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# casus irreducibilis

Let the polynomial

$P(x):=x^{n}+a_{1}x^{{n-1}}+\ldots+a_{n}$ |

with complex coefficients $a_{j}$ be irreducible, i.e. irreducible in the field $\mathbb{Q}(a_{1},\,\ldots,\,a_{n})$ of its coefficients. If the equation $P(x)=0$ can be solved algebraically and if all of its roots are real, then no root may be expressed with the numbers $a_{j}$ using mere real radicals unless the degree $n$ of the equation is an integer power of 2.

# References

- 1 K. Väisälä: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17. Kustannusosakeyhtiö Otava, Helsinki (1950).

Keywords:

irreducible polynomial, roots real

Related:

RadicalExtension, CardanosFormulae, TakingSquareRootAlgebraically, EulersDerivationOfTheQuarticFormula

Major Section:

Reference

Type of Math Object:

Theorem

Parent:

## Mathematics Subject Classification

12F10*no label found*

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## Recent Activity

Jul 5

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

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new question: A trascendental number. by Ron Castillo

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new question: Banach lattice valued Bochner integrals by math ias

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new question: young tableau and young projectors by zmth

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

Jun 13

new question: young tableau and young projectors by zmth

## Comments

## Casus irreducibilis

Jussi,

I'm not sure I understand (or believe) the current statement. Do you mean that if K is a finite extension of Q, and f(x)\in K[x] can be solved algebraically and has real roots, then no radical extension of K contains all the roots of f(x) unless f has degree a power of 2?

Roger

## Re: Casus irreducibilis

In fact, K need not be finite over Q, if, say, one of the coefficients is \pi. I agree that something is fishy here.

Cam

## Re: Casus irreducibilis

> Jussi,

>

> I'm not sure I understand (or believe) the current

> statement. Do you mean that if K is a finite extension of Q,

> and f(x)\in K[x] can be solved algebraically and has real

> roots, then no radical extension of K contains all the roots

> of f(x) unless f has degree a power of 2?

Ah -- the statement is slightly different. Let's suppose that f has rational coefficients for simplicity. Then I think it should be:

"If f\in Q[x] is irreducible, has all real roots, and has degree not a power of 2, then the (real) roots of f are not contained in any *real* radical extension of Q."

This sounds more plausible (maybe real should be replaced with totally real -- not sure), and brings us back to the motivating phenomenon of casus irreducibilis for cubics -- these roots are real numbers which necessitate using complex numbers to express via radicals.

Cam

## Re: Casus irreducibilis

Cam - that is in fact the classical statement of casus irreducibilis. It sounded as though Jussi was trying to generalize it...

Roger

## Re: Casus irreducibilis

Roger,

The contents of the entry may be understood as you formulate. The theorem is proven by induction in the book of V\"ais\"al\"a (a Finnish algebraist and number theorist). I also checked the contents of the theorem in the algebra book of the Finnish function theorist F. Nevanlinna (brother of Rolf Nevanlinna); he proves indirectly the theorem formulated as

"An irreducible equation of odd degree, the roots of which are all real, is not algebraically solvable by adjoining to the coefficient field of the equation mere real radicals."

I have not found the theorem in internet.

Jussi

## Re: Casus irreducibilis

Yes Cam, it's true that K need not be finite over Q.

Jussi

## Re: Casus irreducibilis

The generalisation is not mine...

Jussi

## Re: Casus irreducibilis

> "An irreducible equation of odd degree, the roots of which

> are all real, is not algebraically solvable by adjoining to

> the coefficient field of the equation mere real radicals."

> I have not found the theorem in internet.

So should the entry be only for odd degrees, or is it true for all degrees which are not powers of 2?

By the way, Jussi, fascinating discussion -- I'm glad you brought this up.

Cam

## Re: Casus irreducibilis

> Cam - that is in fact the classical statement of casus

> irreducibilis. It sounded as though Jussi was trying to

> generalize it...

Ah, okay -- my impression had been that the classical version was only for cubics. But it does indeed sound like there's a strict generalization here, what with expanding the coefficient field.

Cam

## Re: Casus irreducibilis

Actually, I've just checked, and you're right. The original statement was regarding cubics. There is a reasonable proof for general irreducible polynomials over the rationals, which I hope to post in the next few days.

But I agree, the article as stated generalizes it even further.

Roger