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Suppose $K$ is splitting field of the polynomial $x^4 - x^2 + 1= 0$ I need to show that Galois Group

$\text{Gal} ({K/\mathbb{Q}})$ is abelian.

For this I first showed that given polynomial is irreducible in $\mathbb{Q}$ by noting that $x^4 - x^2+ 1 = (x^2 + 1 -\sqrt{3}x) (x^2 + 1 + \sqrt{3}x)$

Now, since $\mathbb{Q}$ is a field of characteristic $0$, the extension $K/\mathbb{Q}$ is seperable, therefore $K/\mathbb{Q}$ is a separable extension, since $K$ is splitting field therefore $K/\mathbb{Q}$ is also Normal extension

hence, $\text{Gal}( K/\mathbb{Q})$ = $[K : \mathbb{Q}]$

and since index of $K$ = degree of irreducible polynomial = $4$, thus the group is abelian.

However, I am not sure whether this solution is rigorous enough or not, Can someone please check and tell me the errors in this solution.

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    $\begingroup$ Where did you show that $K=\Bbb{Q}(a)$ for any root $a$ of $x^4-x^2+1$? $\endgroup$
    – reuns
    Apr 22 at 22:22
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    $\begingroup$ I don't really see how that factorization shows $x^4-x^2+1$ is irreducible. I mean it almost does, but $x^4-x^2+1$ has $\binom42$ total quadratic factors over $\Bbb C$. $\endgroup$
    – pancini
    Apr 22 at 23:11
  • $\begingroup$ For different ways to show irreducibility see math.stackexchange.com/q/2414579/254075, but perhaps following up the comment of @pancini and writing down the other 2 factorizations over $\Bbb{C}$ is as good a solution as any. $\endgroup$
    – sharding4
    Apr 23 at 0:27
  • $\begingroup$ Also this doesn't show that the Galois group has order $4$. A priori it could be up to $4!$ $\endgroup$
    – pancini
    Apr 23 at 0:32
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    $\begingroup$ Since the polynomial is palindromic, if $a$ is a root then so is $1/a$. And one can also note even powers of $x$ to conclude that $-a$ is also a root. So if $a$ is a root then the complete set of roots is $a, - a, 1/a,-1/a$. Use this to find splitting field and galois group. $\endgroup$
    – Paramanand Singh
    Apr 23 at 1:26

1 Answer 1

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There is a well known classification of Galois groups of polynomials of degree 4, using the discriminant of the polynomial and its resolvent cubic.

It is easy to see that the discriminant $\delta$ of $f(x)$ is equal to $144$, and $144$ is a square so $\sqrt{144}=12\in\mathbb{Q}$. Also, the resolvent cubic is $g(x):=x^3+2x^2-3x$, which is irreducible over $\mathbb{Q}[t]$. Also, is easy to see that $f$ is irreducible (try the factorization $(x^2+ax+b)(x^2+cx+d)$ for some a,b,c,d rationals). This three things up gives us that $Gal(K/\mathbb{Q})$ is $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.

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