= Solution
Evaluate <Bézout's identity for polynomials> at $T$:
$$
a(T)p(T)+b(T)q(T)=I,\qquad p(T)q(T)=0.
$$
All these <linear operators> commute because they are <polynomials> in the same operator. Thus every $v\in V$ can be written
$$
v=b(T)q(T)v+a(T)p(T)v.
$$
The first summand belongs to $\ker p(T)$ and the second to $\ker q(T)$, since applying the corresponding operator introduces the product $p(T)q(T)=0$.
If $v$ lies in both <kernel of a linear map>[kernels], the same identity gives $v=0+0=0$. Therefore the sum is direct:
$$
\boxed{V=\ker p(T)\oplus\ker q(T).}
$$
This is the <kernel decomposition for coprime polynomials>. It uses both annihilation by $pq$ and the coprimality expressed by the Bézout identity.
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