Newsgroups: sci.math From: pmontgom@cwi.nl (Peter L. Montgomery) Subject: Re: Galois groups of polynomials Nntp-Posting-Host: bark.cwi.nl Date: Sun, 19 Jul 1998 06:12:59 GMT In article <35acf2a3.6967279@news.belgium.eu.net> bogaerm@gib.be writes: >Until a little while I was convinced that the galois group of a >polynomial of degree p was the group S_p of permutations of p elements >(generally speaking, unless the polynomial was rather "special". >Somewhere (I lost the reference) I saw that the Galois group of the >polynomial > x^6+2*x^5+3*x^4+4*x^3+5*x^2+6*x+7 >is _not_ S6. >How can one derive such conclusions ? # This is a Maple script, containing information about how this f splits. f := x^6 + 2*x^5 + 3*x^4 + 4*x^3 + 5*x^2 + 6*x + 7; # The discriminant of f is -2^16 * 7^4. # This is the negative of a perfect square. # Hence sqrt(-1) is in the splitting field. ifactor(discrim(f, x)); # Negative of perfect square # Check how f splits mod p, where p is prime in [1000, 1200]. for p from 1001 to 1199 by 2 do if (isprime(p)) then print(p, Factor(f) mod p); fi; od; # Degrees Primes # 1,1,1,1,1,1 1013 # 1,1,2,2 1021, 1049, 1069, 1097, 1109, 1129, 1193 # 1,1,4 1031, 1039, 1051, 1087, 1091, 1123, 1151, 1171 # 1,5 1009, 1033, 1061, 1181 # 2,2,2 1063, 1087 # 3,3 1093, 1117, 1153 # 6 1103, 1163 # Some degree patterns are missing, notably 2,4 and 1,2,3 # but also 1,1,1,3 and 1,1,1,1,2. # In S6, 90 of the 720 permutations have cycles of lengths # 2 and 4, and another 90 have cycle lengths 1, 1, 4. # Here f splits into factors of degrees 1, 1, 4 # modulo 8 of 27 primes, but never into 2 and 4. # Unless we have encountered a statistical fluke, # the Galois group of f is _not_ S6. # Nor is it A6, since the discriminant is not a perfect square. # According to Maple, it has order 120. galois(f); # If r1 is a root of f, then f(x)/(x - r1) is irreducible # over Q(r1) since f splits into degrees 1,5 modulo some primes. # Even if we start with two distinct roots r1 and r2 of f, # the quotient polynomial f(x)/((x - r1)*(x - r2)) # remains irreducible over Q(r1, r2) since f sometimes # splits into degrees 1,1,4. # If we start with three distinct roots r1, r2, r3 of f, # then Q(r1, r2, r3) will have degree 6 * 5 * 4 = 120 over Q. # If we believe the Galois group has order 120, # then Q(r1, r2, r3) must be the splitting field of f. # That is, f has six linear factors over Q(r1, r2, r3). # To confirm this, I give a fourth root (r4) below. # The last roots can be found by permuting r1, r2, r3. # I also give an expression for sqrt(-1). # Can anyone come up with a more elegant expression for # r4 in terms of r1, r2, r3? Ideally, if this # has the form r4 = h(r1, h2, r3), then we can tell by # inspection whether r3 = h(r1, r2, r4). r1poly := subs(x = r1, f); r2poly := normal((subs(r1 = r2, r1poly) - r1poly)/(r2 - r1)); r3poly := normal((subs(r2 = r3, r2poly) - r2poly)/(r3 - r2)); r4poly := normal((subs(r3 = r4, r3poly) - r3poly)/(r4 - r3)); sym1 := r1 + r2 + r3; sym2 := r1*r2 + r1*r3 + r2*r3; sym3 := r1*r2*r3; sqrtm1 := expand(5 + 7*sym1 - 15*sym1^2 + 27*sym2 - 11*sym1^3 + 18*sym1*sym2 + 3*sym3 - 3*sym2^2 - 3*sym2*sym1^2 + 14*sym1*sym3 - 3*sym2^2*sym1 - 6*sym2*sym3 + 11*sym3*sym1^2 + 3*sym2^3 + 3*sym3^2- 2*sym1*sym2*sym3 + 5*sym3*sym2^2 - 3*sym3^2*sym1 + 3*sym3^2*sym2 + 3*sym3^3)/56; simplify(sqrtm1^2, {r3poly, r2poly, r1poly}); # -1 r4 := (17*r2^3*r1^3+(6*r2^3*r1^3+4*r2^3*r1^2+4*r2^2*r1^3+2*r2^3*r1+6*r2^2*r1 ^2+2*r2*r1^3+4*r2^3+4*r1^3-2*r2^2-18*r2*r1-2*r1^2-8*r2-8*r1+6)*r3^5-9*r2^3*r1^ 2-9*r2^2*r1^3+(11*r2^3*r1^3+7*r2^3*r1^2+7*r2^2*r1^3+3*r2^3*r1+7*r2^2*r1^2+3*r2 *r1^3+9*r2^3+r2^2*r1+r2*r1^2+9*r1^3+3*r2^2-19*r2*r1+3*r1^2+9*r2+9*r1+21)*r3^4-\ 5*r2^3*r1-15*r2^2*r1^2-5*r2*r1^3+(12*r2^3*r1^3+4*r2^3*r1^2+4*r2^2*r1^3+8*r2^2* r1^2+8*r2^3+4*r2^2*r1+4*r2*r1^2+8*r1^3+8*r2^2-12*r2*r1+8*r1^2+16*r2+16*r1+24)* r3^3+11*r2^3-13*r2^2*r1-13*r2*r1^2+11*r1^3+(16*r2^3*r1^3+10*r2^3*r1^2+10*r2^2* r1^3+2*r2^3*r1+20*r2^2*r1^2+2*r2*r1^3+8*r2^3+12*r2^2*r1+12*r2*r1^2+8*r1^3+6*r2 ^2-20*r2*r1+6*r1^2+6*r2+6*r1+8)*r3^2-5*r2^2-29*r2*r1-5*r1^2+(22*r2^3*r1^3+12* r2^3*r1^2+12*r2^2*r1^3-2*r2^3*r1+2*r2^2*r1^2-2*r2*r1^3+16*r2^3-4*r2^2*r1-4*r2* r1^2+16*r1^3-10*r2^2-42*r2*r1-10*r1^2-6)*r3+5*r2+5*r1-25)/112; simplify(r4poly, {r3poly, r2poly, r1poly}); # Zero ;quit; -- Peter-Lawrence.Montgomery@cwi.nl San Rafael, California