proof of casus irreducibilis for real fields

The classical statement of the casus irreducibilisMathworldPlanetmath is that if f(x) is an irreduciblePlanetmathPlanetmathPlanetmath cubic polynomial with rational coefficientsMathworldPlanetmath and three real roots, then the roots of f(x) are not expressible using real radicalsPlanetmathPlanetmath. One example of such a polynomialMathworldPlanetmathPlanetmath is x3-3x+1, whose roots are 2cos(2π/9),2cos(8π/9),2cos(14π/9).

This article generalizes the classical case to include all polynomials whose degree is not a power of 2, and also generalizes the base fieldMathworldPlanetmathPlanetmath to be any real extensionPlanetmathPlanetmathPlanetmath of :

Theorem 1.

Let FR be a field, and assume f(x)F[x] is an irreducible polynomialMathworldPlanetmath whose splitting fieldMathworldPlanetmath L is real with FLR. Then the following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath:

  1. 1.

    Some root of f(x) is expressible by real radicals over F;

  2. 2.

    All roots of f(x) are expressible by real radicals over F using only square roots;

  3. 3.

    FL is a radical extension;

  4. 4.

    [L:F] is a power of 2.

Proof. That (2)(1) is obvious, and (3)(1) since FL is radical, and is real since L. (4) implies that G=Gal(L/F) has order a power of 2. Since G is a 2-group, it has a nontrivial center (this follows directly from the class equationMathworldPlanetmathPlanetmath, or look here ( and thus has a normal subgroupMathworldPlanetmath H of order 2, which corresponds to a subfieldMathworldPlanetmath M of L Galois over F with [L:M]=2. But then Gal(M/F) is also a 2-group, so inductively we see that we can write


where [Ki:Ki-1]=2. Thus each Ki is obtained from Ki-1 by adjoining a square root; it must be a real square root since L. This shows that (4)(2) and (3).

The meat of the proof is in showing that (1)(4). Let the roots of f(x) be α1,,αm, and assume, by renumbering if necessary, that α=α1 lies in a real radical extension K of F but that [L:F] is not a power of 2. Choose an odd prime p dividing [L:F]=|G|, and choose an element τG of order p. Then τ is not the identityPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, so for some i, τ(αi)αi. Also, since f(x) is irreducible, G acts transitively on the roots of f(x), so for some νG, ν(α)=αi. Then σ=ν-1τν does not fix α, since


Let N=Lσ be the fixed field of σ. Then L is Galois over N, and clearly [L:N]=p. But Galois subfields of real radical extensions are at most quadratic, so L cannot lie in a real radical extension of N.

However, αN,αL, and [L:N] is prime. Thus L=N(α)NK (since αK). Additionally, since FF(α)K is a real radical extension of F, we have also that NK is a real radical extension of NF=N. So L lies in the real radical extension NK of N. But this is a contradictionMathworldPlanetmathPlanetmath and thus [L:F] must be a power of 2.

One consequence of this theorem is the fact that if f(x)F[x] has degree not a power of 2, then if f(x) has all real roots, those roots are not expressible in terms of real radicals. If degf=3, we recover the original casus irreducibilis.


Title proof of casus irreducibilis for real fields
Canonical name ProofOfCasusIrreducibilisForRealFields
Date of creation 2013-03-22 17:43:08
Last modified on 2013-03-22 17:43:08
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 8
Author rm50 (10146)
Entry type Theorem
Classification msc 12F10