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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.