proof of casus irreducibilis for real fields

The classical statement of the casus irreducibilis is that if $f(x)$ is an irreducible cubic polynomial with rational coefficients and three real roots, then the roots of $f(x)$ are not expressible using real radicals. One example of such a polynomial is $x^3-3x+1$, whose roots are $2\cos (2\pi /9),2\cos (8\pi /9),2\cos (14\pi /9)$.

This article generalizes the classical case to include all polynomials whose degree is not a power of $2$, and also generalizes the base field to be any real extension of $\mathbb{Q}$:

Theorem 1.

Let $F\mathrm{\subset }\mathrm{R}$ be a field, and assume $f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\in }F\mathit{}\mathrm{[}x\mathrm{]}$ is an irreducible polynomial whose splitting field $L$ is real with $F\mathrm{\subset }L\mathrm{\subset }\mathrm{R}$. Then the following are equivalent:

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

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

3. $F\subset L$ is a radical extension;

4. $[L:F]$ is a power of $2$.

Proof. That $(2)\Rightarrow (1)$ is obvious, and $(3)\Rightarrow (1)$ since $F\subset L$ is radical, and is real since $L\subset ℝ$. $(4)$ implies that $G=\mathrm{Gal}(L/F)$ has order a power of $2$. Since $G$ is a $2$-group, it has a nontrivial center (this follows directly from the class equation, or look here (http://planetmath.org/ANontrivialNormalSubgroupOfAFinitePGroupGAndTheCenterOfGHaveNontrivialIntersection)) and thus has a normal subgroup $H$ of order $2$, which corresponds to a subfield $M$ of $L$ Galois over $F$ with $[L:M]=2$. But then $\mathrm{Gal}(M/F)$ is also a $2$-group, so inductively we see that we can write

$$F=K_0\subset K_1\subset \mathrm{\dots }\subset K_{m-1}=M\subset K_m=L$$

where $[K_i:K_{i-1}]=2$. Thus each $K_i$ is obtained from $K_{i-1}$ by adjoining a square root; it must be a real square root since $L\subset ℝ$. This shows that $(4)\Rightarrow (2)$ and $(3)$.

The meat of the proof is in showing that $(1)\Rightarrow (4)$. Let the roots of $f(x)$ be ${\alpha }_1,\mathrm{\dots },{\alpha }_m$, and assume, by renumbering if necessary, that $\alpha ={\alpha }_1$ 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 $\tau \in G$ of order $p$. Then $\tau$ is not the identity, so for some $i$, $\tau ({\alpha }_i)\ne {\alpha }_i$. Also, since $f(x)$ is irreducible, $G$ acts transitively on the roots of $f(x)$, so for some $\nu \in G$, $\nu (\alpha )={\alpha }_i$. Then $\sigma ={\nu }^{-1}\tau \nu$ does not fix $\alpha$, since

$${\nu }^{-1}\tau \nu (\alpha )={\nu }^{-1}\tau ({\alpha }_i)\ne {\nu }^{-1}({\alpha }_i)=\alpha$$

Let $N=L^{\sigma }$ be the fixed field of $\sigma$. 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, $\alpha \notin N,\alpha \in L$, and $[L:N]$ is prime. Thus $L=N(\alpha )\subset NK$ (since $\alpha \in K$). Additionally, since $F\subset F(\alpha )\subset K$ 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 contradiction and thus $[L:F]$ must be a power of $2$.

One consequence of this theorem is the fact that if $f(x)\in F[x]$ 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 $\mathrm{deg}f=3$, 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