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We know that a cubic equation generates a cubic cyclic extension iff it has a perfect square discriminant. Now I am wondering if there is a similar condition for quintic polynomials. So I am trying to find "are there any conditions on the coefficients of a quintic polynomials under which it generates a cyclic quintic extension?". Or anything that will help in this direction.

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    $\begingroup$ Are you assuming that your base field is $\mathbb{Q}$? $\endgroup$ Jan 21, 2015 at 18:21
  • $\begingroup$ @PaceNielsen Yes I think $\mathbb{Q}$ will be a very good starting point. Unless I am mistaken the result for cubic extension is true for $\mathbb{Q}$. For example see "On cubic Galois field extensions" by Lothar Häberle sciencedirect.com/science/article/pii/S0022314X09002388 $\endgroup$
    – Chunna
    Jan 22, 2015 at 8:10

3 Answers 3

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An answer can be found in the beautiful book by Cox Galois Theory, Theorem 13.2.6 page 372. The result is as follows.

Theorem. Assume that $f \in F[x]$ is monic, separable and irreducible of degree $5$ and that $\textrm{char}(F) \neq 2$. Let $L$ be the splitting field of $f$ over $F$, and set $\textrm{Gal}(L/F)=G \subset S_5$. Then the following holds.

(a) $G \subset A_5$ if and only if the discriminant $\Delta(f)$ is a square in $F;$

(b) $G$ is conjugate to a subgroup of $\textrm{AGL}(1, \, \mathbb{F}_5)$ if and only if the sextic resolvent $\theta_f(x)$ of $f$ has a root in $F;$

(c) $G$ is conjugate to $\langle (12345) \rangle$ (hence cyclic) if and only if $F$ splits completely over $F(\alpha)$, where $\alpha$ is a root of $f.$

In particular, the fact that the discriminat of $f$ is a perfect square is a necessary, but not sufficient condition in order to have a cyclic quintic extension.

Moreover, if $\textrm{char}(F)=0$ then $f$ is solvable by radicals over $F$ if and only if $G$ is isomorphic to a subgroup of $\textrm{AGL}(1, \, \mathbb{F}_5)$, i.e. if and only if its sextic resolvent $\theta_f(x)$ has a root in $F$.

Over the rationals, a more precise criterion useful to compute the Galois group of $f$ (and in particular to check if it is cyclic) can be found in the paper

D. S. Dummit, Solving solvable quintics, Math. Comp. 57 (1991), 387-401,

see in particular Theorem $2$. The conditions given are the following: $G$ is the cyclic group of order $5$ if and only if $\Delta(f)$ is a perfect square, the sextic resolvent $\theta_f(x)$ has a rational root and two quadratic polynomials (constructed in a somewhat complicate way from the coefficients of $f$) are reducible over $\mathbb{Q}$.

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  • $\begingroup$ Thanks Francesco for your answer. But I am sorry I already knew this. Can we find a more precise answer? $\endgroup$
    – Chunna
    Jan 21, 2015 at 10:27
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    $\begingroup$ Then you should have added this in the background of your question :-) Anyway, I added a further reference. $\endgroup$ Jan 21, 2015 at 10:42
  • $\begingroup$ In principle, one could try to look at the generating polynomial of the extension; if it remains irreducible modulo some prime number $p$, then the extension must be cyclic (see, for instance, Theorem 8.23 of jmilne.org/math/CourseNotes/ANTe6.pdf ). $\endgroup$
    – Albertas
    Jan 21, 2015 at 11:41
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Robin French gave a talk at the American Math Society meeting last week, "A new algorithm for Galois groups of quintic polynomials." If I recall correctly, he constructs a resolvent of degree 24, and the degrees of the irreducible factors of that resolvent are all you need to know to get the Galois group of the original irreducible quintic.

EDIT: This degree 24 resolvent is also discussed in Awtrey and Shill, "Absolute resolvents and masses of irreducible quintic polynomials," pages 31-42 of the book, Collaborative Mathematics and Statistics Research, Topics from the 9th Annual UNCG Regional Mathematics and Statistics Conference. Bits of the book are accessible on Google Books.

