# Clarification on Modular Equivalence in Generalized Sumcheck in RZ21

I'm studying generalized sumcheck in RZ21 P.10 and have come across a modular equivalence that I'm trying to understand. The specific step that's unclear to me is the following equivalence:

$$\left(\sum_{\mathsf{h} \in \mathbb{H}} P(\mathsf{h}) \lambda_\mathsf{h}(X)\right)\left(\sum_{s \in \mathcal{S}} \lambda_s(v)^{-1} \lambda_s(X)\right) - \sigma = \left(\sum_{s \in \mathcal{S}} P(s) \lambda_s(v)^{-1}\lambda_s(X)\right) - \sigma \mod t(X)$$

Notation:

1. $$\lambda_{\mathsf{h}}(X)$$ and $$\lambda_{s}(X)$$ are the Lagrange basis polynomials associated with elements.
2. $$\mathcal{S}$$ is a subset of $$\mathbb{H}$$.
3. $$\sum_{s \in \mathcal{S}} P(s) = \sigma$$.

Context: This theorem extends the univariate sumcheck of Aurora to work not only for multiplicative subgroups of finite fields.

Reference: [RZ21], https://eprint.iacr.org/2021/590.pdf, P.10

Note that the whole equation in the paper is a congruence modulo $$t(x)$$, which is the vanishing polynomial of $$\mathbb{H}$$.
The congruence holds because, since $$\mathcal{S} \subseteq \mathbb{H}$$, we can calculate the product in evaluation basis by multiplying evaluations at each point in $$\mathcal{S}$$. The evaluations at points in $$\mathbb{H} \setminus \mathcal{S}$$ vanish because they are zero in the sum over $$\mathcal{S}$$ on the left-hand-side.
• $t(x)$ is the vanishing polynomial of $\mathbb{H}$. Jan 15 at 11:09