Splitting Fields and Constructions

Study Sheet

Splitting Fields and Constructions

Fields that contain every root, cyclotomic polynomials, and the ancient construction problems

Splitting Fields

Concept
The splitting field of a polynomial

A splitting field of f∈F[x]f\in F[x] is a smallest extension EE of FF in which ff factors completely into linear factors; it is generated over FF by the roots of ff. Splitting fields exist and are unique up to isomorphism, and for deg⁡f=n\deg f=n, [E:F]≤n![E:F]\le n!.

Kronecker's construction gives one root at a time; repeating it until every root is present builds the splitting field. Adjoining a single root is often not enough: Q(23)\mathbb Q(\sqrt[3]2) contains one root of x3−2x^3-2 but not the two complex ones. The degree bound n!n! comes from adjoining roots one by one, the kk-th step having degree at most n−k+1n-k+1.

Example
The splitting field of x3−2x^3-2

The roots are 23\sqrt[3]2, 23 ω\sqrt[3]2\,\omega, 23 ω2\sqrt[3]2\,\omega^2, where ω=−1+3 i2\omega=\frac{-1+\sqrt3\,i}2. The splitting field is Q(23,ω)\mathbb Q(\sqrt[3]2,\omega): degree 33 for 23\sqrt[3]2, then degree 22 for ω\omega (it is not real, while Q(23)\mathbb Q(\sqrt[3]2) is), so [E:Q]=6=3![E:\mathbb Q]=6=3! — the largest possible for a cubic.

Example
The splitting field of x4−2x^4-2

The roots are ±24\pm\sqrt[4]2 and ±i24\pm i\sqrt[4]2, so the splitting field is Q(24,i)\mathbb Q(\sqrt[4]2,i). [Q(24):Q]=4[\mathbb Q(\sqrt[4]2):\mathbb Q]=4 (x4−2x^4-2 is Eisenstein at 22), and i∉Q(24)⊂Ri\notin\mathbb Q(\sqrt[4]2)\subset\mathbb R, so the total degree is 4⋅2=84\cdot2=8 — less than 4!=244!=24.

Side note
Where it goes wrong

Adjoining one root does not always split the polynomial: Q(23)\mathbb Q(\sqrt[3]2) is real, so it misses 23 ω\sqrt[3]2\,\omega. But sometimes it does: Q(2)\mathbb Q(\sqrt2) already contains both roots ±2\pm\sqrt2 of x2−2x^2-2, and Q(ω)\mathbb Q(\omega) contains both roots ω,ω2\omega,\omega^2 of x2+x+1x^2+x+1.

Try it
Try it: a product of quadratics

What is the degree over Q\mathbb Q of the splitting field of (x2−2)(x2−3)(x^2-2)(x^2-3)?

Show the answer

The splitting field is Q(2,3)\mathbb Q(\sqrt2,\sqrt3), of degree 2⋅2=42\cdot2=\mathbf4 by the tower law (since 3∉Q(2)\sqrt3\notin\mathbb Q(\sqrt2)).

Roots of Unity and Cyclotomic Polynomials

Concept
Cyclotomic polynomials

The nn-th roots of unity e2πik/ne^{2\pi ik/n} form a cyclic group of order nn; its generators are the primitive nn-th roots. The cyclotomic polynomial Φn(x)\Phi_n(x) has the primitive nn-th roots as its roots. It has integer coefficients, degree φ(n)\varphi(n), is irreducible over Q\mathbb Q, and xn−1=∏d∣nΦd(x)x^n-1=\prod_{d\mid n}\Phi_d(x). So [Q(ζn):Q]=φ(n)[\mathbb Q(\zeta_n):\mathbb Q]=\varphi(n).

Each nn-th root of unity is a primitive dd-th root for exactly one divisor dd of nn (its order in the cyclic group), which is why xn−1x^n-1 factors into the Φd\Phi_d. The field Q(ζn)\mathbb Q(\zeta_n) generated by a primitive root is the splitting field of xn−1x^n-1, because all other roots of unity are powers of ζn\zeta_n.

