x_2,\dotsc,x_n) . $$ Now set $t = (t_1,1,1,\dotsc,1)$ for some $t_1 \in \mathbb{C}$ such that $t_1,t_1^2,\dotsc,t_1^d \neq 1$. (This is easy in $\mathbb{C}$; e.g., $t_1=2$ works. In other fields it's still pretty easy to achieve under a mild condition.) Let this $t$ act on $f$. We have $$ t \cdot f = t_1^d x_1^d p_d(x_2,\dotsc,x_n) + t_1^{d-1} x_1^{d-1} p_{d-1}(x_2,\dotsc,x_n) + \dotsb + p_0(x_2,\dotsc,x_n) . $$ This $t \cdot f \in I$. We can continue on with $t^2 \cdot f$, and so on; each one is in $I$. By linear algebra, we can deduce that each "term" $x_1^k p_k(x_2,\dotsc,x_n) \in I$. This basically just uses invertibility of the Vandermonde matrix.
Now we can repeat the argument with the next variable, to get that each "term" $x_1^{k_1} x_2^{k_2} q(x_3,\dotsc,x_n) \in I$.
Eventually we see that each actual term (monomial) of $f$ is in $I$. This proves that $I$ is a monomial ideal.
It's not that hard to get directly from $f$ to its terms in a single step, rather than this idea of iterating over variables. But I thought it might be simpler to explain this approach.