Graduate Expansion with respect to ##z_1##

[MathLearner123]

I'm reading "From Holomorphic Functions to Complex Manifolds" - Fritzsche & Grauert and I have something that I don't understand very well: If $\nu \in \mathbb{N}_0^n, t \in \mathbb{R}^n_+$ and $z \in \mathbb{C}^n$, write $\nu = (\nu_1, \nu'), t = (t_1, t')$ and $z = (z_1, z')$. \ An element $f=\sum_{\nu \geq 0} a_\nu \mathbf{z}^\nu \in \mathbb{C} [\![ z ]\!]$ (formal power series ring) can be written in the form $f=\sum_{\lambda = 0}^\infty f_\lambda z_1^\lambda$ where $f_\lambda(z')=\sum_{\nu' \ge 0} a_{(\lambda, \nu')} (z')^{\nu'}$. Why can we write this? Changing the order of factors doesn't affect the formal power series? We have that $f=\sum_{\nu \ge 0} a_\nu z^\nu = \sum_{(\nu_1, \nu')\ge 0} a_{(\nu_1, \nu')}z_1^{\nu_1} (z')^{\nu'} = \sum_{\nu_1 \ge 0} \sum_{\nu' \ge 0} a_{(\nu_1, \nu')}z_1^{\nu_1} (z')^{\nu'} = \sum_{\nu_1 \ge 0} \left(\sum_{\nu' \ge 0} a_{(\nu_1, \nu')} (z')^{\nu'}\right)z_1^{\nu_1} = \sum_{\lambda \ge 0} \left(\sum_{\nu' \ge 0} a_{(\lambda, \nu')} (z')^{\nu'}\right)z_1^{\lambda} = \sum_{\lambda \ge 0} f_\lambda z_1^{\lambda}$. But taking out that common factor $z_1^{\nu_1}$ doesn't change the formal power series? And another thing that I dont't understand: In what order the terms appear in the expansion $f = \sum_{\nu \ge 0} a_\nu z^\nu$. For example, in $\mathbb{C}[\![z_1,z_2,z_3]\!]$ both terms $a_{(1,2,3)} z^{(1,2,3)}$ and $a_{(1,3,2)} z^{(1,3,2)}$ have "order" $1+2+3=6$ but who comes first in the expansion? I need to consider some order on $\mathbb{N}_0^3$ or something?

 

[pasmith]

The order of summation does not matter in formal power series (we aren't actually summing them, so issues of convergence do not arise). Hence<br /> \sum_{n=0}^\infty \sum_{m=0}^\infty a_{nm}z_1^nz_2^m = \sum_{n=0}^\infty\left(\sum_{m=0}^\infty a_{nm}z_2^m\right)z_1^n = \sum_{n=0}^\infty f_n(z_2) z_1^n.