Example
Small cyclotomic polynomials

Φ1=x−1\Phi_1=x-1, Φ2=x+1\Phi_2=x+1, Φ3=x2+x+1\Phi_3=x^2+x+1, Φ4=x2+1\Phi_4=x^2+1, Φ6=x2−x+1\Phi_6=x^2-x+1, Φ8=x4+1\Phi_8=x^4+1, Φ12=x4−x2+1\Phi_{12}=x^4-x^2+1. Their degrees 1,1,2,2,2,4,41,1,2,2,2,4,4 are φ(1),φ(2),φ(3),φ(4),φ(6),φ(8),φ(12)\varphi(1),\varphi(2),\varphi(3),\varphi(4),\varphi(6),\varphi(8),\varphi(12).

Example
Factoring x6−1x^6-1

The divisors of 66 are 1,2,3,61,2,3,6, so x6−1=Φ1Φ2Φ3Φ6=(x−1)(x+1)(x2+x+1)(x2−x+1)x^6-1=\Phi_1\Phi_2\Phi_3\Phi_6=(x-1)(x+1)(x^2+x+1)(x^2-x+1), and the degrees add up: 1+1+2+2=61+1+2+2=6.

Side note
Where it goes wrong

Not every root of xn−1x^n-1 is primitive: ζ62=ω\zeta_6^2=\omega is a primitive cube root, not a primitive sixth root. And Φn\Phi_n is not xn−1+⋯+1x^{n-1}+\cdots+1 in general — that formula is Φn\Phi_n only when nn is prime.

Try it
Try it: a cyclotomic degree

What is [Q(ζ12):Q][\mathbb Q(\zeta_{12}):\mathbb Q]?

Show the answer

φ(12)=12⋅12⋅23=4\varphi(12)=12\cdot\frac12\cdot\frac23=\mathbf4, the degree of Φ12=x4−x2+1\Phi_{12}=x^4-x^2+1.

Constructible Numbers

Concept
What ruler and compass can build

Starting from 00 and 11, the numbers obtainable by ruler and compass are the constructible numbers: they form a field and are closed under square roots, and nothing else can be reached. Hence a constructible number lies in a tower of quadratic extensions, and its degree over Q\mathbb Q is a power of 22.

Each ruler-and-compass step intersects lines and circles, which means solving equations of degree at most 22. So each step adds at most a square root, and the degree over Q\mathbb Q can only grow by factors of 22. That single fact settles three problems that stood open for two thousand years: doubling the cube, trisecting every angle, and squaring the circle are all impossible.

Example
Doubling the cube

A cube of volume 22 needs side 23\sqrt[3]2, whose minimal polynomial x3−2x^3-2 has degree 33. Since 33 is not a power of 22, 23\sqrt[3]2 is not constructible.

Example
Trisecting 60∘60^\circ

Trisecting 60∘60^\circ means constructing cos⁡20∘\cos20^\circ. By cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta with cos⁡60∘=12\cos60^\circ=\frac12, c=cos⁡20∘c=\cos20^\circ satisfies 8c3−6c−1=08c^3-6c-1=0. This cubic has no rational root, so it is irreducible and cc has degree 33: the 60∘60^\circ angle cannot be trisected.

Example
A number that IS constructible

2+3\sqrt{2+\sqrt3} is built by two square roots: 3\sqrt3, then 2+3\sqrt{2+\sqrt3}. Its minimal polynomial is x4−4x2+1x^4-4x^2+1, of degree 4=224=2^2, consistent with the rule.

Side note
Where it goes wrong

Degree a power of 22 is necessary, not sufficient, in general — and an angle being impossible to trisect does not mean NO angle can be trisected: 90∘90^\circ trisects easily into 30∘30^\circ. The theorem says there is no method that works for every angle.

