Chebotarev Referee Audit

Referee E's audit method is to independently recalculate the entire Chebotarev argument from scratch, not just double-checking the derivation steps of Document 16. The discriminant of $f_2(x)=x^3+x^2+8x-16$ is $-11136=-2^7\cdot3\cdot29$; irreducible plus a non-square discriminant gives $\operatorname{Gal}(L/\mathbf Q)=S_3$, and the unique quadratic subfield is $F_0=\mathbf Q(\sqrt{-174})$. Let $K=\mathbf Q(\zeta_{24},\sqrt{29})$, $[K:\mathbf Q]=16$; from $\sqrt{-174}=\sqrt{-6}\sqrt{29}$ and $\mathbf Q(\sqrt{-6})\subset\mathbf Q(\zeta_{24})$, we get $F_0\subset K$; because $K$ is abelian and the only nontrivial proper Galois subfield of $L$ is $F_0$, we have $L\cap K=F_0$, deducing $[LK:\mathbf Q]=48$. The support condition is expressed in a cleaner form: taking the identity on $K$ and a 3-cycle on $L$; the 3-cycle fixes $F_0$, so the two are compatible; the conjugacy class size is $2$, and the exact density is $\delta(\mathcal P)=2/48=1/24$—completely consistent with the independently calculated result in Document 16, the two calculations corroborating each other. Finally, the document confirms that v0.4 additionally ran a completely independent small polynomial-mod-$q$ verifier, scanning $q<10^7$, and once again explicitly states: this only verifies the implementation and density trend, and cannot replace the theorem itself.