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Deducing that an element is constructable

  1. Nov 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Let c be a primitive 16th root of unity in the field of complex numbers. Show that c is a constructible number.

    2. Relevant equations


    3. The attempt at a solution
    I showed that c^2 is a primitive 8th root of unity and c^4 is a primitive 4th root of unity, so if I can show that one of these is constructible then I can use the fact that the square roots of constructible numbers are constructible to show that c is constructible. That being said, I am quite out of my area of knowing here, I don't have any clue how to go about showing any of these are constructible. Well, if 4 is a primitive 4th root of unity is constructible, that would mean that some kind of diamond is constructible in the complex plane? Am I on the right track here?
     
  2. jcsd
  3. Nov 16, 2016 #2

    fresh_42

    Staff: Mentor

    Do you already have the connection between type and degree of extensions and constructible numbers? Or are you supposed to do this?
     
  4. Nov 17, 2016 #3
    I'm allowed to use the fact that every extension that has a degree that's NOT a power of 2 is NOT constructible. It's my understanding that if an extension has a degree that is a power of 2 that it is construnctible, but we are not allowed to use this. I think it has something to do using some geometry stuff, like showing that if some cos(2pi/something) is constructible then the 16-gon is constructible (I'm not sure I'm making complete sense).
     
  5. Nov 17, 2016 #4

    fresh_42

    Staff: Mentor

    So you have
    ##[\mathbb{Q}:\mathbb{Q}(c)] \neq 2^n \Rightarrow c \text{ not constructible }## which is equivalent to ##c \text{ constructible } \Rightarrow [\mathbb{Q}:\mathbb{Q}(c)] = 2^n##.
    (You should add the extension to be normal, too, simply for completeness.) Too bad, the other direction is what you needed.

    Your idea to resolve the problem into steps by two is correct, because square roots are constructible as I think you may also use (or simply show). The fun part is, this apparently trivial sentence bears already the answer you need: consider the Galois group. What can you say about it? I've already written the key word here.
     
  6. Nov 17, 2016 #5
    The Galois group is isomorphic to the group of units Z*_16, which I could break down into a direct product of cyclic groups since it is finite and abelian. I don't understand how looking at the Galois group is going to help me prove that c is constructible. Would looking at subgroups of the Galois group i.e. Gal(Q(c^2):Q) and Gal(Q(c^4):Q) be helpful? You said you already have written a key word huh? I must be pretty dense.

    And yes, we can use the fact that square roots of a constructible number is constructible
     
  7. Nov 17, 2016 #6

    LCKurtz

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    Probably a silly question, but why can't you just show a construction that works?
     
  8. Nov 17, 2016 #7

    fresh_42

    Staff: Mentor

    How many elements does ##\mathbb{Z}_{16}^*## have? And, yes, it's finite and Abelian. What do we know about those groups? You overlooked the keyword I used, and what is even more important, disregarded the reason why this term once has been chosen:
    solvable!
    With that you only need to add the fundamental theorem of Galois theory and voilà: done (without drawing circles).
     
  9. Nov 17, 2016 #8
    Eh, I'm doing this in a Galois theory class so I'm trying to find an algebraic method of showing it's constructible
     
  10. Nov 17, 2016 #9
    Ah, we're learning about solvability tomorrow actually :-) I'll get back to you on this haha.
     
  11. Nov 17, 2016 #10

    fresh_42

    Staff: Mentor

    That's what the algebraic method describes: Dividing something into two equal parts, and proceed until all ##16## are done.
    And it was your initial idea:
    Basically an induction (or recursion).
     
  12. Nov 18, 2016 #11
    So should I show that c^8 is a second root of unity now? I'm still a bit confused on how this is going to show that c is constructible.
     
  13. Nov 18, 2016 #12

    fresh_42

    Staff: Mentor

    If the concept of solvable groups and derived series is yet to come, then you might follow @LCKurtz advise and simply construct ##c##.
    Draw the complex plane, a unit circle and start to cut the angles in halves. Begin with the full circle as your first.

    This geometric process can also be done algebraically.
     
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