I'm studying Sturmfels's "Algorithms in Invariant theory", and in particolar this result (page 30)
Lemma. Let $p_1,\ldots,p_m$ be algebraically independent elements of $\mathbb{C}[x_1,\ldots,x_n]$ which are homogeneous of degrees $d_1,\ldots,d_m$. Then the Hilbert series of $R=\mathbb{C}[p_1,\ldots,p_m]$ equals (where $z$ is a formal variable) $$\frac{1}{(1-z^{d_1})\cdots(1-z^{d_m})}.$$
Proof. Since $R_d=\langle p_1^{i_1}\cdots p_m^{i_m}\mid \sum_{k=1}^m i_kd_k=d \rangle$, then $$dim(R_d)=|A_d|=|\{(i_1,\ldots,i_m)\in\mathbb{N}^m\mid \sum_{k=1}^m i_kd_k=d\}\mid.$$ Thus \begin{equation} \begin{split} \frac{1}{(1-t^{d_1})\cdots(1-t^{d_m})}&=\frac{1}{(1-t^{d_1})}\cdots\frac{1}{(1-t^{d_m})} = \\ & = (\sum_{i_1=0}^{\infty}z^{i_1 d_1})(\sum_{i_2=0}^{\infty} z^{i_2 d_2})\cdots(\sum_{i_m=0}^{\infty}z^{i_m d_m}) = \\ &\overset{\star}{=} \sum_{d=0}^{\infty} \sum_{(i_1,i_2,\ldots,i_m)\in A_d} z^d= \sum_{d=0}^{\infty}|A_d|z^d \end{split} \end{equation}
The $\star$ confuses me a lot, and I'd like to see why this equality holds.
My idea. Let me call LHS and RHS the left and right-hand side of $\star$ identity. I can re-write the LHS as $$\sum_{(i_1,i_2,\ldots,i_m)\in \mathbb{N}^m} z^{i_1d_1+\ldots+i_md_m}, $$ and then setting $d=i_1d_1+\ldots+i_md_m$, we got $$\sum_{d=0}^{\infty} \sum_{(i_1,\ldots,i_m)\in A_d} z^d?$$ I don't know, from my point it looks like cheating, but it's the best I came up with.
I'm very sorry to bother you with this, but I keep getting confused about that passage and I wanna now if my justification is it right, or there are some re-ordering involved. Thanks in advance.
Your justification is entirely correct. Indeed the left hand side equals your expression $$\sum_{(i_1,i_2,\ldots,i_m)\in \mathbb{N}^m} z^{i_1d_1+\ldots+i_md_m},$$ and now we want to find the coefficients $c_d$ of this power series, i.e. the numbers $c_d$ such that $$\sum_{(i_1,i_2,\ldots,i_m)\in \mathbb{N}^m} z^{i_1d_1+\ldots+i_md_m}=\sum_{d\geq0}c_dz^d.$$ It should be clear that $c_d$ is precisely the number of solutions to $$i_1d_1+\cdots+i_md_m=d,$$ with $(i_1,i_2,\ldots,i_m)\in \mathbb{N}^m$. And this is precisely $|A_d|$ by definition. So we conclude that $$\sum_{(i_1,i_2,\ldots,i_m)\in \mathbb{N}^m} z^{i_1d_1+\ldots+i_md_m}=\sum_{d\geq0}|A_d|z^d.$$