Sum-check with explicit degree bounds

A sum-check instance consists of a supplied finite field $F$, a fixed formal polynomial $g\in F[X_1,\dots ,X_n]$ with $n\ge 1$, trusted nonnegative integer individual degree bounds $d_i$ satisfying ${\mathrm{deg}}_{X_i}g\le d_i$ for every $i$, a claimed value $H\in F$, and trusted access to point evaluation of that same $g$ on $F^n$. Zero polynomials satisfy every nonnegative bound. The claim is $H={\sum }_{b\in \{0,1{\}}^n}g(b)\mathrm{.}$ The field order is as in Finite fields and their order, the formal ring as in Polynomial rings in finitely many commuting indeterminates by iteration, and correctness uses the honest/universal-strategy quantifiers of Completeness and soundness of an interactive protocol.

Set $C_0=H$. For each $i=1,\dots ,n$, in this order:

1. Receive a coefficient list $(a_0,\dots ,a_m)$ specifying $h_i(T)={\sum }_{j=0}^m{a}_j{T}^j$, with $0\le m\le d_i$. Reject a missing, malformed, non-field, or overlength message. The single coefficient $0$ represents zero; trailing zero coefficients are allowed up to the length cap.
2. Check $h_i(0)+h_i(1)=C_{i-1}$; reject on failure.
3. Only after that message is fixed and checked, draw a fresh independent uniform $r_i\in F$ and put $C_i=h_i(r_i)$. Send $r_i$ to the prover if $i<n$; sending it when $i=n$ is optional.

Finally compute $g(r_1,\dots ,r_n)$ through the trusted evaluator and accept exactly when it equals $C_n$ and every earlier check passed. In particular $n=1$ still has a coefficient message, a fresh challenge, and a terminal comparison. The evaluator's answer is not an unchecked prover assertion.

For a fixed challenge prefix, the honest round polynomial is $q_i(T)={\sum }_{b\in \{0,1{\}}^{n-i}}g(r_1,\dots ,r_{i-1},T,b)\mathrm{.}$ The cube of dimension zero has one empty tuple, so $q_n(T)=g(r_1,\dots ,r_{n-1},T)$. Write $S_i={\sum }_{b\in \{0,1{\}}^{n-i}}g(r_1,\dots ,r_i,b),S_0={\sum }_{b\in \{0,1{\}}^n}g(b)\mathrm{.}$ A scalar claim at stage $i$ is true when $C_i=S_i$. The honest prover sends $h_i=q_i$. The input $H$ is not counted as an additional prover message.