As part of the proof of the Primary Decomposition Theorem, we first show that if $f(x)=a(x)b(x)$ with $gcd(a,b)=1$ and $f(T)=0$ where $T:V\to V$ is a linear transformation, then we have that $V=Ker(a(T)) \oplus Ker(b(T))$.
I get that $f(x)=a(x)b(x) \implies a(x)s(x)+b(x)t(x)=1$ for some polynomials $s,t \in \Bbb F$, so that $a(T)s(T)v+b(T)t(T)v=v$. But how does it follow that \begin{align} \ & a(T)(b(T)t(T))v=0 \\ \ & b(T)(a(T)s(T))v=0 \\ \end{align}
and hence that $V=Ker(a(T))+Ker(b(T))$?
Is it because $a(T)(b(T)t(T))v=(a(T)b(T))t(T)v=f(T)(t(T)v)=0$?
Then what about $b(T)(a(T)s(T))v=0$? Does $b(T)$ necessarily commute with $a(T)$? Supposing we have established this, why then does this imply that $V=Ker(a(T))+Ker(b(T))$?
Thanks
Yes $a(T)$ and $b(T)$ do commute: $$f(T)=a(T)b(T)=b(T)a(T)$$ As you have shown correctly, $\gcd(a, b)=1$ implies that for any $v\in V$, $$v=a(T)s(T)(v)+b(T)t(T)(v)$$ where $a(T)s(T)(v)\in\ker(b(T))$ and $b(T)t(T)(v)\in\ker(a(T))$. So $$V=\ker(a(T))+\ker(b(T)).$$ Now if $v\in\ker(a(T))\cap \ker(b(T))$, then $$v=a(T)s(T)(v)+b(T)t(T)(v)=s(T)a(T)(v)+t(T)b(T)(v)=0$$ and therefore $\ker(a(T))\cap \ker(b(T))=\{0\}$ which proves that $V=\ker(a(T))\oplus\ker(b(T)).$