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Let $f(x)=x^4-x^2-1\in \mathbb{Q}[x]$. Find a splitting field and then prove that the Galois group of $f$ is isomorphic to the dihedral group of order 8.

My approach:

First I need to show that the polynomial $f$ is irreducible. To do that I use rational root test.(rational roots test require polynomial to live in $\mathbb{Z}[x]$, but $f$ has no pure rational coefficient so I think its okay to apply here.) That tells me $f$ is irreducible over $\mathbb{Q}$.

Now, I just use quadratic formula to solve for the roots and they are

$$\pm\sqrt{\dfrac{-1\pm \sqrt{5}}{2}}$$ Denoting them by $\alpha_1=\sqrt{\dfrac{-1+ \sqrt{5}}{2}},\alpha_2=-\sqrt{\dfrac{-1+ \sqrt{5}}{2}},\alpha_3=\sqrt{\dfrac{-1- \sqrt{5}}{2}},\alpha_4=-\sqrt{\dfrac{-1- \sqrt{5}}{2}}$ then my splitting field is $\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$, here is where I am not sure. I think it is okay to use one of the roots to adjoin since they all look very similar, namely $\mathbb{Q}(\alpha_1,\alpha_3)$ should suffice, since the other two can be obtained by negating the sign. But I don't know how to argue that formally.

To find the Galois group, let $\tau_i\in $ Aut$(\mathbb{Q}(\alpha_1))$, then let $\tau_0$ be identity, $\tau_1(\alpha_1)=-\alpha_1=\alpha_2$ and $\tau_1(\alpha_3)=\alpha_4$. And $\tau_2(\alpha_1)=-\alpha_1=\alpha_2$ and $\tau_2(\alpha_3)=\alpha_3$. And $\tau_3(\alpha_1)=-\alpha_1=\alpha_2$ and $\tau_3(\alpha_3)=\alpha_4$. I think thats all the automorphisms. If my generators are correct.

Is this reasonable? Thanks!

Remu X
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    You are not quite done with irreducibility. The rational root test can ever only eliminate the presence/absence of linear factors, but a quartic may factor as a product of two irreducible quadratics. Then, there is the possibility of hidden relations tying, say, $\alpha_1$ and $\alpha_3$. Here is a more general answer. Not sure, if it has enough details for you. The exact duplicate Derek Allums found OTOH... – Jyrki Lahtonen May 15 '24 at 04:30
  • @DerekAllums I seen that one but I cant understand why they include the multiplicative inverse. – Remu X May 15 '24 at 04:35
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    I used my dupehammer, because I find the answer at the target quite complete. If there are steps you are uncomfortable with, please ask (either here by editing the question or commenting), or by commenting at the target thread. – Jyrki Lahtonen May 15 '24 at 04:35
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    Calculate the product $\alpha_1\alpha_3$, and you will see where the multiplicative inverse comes from. That will make proving irreducibility easier also. Consequently you also get that the splitting field will be a degree $8$ extension. After all, $\alpha_1$ is real. – Jyrki Lahtonen May 15 '24 at 04:36
  • @JyrkiLahtonen ah I see thank you, so to complete the irreducible part I should just say by rational root test it has no linear factor, and suppose $f=pq$ for some quadratic $p,q$, then I must have some sort of contradiction here I dont see, could you give a hint? Thanks – Remu X May 15 '24 at 04:41
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    No, because you need to prove that no such factorization is possible at all when $p,q$ are required to have rational coefficients! But you have a complete list of roots, so one way of seeing irreducibility is to check that none of $(x-\alpha_1)(x-\alpha_j)$, $j=2,3,4,$ have rational coefficients. After all $p$ or $q$ needs to be one of those (by uniqueness of factorization in the splitting field). That is easy. There are also other methods, but the rational root test is decisive only up to cubic polynomials. – Jyrki Lahtonen May 15 '24 at 04:46
  • @JyrkiLahtonen Thank you this helps a lot! – Remu X May 15 '24 at 04:46

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