# Prove that the Galois group of $x^5+x^4-4x^3-3x^2+3x+1$ is cyclic of order $5$

I have been asked to prove that the Galois group of $x^5+x^4-4x^3-3x^2+3x+1$ (presumably over $\mathbb{Q}$ is cyclic of order $5$. Unfortunately, I have no idea where to start.

The hint says to show that this is the minimal polynomial of $\zeta_{11} + \zeta_{11}^{-1}$. Which I guess I could do by plugging it into the polynomial, but there is most likely a better way to do it. Anyway, even if I could establish that, I wouldn't know any of the other roots so it's not like I could find any of the automorphisms. I guess it would kind of help because it would show that the Galois group had order divisible by 5. Except a lot of them are. I think I am completely missing the point of the hint.

The section was about how you could reduce polynomials mod $p$, and the Galois group would have to contain a permutation with cycle type $(n_1 n_2 ... n_k)$, where $n_1$, ..., $n_k$ are the degrees of the irreducible factors (mod $p$). But this is not very relevant because since the group is cyclic of prime order, all the permutations would be 5-cycles. And lots of other subgroups of $S_5$ have 5-cycles too.

So at this point I am kind of stuck. Can you give me a hint as to how to solve the problem? Thanks!

P.S. We are allowed to use the computer program SAGE for this assignment, but not the command that calculates the Galois group. Also, I tried to use SAGE to calculate the roots but it gave me numerical answers.

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Suppose you have shown that the minimal polynomial of $\zeta_{11}+\zeta_{11}^{-1}$ is the given polynomial.

What does that mean $[\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1}):\mathbb{Q}]$ is?

For any $n$, we know that $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is Galois, and $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^\times$ (or, if this has not come up in class, you should prove it separately to make your argument complete). In particular, $$\text{Gal}(\mathbb{Q}(\zeta_{11})/\mathbb{Q})\cong(\mathbb{Z}/11\mathbb{Z})^\times\cong \mathbb{Z}/10\mathbb{Z},$$ the cyclic group of order 10.

Note that $\mathbb{Q}(\zeta_{11})\supset\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1})$.

The cyclic group of order $n$ has a single subgroup of size $d$, for each $d\mid n$. Therefore it also has a single quotient group of size $d$, for each $d\mid n$.

We know that $[\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1}):\mathbb{Q}]=|\text{Gal}(\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1})/\mathbb{Q})|$.

What does that mean $\text{Gal}(\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1})/\mathbb{Q})\cong\text{Gal}(\mathbb{Q}(\zeta_{11})/\mathbb{Q})\bigg/\text{Gal}(\mathbb{Q}(\zeta_{11})/\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1}))$ is?

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Hello and thanks :) If $\zeta_{11} + \zeta_{11}^{-1}$ is the root of that minimal polynomial, then $Gal(\mathbb{Q}(\zeta_{11}+\zeta_{11}^{-1})/\mathbb{Q})$ has to be $\mathbb{Z}_5$ (because that's the only subgroup of order $5$). But how do I know $\mathbb{Q}(\zeta_{11} + \zeta_{11}^{-1})$ contains all the roots? – badatmath May 18 '11 at 2:16
@badatmath: We know (or you should show that) $\mathbb{Q}(\zeta_{11})/\mathbb{Q}$ is Galois, as it is the splitting field of $x^{11}-1$. Because $\text{Gal}(\mathbb{Q}(\zeta_{11})/\mathbb{Q})\cong\mathbb{Z}/10\mathbb{Z}$ is abelian, every subgroup is normal. What does the Fundamental Theorem of Galois Theory say about the correspondence between subgroups and subfields when the subgroup is normal? – Zev Chonoles May 18 '11 at 2:20
@Zev: Oh, okay! I'm a dumbass. Thanks. :D – badatmath May 18 '11 at 2:24
@badatmath: No problem! Also, you're not a dumbass, you just needed a hint in the right direction. We've all been in that position :) – Zev Chonoles May 18 '11 at 2:30
@badmath: We all learn by experience. Dont worry. As Neumann says, In Mathematics, we don't understand anything, we just get used to them. – user9413 May 18 '11 at 5:25
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The simplest way to show that the minimal polynomial is this one is just to plug it in and develop using the binomial theorem ; just remember that $\zeta_n \zeta_m = \zeta_{n+m}$ and that $\zeta_{n} \zeta_m^{-1} = \zeta_{n-m}$ when you're computing, and then prove its irreducibility by noticing that $1$ and $-1$ are not roots of it (rational root theorem), plus it cannot factor as a product of degree $2$ and $3$ (use linear algebra to do this part).

Now $[\mathbb Q(\zeta_{11}+\zeta_{11}^{-1}) : \mathbb Q] = 5 = |\mathrm{Gal}(\mathbb Q(\zeta_{11}+\zeta_{11}^{-1}) \backslash \mathbb Q)|$ and the $5$ comes from the fact that the dimension of the extension divides $5$ (obviously) but is not $10$ since $\mathbb Q(\zeta_{11}) \neq \mathbb Q(\zeta_{11} + \zeta_{11}^{-1})$ . Since the only group of order $5$ is the cyclic one, you're done. (Any non-trivial element of the group has order dividing $5$, hence $5$, so that the group is cyclic.)

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Ah, I was afraid of that. Thanks. – badatmath May 18 '11 at 2:44
Afraid of what? =P – Patrick Da Silva May 18 '11 at 2:58
having to plug that in. – badatmath May 18 '11 at 4:07