Clarification Regarding to prove diagonalisable operator iff Minimal Polynomial Splits and

Question

I wanted to prove that A is diagonalisable over F iff minimal Polynomial of A splits over F and Have square free terms.
In that proof, I had following doubts
If A is diagonalisable then it is clear that A Has characteristics polynomial which splits over F.
As Minimal Polynomial of A divide characteristics polynomial of A.
Then it also splits. Only thing remains to show that It has the square free term.
As A is diagonalisable it has eigenvalues then I wanted to show that $m_T(x)=(x-a_1)(x-a_2).....(x-a_k)$ .
where $a_1,a_2,.....a_k $be eigenvalues . TO show this minimal polynomial I wanted to show that it Annihilates everything in Vector space.So take $v_i\in V$ such that $T(v_i)=a_iv_i$ Now consider above polynomial as
$((T-a_1)(T-a_2).....(T-a_i)......(T-a_k)v_i=?$ Now I know that polynomial is commutative ring But How to think for this Because this is the operator. Is this also commutative.
If this I know it annihilates everything in V.
Any Help will be appreciated