From Galois Theory (Rotman): I wrote down the whole proof, but my question is only about the third paragraph.

There exists a quantic polynomial $f(x) \in \Bbb{Q}[x]$ that is not solvable by radicals.

Proof If $f(x)=x^5 -4x + 2$, then $f(x)$ is irreducible over $\Bbb{Q}$, by Eisenstein's criterion. Let $E/\Bbb{Q}$ be the splitting field of $f(x)$ contained in $\Bbb{C}$, and let $G=Gal(E/\Bbb{Q})$. If $\alpha$ is a root of $f(x)$, then $[\Bbb{Q}(\alpha): \Bbb{Q}]=5$, and so

$$[E : \Bbb{Q}] = [ E : \Bbb{Q}(\alpha)][\Bbb{Q}(\alpha) : \Bbb{Q}] = 5[E : \Bbb{Q}(\alpha)].$$

By Theorem 56 $|G|=[E: \Bbb{Q}]$ is divisible by 5.

We now use some calculus; $f(x)$ has exactly two critical points, namely, $\pm (4/5)^{1/4}$ ~ $\pm .946$, and $f((4/5)^{1/4}) < 0$ and $f(-(4/5)^{1/4}) > 0$; it follows easily that $f(x)$ has exactly thre real roots (they are, approximately, -1.5185, 0.5085, and 1.2435; the complex roots are $-.1168 \pm 1.4385i.$)

Regarding $G$ as a group of permutations on the 5 roots, we note that $G$ contains a 5-cycle (it contains an element of order 5, by Cauchy's theorem, and the only elements of order 5 in $S_5$ are 5-cycles). The restriction of complex conjugation, call it $\sigma$, is a transposition, for $\sigma$ interchanges the two complex roots while it fixes the three real roots.

I was a bit confused here about how we can assume that $\sigma$ restricted to the complex conjugation is a transposition. Could we not exchange a complex number with a real number? In other words, how do we know that complex numbers must be sent to other complex numbers?

By Theorem G.39, $S_5$ is generated by any transposition and any 5-cycle, so that

$$G = Gal(E/\Bbb{Q}) \cong S_5$$

is not a solvable group, by Theorem G.34, and Theorem 74 shows that $f(x)$ is not solvable by radicals.

  • 1
    $\begingroup$ The point is that conjugation swaps 2 roots and leave 3 fixed - a transposition - so there is a transposition in the Galois group. There are other transpositions, and other elements of order 2 which aren't transpositions - but a 5-cycle and a transposition generate all these. $\endgroup$ – Mark Bennet Jun 30 '13 at 21:08

Conjugation maps $a + i b$ to $a - i b$, where $a, b$ are real numbers. So the complex numbers fixed by conjugation, that is, those complex numbers for which $a + i b = a - i b$, are exactly those for which $b = 0$, that is, the real numbers.

So the three real roots are fixed, and the two complex, non-real ones cannot be fixed, so they must be exchanged by conjugation. (Since the coefficients of $f(x)$ are real, the conjugate of a root of $f(x)$ must be a(nother) root of $f(x)$.)


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