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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.

night_crawler
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    Where did you show that $K=\Bbb{Q}(a)$ for any root $a$ of $x^4-x^2+1$? – reuns Apr 22 '22 at 22:22
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    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 $\Bbb C$. – pancini Apr 22 '22 at 23:11
  • For different ways to show irreducibility see https://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. – sharding4 Apr 23 '22 at 00:27
  • Also this doesn't show that the Galois group has order $4$. A priori it could be up to $4!$ – pancini Apr 23 '22 at 00:32
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    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. – Paramanand Singh Apr 23 '22 at 01:26
  • @ParamanandSingh So, the splitting field is $\mathbb{Q(a)}$ , but my question is why is it necessary to express the splitting field in this form? After proving this
    $K = \mathbb{Q(a)}$ and $f$ is irreducible can I directly say order of Group is $4$     – night_crawler Apr 23 '22 at 01:39
  • @reuns: Why is it necessary to express the splitting field in that fomat? – night_crawler Apr 23 '22 at 01:41
  • Well the splitting field has to be some extension of rationals and thus is obtained by adjoining roots to $\mathbb {Q}$. In this particular case it is sufficient to adjoin any one root. But in other cases more roots may need to be adjoined. Writing out splitting field explicitly is essential to determine the degree of this field extension (which is same as order of Galois group). And yes if the polynomial is irreducible then $\mathbb {Q} (a) $ is of degree $4$ over $\mathbb {Q} $. And so Galois group is of order $4$. – Paramanand Singh Apr 23 '22 at 02:22
  • Ideas expressed in my last comment are pretty standard and available in any text on Galois and field theory. Please revisit your textbook to study proofs of those. – Paramanand Singh Apr 23 '22 at 02:25

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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 $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.

Guillermo García Sáez
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