# Galois Group of splitting field is Abelian

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.

• Where did you show that $K=\Bbb{Q}(a)$ for any root $a$ of $x^4-x^2+1$? Apr 22 at 22:22
• 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 $\Bbb C$. Apr 22 at 23:11
• 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. Apr 23 at 0:27
• Also this doesn't show that the Galois group has order $4$. A priori it could be up to $4!$ Apr 23 at 0:32
• 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. Apr 23 at 1:26

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