Consider the general polynomial equation :

$$ P(X) = X^4 + a_1 X^3 + a_2 X^2 + a_3 X + a_4 $$

We have $ \operatorname{Gal}( \mathbb{Q}(r_1, ..., r_4)/ \mathbb{Q}(\sigma_1, ..., \sigma_4) ) \cong S_4 $

With $ r_i$ the roots of P and $ \sigma_i $ the elementary symmetric polynomials.

Although for $ \Phi_5(X) = 1 + X + X^2 + X^3 + X^4 $

We have $ \: \operatorname{Gal}( \mathbb{Q}(\zeta_5)/ \mathbb{Q} ) \cong \mathbb{Z}/4\mathbb{Z} $

$ \bullet $ Why are these two groups not isomorphic? Can't we just evaluate the $ a_i$ ? Or it is because they are transcendental ? I don't get the intuition.

Thanks for any help with this.

  • 5
    $\begingroup$ This is because the $a_i$s are algebraically independent. $\endgroup$
    – Bernard
    Commented May 31, 2018 at 17:23
  • $\begingroup$ Yes in the theorem the $ a_i $ are transcendental over $ \mathbb{Q} $ or another field K and they are algebraically independent. So perhaps we could say that the polynomial P(X) represents just a particular case of equations, and not the general case. So we have just found one type of non solvable equation for any degree $ n \in \mathbb{N} $ $\endgroup$
    – Psylex
    Commented May 31, 2018 at 18:05
  • 1
    $\begingroup$ On the contrary: $P(X)$ represents the general quartic equation, in the sense that there's no relation whatsoever between its coefficients. $\endgroup$
    – Bernard
    Commented May 31, 2018 at 18:09
  • $\begingroup$ Ok thanks that's what they mean by general. $\endgroup$
    – Psylex
    Commented May 31, 2018 at 18:13

1 Answer 1


Some intuition, rather than a completely rigorous field-theoretic answer:

One way to think about this is that in the general case, there are very few relations between the roots (that is, polynomials $ p \in \mathbb{Q}(\sigma_1,\dotsc,\sigma_4)[X_1,\dotsc,X_4]$ so that $p(r_{1},\dotsc,r_{4}) \in \mathbb{Q}(\sigma_1,\dotsc,\sigma_4)$; by one of Galois' original results, these are precisely those that have the same values for any permutation in the Galois group). Indeed, there are essentially none apart from Vietà's formulae. On the other hand, there are lots of relations $\in \mathbb{Q}$ between the roots of $\Phi_5$, like $X_1^k - X_{k} = 0$ for $k=2,3,4$, among others. This essentially forces the Galois group to be small, because all of these relations still have to be preserved for any of its permutation of the roots. Indeed, $\zeta_5$ is itself a Galois resolvent of $\Phi_5$, and it is then straightforward to construct the other roots as rational functions of $\zeta_5$, simply by taking powers. But then the cyclic structure of the multiplicative group modulo $5$ forces the only allowable permutations to be cyclic.


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