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Determine the Galois group over $\mathbb{Q}$ of the polynomial $f(X)$, where $$f(X)=X^6-12X^4+15X^3-6X^2+15X+6.$$

Firstly, based on Eisenstein's criterion, pick prime $p=3$, we can show that $f$ is irreducible over $\mathbb{Q}$, but how can we find the Galois group over $\mathbb Q$ of $f$?

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With the aid of Dedekind's theorem we can relatively quickly conclude that the Galois group of this polynomial is the full symmetric group $S_6$.

Because the Galois group $G$ can be viewed as a group of permutations of the six roots of $f(X)$ we only need to identify the subgroup of $G$.

  1. Because $f(X)$ is irreducible, $G$ is a transitive subgroup.
  2. Modulo $2$ we see that $$f(X)=X(X^5+X^2+1).$$ With the latter factor a "known" irreducible, Dedekind tells us that $G$ contains a 5-cycle. This means that the point stabilizer $Stab_G(\alpha)$ of some root $\alpha$ is transitive on the other five roots. Because the various point stabilizers are all conjugate to each other, all of them have transitive stabilizers.
  3. Modulo $p=13$ we get (thanks to Mathematica) $$f(X)=(X+3)(X^2+4X+10)(X^3+6X^2+2x+8),$$ so Dedekind tells us that $G$ contains an element $\sigma$ that is a disjoint product of a 2-cycle and a 3-cycle.
  4. The power $\sigma^3$ is thus a 2-cycle. So $Stab_G(\alpha)$ contains a 2-cycle as well as a 5-cycle. It is a standard exercise to show that a subgroup of $S_5$ containing a 5-cycle and a 2-cycle must be all of $S_5$.
  5. By the previous bullet $Stab_G(\alpha)=S_5$. As $G$ is also transitive, we can conclude that $G$ is all of $S_6$.
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  • $\begingroup$ When we get the whole of $S_n$ Dedekind often takes us the distance. When we have a proper subgroup we need something extra. $\endgroup$ – Jyrki Lahtonen Mar 26 '17 at 17:48
  • $\begingroup$ Tip: Look for an integer $n$ such that $f(n)$ has a small prime factor $p$. Then modulo $p$ you are guaranteed at least one linear factor. Unless there are multiple factors (e.g. when $p$ is a factor of the discriminant), then Dedekind will say something about the point stabilizer within the Galois group. $\endgroup$ – Jyrki Lahtonen Mar 26 '17 at 17:56

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