696.e1 Chebotarev Support Family

Define the prime family $\mathcal P=\{q\text{ prime}: q\equiv1\pmod{24},\ (\frac q{29})=1,\ f_2\bmod q\text{ irreducible}\}$. The document first proves why this set of congruence conditions is effective: $q\equiv1\pmod{24}$ gives $q\equiv1\pmod8$ and $q\equiv1\pmod3$, making $2,3$ split in $\mathbb Q(\sqrt q)$; adding $(\frac q{29})=1$ makes $29$ split as well—all three primes of the conductor, $2,3,29$, split. Next, it performs the actual Chebotarev computation: let $L$ be the Galois closure of $f_2$ ($\mathrm{Gal}(L/\mathbb Q)=S_3$, resolvent $F_0=\mathbb Q(\sqrt{-174})$), and $K=\mathbb Q(\zeta_{24},\sqrt{29})$. Since $\sqrt{-174}=\sqrt{-6}\sqrt{29}$ and $\mathbb Q(\sqrt{-6})\subset\mathbb Q(\zeta_{24})$, it can be proven that $L\cap K=F_0$, hence $[LK:\mathbb Q]=48$; taking a 3-cycle element of $S_3$ (trivial on $F_0$, which is a valid element of the fiber-product Galois group), its conjugacy class size is $2$, yielding an exact density of $\frac2{48}=\frac1{24}$. The document then proves a very neat corollary—"Automatic ordinary": $q$ being inert in the cubic field is equivalent to $\mathrm{Frob}_q$ being an order-3 element of $S_3\simeq GL_2(\mathbb F_2)$, whose characteristic polynomial $X^2+X+1$ gives trace $\equiv1\pmod2$, meaning $a_q$ is odd; if $q\ge5$ and supersingular, then $q\mid a_q$ combined with the Hasse bound would force $a_q=0$, a contradiction—therefore $q\in\mathcal P\Rightarrow q$ is good ordinary for $E$. Finally, the document provides the first concretely verified example: $q=241$ ($241\equiv1\pmod{24}$, $241\bmod29=9$ is a quadratic residue, $f_2\bmod241$ has no roots and is thus irreducible), direct point counting yields $a_{241}(E)=-7$, so the first explicit twist parameter is $d=241$.