# 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 $2\cos(2\pi/9),2\cos(8\pi/9),2\cos(14\pi/9)$.

###### Theorem 1.

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

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.

$F\subset L$ is a radical extension;

4. 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 $L\subset\mathbb{R}$. $(4)$ implies that $G=\operatorname{Gal}(L/F)$ 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 $\operatorname{Gal}(M/F)$ is also a $2$-group, so inductively we see that we can write

 $F=K_{0}\subset K_{1}\subset\ldots\subset K_{m-1}=M\subset K_{m}=L$

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 $L\subset\mathbb{R}$. 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 $\alpha_{1},\ldots,\alpha_{m}$, 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 $[L:F]=\lvert G\rvert$, and choose an element $\tau\in G$ of order $p$. Then $\tau$ is not the identity     , so for some $i$, $\tau(\alpha_{i})\neq\alpha_{i}$. 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

 $\nu^{-1}\tau\nu(\alpha)=\nu^{-1}\tau(\alpha_{i})\neq\nu^{-1}(\alpha_{i})=\alpha$

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 $\deg f=3$, we recover the original casus irreducibilis.

## References

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Title proof of casus irreducibilis for real fields ProofOfCasusIrreducibilisForRealFields 2013-03-22 17:43:08 2013-03-22 17:43:08 rm50 (10146) rm50 (10146) 8 rm50 (10146) Theorem msc 12F10