Polynomial Modulo Soundness
We want to proof a polynomial modular multiplication (PMM)
$$ A_i(X) ⋅ B_i(X) = R_i(X) \mod P(X) $$
Where $A,B,R$ are low degree.
We can do this using an auxiliary low-degree polynomial $Q_i(X)$ and evaluating the following in a random point $z$
$$ A_i(z) ⋅ B_i(z) = Q_i(z) ⋅ P(z) + R_i(z) $$
To verify many such PMMs we can take a linear combination in a random $α$
$$ \sum_i α^i ⋅ A_i(X) ⋅ B_i(X) = P(z) ⋅ \sum_i α^i ⋅ Q_i(z) + \sum_i α^i ⋅ R_i(z) $$
Question Can the prover pre-sum the quotients? I.e. have an auxiliary low-degree polynomial $Q(X)$ and check:
$$ \sum_i α^i ⋅ A_i(X) ⋅ B_i(X) = P(z) ⋅ Q(z) + \sum_i α^i ⋅ R_i(z) $$
Given $A_i, B_i, R_i, Q$ and $P$ low-degree polynomials that satisfy the above equation. We want to show the only solution is
$$ Q(X) = \sum_i α^i ⋅ Q_i(X) $$
where
$$ Q_i(X) = \frac{A_i(X) ⋅ B_i(X) - R_i(X)}{P(X)} $$
We know that
$$ Q(X) = \frac{\sum_i α^i ⋅ (A_i(X) ⋅ B_i(X) - R_i(X) )}{P(X)} $$
So this amounts to proving
$$ \frac{\sum_i α^i ⋅ (A_i(X) ⋅ B_i(X) - R_i(X) )}{P(X)} = \sum_i α^i ⋅ \frac{A_i(X) ⋅ B_i(X) - R_i(X)}{P(X)} $$
Let's assume that between $A_i(X) ⋅ B_i(X) - R_i(X)$ there are enough degrees of freedom that these can be arbitrary polynomials $D_i(X)$ of degree $2⋅d$:
$$ \frac{\sum_i α^i ⋅ D_i(X)}{P(X)} = \sum_i α^i ⋅ \frac{D_i(X)}{P(X)} $$