On the affine plane $\mathbb C^2$ we have the action of the torus $T=(\mathbb C^\times)^2$ given by rescaling: $$(t_1,t_2)\cdot (a,b)=(t_1a,t_2b)\in\mathbb C^2.$$ This action extends to the Hilbert scheme $\textrm{Hilb}_n(\mathbb C^2;d)$ consisting of ideals $I\subset \mathbb C[x,y]$ such that, for $k>>0$, one has $\textrm{codim}\,I_{k}=dk+n$. Here, the codimension of $I_k$ is the number of linearly independent polynomials in the vector space $I\cap \mathbb C[x,y]_k$. Just to be concrete: if $d=0$, we are dealing with $\textrm{Hilb}^n(\mathbb C^2)$.
I would like to understand the statement asserting that $T$-fixed ideals are exactly monomial ideals (i.e. ideals spanned by monomials).
A crucial step seems to be the following assertion:
Monomials $x^iy^j$ are eigenvectors of the torus action with distinct eigenvalues.
Could anyone please explain to me the latter sentence (which I cannot make sense of), and its relation with $T$-fixed ideals?
Thank you very much.
Recall that the character lattice of algebraic characters of $T^2$ is just $\mathbb{Z}^2$, where $(n,m)$ corresponds to $(z,w) \mapsto z^nw^m$.
Now, thinking about $T^2$'s dual action on all of $\mathbb{C}[x,y]$, up to inverting our action, when we diagonalize on the character lattice, we get each weight appearing with multiplicity one; namely, the monomials $x^ny^m$ are a basis for $\mathbb{C}[x,y]$, and have weight $(n, m)$.
Now, if we have an ideal $I$ fixed by the action, in particular it's a subrepresentation of $\mathbb{C}[x,y]$. It's a general fact that a subrepresentation of a diagonalizable representation is again diagonalizable, in particular, your $I$ is as a vector space a direct sum of some of these weight spaces, whence a monomial ideal (as a vector space it's spanned by monomials!).
*Edit: a character is just a group homomorphism $T^2 \rightarrow \mathbb{C}^\times$. We started with $T$ acting on $\mathbb{C}^2$ by scaling, so you get a 'dual' representation on all (regular) functions on it, $\mathbb{C}[x,y]$: $$t \cdot f := f \circ t$$ (this is okay since $T^2$ is abelian).
Given a representation $V$ of a torus, the analogous thing to finding an eigenvector with eigenvalue $\lambda \in \mathbb{C}$ for a single operator is finding an 'eigenvector' with for the whole torus, by which we mean some $v \in V$ such that $t \cdot v = \lambda(t) v$, for some weight $\lambda: T^2 \rightarrow \mathbb{C}^*$ (this is the 'exponential' of the story on lie algebras, where you really are just simultaneously diagonalizing a family of commuting operators $Lie(T^2)$).
Now, for $t = (z,w)$, $t \cdot x^n y^m = (zx)^n (wy)^m = z^n w^m x^n t^m$, so $x^ny^m$ is an eigenvector with weight $\lambda: (z,w) \mapsto z^n w^m$.
So, the monomials form an 'eigenbasis' for $\mathbb{C}[x,y]$ as a $T^2$ representation. Since they correspond to different weights, each weight space is 1 dimensional.
How does this help us? Well the key fact is, if $V$ is a diagonalizable $T^2$ representation, i.e. has an eigenbasis, then so does any subrepresentation (i.e. vector subspace closed under the action of $T^2$). Now, your ideal stable under $T^2$ would be such a subrepresentation, so it has a basis consisting of eigenvectors, and since the weight spaces were one dimensional, these have to be monomials, which is what you wanted.
Let's prove the key fact, for a reasonable algebraic group $G$. Let's write $V^\lambda$ for the space of eigenvectors of $V$ with weight $\lambda$, i.e. $V = \oplus_{\lambda \in \text{Hom}(G, \mathbb{C}^*) } V^\lambda$. The claim is that for $W$ a subrepresentation, the inclusion $\oplus W \cap V^\lambda \rightarrow W$ is an isomorphism.
Well, take any $w \in W$, write it as $v_{\lambda_1} + \ldots + v_{\lambda_n}$. For $G$ reasonable, the locus of $g$ for which $\lambda_1(g) \neq \ldots \neq \lambda_n(g)$ should be nonempty (e.g. the locus where $\lambda_i = \lambda_j$ is closed, so if $G$ is irreducible the desired locus is the complement of a finite union of subvarieties of positive codimension).
From here, we know that $w, gw, \ldots, g^{n-1}w$ are all in $W$. By the Vandermonde determinant these are a basis for $v_{\lambda_1}, \ldots, v_{\lambda_n}$, as desired.