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[parent] casus irreducibilis (Theorem)

Let the polynomial $$P(x) := x^n+a_1x^{n-1}+\ldots+a_n$$ 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.

Bibliography

1
K. V¨AISÄLÄ: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17.    Kustannusosakeyhtiö Otava, Helsinki (1950).




"casus irreducibilis" is owned by pahio.
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See Also: radical extension, Cardano's formulae, taking square root algebraically, Euler's derivation of the quartic formula

Keywords:  irreducible polynomial, roots real

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proof of casus irreducibilis for real fields (Theorem) by rm50
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Cross-references: numbers, real, roots, equation, field, coefficients, complex, polynomial
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This is version 9 of casus irreducibilis, born on 2005-06-20, modified 2008-03-03.
Object id is 7171, canonical name is CasusIrreducibilis.
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AMS MSC12F10 (Field theory and polynomials :: Field extensions :: Separable extensions, Galois theory)

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Casus irreducibilis by rm50 on 2007-12-28 17:42:41
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
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