2
$\begingroup$

Suppose that our polynomial is $x^5-1$, thus the splitting field is $\mathbb{Q}(\gamma)$ where $\gamma$ is a primitive 5-th foot of unity. Then our basis for the extension field will be:

$\{1, \gamma, \gamma^2, \gamma^3, \gamma^4\}$

How would one find a Galois group for this extension? I can see that 1 needs to be fixed, since it is a member of $\mathbb{Q}$, then we can create the maps:

$\sigma_1 = id$

$\sigma_2: \gamma \rightarrow \gamma^2$

$\sigma_3: \gamma \rightarrow \gamma^3$

$\sigma_4: \gamma \rightarrow \gamma^4$

$\sigma_5: \gamma^2 \rightarrow \gamma^3$

$\sigma_6: \gamma^3 \rightarrow \gamma^4$

Does that seem right?

$\endgroup$
3
  • 1
    $\begingroup$ Notice that $\sigma_4(\gamma^2) = \sigma_4(\gamma)^2 = \gamma^8 = \gamma^5 \cdot \gamma^3 = \gamma^3 = \sigma_5(\gamma^2)$. So $\sigma_5$ is not a new automorphism. In general, an automorphism is determined by where it sends a basis, and a cyclotomic extension has a particularly nice basis that puts a lot of restriction on the automorphisms that can appear. $\endgroup$
    – bzc
    May 10, 2016 at 2:51
  • 1
    $\begingroup$ it is clear that once you defined $\sigma(\gamma)$, it determinates the whole automorphism. $\endgroup$
    – reuns
    May 10, 2016 at 3:11
  • 1
    $\begingroup$ The basis, rather, is $\{1,\gamma,\gamma^2,\gamma^3\}$. That is, the splitting field is quartic, not quintic. $\endgroup$
    – Lubin
    May 10, 2016 at 4:26

1 Answer 1

2
$\begingroup$

In general for cyclotomic extensions, $K=\Bbb Q(\zeta_n)$, your automorphism is totally determined by what it does on $\zeta$ since that generates the extension. This field is just $\Bbb Q[x]/(\Phi_n(x))$ with $\Phi_n(x)$ the $n^{th}$ cyclotomic polynomial. So an automorphism is just sending one primitive root to another, so there are at most $\varphi(n)$ such automorphisms. However, raising $\zeta_n^k$ for $k\in\Bbb Z/n\Bbb Z^*$ gives $\zeta_n'$ another primitive $n^{th}$ root of $1$, so each $k\in\Bbb Z/n\Bbb Z^*$ gives rise to an automorphism of the group. This induces an injection

$$\Bbb Z/n\Bbb Z^*\to \text{Gal}(K/\Bbb Q)$$

but then since the target has cardinality at most the cardinality of the domain, this must be an isomorphism.

In your case, this exactly reduces to $\gamma\mapsto \gamma^k, \quad 1\le k\le 4$ are all the automorphisms of your field.

Your $\sigma_5$ note has $\sigma_5(\gamma^1)=\sigma_5(\gamma^6)=\sigma_5(\gamma^2)^3=\gamma^{3\cdot 3}=\gamma^4$ so this is the same as $\sigma_4$. And similarly your $\sigma_6$ has $\sigma_6(\gamma^1)=\sigma_6(\gamma^6)=\sigma_6(\gamma^3)^2=\gamma^{2\cdot 4}=\gamma^3$ so it's the same as $\sigma_3$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .