# Thread: Splitting Field for x^3+1

1. ## Splitting Field for x^3+1

What is the splitting field for x^3+1? The splitting field is obtained by adjoining the roots of this polynomial to Q, but I can't get a reasonable answer simply by doing that. Compare with x^3-1, whose splitting field is obtained by adjoining (-1)^(2/3), which gives a degree two extension. Doing similarly in my case creates a splitting field that, to me, seems "too big", namely the splitting field for x^6-1:

The roots of x^3+1 are -1, (-1)^(1/3) and (-1)^(5/3). Adjoining -1 accomplishes nothing, so lets try (-1)^(1/3) =: u. Its minimal polynomial has degree two, so elements of K:=Q(u) are of the form a+bu. Hence u is in K as is its inverse 1-u. Also, all powers of u are in K, thus for example u^2 and u^4 are in K, i.e. the roots of x^3-1 are in K, and consequently the roots of x^6-1 are in K. Hence, spl(x^3+1) = spl(x^6-1).

What am I doing wrong?

2. ## Re: Splitting Field for x^3+1

nothing. the degree of an extension matches the degree of a polynomial only if the polynomial is IRREDUCIBLE.

let's look at x6 - 1:

this isn't irreducible, it factors as:

x6 - 1 = (x3 - 1)(x3 + 1) = (x - 1)(x + 1)(x2 + x + 1)(x2 - x +1).

so one thing is clear from this factorization, if we adjoin the roots of x2 - x + 1 to Q, we get at least 4 of the roots of x6 - 1.

the 6-th roots of 1 form a cyclic subgroup of the multiplicative group of Q(u)*, where u is a root of x2 - x + 1 (the other root is u5 = 1/u.

it turns out that u + 1/u = 1. you can use trigonometry to prove this: u = 1/2 + (√3/2)i)

we obviously can't have a subgroup of a group of order 6 of order 4: if we have at least 4 elements of it, we have them all.

also, if u is a 6-th root of 1, clearly u2 is a 3-rd root of 1, so u2 is a root of x3 - 1, so we can write:

x6 - 1 = (x - 1)(x + 1)(x - u2)(x - 1/u2)(x - u)(x - 1/u)

(it turns out that u2 + 1/u2 = -1. this can be proved using some complex number algebra, since u2 = -1/2 + (√3/2)i).

another way to look at it:

suppose u is a non-rational root of x3 + 1. this means that u3 = -1.

but -1 is a SQUARE root of 1 (remember every positive number has TWO square roots: the positive one, and the negative one. we often forget about the negative one).

so if u3 = -1, then:

u6 = (u3)2 = (-1)2 = 1, that is: u is a 6-th root of unity.

6 = 3*2, so it is NOT surprising that including the "3 part" (the 6-th roots that come from the cubic x3 + 1), we get everything, because Q already contains the "2-part"

(that is: the factor x2 - 1 of x6 - 1 splits completely over Q).

what IS sort of surprising is this, we have:

Q < Q(u2) ≤ Q(u)

but since [Q(u):Q] = 2, we must have either Q(u2) = Q, or Q(u2) = Q(u).

it turns out that Q(u2) = Q(u). the easiest way to prove this is to show we can write u in terms of u2:

so convince yourself that u = -1/u2, which will prove Q(u) ≤ Q(u2).

Thank you!