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Did not realize this question was here. I went through the chapter in Cox, Galois Theory where he says how Gauss produced polynomials solved by cyclotomic elements... In particular, all have five real roots, each twice a cosine of $2 k \pi / p$ or a sum of such cosines, where $p \equiv 1 \pmod 5$ is prime. A few of these are in wikipedia primes up to 71. For primes up to 100, also in Reuschle (1875).

x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1   p  11 p.root  2 exps 10^k  d  =  11^4
  x^5 + x^4 - 12 x^3 - 21 x^2 + 1 x + 5   p  31 p.root  3 exps 6^k  d  =  5^2 31^4
  x^5 + x^4 - 16 x^3 + 5 x^2 + 21 x - 9   p  41 p.root  6 exps 3^k  d  =  3^6 41^4
  x^5 + x^4 - 24 x^3 - 17 x^2 + 41 x - 13   p  61 p.root  2 exps 21^k  d  =  29^2 61^4
  x^5 + x^4 - 28 x^3 + 37 x^2 + 25 x + 1   p  71 p.root  7 exps 23^k  d  =  23^2 71^4
  x^5 + x^4 - 40 x^3 + 93 x^2 - 21 x - 17   p  101 p.root  2 exps 32^k  d  =  17^2 101^4
  x^5 + x^4 - 52 x^3 - 89 x^2 + 109 x + 193   p  131 p.root  2 exps 18^k  d  =  79^2 131^4
  x^5 + x^4 - 60 x^3 - 12 x^2 + 784 x + 128   p  151 p.root  6 exps 23^k  d  =  2^18 151^4
  x^5 + x^4 - 72 x^3 - 123 x^2 + 223 x - 49   p  181 p.root  2 exps 17^k  d  =  7^2 149^2 181^4
  x^5 + x^4 - 76 x^3 - 359 x^2 - 437 x - 155   p  191 p.root  19 exps 11^k  d  =  5^2 11^2 191^4
  x^5 + x^4 - 84 x^3 - 59 x^2 + 1661 x + 269   p  211 p.root  2 exps 26^k  d  =  31^2 67^2 211^4
  x^5 + x^4 - 96 x^3 - 212 x^2 + 1232 x + 512   p  241 p.root  7 exps 11^k  d  =  2^16 11^2 241^4
  x^5 + x^4 - 100 x^3 - 20 x^2 + 1504 x + 1024   p  251 p.root  6 exps 2^k  d  =  2^18 5^4 251^4
  x^5 + x^4 - 108 x^3 - 401 x^2 - 13 x + 845   p  271 p.root  6 exps 12^k  d  =  5^2 13^4 271^4
  x^5 + x^4 - 112 x^3 - 191 x^2 + 2257 x + 967   p  281 p.root  3 exps 6^k  d  =  193^2 281^4
  x^5 + x^4 - 124 x^3 + 535 x^2 - 413 x - 539   p  311 p.root  17 exps 11^k  d  =  7^4 13^2 311^4
  x^5 + x^4 - 132 x^3 - 887 x^2 - 1843 x - 1027   p  331 p.root  3 exps 13^k  d  =  13^2 31^2 331^4
  x^5 + x^4 - 160 x^3 + 369 x^2 + 879 x - 29   p  401 p.root  3 exps 26^k  d  =  29^2 401^4 433^2
  x^5 + x^4 - 168 x^3 + 219 x^2 + 3853 x - 3517   p  421 p.root  2 exps 32^k  d  =  223^2 239^2 421^4
  x^5 + x^4 - 172 x^3 - 724 x^2 + 1824 x + 1728   p  431 p.root  7 exps 47^k  d  =  2^20 3^4 431^4
  x^5 + x^4 - 184 x^3 - 129 x^2 + 4551 x + 5419   p  461 p.root  2 exps 13^k  d  =  163^2 461^4 491^2
  x^5 + x^4 - 196 x^3 + 59 x^2 + 2019 x + 1377   p  491 p.root  2 exps 32^k  d  =  3^4 17^2 229^2 491^4
  x^5 + x^4 - 208 x^3 - 771 x^2 + 4143 x + 2083   p  521 p.root  3 exps 24^k  d  =  61^2 521^4 577^2
  x^5 + x^4 - 216 x^3 + 1147 x^2 - 805 x - 2629   p  541 p.root  2 exps 11^k  d  =  11^2 311^2 541^4
  x^5 + x^4 - 228 x^3 + 868 x^2 + 3056 x - 7552   p  571 p.root  3 exps 2^k  d  =  2^22 31^2 571^4
  x^5 + x^4 - 240 x^3 + 1755 x^2 - 3731 x + 2399   p  601 p.root  7 exps 17^k  d  =  5^2 13^2 17^2 601^4
  x^5 + x^4 - 252 x^3 + 2095 x^2 - 5785 x + 5069   p  631 p.root  3 exps 24^k  d  =  89^2 631^4
  x^5 + x^4 - 256 x^3 - 564 x^2 + 5328 x - 5120   p  641 p.root  3 exps 21^k  d  =  2^16 5^2 61^2 641^4
  x^5 + x^4 - 264 x^3 - 185 x^2 + 16837 x + 4851   p  661 p.root  2 exps 32^k  d  =  3^16 7^2 661^4
  x^5 + x^4 - 276 x^3 - 1299 x^2 + 5329 x + 15581   p  691 p.root  3 exps 11^k  d  =  379^2 397^2 691^4
  x^5 + x^4 - 280 x^3 + 2047 x^2 - 3791 x + 1699   p  701 p.root  2 exps 23^k  d  =  17^2 19^2 23^2 701^4
  x^5 + x^4 - 300 x^3 - 2313 x^2 - 3761 x - 571   p  751 p.root  3 exps 11^k  d  =  41^2 631^2 751^4
  x^5 + x^4 - 304 x^3 + 2831 x^2 - 8925 x + 8775   p  761 p.root  6 exps 3^k  d  =  3^4 5^2 23^2 761^4
  x^5 + x^4 - 324 x^3 - 3471 x^2 - 12431 x - 13603   p  811 p.root  3 exps 12^k  d  =  7^4 47^2 811^4
  x^5 + x^4 - 328 x^3 - 1215 x^2 + 3573 x + 2179   p  821 p.root  2 exps 32^k  d  =  37^4 109^2 821^4
  x^5 + x^4 - 352 x^3 - 2361 x^2 + 4257 x + 9967   p  881 p.root  3 exps 29^k  d  =  29^2 881^4 953^2
  x^5 + x^4 - 364 x^3 - 2988 x^2 - 1392 x + 9856   p  911 p.root  17 exps 22^k  d  =  2^18 7^2 11^2 911^4
  x^5 + x^4 - 376 x^3 + 3877 x^2 - 13445 x + 15271   p  941 p.root  2 exps 12^k  d  =  191^2 941^4
  x^5 + x^4 - 388 x^3 + 1476 x^2 + 8304 x + 7168   p  971 p.root  6 exps 2^k  d  =  2^20 7^2 13^2 971^4
  x^5 + x^4 - 396 x^3 + 2101 x^2 + 8039 x - 1819   p  991 p.root  6 exps 30^k  d  =  107^2 991^4 1399^2

