How to calculate the Galois group of $x^5+15x+12$?

Question

How to calculate the Galois group of $x^5+15x+12$ over the field $\Bbb Q$? Using the Tchebotarov Density theorem which states that "the density of primes $p$ for which $f(x)$ splits into type $T$ modulo $p$ is precisely $d_T=\frac{n_T}N$[ where $N=|G|$ order of the Galois group $G$ and $n_T$ is the number of the cycle type $T$ in $G$], I guess that $G=F_{20}$ the Frobenius group of order $20$.

Answer 1

Hint The polynomial $f(x) := x^5 + 15 x + 12$ is Einstein at $3$, so is irreducible, and thus $\operatorname{Gal}(f)$ is a transitive subgroup of $S_5$, namely, one of $S_5, A_5, F_{20}, D_{10}, C_5$. The discriminant of $f$ is $\Delta := 2^{10} 3^4 5^5$, which is not a square, so $\operatorname{Gal}(f) \not\leq A_5$, leaving only the possibilities $S_5$ and $F_{20}$. To distinguish between these possibilities, compute and factor Cayley's resolvent $R$ of $f$.

Recall that an irreducible quintic $f$ is solvable by radicals iff Cayley's resolvent $R(x) = P(x)^2 - 2^{10} \Delta x$ has a rational root; for polynomials $x^5 + a x + b$, $P(x) = x^3 - 20 a x^2 + 240 a^2 x + 320 a^3$. Evaluating for our polynomial ($a = 15, b = 12$) and factoring gives $R(x) = (x - 180) q(x)$ for some quintic polynomial $q$. Thus, $f$ is solvable by radicals, so $\operatorname{Gal}(f) \cong F_{20}$.

Answer 2

Here is a supplement of Travis' answer, which I prove an irreducible quintic over $\mathbb{Q}$ is solvable by radical iff its Cayley resolvent has a rational root.

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Consider formal variables $x_1,\cdots,x_5$. Denote $$\theta_1 = x_1^2 x_2 x_5 + x_1^2 x_3 x_4 + x_2^2 x_1 x_3 + x_2^2 x_4 x_5 + x_3^2 x_1 x_5 + x_3^2 x_2 x_4 + x_4^2 x_1 x_2 + x_4^2 x_3 x_5 + x_5^2 x_1 x_4 + x_5^2 x_2 x_3$$

The stabilizer of $\theta_1$ under the action of $S_5$ is a group $M$, you can check it has order $20$ (isomorphic to $F_{20}$). The orbit of $\theta_1$ by $S_5$ consists of six elements, denote them by $\{\theta_1, \cdots, \theta_6\}$. Note that normalizer of $M$ in $S_5$ is itself.

When $x_i$ are substituted as roots of an irreducible quintic $f$, the Cayley resolvent is defined to be the polynomial $$f_{20}(x):=(x-\theta_1)\cdots(x-\theta_6)$$ evidently it is in $\mathbb{Q}[x]$.

If the Galois group $G$ is $S_5$ or $A_5$, then $G$ acts transitively on $\{\theta_1, \cdots, \theta_6\}$. If one of $\theta_i$ is rational, then all roots of $f_{20}$ are the same. But Cayley resolvent of an irreducible polynomial with Galois group containing $A_5$ over characteristic $0$ field has distinct roots, contradiction. Therefore $f_{20}$ has no rational root.

If the Galois group $G$ is $C_5$, $D_5$ or $F_{20}$, we choose the a conjugate of $G$ which lies in our $M$ (possible because $M$ is self-normalizing), then $G$ fixes $\theta_1$, so $\theta_1\in \mathbb{Q}$. This completes the proof.