This arose recently in an online discussion about roots and irreducibility.

Let $f(X) = X^4 - 4X + 2$. $f(X)$ has two real roots and two complex roots, which means that complex conjugation $\sigma$ acts on the roots by transposition, i.e. it has cycle type $1,1,2$. In particular, $\sigma$ is an odd permutation.

My question is: Can we see this fact directly, without using analysis to determine the number of real roots of $f$?

Said another way: Is there a straightforward algebraic reason why $\sigma$ induces an odd permutation on the roots, that does not go through something like the Intermediate Value Theorem to show that there is a real root?

More generally, how would I determine the cycle type of a given automorphism (of some parent field) acting on the roots of an irreducible polynomial?

An example of this kind of thing is the discriminant: the discriminant of an irreducible polynomial is a square if and only if the Galois group consists only of even permutations of the roots.

Another example is the resolvent cubic of a quartic, which is irreducible if and only if the Galois group contains a $3$-cycle.

I suspect that there should be a similar computation that gives us information about the cycle type of complex conjugation.

In case it's useful, here is a computation of the Galois group of $f$ over $\mathbb{Q}$:

$f(X)$ is irreducible, as is its resolvent cubic, $X^3 - 8X - 16$, so the Galois group has order divisible by $3$ and $4$. Furthermore, its discriminant $-2^8 \cdot 19$ is not a square, so the Galois group is not contained in $A_4$, and must therefore be $S_4$.

  • $\begingroup$ $f(X)$ is patently reducible if it has real roots. Do you mean irreducible over $\mathbb Q$? $\endgroup$ – TonyK Jun 12 '15 at 19:54
  • $\begingroup$ @TonyK Yes—all quartics are reducible over $\mathbb{R}$. $\endgroup$ – Slade Jun 12 '15 at 20:16
  • $\begingroup$ Well, exactly. Why don't you say so? $\endgroup$ – TonyK Jun 12 '15 at 21:21
  • $\begingroup$ @TonyK: Because the default assumption in Galois theory is that polynomials are studied under $\mathbb{Q}$? $\endgroup$ – Charles Jun 12 '15 at 23:30
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    $\begingroup$ Complex conjugation is an odd permutation if and only if the discriminant of the polynomial is negative. (BTW, you may have made a typo, because the polynomial $x^4 - 2x + 4$ you wrote down has no real roots.) $\endgroup$ – Epargyreus Jun 14 '15 at 8:26

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