My Math Forum Fixed fields and Galois subgroups.
 User Name Remember Me? Password

 Abstract Algebra Abstract Algebra Math Forum

 May 9th, 2013, 05:58 PM #1 Member   Joined: May 2012 Posts: 56 Thanks: 0 Fixed fields and Galois subgroups. For the group $Q(\zeta_7)$. I found all the permutations that are possible for $\zeta_7$. Here is what I have: Since $Aut(=<\zeta_7=>)= Z^{\times}_7 = \{1, 2, 3, 4, 5, 7\}$, we know that we can have 6 automorphisms. Let $\zeta= \zeta_7$ $\zeta \rightarrow \zeta$ $\zeta \rightarrow \zeta^2$ $\zeta \rightarrow \zeta^3$ $\zeta \rightarrow \zeta^4$ $\zeta \rightarrow \zeta^5$ $\zeta \rightarrow \zeta^6$ I computed the orders of these automorphisms (since I want to know the subgroup orders in the Galois group in order to determine the fixed fields). So... $\zeta \rightarrow \zeta$ has order 1 $\zeta \rightarrow \zeta^2 \rightarrow \zeta^4 \rightarrow \zeta$ Order=3 $\zeta \rightarrow \zeta^3 \rightarrow \zeta^2 \rightarrow \zeta^^6 \rightarrow^4 \rightarrow \zeta^5 \rightarrow \zeta$ Order = 6 $\zeta \rightarrow \zeta^4 \rightarrow \zeta^2 \rightarrow \zeta$ Order = 3 $\zeta \rightarrow \zeta^5 \rightarrow \zeta^4 \rightarrow \zeta^6 \rightarrow \zeta^2 \rightarrow \zeta^3 \rightarrow \zeta$ Order = 6 $\zeta \rightarrow \zeta^6 \rightarrow \zeta$ Order = 2 So now I need to find the fixed subgroups. I know that $< \zeta>$ will correspond to $Q(\zeta)$ since it fixes everything. I know that the other fields I need to look at are $Q(\zeta^2)$, $Q(\zeta^3)$, $Q(\zeta^4)$, $Q(\zeta^5)$, and $Q(\zeta^6)$. There are two reasons why I was stuck: 1) Since, for an intermediate field K and Galois subgorup H of G, $[ K : Q]= [G : H]$, finding [K:Q] would help reduce the possibilities, right? But I'm kind of confused about how we would be able to find that. 2) Even if we did find [K:Q] for each K, that would only reduce the possibilites and not tell us exactly which subgorup corresponds to which intermediate field, right? So we have to check which subgroup fixes $\zeta^k$ for some k between 1 and 6, right? But I don't really get anywhere with this method. I'm probably doing something wrong, but I'm not sure what. For example, for $\zeta^3$, I tried to see which automorphism would give me $\zeta^3$ back but couldn't really find any that would. So I'm probably missing something, but I'm not sure why... Thank you in advance

 Tags fields, fixed, galois, subgroups

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post jezza10181 Abstract Algebra 1 December 15th, 2013 02:34 PM gaussrelatz Algebra 1 October 10th, 2012 11:30 PM DanielThrice Abstract Algebra 1 November 25th, 2010 02:28 PM TTB3 Abstract Algebra 6 June 27th, 2009 06:57 AM bjh5138 Abstract Algebra 2 August 14th, 2008 12:03 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.