What is the smallest degree monic polynomial with a/b as a root?

The discriminant of f(x) := 2x^3 + 2x + 1 is Δ = -172 < 0, so the polynomial does indeed have two distinct complex conjugate roots, and one real root. Call the real root c. Since the constant term of f is nonzero, a, b, and c are nonzero. Let E be a splitting field of f(x) over the rationals.

If f had any rational roots, they would be of the form ±g/h, where g divides the constant term of f (1) and h divides its leading coefficient (2). This gives as candidates only ±1, ±1/2. None of these are roots, so f has no rational roots and so is irreducible over Q. Therefore, [Q(a):Q] = 3 and so 3 divides [E:Q]. Also, Δ is not a square in the rationals, so [Q(√Δ):Q] = 2 and 2 must divide [E:Q]. However, E/Q is Galois and generated by a, b, and c, and Gal(E/Q) permutes the set {a, b, c}, so Gal(E/Q) is isomorphic to a subgroup of S_3. Therefore, we must have [E:Q] = 6 and Gal(E/Q) ≅ S_3.

For each permutation η of {a, b, c}, let σ[η] be the map in Gal(E/Q) restricting to this permutation. We have σ[(a b)](a/b) = b/a. If we write a = r exp(iθ), where r > 0, 0 ≤ θ < 2π, then b = r exp(-iθ), so a/b = exp(2iθ) and b/a = exp(-2iθ). Therefore, if a/b = b/a, θ must be a multiple of π/2, so a is pure imaginary, a = ±ri; but this is impossible as then

f(a) = 1 ± 2i (r – r^3) ≠ 0.

Since a, b, and c are all distinct, we have σ[(a c)](a/b) = c/b ≠ a/b and σ[(b c)](a/b) = a/c ≠ a/b. Finally, if σ[(a b c)](a/b) = b/c was equal to a/b, we would have b^3 = abc = -1/2; this means that θ must be π/3 or 5π/3, but this is impossible as then

f(a) = (1 – 2r^3) + 2r exp(iθ) ≠ 0.

Therefore, a/b is fixed by no subgroup of Gal(E/Q), so since E/Q is Galois, E = Q(a/b) and the minimal polynomial, p(x), of a/b over the rationals will have the conjugates of a/b as roots:

p(x) = (x – a/b)(x – b/a)(x – a/c)(x – c/a)(x – b/c)(x – c/b)

= (x^2 – (a^2 + b^2)/(ab) x + 1) (x^2 – (a^2 + c^2)/(ac) x + 1) (x^2 – (b^2 + c^2)/(bc) x + 1). (*)

For nonnegative integers i, j, and k, write S[ijk] for the symmetrization of a^i b^j c^k, so that, for example,

S[110] = ab + ac + bc,

S[210] = a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2.

Also, write

T = S[1] = a + b + c,

U = S[110] = ab + ac + bc,

V = S[111] = abc.

Then expanding (*) gives

p(x) = x^6

– (S[210] / V) x^5

+ (3 + (S[411] + S[321] + S[330]) / V^2) x^4

– ((2 S[210] / V) + (S[420] + 2 V^2) / V^2) x^3

+ (3 + (S[411] + S[321] + S[330]) / V^2) x^2

– (S[210] / V) x

+ 1. (**)

Expanding the products TU, T^3, U^3, and T^2 U^2 gives the identities

S[210] = TU – 3V,

S[321] = V S[210],

S[3] = T^3 – 3 S[210] – 6V,

S[411] = V S[3],

S[330] = U^3 – 3 S[210] V – 6 V^2,

S[420] = T^2 U^2 – 2 S[330] – 8 S[321] – 2 S[411] – 15 V^2.

In our case we have

T = 0, U = 1, V = -1/2,

giving

S[210] = 0*1 – 3*(-1/2) = 3/2,

S[321] = (-1/2)*(3/2) = -3/4,

S[3] = -3*(3/2) – 6*(-1/2) = -3/2,

S[411] = (-1/2)*(-3/2) = 3/4,

S[330] = 1^3 – 3*(3/2)*(-1/2) – 6*(-1/2)^2 = 7/4,

S[420] = – 2*(7/4) – 8*(-3/4) – 2*(3/4) – 15*(-1/2)^2 = -11/4

and substituting these values into (**) gives

p(x) = x^6 + 3 x^5 + 10 x^4 + 15 x^3 + 10 x^2 + 3 x + 1.

Since this polynomial has integer coefficients, it is the solution to the problem.

1

because of the fact of fact which you haven’t any longer gotten an answer yet, i’ll show out that Wolfram Alpha can a minimum of compute the minimum polynomial in question: x^6 + 3x^5 + 10x^4 + 15x^3 + 10x^2 + 3x + a million. Naively plugging in undesirable expressions for the roots in could coach that the ratio is a root, and there are deterministic algorithms for proving the irreducibility of polynomials over Q, yet this could be a foul and backwards attitude.

Since you haven’t gotten an answer yet, I’ll mention that Wolfram Alpha can at least compute the minimal polynomial in question: x^6 + 3x^5 + 10x^4 + 15x^3 + 10x^2 + 3x + 1. Naively plugging in awful expressions for the roots in would prove that the ratio is a root, and there are deterministic algorithms for proving the irreducibility of polynomials over Q, but this is a terrible and backwards approach.

Source(s): http://www.wolframalpha.com/input/?i=roots+of+2x^3…
http://www.wolframalpha.com/input/?i=minimal+polyn…