Schweitzer 2018

Write $f(x)=x^4+x^3+2x^2-4x+3$.

Step 0: what must be shown. The group $(\mathbb Z/13)^\times$ is cyclic of order 12, so its fourth powers form the subgroup $H=\{1,3,9\}$ of order 3. Also $0=0^4$. The value $f(a)$ is never $0$ (by the rational root test, an integer root would divide 3, and $f(\pm1),f(\pm3)$ are $3,9,117,87$). If $13\mid d$, then $d\equiv0$ is a fourth power. Otherwise $d$ is a product of primes $p\ne13$ dividing $f(a)$, and since $H$ is closed under multiplication, it suffices to prove: every prime $p\ne13$ that divides some $f(a)$ is in $H$ modulo 13.

Step 1: the roots of $f$. Let $\zeta=e^{2\pi i/13}$. The number 2 generates $(\mathbb Z/13)^\times$, so the cosets of $H$ are $2^jH$, $j=0,1,2,3$. Put $$\eta_j=\sum_{h\in H}\zeta^{\,2^jh}.$$ For $r$ prime to 13, the automorphism $\sigma_r:\zeta\mapsto\zeta^r$ permutes the $\eta_j$ (it sends the coset $2^jH$ to $r2^jH$), and it fixes each $\eta_j$ exactly when $r\in H$. So the coefficients of $\prod_j(x-\eta_j)$ are fixed by every $\sigma_r$; they are rational algebraic integers, hence integers. Integers are pinned down by approximations to within $\tfrac12$, and a numerical evaluation gives $$\prod_{j=0}^3(x-\eta_j)=x^4+x^3+2x^2-4x+3=f(x).$$ The same kind of computation gives the discriminant $\Delta=\prod_{i<j}(\eta_i-\eta_j)^2=3^2\cdot13^3$. In particular the four $\eta_j$ are distinct.

Step 2: Frobenius. Let $p\ne13$ be a prime with $p\mid f(a)$, and let $\mathfrak P$ be a maximal ideal of $\mathbb Z[\zeta]$ containing $p$. For $z=\sum c_i\zeta^i\in\mathbb Z[\zeta]$, the binomial theorem in characteristic $p$ and Fermat’s little theorem give $$z^p\equiv\sum c_i^p\zeta^{ip}\equiv\sum c_i\zeta^{ip}=\sigma_p(z)\pmod{p}.$$ Since $0=f(a)=\prod_j(a-\eta_j)$ lies in the prime ideal $\mathfrak P$, we have $a\equiv\eta_j\pmod{\mathfrak P}$ for some $j$. Then $$\sigma_p(\eta_j)\equiv\eta_j^{\,p}\equiv a^p\equiv a\equiv\eta_j\pmod{\mathfrak P}.$$ If $p\notin H$ (mod 13), then $\sigma_p(\eta_j)=\eta_k$ with $k\ne j$, so $\eta_k-\eta_j\in\mathfrak P$, hence $\Delta\in\mathfrak P$ and $p\mid\Delta=3^2\cdot13^3$. That forces $p=3$, but $3\in H$. So in every case $p$ is $1$, $3$ or $9$ modulo 13, as needed.