This secret sharing scheme takes place in the polynomial ring over a finite field $F_p[x]$, $p$ being a large prime, in the following simplified way:
The dealer chooses a private polynomial $f(x)$.
Every participant $P_i, i \in N$, has a public polynomial $m_i(x)$, with $deg(m_i(x)) \leq deg(f(x))$.
The dealer computes the private secret share of $P_i$ as $s_i(x) = f(x) \mod m_i(x)$ and sends it to $P_i$.
I need a way for $P_i$ to verify that the private secret share $s_i(x)$ was well calculated.
Suppose there is a polynomial commitment scheme that allows you to commit to a certain polynomial $p(x)$ as $p(\beta)G$.
$\beta$ is a value unknown to everyone (as well as $p(\beta)$), including the dealer, and $G$ is a point generator of an Elliptic Curve over $F_p$.
This polynomial commitment scheme also allows to produce a proof evaluation of a certain committed polynomial, i.e., you can prove that a certain point (a, b) belongs to the committed polynomial.
With this, my idea to verify the private secret share $s_i(x)$ is the following:
Since $f(x) = m_i(x)k_i(x) + s_i(x)$, the dealer commits to $f(x)$ and $k_i(x)$ as $f(\beta)G$ and $k_i(\beta)G$, respectively, and sends them to $P_i$.
$P_i$ asks for an evaluation of a random $z$.
The dealer sends $f(z)$ and $k_i(z)$ to $P_i$, along with the proofs that they belong to the committed polynomials.
$P_i$ verifies that $f(z) = k_i(z)m_i(z) + s_i(z)$.
My thinking was that two polynomials $p(x)$ and $p'(x)$, with degrees $n$ and $m$, can have at most $max(n, m)$ intersections, so if the dealer commits to polynomials $f'(x)$ and/or $k'(x)$ instead of $f(x)$ and/or $k(x)$, the probability that the equation holds for those committed polynomials, is negligible.
I don't know how to prove or disprove this, so I would appreciate if anyone can help me.