# How to determine the Galois Group of a polynomial with "ugly" roots? [duplicate]

This is related to my previous question.

Suppose I have the polynomial $p(x) = x^2 - 2 \in \Bbb Q[x]$. This is irreducible over $\Bbb Q$ so we need to move up to the extension field $K/\Bbb Q = \Bbb Q(\sqrt{2})$. Now $p$ splits over $K$ into $(x - \sqrt{2})(x+\sqrt{2})$ and it can be shown that the Galois Group $\operatorname{Gal}(K/\Bbb Q) \cong \Bbb Z/2\Bbb Z$ since it contains only the identity automorphism and one involution that sends $\sqrt 2$ to its conjugate $-\sqrt 2$.

My question is this; what if I have a polynomial with ugly roots? Say, for instance,

$$p(x) = x^5 - x^3 - 2x^2 - 2x - 1$$

This was shown in my previous question to be irreducible over $\Bbb Q$ so the splitting field of $p$ is simply $\Bbb Q$ adjoin the roots of $p$. These roots are, however,

$$\alpha \approx 1.73469\ \ \ \beta \approx 0.79645e^{0.49411\pi i}\ \ \ \gamma \approx 0.9533e^{5.6494\pi i}$$

and the conjugates $\overline \beta, \overline \gamma$ (do we get the conjugates for free when we adjoin complex roots or do we explicitly have to adjoin them?).

Then the splitting field $F$ of $p$ is $\Bbb Q(\alpha, \beta, \gamma)$ and their conjugates (depending on the answer to the question just asked).

At this point I'm not sure where to go; I've been trying to list automorphisms that fix $\Bbb Q$, obviously the identity is included, it seems also that the autmorphisms

$$\varphi_1 : \begin{cases} \beta \to \overline \beta \\ \gamma \to \gamma \end{cases}\ \ \ \text{and}\ \ \ \varphi_2 : \begin{cases} \beta \to \beta\\ \gamma \to \overline \gamma \end{cases}$$

are in the galois group. Also, $\varphi_2 \circ \varphi_1 = \psi$ should be in the group and this sends both $\beta$ and $\gamma$ to their conjugates. Any composition of $\psi$ with either of the others I've listed gives the other. I'm stuck for finding more now though. Since conjugation is an involution automorphism these are self inverse.

I'm not sure where to go from here! Is what I've done thus far even correct or have I totally misunderstood the concepts I'm trying to understand here?

• Since $p$ is irreducible, $\textrm{Gal}(p)$ is a transitive subgroup of $S_5$, but for $p$ prime every transitive subgroup of $S_p$ contains a $p$-cycle. If $\varphi_1$ is an automorphism (I didn't think this is immediately), then it is a transposition in $\textrm{Gal}(p)$. For $p$ prime, a $p$-cycle and a transposition together generate all of $S_p$, so we can conclude (given that the claim about $\varphi_1$ is correct) that $\textrm{Gal}(p) \cong S_5$. Apr 9, 2017 at 21:16
• It seems the claim about $\varphi_1$ is false: Consulting a CAS gives that $\textrm{Gal}(p) \cong D_{10}$. Since $p$ has two pairs of nonreal conjugate roots, complex conjugation is a product of two disjoint transpositions in $\textrm{Gal}(p)$, and such a product and the guaranteed $5$-cycle together generate a copy of $D_{10}$. Apr 9, 2017 at 21:21
• @Travis Thanks a lot. Do you have a source for your claim about the pairs of non-real conjugate roots? I'd like to read about it.. if you can recommend something! Apr 9, 2017 at 21:25
• It's more or less immediate from the definition: The Galois group can be regarded as a group of permutations on the roots. We can make a little more cheap progress without computer assistance: The only transitive subgroups of $S_5$ that contain such a product are $D_{10}, \Bbb Z_5 \rtimes \Bbb Z_4$, $A_5$, and $S_5$. Computing gives that the discriminant is $2209 = 47^2$, so $\textrm{Gal}(p) \leq A_5$, so the only possibilities are $D_{10}$ and $A_5$. Distinguishing between these cases manually might be more difficult. Apr 9, 2017 at 21:35