A random subsum checks all quadratic equations at once

Put $d_j=A_j\cdot (u\otimes u)+b_j$. Since $u$ fails at least one equation, $d\mathrm{\ne }0$. Linearity gives $A(z)\cdot (u\otimes u)+b(z)=z\cdot d$, so the combined equation fails exactly when $z\cdot d=1$; by [F3] this occurs with probability exactly $1\mathrm{/}2$.

Fix any requested tensor coordinate $q$. Let $E=\{y:g(y)\mathrm{\ne }G(y)\}$, so $\mathrm{\mid }E\mathrm{\mid }\mathrm{/}{2}^{N^2}=\delta$. Both $y$ and $q+y$ are uniform, and by [F4] the corrector returns $g(y)+g(q+y)$. Unless one of these two points lies in $E$, this equals $G(y)+G(q+y)=G(q)$ by linearity of the intended oracle in [F2]; a union bound therefore gives correction failure probability at most $2\delta$, uniformly for every $q$, including $q=0$.

For the test, condition on each $z$ and set $q=A(z)$. By [F2] and [F1], on the event from step 1.1 the ideal value $G(q)=A(z)\cdot (u\otimes u)$ differs from $b(z)$; whenever the correction in step 1.2 returns $G(q)$, the test rejects. Its failure probability conditional on each $z$ is at most $2\delta$, so averaging gives $\mathrm{Pr}[\text{reject}]\ge \mathrm{Pr}[z\cdot d=1]-2\delta =\frac{1}{2}-2\delta$, without any independence assumption between rejection and correction errors.

The test samples $M$ bits for $z$ and $N^2$ bits for $y$, computes both query locations before reading $g$, and makes exactly two symbol queries; thus it is nonadaptive and uses one combined equation instead of querying all $M$ original equations. If $M=0$, its premise is impossible; if $N=0$, the tensor domain is a singleton and the corrector queries that same coordinate twice, as allowed by [F4].