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Apparently, nice necessary and sufficient conditions are known for a Galois group of a degree 5 polynomial to have order divisible by 3. What are these conditions?

The possible Galois groups for an irreducible quintic are $S_5$, $A_5$, $D_5$, $C_5$, and $F_{20}$. These only groups with orders divisible by $3$ are $S_5$ and $A_5$, so the problem reduces to characterizing when when a polynomial has one of those as its Galois group.

Alternatively, we want to know if the Galois group contains a 3-cycle. There is a theorem by Dedekind that about reduction mod $p$ can would provide sufficient conditions here: some reduction mod $p$ factors as two linear terms and a cubic. This condition may even be necessary -- I vaguely recall a result to the effect that all of the cycle types in the group appear as mod $p$ reductions, or something like that. Is this true? Regardless, can we find simpler conditions?

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  • $\begingroup$ Are you restricting attention to quintics with rational coefficients? $\endgroup$ – hardmath Jul 22 '15 at 17:55
  • $\begingroup$ @hardmath Yes. $\bf{}$ $\endgroup$ – Potato Jul 22 '15 at 17:56
  • $\begingroup$ Have you heard of the resolvent polynomials and discriminants? $\endgroup$ – GiantTortoise1729 Jul 22 '15 at 18:00
  • $\begingroup$ @GiantTortoise1729 The discriminant will only tell us whether it's in $A_n$, which doesn't seem helpful. I don't recall the theory of the resolvent from memory, but possibly it would help. $\endgroup$ – Potato Jul 22 '15 at 18:01
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The Galois group of an irreducible quintic polynomial has its order divisible by 3 if and only it has a root which is not expressible by radicals (i.e. the polynomial is not solvable by radicals). This is simply because $A_5$ and $S_5$ are the only transitive subgroups of $S_5$ that are not solvable.

To answer your second question, yes all cycle types in the group appear as mod $p$ reductions. This follows from the Frobenius density theorem.

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  • $\begingroup$ Ah, OK. I was thinking of some condition in terms of the coefficients but I think this was probably what was intended. After all, there's no way to characterize this in terms of the coefficients easily! $\endgroup$ – Potato Jul 22 '15 at 18:07
  • $\begingroup$ Brandon, did you mean Chebotarev's density theorem? $\endgroup$ – Jyrki Lahtonen Jul 23 '15 at 9:50
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    $\begingroup$ Jyrki, I meant the Frobenius density theorem (it is referred to as the "theorem of Frobenius" in the second paragraph of the section I linked to). It certainly follows from Chebotarev, as Chebotarev generalizes Frobenius' theorem, but the full power of Chebotarev is not necessary and the proof of Frobenius density is much simpler, see mathoverflow.net/questions/136025/frobenius-density-theorem. $\endgroup$ – Brandon Carter Jul 23 '15 at 16:27
  • $\begingroup$ Thanks. Feeling a bit more educated :-) $\endgroup$ – Jyrki Lahtonen Jul 24 '15 at 19:18
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As mentioned in the question, the order of the Galois group is divisible by three if and only if it is isomorphic to either $A_5$ or $S_5$, i.e. if and only if it is not solvable.

According to Dummit, "Solving Solvable Quintics", a quintic polynomial $$ x^5+px^3+qx^2+rx+s $$ with rational coefficients has solvable Galois group if and only if the associated sextic polynomial $$ x^6 + ax^5 + bx^4 + cx^3 +dx^2 + ex +f $$ has a rational root, where \begin{align*} a \;&=\; 8r \\[12pt] b \;&=\; 2pq^2-6p^2r+40r^2-50qs, \\[12pt] c \;&=\; −2q^4 + 21pq^2r − 40p^2 r^2 + 160r^3 − 15p^2qs − 400qrs + 125ps^2, \\[12pt] d \;&=\; p^2q^4 − 6p^3q^2r − 8q^4r + 9p^4r^2 + 76pq^2r^2 − 136p^2r^3 + 400r^4 − 50pq^3s \\ &\qquad + 90p^2qrs − 1400qr^2s + 625q^2s^2 + 500prs^2, \\[12pt] e \;&=\; −2pq^6 + 19p^2q^4 r − 51p^3 q^2 r^2 + 3q^4 r^2 + 32p^4 r^3 + 76pq^2 r^3 − 256p^2 r^4 \\ &\qquad + 512r^5 − 31p^3 q^3 s − 58q^5 s + 117p^4 qrs + 105pq^3 rs + 260p^2 qr^2 s \\ &\qquad − 2400qr^3 s − 108p^5 s^2 − 325p^2 q^2 s^2 + 525p^3 rs^2 + 2750q^2 rs^2 \\ &\qquad − 500pr^2 s^2 + 625pqs^3 − 3125s^4, \\[12pt] f\;&=\; q^8 − 13pq^6 r + p^5 q^2 r^2 + 65p^2 q^4 r^2 − 4p^6 r^3 − 128p^3 q^2 r^3 + 17q^4 r^3 \\ &\qquad + 48p^4 r^4 − 16pq^2 r^4 − 192p^2 r^5 + 256r^6 − 4p^5 q^3 s − 12p^2 q^5 s \\ &\qquad + 18p^6 qrs + 12p^3 q^3 rs − 124q^5 rs + 196p^4 qr^2 s + 590pq^3 r^2 s \\ &\qquad − 160p^2 qr^3 s − 1600qr^4 s − 27p^7 s^2 − 150p^4 q^2 s^2 − 125pq^4 s^2 \\ &\qquad − 99p^5 rs^2 − 725p^2 q^2 rs^2 + 1200p^3 r^2 s^2 + 3250q^2 r^2 s^2 − 2000pr^3 s^2 \\ &\qquad − 1250pqrs^3 + 3125p^2 s^4 − 9375rs^4. \end{align*}

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  • $\begingroup$ It would help future Readers if make explicit the connection between solvability and the condition of Galois group divisible by three. This is brought out in the other Answer, but someone browsing here might not think to look there. $\endgroup$ – hardmath Jul 22 '15 at 19:02

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