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Is this field extension normal?

  1. Nov 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the field extension Q(c):Q, where c is a primitive nth root of unity. Is this extension normal?

    2. Relevant equations


    3. The attempt at a solution
    I believe this polynomial splits in x^n - 1,where it's n roots are exactly the powers of c. Thus this extension is normal because Q(c) is a finite splitting field of the polynomial x^n-1 over Q. Is this correct?
     
  2. jcsd
  3. Nov 18, 2016 #2

    fresh_42

    Staff: Mentor

    Yes. Only that ##x^n-1## is divided by the minimal polynomial of ##\mathbb{Q}(c)##. "splits in ##x^n-1##" doesn't make sense. And you get less than ##n## factors, since ##(x-1)## is one factor (eventually with others) which you don't need.
    You already know, which powers of ##c## are primitive, so you have two ways to write down the minimal polynomial: either by all the factors that give you all the necessary roots in terms of powers of ##c##, or explicitly with the roots in ##\mathbb{C}##. Both ways show you the decomposition. The way using complex numbers has the advantage, that one easily sees, that and which powers of ##c## are involved.
     
  4. Nov 18, 2016 #3
    My bad, I should have said x^n - 1 splits in Q(c). x^n-1 is divided by the minimal polynomial of Q(c)? The minimal polynomial of Q(c) will not b the nth cyclotomic polynomial because n is not primitive, do I need to know more about it's minimal polynomial. But the nth cyclotomic polynomial will be a factor of the minimal polynomial in c because all of it's roots are powers of c.
     
  5. Nov 18, 2016 #4

    fresh_42

    Staff: Mentor

    Well the ##n-##th roots of unity are all complex numbers, because at latest ##x^n-1## splits in ##\mathbb{C}##.
    You know, that they divide the unit circle in equidistant parts. Since ##1^n-1=0\, , \,(x-1)\,|\,(x^n-1)## and our first point is ##1##. The others are therefore ##\exp(\frac{k}{n}\,\cdot\,2\pi\,i)=\exp(\frac{1}{n}\,\cdot\,2\pi\,i)^k=c^k## with ##k=1,\ldots,n-1##. Split. You know from previous posts, that you need a primitive root ##c## as given to get all others as a power of it.

    I'm not sure how you define the cyclotomic polynomial. In my books, it is the (irreducible) minimal polynomial (which of course has to divide ##x^n-1##).

    We don't say ##n## is primitive or not. A root of ##x^n-1## can be primitive or not: ##\exp(\frac{1}{n}\,\cdot\,2\pi\,i)^1## is, ##\exp(\frac{1}{n}\,\cdot\,2\pi\,i)^0## is not. ##c## is primitive simply means, that you can hop around the circle starting at ##c^0## by steps of ##\cdot c## and hit all other roots before you hit one twice. Like ##(c^0,c^3,c^2,c^1)## hops in steps of ##c^3## and hits all of the ##4-##th roots of unity, whereas ##(c^0,c^2)## in steps of ##c^2## does not.
     
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