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same rules but degree 7

$$ x^7 + x^6 - 12 x^5 - 7 x^4 + 28 x^3 + 14 x^2 - 9 x + 1, \; \; p = 29, \; \; r = 2, \; \; 12^k $$ $$ x^7 + x^6 - 18 x^5 - 35 x^4 + 38 x^3 + 104 x^2 + 7 x - 49, \; \; p = 43, \; \; r = 3, \; \; 7^k $$ $$ x^7 + x^6 - 30 x^5 + 3 x^4 + 254 x^3 - 246 x^2 - 245 x + 137, \; \; p = 71, \; \; r = 7, \; \; 14^k $$ $$ x^7 + x^6 - 48 x^5 + 37 x^4 + 312 x^3 - 12 x^2 - 49 x - 1, \; \; p = 113, \; \; r = 3, \; \; 35^k $$
$$ x^7 + x^6 - 54 x^5 - 31 x^4 + 558 x^3 - 32 x^2 - 1713 x + 1121, \; \; p = 127, \; \; r = 3, \; \; 24^k $$ $$ x^7 + x^6 - 84 x^5 - 217 x^4 + 1348 x^3 + 3988 x^2 - 1433 x - 1163, \; \; p = 197, \; \; r = 2, \; \; 20^k $$ $$ x^7 + x^6 - 90 x^5 + 69 x^4 + 1306 x^3 + 124 x^2 - 5249 x - 4663, \; \; p = 211, \; \; r = 2, \; \; 10^k $$ $$ x^7 + x^6 - 102 x^5 - 195 x^4 + 1850 x^3 + 978 x^2 - 8933 x + 5183, \; \; p = 239, \; \; r = 7, \; \; 23^k $$ $$ x^7 + x^6 - 120 x^5 - 711 x^4 - 784 x^3 + 1956 x^2 + 2863 x - 343, \; \; p = 281, \; \; r = 3, \; \; 61^k $$ $$ x^7 + x^6 - 144 x^5 + 399 x^4 + 2416 x^3 - 10808 x^2 + 10831 x - 1237, \; \; p = 337, \; \; r = 10, \; \; 38^k $$

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