Try it
Try it: constructible or not?

Is 34\sqrt[4]3 constructible?

Show the answer

Yes: 34=3\sqrt[4]3=\sqrt{\sqrt3}, two square roots. (Its degree is 44, a power of 22.) By contrast 33\sqrt[3]3 has degree 33 and is not constructible.

Constructible Regular Polygons

Concept
Gauss and Wantzel

A regular nn-gon is constructible exactly when cos⁡2πn\cos\frac{2\pi}n is constructible, which happens exactly when φ(n)\varphi(n) is a power of 22 — equivalently, when nn is a power of 22 times a product of DISTINCT Fermat primes 3,5,17,257,655373,5,17,257,65537.

cos⁡2πn\cos\frac{2\pi}n generates a field of degree φ(n)2\frac{\varphi(n)}2 over Q\mathbb Q (for n≥3n\ge3), so the degree test says φ(n)\varphi(n) must be a power of 22. Gauss proved the converse at age 19, constructing the regular 1717-gon.

Example
The heptagon is impossible

φ(7)=6\varphi(7)=6, not a power of 22. Concretely, cos⁡2π7\cos\frac{2\pi}7 has minimal polynomial 8x3+4x2−4x−18x^3+4x^2-4x-1, of degree 33. So the regular 77-gon cannot be constructed.

Example
The pentagon is possible

φ(5)=4\varphi(5)=4, and cos⁡2π5=5−14\cos\frac{2\pi}5=\frac{\sqrt5-1}4 has minimal polynomial 4x2+2x−14x^2+2x-1, of degree 22: one square root suffices.

Example
Which n≤20n\le20 work?

Checking φ(n)\varphi(n) for 3≤n≤203\le n\le20, the powers of 22 occur for n=3,4,5,6,8,10,12,15,16,17,20n=3,4,5,6,8,10,12,15,16,17,20 — eleven polygons. The others (7,9,11,13,14,18,197,9,11,13,14,18,19) have φ(n)\varphi(n) with an odd factor greater than 11.

Side note
Where it goes wrong

The Fermat primes must be DISTINCT: 9=3⋅39=3\cdot3 fails, since φ(9)=6\varphi(9)=6. So the regular 99-gon is not constructible even though 33 is a Fermat prime and the triangle is constructible.

Try it
Try it: two polygons

Are the regular 5151-gon and 4545-gon constructible?

Show the answer

51=3⋅1751=3\cdot17, distinct Fermat primes, and φ(51)=2⋅16=32\varphi(51)=2\cdot16=32: yes. 45=9⋅545=9\cdot5 and φ(45)=24\varphi(45)=24: no.

Why It Works

Adjoin a root α1\alpha_1 of ff: degree at most nn. Over F(α1)F(\alpha_1), f=(x−α1)gf=(x-\alpha_1)g with deg⁡g=n−1\deg g=n-1; adjoin a root of gg: degree at most n−1n-1. Continuing, the tower law gives [E:F]≤n(n−1)⋯1=n![E:F]\le n(n-1)\cdots1=n!.

Each new point of a construction is an intersection of lines and circles whose equations have coefficients in the current field KK. Solving such a system needs at most a quadratic equation, so the new coordinates lie in KK or in K(d)K(\sqrt d) with d∈Kd\in K, an extension of degree 11 or 22. A constructible number therefore lies in a field KmK_m with [Km:Q]=2m[K_m:\mathbb Q]=2^m, and by the tower law its own degree divides 2m2^m.

Every root of xn−1x^n-1 has some order dd in the group of nn-th roots of unity, with d∣nd\mid n, and is then a primitive dd-th root, a root of Φd\Phi_d. Each root appears exactly once on both sides, and both sides are monic, so they are equal. Comparing degrees gives n=∑d∣nφ(d)n=\sum_{d\mid n}\varphi(d).