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Math Help - Finite Subfield

  1. #1
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    Finite Subfield



    (a) I want to use a result known as the "Subfield Test", but unfortunately I don't see how to make use of the fact that char(F)=p in order to simplify things.

    To use the Subfield Test I must show that for any x,y (y \neq 0) in K, x-y and xy^{-1} belong to K.

    Let x , y \in K, so by definition x^p =x and y^p=y. But how do we deduce that ( x-y)^p = (x-y) or (xy^{-1})^p = xy^{-1}?

    The only thing we are told is the characteristic of F is a prime (which is the least positive integer p such that px=0 for all x in F). Is there any way we can somehow use Fermat's little theorem?

    (b) Any clues to get me started is really appreciated.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by demode View Post


    (a) I want to use a result known as the "Subfield Test", but unfortunately I don't see how to make use of the fact that char(F)=p in order to simplify things.

    To use the Subfield Test I must show that for any x,y (y \neq 0) in K, x-y and xy^{-1} belong to K.

    Let x , y \in K, so by definition x^p =x and y^p=y. But how do we deduce that ( x-y)^p = (x-y) or (xy^{-1})^p = xy^{-1}?

    The only thing we are told is the characteristic of F is a prime (which is the least positive integer p such that px=0 for all x in F). Is there any way we can somehow use Fermat's little theorem?

    (b) Any clues to get me started is really appreciated.
    For (a), you want to use the binomial theorem. To get the result, you will need to prove that p| \frac{p!}{n!(p-n)!}. For the second bit, you should remember that x^p=x for all x \in K and that xy=yx, so (xy)^p=\ldots.

    For (b), what is the only infinite cyclic group? Can this be a field? Alternatively, you could prove that every infinite field contains a copy of the rational numbers, but that might be a bit like overkill...
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    For (a), you want to use the binomial theorem. To get the result, you will need to prove that p| \frac{p!}{n!(p-n)!}. For the second bit, you should remember that x^p=x for all x \in K and that xy=yx, so (xy)^p=\ldots
    But why does proving that p dicvides the coefficients of x^{p-n}y^n show that (x-y)^p=(x-y)?

    I'm not sure how to prove this formally, but since \frac{p!}{n!\,(p-n)!} is the same as

    \frac{p(p-1)(p-2) \cdots (p-n+1)}{n(n-1)(n-2) \cdots 1}

    we can observe that this number must be divisible by p. This is okay?

    For (b), what is the only infinite cyclic group? Can this be a field? Alternatively, you could prove that every infinite field contains a copy of the rational numbers, but that might be a bit like overkill...
    \mathbb{Z} is the only infinite cyclic group, and it has two generators 1 and -1. But \mathbb{Z} is NOT a field.

    Only I think \mathbb{Z}_p, the ring of integers modulo p is a field with characteristic p (where p is a prime).
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by demode View Post
    But why does proving that p dicvides the coefficients of x^{p-n}y^n show that (x-y)^p=(x-y)?

    I'm not sure how to prove this formally, but since \frac{p!}{n!\,(p-n)!} is the same as

    \frac{p(p-1)(p-2) \cdots (p-n+1)}{n(n-1)(n-2) \cdots 1}

    we can observe that this number must be divisible by p. This is okay?
    Yes, that is fine. You want to do this as then you can expand (x-y)^p using the binomial theorem...


    Quote Originally Posted by demode View Post
    \mathbb{Z} is the only infinite cyclic group, and it has two generators 1 and -1. But \mathbb{Z} is NOT a field.

    Only I think \mathbb{Z}_p, the ring of integers modulo p is a field with characteristic p (where p is a prime).
    It is a cyclic group though, which is what you are looking for. Also, as a group it only have one generator (inverses are implicit).
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  5. #5
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    I must use the fact that the characteristic is not 2. So here's what I've come up with:

    We have 1+1 \ne 0, that is, 1 \ne -1. Now let's consider the assumption that F^* = F\setminus \{0\} is cyclic, F^* = < a > say. This means that every non-zero element of F is a power of a. And we know that a ^0 = 1 and we recall that -1 \ne 1.

    So, could you please help me finish it off from here? Now I just need to show that a has finite order, that is, there exists an integer n with a^n=1. Then F^* is finite, and so is F. But I don't know how to do this bit...

    I know that -1 \in F^*, so it is a power of a. What else do I need to say to complete the proof?
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  6. #6
    MHF Contributor Swlabr's Avatar
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    You do not need to re-write the proof. It works. You just need to work out where you used the fact that the field has characteristic greater than 2. Perhaps the best way to think about this is to work out what the field \mathbb{Z}/2\mathbb{Z} of characteristic 2 looks like? What is it's multiplicative group?
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