Re: Galois group of a given quartic equation
The Doc
ramified in cyclic quartic extensions of Q. The conclusion is that if
a, b, and c are positive integers,
f(x) = x^4 - 2c*x^3 + (c^2 - a^2 - b^2)*x^2 + 2c*a^2 * x - a^2 * c^2
can never have Galois group Z4. It would be interesting if a more
elementary argument could be found. As before, we assume that
gcd(a, b, c) = 1.
From previous considerations, we have reduced the possible occurrences
of Z4 as Galois group of f(x) to cases where the discriminant
D = (4*a*b*c)^2 * ((c^2 - a^2 - b^2)^3 - 27*(abc)^2)
and therefore the factor
D1 = (c^2 - a^2 - b^2)^3 - 27*(abc)^2,
is twice a square. The examples
a = 32 b = 155 c = 247, D1 = 2 * 288560^2
a = 187 b = 224 c = 617, D1 = 2 * 62368976^2
give f(x) having discriminants that are twice squares, and Galois
groups S4 and D4, respectively.
If K is the splitting field of f(x), and K/Q is cyclic of degree 4,
from previous arguments we know 2 must be the unique prime which is
totally ramified in K/Q. For further consideration of the possibility
of K/Q being cyclic of degree 4 when D is twice a square, we can use
the criterion Mr. Holing mentioned in <kh1vsgdip24j@legacy> to
distinguish the cases Z4 and D4, because we now know the field
E = Q(sqrt(D))
is Q(sqrt(2)). If f(x) has Galois group Z4, it must factor over E.
Clearly the factors are irreducible quadratics over E, and are
algebraically conjugate. And since here f(x) is monic, the factors
are monic, and the other coefficients are algebraic integers. So in
our situation the factorization is of the form
f(x) = (x^2 + (u + s*v)*x + u' + s*v')(x^2 + (u - s*v)*x + (u' - s*v')
where s = sqrt(2) and u, v, u' and v' are rational integers. Equating
coefficients,
coeff of x^3: 2u = -2c
coeff of x^2: 2u' + u^2 - 2v^2 = c^2 - a^2 - b^2
coeff of x: 2uu' - 4vv' = 2a^2 * c
constant term: (u')^2 - 2(v')^2 = -a^2 * c^2
Substituting -c for u in the expression for the coefficient of x^2
gives
a^2 + b^2 = 2v^2 - 2u',
so if K/Q is cyclic of degree 4, a^2 + b^2 must be even (in the D4
example above, this sum is odd). Now since K/Q is cyclic of degree 4,
and 2 is totally ramified in K/Q, we have from previous considerations
f(x) == x^4 or (x+1)^4 (mod 2).
This requires the coefficient c^2 - a^2 - b^2 of x^2 to be even.
Since we are assuming
gcd(a, b, c) = 1
the only possibility is that a and b are both odd, and c is even.
Since any odd square is congruent to 1 modulo 8, we have
c^2 - a^2 - b^2 == 2 (mod 4), so
(c^2 - a^2 - b^2)^3 == 8 (mod 16)
If c == 2 (mod 4) then 27*(abc)^2 == 4 (mod 8), which would force
D1 == 4 (mod 8)
which is inconsistent with being twice a square. Therefore, c must be
divisible by 4. A similar argument shows that
if 8|c then D1/8 == -1 (mod 8)
which is incinsistent with D1 being twice a square. Therefore, c is
divisible by 4 but not by 8. So, -a^2 * c^2 is divisible by 16 but
not by 32.
Now, using
r1*r2*r3*r4 = -a^2 * c^2
where r1,r2, r3 and r4 are the zeroes of f(x), we have as in a
previous argument, that if P is the unique prime of K containing 2,
v_P(r1) = v_P(r2) = v_P(r3) = v_P(r4) = v_P(2).
Thus, r1/2, r2/2, r3/2 and r4/2 are algebraic integers (but not in P).
This causes no obvious difficulty with the coefficients of x or x^3.
But with the coefficient of x^2, we have
r1*r2 + r1*r3 + r1*r4 + r2*r3 + r2*r4 + r3*r4 = c^2 - a^2 - b^2.
Each term on the left side is 4 times an algebraic integer, hence so
is the sum. But as previously noted,
c^2 - a^2 - b^2 == 2 (mod 4)
so the coefficient of x^2 in f(x) is not 4 times an algebraic integer.
This (finally!) eliminates the last possibility for f(x) to have
Galois group Z4.
Peter L. Montgomery
>This continues a line of argument using the primes that are totally
>ramified in cyclic quartic extensions of Q. The conclusion is that if
>a, b, and c are positive integers,
>
>f(x) = x^4 - 2c*x^3 + (c^2 - a^2 - b^2)*x^2 + 2c*a^2 * x - a^2 * c^2
>
>can never have Galois group Z4. It would be interesting if a more
>elementary argument could be found.
I observe that f(x) = 0 is equivalent to
a^2/x^2 + b^2/(x - c)^2 - 1 = 0 .
If a/x = sin(theta) and b/(x - c) = cos(theta), then
x = a*csc(theta) and x - c = b*sec(theta), so we are looking
for solutions to a*csc(theta) - b*sec(theta) = c.
Change our variable to t = tan(theta/2):
a*(1 + t^2)/(2*t) - b*(1 + t^2)/(1 - t^2) = c .
Clearing fractions gives
a*t^4 + (2*b - 2*c)*t^3 + (2*b + 2*c)*t - a = 0 .
This polynomial is not monic, but may be easier to analyze
than the original. The two have identical Galois groups.
--
If Scott Peterson had not been convicted, he would be Scott free.
pmon...@cwi.nl Microsoft Research and CWI Home: Bellevue, WA