Suppose there exists co-prime ploynomials $p,q\in F[X]$ such that $(p \cdot q)(T)=0$, Prove: $V=Im(p(T)) \oplus Im(q(T))$ Let $V$ be a Vector space over filed $F$, and let  $T : V \to V$ be a linear transformation
suppose there exists co-prime ploynomials $p,q\in F[X]$ such that $(p \cdot q)(T)=0$
Prove that $V=Im(p(T)) \oplus Im(q(T))$
With this few information about T I am having hard time dealing with this question, thought of using the Primary decomposition of V generated by $p \cdot q$ but couldn't go further.
I would appriciate any hint or directions.
Thanks.
 A: If you are allowed to use the primary decomposition, then you only have to show that $ker(p(T))=im(q(T))$ (and the analogous equality with $p$ and $q$ exchanges). Since $pq(T)=qp(T)=0$, it is clear that $im(q(T))\subset ker(p(T))$. On the other hand, using the polynomials $h$ and $k$ such that $kp+hq=1$ as in the comment of @PedroTamaroff, you see that for $v\in ker(q(T))$ you get $v=p(T)(k(T)(v))+k(T)(q(T)(v))=p(T)(k(T)(v))$, so $v\in im(p(T))$. 
A: You have to show that every vector $v\in V$ can be represented as $p(T)(x)+q(T)(y)$, for some $x,y\in V$, and that $\operatorname{Im}(p(T))\cap\operatorname{Im}(q(T))=\{0\}$.
Since $p$ and $q$ are coprime, you can write
$$
1=p(x)p'(x)+q(x)q'(x)
$$
for some $p',q'\in F[x]$. Then, for every $v\in V$,
$$
v=p(T)(p'(T)(v))+q(T)(q'(T)(v))
$$
so you can take $x=p'(T)(v)$ and $y=q'(T)(v)$.
Suppose now $v=p(T)(x)=q(T)(y)\in\operatorname{Im}(p(T))\cap\operatorname{Im}(q(T))$.
Then, by assumption, $q(T)(v)=q(T)(p(T)(x))=0$ and $p(T)(v)=p(T)(q(T)(y))=0$. Now apply $v=p(T)(p'(T)(v))+q(T)(q'(T)(v))$ to show $v=0$.
