17
$\begingroup$

From Hungerford, section V, chapter 4 exercise 9:

Let $x^4+ax^2+b$ in $K[x]$ (with char $K\neq $2) be irreducible with Galois group $G$.

(a) If $b$ is a square in $K$, then $G = \mathbb{Z}_2\times\mathbb{Z}_2$.
(b) If $b$ is not a square in $K$ and $b(a^2-4b)$ is a square in $K$, then $G = \mathbb{Z}_4$.
(c) If neither $b$ nor $b(a^2-4b)$ is a square in $K$, then $G = D_4$ (dihedral group).

What I've tried

With (a): Let be $u_1,u_2,u_3,u_4$ the roots of the quartic, then $b=u_1u_2u_3u_4$ is in $K$ and there exists $a\in K$ with $a^2=b$. On the other hand, $G = \mathbb{Z}_2\times\mathbb{Z}_2$ if and only if $\alpha=u_1u_2+u_3u_4$, $\beta=u_1u_3+u_2u_4$, $\gamma=u_1u_4+u_2u_3$ are in $K$, but I don't find the way to rely those results.

With (b) I know that $b=u_1u_2+u_1u_3+u_1u_4+u_2u_3+u_2u_4+u_3u_4$ but again I dont know how to continue.

Any help? Thanks.

$\endgroup$
9
  • $\begingroup$ Not precisely the same but it has been usefull: math.stackexchange.com/questions/204709/… $\endgroup$
    – Lotus
    Commented Jan 24, 2014 at 19:12
  • 1
    $\begingroup$ Is it possible that the polynomial is meant to be a biquadratic, so it should read $$x^4+ax^2+b?$$ Then the roots come in pairs of negatives of each other. That constrains the Galois group to be a 2-group, and gives a better match with the listed alternatives. $\endgroup$ Commented Feb 6, 2014 at 22:01
  • 3
    $\begingroup$ @Lotus this should help for cubics and quartics math.uconn.edu/~kconrad/blurbs/galoistheory/cubicquartic.pdf $\endgroup$
    – ir7
    Commented Feb 6, 2014 at 22:13
  • $\begingroup$ oh sure, it was $x^4+ax^2+b$ sorry, edited $\endgroup$
    – Lotus
    Commented Feb 6, 2014 at 22:47
  • 1
    $\begingroup$ @JyrkiLahtonen You are right but, albeit what you say is easy to proof, it is not evident at first sight, so in my opinion a better statement for the problem would have been "Assume $K$ is infinite, then 1. + 2. + 3. Now assume $K$ is finite: prove that in this case only 2. can hold". Note that in characteristic 2 you use the Binomial Theorem to factor. In any case, IMO this is regarding the problem from the other end of the perspective, because in a finite field the Galois group MUST be cyclic by a well known theorem so that the fact that only 2. can hold is more a consequence than a cause. $\endgroup$
    – brad
    Commented Jul 4, 2022 at 11:27

1 Answer 1

23
+50
$\begingroup$

There are two numbers $\alpha,\beta$ (in some algebraic closure of $K$) such that the identity $P=X^4+aX^2+b=(X^2-\alpha^2)(X^2-\beta^2)$ holds. Then the set $R$ of all roots of $P$ is $\lbrace \pm \alpha, \pm \beta\rbrace$. Since $P$ is irreducible, $G$ can be identified to a transitive subgroup of ${\mathfrak S}(R)$, the group of permutations of $R$.

Also, any $\sigma\in G$ obviously satisfies $\sigma(-\alpha)=-\sigma(\alpha)$ and $\sigma(-\beta)=-\sigma(\beta)$. The subgroup $H$ of permutations satisfying those two conditions consists of eight elements: in cycle notation,

$$ \begin{array}{lcl} H &=\lbrace& {\sf id},(\alpha,-\alpha)(\beta,-\beta),\\ & & (\alpha,\beta)(-\alpha,-\beta),(\alpha,-\beta)(-\alpha,\beta), \\ & & (\alpha,\beta,-\alpha,-\beta), (\alpha,-\beta,-\alpha,\beta), \\ & & (\alpha,-\alpha), (\beta,-\beta) \rbrace \end{array} $$

There are exactly three transitive subgroups of $H$, namely $H$ itself and

$$ \begin{array}{lcl} H_1 &=&\lbrace {\sf id},(\alpha,-\alpha)(\beta,-\beta), (\alpha,\beta)(-\alpha,-\beta),(\alpha,-\beta)(-\alpha,\beta) \rbrace \\ H_2 &=& \lbrace {\sf id},(\alpha,-\alpha)(\beta,-\beta), (\alpha,\beta,-\alpha,-\beta), (\alpha,-\beta,-\alpha,\beta) \rbrace \end{array} $$

In case (a), $(\alpha\beta)^2=b$ is a square in $K$, so $\gamma_1=\alpha\beta$ is in $K$. Now if $\tau_1=(\alpha,-\beta,-\alpha,\beta)$, we have $\tau_1(\gamma_1)=-\gamma_1$, so $\tau_1\not\in G$ and this forces $G=H_1$.

In case (b), $(\alpha\beta(\alpha^2-\beta^2))^2=b(a^2-4b)$ is a square in $K$, so $\gamma_2=\alpha\beta(\alpha^2-\beta^2)$ is in $K$. Now if $\tau_2=(\alpha,\beta)(-\alpha,-\beta)$, we have $\tau_2(\gamma_2)=-\gamma_2$, so $\tau_2\not\in G$ and this forces $G=H_2$.

Finally, in case (c) we have $\gamma_1\not\in K$ and $\gamma_2\not\in K$. By the fundamental theorem of Galois theory, $\gamma_1$ is not fixed by all the elements of $G$, so $G\neq H_2$. Similarly, $\gamma_2$ is not fixed by all the elements of $G$, so $G\neq H_1$. The only possibility left is then $G=H$.

$\endgroup$
2
  • 1
    $\begingroup$ Observe that $\tau_1$ does not fix $\gamma_2$, so in case (b) you cannot have $G=H_2$. Instead of $\alpha \beta (\alpha-\beta)$ you probably mean $\alpha \beta (\alpha^2-\beta^2)$. Observe that $(\alpha \beta (\alpha^2-\beta^2))^2 = b(a^2-4b)$. $\endgroup$
    – Martino
    Commented Nov 4, 2018 at 17:14
  • $\begingroup$ @Martino Corrected, thanks. $\endgroup$ Commented Nov 5, 2018 at 5:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .