Let $K$ be a field, and $a_1(x), \dots, a_r(x) \in K[x]$, $r \geq 2$, be pairwise relatively prime non-constant polynomials. I am trying to prove the following proposition.
Proposition. For all $c(x)\in K[x]$ such that $\deg c(x) < \sum\limits_{i = 1}^{r}{\deg a_i(x)}$ there exist uniquely determined polynomials $u_i(x) \in K[x]$ such that $$c(x) = \sum\limits_{i = 1}^{r}{u_i(x)\prod\limits_{j = 1 \\ j \neq i}^{r}{a_j(x)}}$$ and $\deg u_i(x) < \deg a_i(x)$ for $i = 1, \dots, r$.
(In some way, this is a specification and generalisation of polynomial Bézout's identity, i.e. $c = xa + yb$. I think I need this result for proving results related to Hensel lifting for factorising polynomials over $\mathbb Z[x]$).
Effort
If one takes $A_i(x) := \prod\limits_{j = 1 \\ j \neq i}^{r}{a_j(x)}$, then the $A_i$ all together have $\gcd$ equal to $1$. Thus one can write $$1 = \sum\limits_{i = 1}^{r}\tilde{u}_i(x)A_i(x)$$ for some $\tilde u_i(x)$. When multiplied by $c(x)$, one does get the necessary form $$c(x) = \sum\limits_{i = 1}^{r}[c(x)\tilde{u}_i(x)]A_i(x)$$
but the degree bounds are probably not fulfilled, and uniqueness is still open. For the degree bounds, maybe one ought to begin by dividing $c(x)\tilde u_1(x)$ with remainder by $a_1(x)$, giving $c(x)\tilde u_1(x) = q_1(x)a_1(x) + r_1(x)$. Substituting this back gives $$c(x) = [q_1(x)a_1(x) + r_1(x)]A_1(x) + \sum\limits_{i = 2}^{r}[c(x)\tilde{u}_i(x)]A_i(x),$$ or, after regruping, $$c(x) = r_1(x) A_1(x) + [q_1(x)a_2(x) + c(x)\tilde{u}_2(x)]A_2(x) + \sum\limits_{i = 3}^{r}[c(x)\tilde{u}_i(x)]A_i(x).$$
If I were to take $u_1(x) := r_1(x)$, then $\deg u_1(x) < \deg r_1(x)$ is satisfied by the definition of $r_1(x)$ as remainder. Were we to continue this process of division by remainder and regrouping, then $c(x)\tilde{u}_2(x) = q_2(x)a_2(x) + r_2(x)$, therefore $q_1(x)a_2(x) + c(x)\tilde{u}_2(x) = (q_1(x) + q_2(x))a_2(x) + r_2(x)$ whence
$$c(x) = r_1(x) A_1(x) + r_2(x) A_2(x) + [[q_1(x) + q_2(x)]a_3(x) + c(x)\tilde{u}_3(x)]A_3(x) + \sum\limits_{i = 4}^{r}[c(x)\tilde{u}_i(x)]A_i(x).$$
Now take $u_2(x) := r_2(x)$, and progress forward.
Question
Moving forward like above, one can prove the degree bounds for $u_1(x), \dots, u_{r-1}(x)$. For the last summand, we likely have
$$\underbrace{\left[\left(\sum\limits_{i = 1}^{r-1}{q_i}(x)\right)a_{r}(x) + c(x)\tilde u_r(x)\right]}_{=:\ u_r(x)} A_r(x) \tag{*}$$
where the $q_i(x)$ are quotients, $c(x)\tilde u_i(x) = q_i(x) a_i(x) + r_i(x)$.
- Question: The hope is that $\deg u_r(x) < \deg a_r(x)$. But is this the case?
A good place to start is $r = 2$.
The question for $r := 2$
Let us denote $b(x) := a_1(x)$ and $a(x) := a_2(x)$. The question is then to find unique $u_1(x)$ and $u_2(x)$ such that $c(x) = u_1(x) a(x) + u_2(x) b(x)$ satisfying the bounds $\deg u_1(x) < \deg b(x)$ and $\deg u_2(x) < a(x)$.
Also $A_1(x) = b(x)$, $A_2(x) = a(x)$, and $c(x)u_1(x) = q_1(x)b(x) + r_1(x)$. As before, $u_1(x) := r_1(x)$. The last (or second) summand here is
$$[q_1(x) a(x) + c(x)\tilde u_2(x)] b(x)$$
- And the question, in this case, is whether $q_1(x)a(x) + c(x) \tilde u_2(x) =: u_2(x)$ satisfies $\deg u_2(x) < \deg a(x)$.
I am not sure what to do here, even for $r = 2$.
For $r = 2$, uniqueness comes directly from the degree bounds, so starting from the degree bounds is probably a good idea.