When I read the definition of a radical extension, I thought: What is the difference with a algebraic extension?

Could you give me an example of an algebraic extension which is not a radical extension?


  • 3
    $\begingroup$ The difference is that in a radical extension you're only adjoining roots of polynomials of the form $x^n - a$ rather than arbitrary polynomials. $\endgroup$ – Qiaochu Yuan Oct 12 '15 at 6:06
  • $\begingroup$ 'Radical' is a dangerous name for an extension. Some of us think in purely inseparable (following Bourbaki, EGA, etc.). $\endgroup$ – Heinrich Oct 12 '15 at 11:02
  1. The only way for a cubic extension of the rationals to be radical is for it to be a pure cubic extension, that is, it has to be ${\bf Q}(r^{1/3})$ for some rational $r$ (indeed, we may take $r$ to be an integer). But most cubic extensions are not pure cubics. Maybe the simplest example would be a cyclic cubic field, such as the one generated over the rationals by $\cos(2\pi/7)$ or $\cos(2\pi/9)$. These can't be pure cubic, since they are normal extensions, which pure cubics aren't.

  2. Let $K$ be the splitting field of $f(x)=x^5-x-1$ over the rationals. It can be proved that the zeros of $f$ can't be expressed in radicals at all. That means $K$ isn't even contained in a radical extension.

  • $\begingroup$ Thank you very much. I understand your explanations It took so long, because I was busy with your original answer, the one with the number (11+2i)^1/3 + (11-2i)^1/3. I could not prove that Q(s) was a cubic extension, where s is that number, and that it is not Q(r^1/3) for some rational r. I first took a simpler example, t=(2+2i )^1'/3+(2-2i)^1/3, but Q(t) is a quadratic extension. Maybe you can give me a clue about Q(s). $\endgroup$ – ericj Oct 21 '15 at 11:05
  • $\begingroup$ I wish you had asked me about that. I edited out that original answer, because it doesn't work. $(-1-2i)^3=11+2i$, so $$(11+2i)^{1/3}+(11-2i)^{1/3}=(-1-2i)+(-1+2i)=-2$$ and it's not a cubic extension at all, the element is already in the rationals. $\endgroup$ – Gerry Myerson Oct 21 '15 at 12:22
  • $\begingroup$ OK. Exactly the same I saw with (2+2i)^1/3 + (2-2i)^1/3. $\endgroup$ – ericj Oct 21 '15 at 12:37

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