Let $f=T^{3}+aT+b\in\mathbb{Z}[T]$ irreducible. To prove that if $f$ has cyclic Galois group and primes coefficients $a,b$, then $a=b$,

Edit: added irreducible. (which I didn't).

The discriminant $\Delta=-2^{2}a^{3}-3^{3}b^{2}$ is a square, $r^{2}\in\mathbb{Z}^{2}$. Then $a<0$. If $(a,b)$ is a solution, then $(a,-b)$ is also a solution so we can take $(a,b)=(-p,-q)$ with with $p,q$ prime.

Not all cyclic polynomials $T^3-aT-a$ have prime coefficient: for example, $a=9,27,49,63...$. Furthermore, one has cyclic cubics with only one prime coefficient: $T^3-31T-62$. However, if both coefficients are prime, they seem to be equal and furthermore this prime is $p\equiv 1\mod 3$.

Let $a=-p$, $b=q$ so the discriminant is the square $\Delta=r^2=4p^3-27q^2$. Suppose $p^2\mid r^2$, so $m^2p^2=4p^3-27q^2$ with $m\in \mathbb{Z}$. Then $p^2(4p-m^2)=27q^2$. The case $p=3$ can be eliminated. Since $m^2\neq 4p$, then $p\mid q$ so $p=q$. Hence, $r^2=p^2(4p-27)=m^2p^2$ and $m^2=4p-27$ must be a square. This has lots of solutions in terms of $(m,p)$.

I am not able to eliminate the case $p\neq q$ "by hand". Perhaps one needs more results from the cyclicity of the group than only knowing that the discriminant is a square?

My inspiration dried out there.

Evidence of this by numerical research: the integral coefficients $a,b$ such that $f$ is cubic cyclic is printed, and polynomials with prime coefficients are exclusively on the $x=y$ line (green spots), and exclusion multiple roots/zero discriminant $4x^3+27y^2=0$ in red. enter image description here


2 Answers 2


The Galois group of $x^3 - px + q$ is cyclic if and only if the discriminant $D = 4p^3 - 27q^2 = x^2$ is a square. This is equivalent to $$ p^3 = \frac{x^2 + 27q^2}4 = \alpha \alpha', \quad \text{where} \quad \alpha = \frac{x + 3q\sqrt{-3}}2. $$ If $\gcd(\alpha,\alpha') = 1$ in ${\mathbb Z}[\zeta_3]$, both factors must be cubes up to units, and this leads to a contradiction quickly (use the fact that $q$ is prime).

Thus $\alpha$ and $\alpha'$ must have a factor in common, and this can only be $p$. The case $p = 3$ is quickly eliminated, leaving $p = q$ as the only possibility.

  • $\begingroup$ Once I fully understand this I will eventually complete the details. $\endgroup$
    – NevD
    May 10, 2017 at 19:52

I post this verbatim as some parts are not totally clear for me.

If $\alpha$ and $\alpha'$ are coprime and since their product is the cube $p^3$, each factor must be a cube (I admit this part *): $u\alpha=(a+ b\sqrt{-3})^{3}$ with $u$ a unit in $\mathbb{Z}[\zeta_{3}]$ ( these are $\pm1,\pm\zeta_{3},\pm\zeta_{3}^{2}$, with $u^{3}=1$).

Then expanding $u\alpha=a^{3}+3a{}^{2}b\sqrt{-3}-9ab{}^{2}-3b{}^{3}\sqrt{-3}=u(\frac{r}{2}+\frac{3q}{2}\sqrt{-3})$ and separating terms gives:

$$a^{3}-9ab^{2}=\frac{r}{2}, \quad 2(a^{2}b-b^{3})=q$$ Since $q$ is prime, the last equation gives $a=\pm b$, so with the first equation $\alpha=-16a^{3}(1+\sqrt{-3})$. Multiplying by $\alpha'=-16a^{3}(1-\sqrt{-3})$ gives $p^{3}=2^{10}a^{6}\in \mathbb{Q}$, impossible since $p$ is prime.

Then there exists $\beta$ such that $\beta\mid\alpha$ and $\beta\mid\alpha'$. Let $\gamma=c+d\sqrt{-3}$ and write $\alpha=\beta\gamma$ and $\alpha'=\beta\gamma'$, so $p^{3}=\beta^{2}\gamma\gamma'=\beta^{2}(c^{2}+3d^{2})$, an equation in $\mathbb{Q}$ so $\beta=p$ and $p=c^2+3d^2$.

Then $p$ divides $\alpha=\frac{1}{2}(r+3q\sqrt{-3})$ so divides $2\alpha-r=3q\sqrt{-3}$ so elminating $p=3$ (not cyclic), we have $p$ divides $q$, so $p$=$q$.

Furthermore, $p\equiv c^2\mod 3$. Since in $\mathbb{F}_3$, whether $x^2\equiv 0\mod 3$ or $x^2\equiv 1\mod 3$, so all the "good" primes are $p\equiv 1\mod 3$.

(*) After a bit of research, I understand that $\mathbb{Z}[\zeta_3]$ is a UFD and $\alpha$ and $\alpha'$ being coprime have no common factor, hence the result. This opens a new area for me. Thanks.


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