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Let $K=\mathbb{Q}$ and consider a cyclic extension $L$.(For example say, the splitting field of the polynomial say $f=x^3+x^2-2x-1$). Now consider a cyclotomic extension of $\Phi_3=X^2+X+1$. Let us call this field $F$. Now we know that $[F:K]=6$ by the Tower Law. But in particular, if we consider the cyclotomic extension of $\mathbb{Q}$ which we shall call $M$,we have that $[F:M]=3$(by a suitable consideration of the Tower Law.) But we do know $F/M$ is a Galois extension; we adjoined all roots of $f$.But then $Gal(F/M)=C_3$ and we know this cannot be true because cyclic extensions of degree $N$ of a field which contains the $N$-th roots of unity are Kummer by Kummer Theory. Where did I make a mistake in my reasoning?

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    $\begingroup$ Exactly what is $M$? Do you mean $K(\omega)$, where $\omega^2+\omega+1=0$? If so, now there is no difficultly: once you have the cube roots of unity there, the extension becomes Kummer. That’s the message of the Cubic Formula. $\endgroup$ – Lubin May 6 '16 at 15:37
  • $\begingroup$ Yes, that is what I meant $\endgroup$ – daruma May 6 '16 at 15:41
  • $\begingroup$ And is your confusion resolved now? $\endgroup$ – Lubin May 6 '16 at 15:45
  • $\begingroup$ But $f$ is not of the form $X^n+a$? $\endgroup$ – daruma May 6 '16 at 15:58
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    $\begingroup$ Oh yes, that was very dumb. It's a splitting polynomial of some other polynomial as well. Thanks. My confusion has being resolved $\endgroup$ – daruma May 6 '16 at 16:59

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