1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this a Galois extension?

  1. Nov 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Let Q < L < Q(c) where c is a primitive nth root of unity over Q. Is [L:Q] a Galois extension?

    2. Relevant equations


    3. The attempt at a solution
    L must be equal to Q(d) where d is a non primitive nth root of unity. [Q(d):Q] is not a galois extension because the minimal polynomial of d over Q is x^d -1, and this polynomial has d roots, not all of which are in Q. Is this correct?
     
  2. jcsd
  3. Nov 18, 2016 #2

    fresh_42

    Staff: Mentor

    What does it mean for ##\mathbb{Q}\subsetneq \mathbb{Q}(d)## to be a Galois extension? Can the minimal polynomial of ##d## have other roots than powers of ##d##? By the way ##x^d-1## is not the minimal polynomial, but only a factor of this: ##x-1## divides ##x^d-1## and the result is still in ##\mathbb{Q}[x]##. So why should we call such a polynomial minimal though?
     
  4. Nov 19, 2016 #3
    Q<Q(d) is a galois extension if Q(d) is a finite normal and separable extension. We know it is separable because Char(Q)=0. If Q(d) is a finite and normal extension then it is a splitting field for the minimal polynomial of d in Q. The minimal polynomial of d will be a factor of x^d-1, but I'm not sure if the roots of it's minimal polynomial are all powers of d; if they are, then I suspect this is a Galois extension.
     
  5. Nov 19, 2016 #4

    fresh_42

    Staff: Mentor

    You don't need the powers. Only the definition of normal. Firstly, how did you find ##d##? And why has it to be of the claimed form, i.e. its minimal polynomial a divisor of ##(x^d-1)##? Btw, it's a bit confusing to denote the element by ##d## as well as the degree of ##x^d-1##. One is a complex and one a natural number. And then, does this situation differ at any point from the one you started at?
     
  6. Nov 21, 2016 #5
    my bad, I'm more lost than I thought I suppose. I was thinking something like if the degree of d is n then the minimal polynomial of d will be a divisor of x^n-1. Am I way off track here?
     
  7. Nov 21, 2016 #6

    fresh_42

    Staff: Mentor

    You should proceed step by step.

    What we have is ##\mathbb{Q} \subseteq \mathbb{Q}(c)## with a primitive ##h-##th root of unity, i.e. ##c^h=1##. (I choose ##h## because you burnt ##n## in your question.) Let us define the minimal polynomial of ##c## by ##m_c(x)##. Then ##m_c(x)\,\vert \,(x^h-1)## and ##m_c(x)## is the product of all ##(x-c^k)## with ##(k,h)=1## being coprime. The Galois group of ##\mathbb{Q}(c)## over ##\mathbb{Q}## is isomorphic to the cyclic group ##\mathbb{Z}_h^* = \mathbb{Z}_{\varphi(h)}##.

    Now we assume a field ##\mathbb{Q} \subseteq L \subseteq \mathbb{Q}(c)## and we know, that its Galois group is a cyclic subgroup ##\mathbb{Z}_n## of ##\mathbb{Z}_{\varphi(h)}##, i.e. ##n\cdot m = \varphi(h)## for some ##m##. We may further assume that ##\mathbb{Z}_n## is generated by a (single) automorphism of order ##n##, which is also an automorphism ##\sigma## of ##\mathbb{Q}(c)## and therefore ##\sigma^{\varphi(h)}=(\sigma^m)^n = 1##. So ##\mathbb{Z}_n=\{\sigma^m,\sigma^{2m},\ldots,\sigma^{(n-1)m},\sigma^{nm}=1\}## is the Galois group of ##L##. This means in return that ##L## consists of all elements ##a##, that can be substituted by ##\sigma^{m}(a)##.

    This is what we know from the given situation. And it might be helpful to have examples like ##h=12, 16## or ##24## in mind. Now from here we can either list all elements ##a \in L## given a basis of ##\mathbb{Q}(c)## over ##\mathbb{Q}##, i.e. powers of ##c##, or construct a primitive element ##d## (and show that ##L=\mathbb{Q}(d)##).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Is this a Galois extension?
  1. Galois fields (Replies: 3)

  2. Galois Extensions (Replies: 1)

  3. Galois extensions (Replies: 7)

Loading...