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Given a = $e^\frac{2i\pi}{18}$, field extension $|\Bbb Q(a):\Bbb Q| = 6$, I was wondering what the basis for $\Bbb Q(a)$ being, or how the elements are represented. I found the minimal polynomial to be $x^6-x^3+1$, which is degree 6 in $\Bbb Q$, so I'm guessing the basis for $\Bbb Q(a)$ would $\{1,a,a^2,a^3,a^4,a^5\}$. Does that make any sense? Normally the bases I have dealt with were degree $2$ so the elements were just like $a+bi$ or something with basis $\{1,i\}$

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  • $\begingroup$ Since $f(x) = x^6-x^3+1$ is irreducible and $f(a) = 0$ you know $\mathbb{Q}[a] = \{ \sum_{n=0}^5 c_n a^n, c_n \in \mathbb{Q}\}$. For proving that $\mathbb{Q}(a) = \mathbb{Q}[a]$ you need to show that $\mathbb{Q}[x]/f(x)$ is a field. With $\phi(p(x)) = p(a)$ you have an isomorphism $\phi : \mathbb{Q}[x]/f(x) \to \mathbb{Q}[a]$, ie. $\mathbb{Q}[a]$ is a field. $\endgroup$ – reuns Apr 7 '17 at 5:07
  • $\begingroup$ Oh I see. Thank you very much. This might be much, but how would I attempt to find the galois group for Q(a):Q? If you don't want to answer thats fine. $\endgroup$ – Tesuji Apr 7 '17 at 5:58
  • $\begingroup$ The Galois group of cyclotomic extensions is easy : $\sigma(\zeta_9)=\zeta_9^k$ where $gcd(k,9) = 1$. There are $\varphi(9) = 6$ such automorphisms. $\endgroup$ – reuns Apr 7 '17 at 6:18
  • $\begingroup$ Oh man we didn't learn about the zeta stuff yet. Is there a link you can provide me to read up on this? Despite your comment, I can't list the different $\sigma$ elements, since our professor wants a product table and then show what group this is isomorphic too.. If you can't, regardless, thank you very much for your help. $\endgroup$ – Tesuji Apr 7 '17 at 6:34
  • $\begingroup$ $\zeta_9 = a = e^{2 i \pi / 9}$ and $\sigma$ is a ring morphism $\mathbb{Q}[a] \to \mathbb{Q}[a]$, so $\sigma(a) = a^k$ means $\sigma(\sum_{n=0}^5 c_n a^n) =\sum_{n=0}^5 c_n \sigma(a^n)= \sum_{n=0}^5 c_n a^{nk}$ $\endgroup$ – reuns Apr 7 '17 at 6:39
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Your finding of the minimal polynomial and the basis are fine, it proves that you have well understood the setup. A trick to retrieve the Galois group is to find another root for $x^6-x^3+1$. Some trial and error gives you that $-a^2$ is such a root. So there is an element $\sigma$ of the Galois group that maps $a \mapsto -a^2$, so we can then ask what is the image $\sigma(-a^2)$? It's $--a^4$ and we verify that it is indeed another root! we can go on and find the three other roots: $\sigma(-a^4) = -a^8 = a^2-a^5$, $\sigma(a^2-a^5) = \sigma(a^2)-\sigma(a^5) = -a^4 + a^{10} = -a^4 -a$ and $\sigma(-a^4 -a) = \sigma(-a^4) - \sigma(a) = a^2 + a^5 + a^2$. If we change basis to these roots $$ \{a, -a^2,-a^4,a^2-a^5,a^4-a, a^5\} $$ we have an easy expression for $\sigma$ as a matrix.
When we revert this matrix to the original basis $\{1,a,a^2,a^3,a^4,a^5\}$ this yields the matrix $$ \sigma = \begin{pmatrix}1&0&0&1&0 &0\\ 0&0&0&0&0 &1\\ 0&-1&0&0&-1&0\\ 0&0&0&-1 & 0&0\\ 0&0&1&0&0&0\\ 0&0&0&0&1&0 \end{pmatrix} $$ This matrix generates the Galois group, a cyclic group of order $6$.

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