I am trying to understand a proof from N. Jacobson book ``Lectures in Abstract Algebra, II. Linear Algebra'', Chapter IV.8 (p.130, but I've attached the screen for convinience).
Let $V$ be a finite-dimensional vector space over a field $\mathsf{F}$ (no nessessary algebraic closed and it may be also be a skew-field I guess). Take a linear map $A:V \to V$, then $A\in \mathrm{Mat}_{n \times n}(\mathsf{F})$. Hence we there is a minimal $N >0$ such that $A^0=\mathbf{id}_V, A^1,\ldots, A^N$ are liniear depending vectors of $\mathrm{Mat}_{n\times n}(\mathsf{F})$, and we thus get a polynomial $\mu(t) \in \mathsf{F}[t]$ which is exactly the minimal polynomial of $A$.
Assume further, that $\mu(t)$ can be decomposed as follows $\mu(t) = \pi_1^{n_1}(t)\cdots \pi_\ell(t)^{n_\ell}$, where all $\pi_i^{n_i}(t)$ are assumed to be relativly prime. It is clear that $V = \{v\in V: \mu(A)(v) = 0\}$.
Next, set $V_i:=\{v\in V: \pi_i^{n_i}(v) = 0\}$. We want to show that $V \cong \bigoplus_{i=1}^\ell V_i$. To do so we remard that polynomials $\mu_i(t):= \mu(t)/\pi_i^{n_k}(t)$ are relatively prime and hence there exist some $\phi_1,\ldots, \phi_\ell \in \mathsf{F}[t]$ such that $$ \phi_1(t)\mu_1(t) + \cdots + \phi_\ell(t) \mu_\ell(t)=1. $$ then $$ \phi_1(A)\mu_1(A) + \cdots + \phi_\ell(A) \mu_\ell(A)=\mathbf{id}_V. $$
Then, Jacobson wrote ``Since $\mu_i(t)\mu_j(t)$ is divisible by $\mu(t)$ if $i\ne j$, $$ \phi_i(A)\mu_i(A)\phi_j(A)\mu_j(A)=0. $$
And this is my trouble. I don't understand why? Can somebody explane me?
It is clear, because of $\mu(A)(v)=0$ for any $v\in V$ and hence $\mu(A)=0$. Since $\mu_i(A)\phi(A)\mu_j(A)\phi_j(A)$ is divisble by $\mu(A)$ it also must be zero.