I have a little trouble with a proof involving PI algebras, as you can see here at page 6 - Giambruno Zaicev - Polynomial Identities and Asymptotic methods.
First of all let me recall some facts.
Let $\mathbb{F}$ be a field and $X$ a set. The free associative algebra on X over $\mathbb{F}$ is the algebra $\mathbb{F}\langle{X}\rangle$ of polynomials in the non-commuting indeterminates $x \in X$.
If $f\in \mathbb{F}\langle{X}\rangle$, we write $f=f(x_1,...,x_n)$ to indicate that $x_1,...,x_n \in X$ are the only indeterminates occurring in $f$.
We define $\deg u$, the degree of a monomial $u$, as the length of the word $u$. Also $\deg_{x_i} u$, is the number of occurrences of $x_i$ in $u$.
Accordingly, the degree $\deg f$ of a generic polynomial $f=f(x_1,...,x_n)$ is the maximum degree of a monomial in $f$; instead $\deg_{x_i} f$, the degree of $f$ in $x_i$, is the maximum of $\deg_{x_i} u$, for $u$ monomial in $f$.
DEFINITION Let $A$ be an $\mathbb{F}$-algebra and $f=f(x_1,...,x_n) \in \mathbb{F}\langle{X}\rangle$. We say that $f$ is a polynomial identity of $A$ if $f(a_1,...,a_n)=0$ for all $a_1,...,a_n \in A$ (shortly we can say that $f$ is a PI for $A$).
DEFINITION Let $f=f(x_1,...,x_n) \in \mathbb{F}\langle{X}\rangle$. We say that $f$ is multihomogeneous of multidegree $(a_1,...,a_n)$ if $f$ is a sum of monomials of degree $a_1$ in $x_1$,..., $a_n$ in $x_n$. In other words, if $f=\sum_{j=1}^{m}u_j$ where $u_j$'s are monomials of $f$ then $deg_{x_i}u_j=a_i\, \forall i=1,...,n\,\,j=1,...,m$.
Example $f(x_1,x_2,x_3,x_4)=x_2x_4x_1x_2x_1x_2+x_4{x_2}^2x_1x_2x_1$ is multihomogeneous of multidegree $(2,3,0,1)$.
If $f=f(x_1,...,x_n) \in \mathbb{F}\langle{X}\rangle$, we can always write $$f=\sum_{a_1\geq 0,...,a_n\geq 0} f^{(a_1,...,a_n)}$$ where $f^{(a_1,...,a_n)}$ is the sum of multihomogeneous monomials of multidegree $(a_1,...,a_n)$ of $f$, ie the sum of monomials in $f$ where $x_1,...,x_n$ appear at degree $a_1,...,a_n$ respectively. The polynomials $f^{(a_1,...,a_n)}$ wich are nonzero ar called multihomogeneous components of $f$. Now the proposition I'm trying to understand:
Proposition Let $\mathbb{F}$ be an infinite field. If $f=f(x_1,...,x_n) \in \mathbb{F}\langle{X}\rangle$ is a PI for the algebra A, then every multihomogeneous component of $f$ is still a PI for the algebra A.
proof. For every variable $x_t$, $1\leq t\leq n$, we can decompose $f=\sum_{i=0}^{m}f_i$ where $f_i$ is the sum of all monomials of $f$ in which $x_t$ appears at degree $i$ and $m=\deg_{x_t} f$ is the degree of $f$ at $x_t$.
Let $\alpha_0,...,\alpha_m$ be distinct elements of $\mathbb{F}$. Clearly for every $j=o,...,m$ $$f(x_1,...,\alpha_j x_t,...,x_n)$$ is still a PI for A. Since each $f_i$ is homogeneous in $x_t$ of degree $i$, $$f_i(x_1,...,\alpha_j x_t,...,x_n)={\alpha_j}^i f_i(x_1,..., x_t,...,x_n).$$ Hence $$f(a_1,...,\alpha_j a_t,...,a_n)=\sum_{i=0}^{m}{\alpha_j}^i f_i(a_1,...,a_n)=0$$ for all $j=0,...,m$ and for all $a_1,...,a_n \in A$.
Write the Vandermonde matrix $$\Delta=\begin{bmatrix} 1 & 1 & \cdots & 1\\ \alpha_0 & \alpha_1 & \cdots & \alpha_m \\ \vdots & \vdots & & \vdots \\ {\alpha_0}^m & {\alpha_1}^m & \cdots & {\alpha_m}^m \end{bmatrix}$$ then the previous says that for every $a_1,...,a_n \in A$, if we write $f_i(a_1,...,a_n)=\overline{f_i}$ then $$(\overline{f_0},...,\overline{f_m})\Delta=0$$ is a square linear system of equations which has a unique solution $(\overline{f_0},...,\overline{f_m})=(0,...,0)$ because $\det \Delta=\prod_{0\leq i<j\leq m}(\alpha_i-\alpha_j)$ is non-zero. Then for every variable $x_t$, $1\leq t\leq n$, if $f=\sum_{i=0}^{m}f_i$ where $f_i$ is the sum of all monomials of $f$ in which $x_t$ appears at degree $i$, then $f_0,...,f_m$ are PI for A.
The author concludes the proof saying that by an obvious induction argument one can show that the theorem holds.
It's everything clear to me, but this obvious induction argument that I have to apply to a generic multihomogeneous component of $f$. I don't really know where to apply the inductive step (maybe on the number of multihomogeneous components of $f$?).
Any suggestion would be really appreciated! Thanks a lot!!
As for the next step, fix now two variables: $x_{t_1},\ x_{t_2}$, and loop over their common degrees occuring in $f$: $$f=\sum_{i_1,i_2\ge 0}f_{i_1,i_2}$$ where $f_{i_1,i_2}$ is the sum of monomials $m$ of $f$ where $\deg_{x_{t_1}}m=i_1$ and $\deg_{x_{t_2}}m=i_2$.
Note that for a fixed $i_1$, we get $\sum_{i_2\ge 0}f_{i_1,i_2}=f_{i_1}=$ the sum of monomials of $f$ where $\deg_{x_{t_1}}=i_1$.
By the above, we already know that $f_{i_1}$ is a PI for $A$, so we can apply it once again with $x_{t_2}$ to arrive that each $f_{i_1,i_2}$ is again PI for $A$.