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Math Help - Question on proof regarding structure of finite fields

  1. #1
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    Question on proof regarding structure of finite fields

    I am reading the proof of the fact that as a group under multiplication, the set of nonzero elements of \mbox{GF}(p^n) (the Galois field/finite field of order p^n) is isomorphic to \mathbb{Z}_{p^n-1} and is therefore cyclic.

    During the course of the proof, the author reminds the reader that the operation is componentwise addition. I do not see why this is true...we are looking at the multiplicative group of the field. Can someone explain? (I will add more detail if necessary).
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    Re: Question on proof regarding structure of finite fields

    Quote Originally Posted by AlexP View Post
    I am reading the proof of the fact that as a group under multiplication, the set of nonzero elements of \mbox{GF}(p^n) (the Galois field/finite field of order p^n) is isomorphic to \mathbb{Z}_{p^n-1} and is therefore cyclic.

    During the course of the proof, the author reminds the reader that the operation is componentwise addition. I do not see why this is true...we are looking at the multiplicative group of the field. Can someone explain? (I will add more detail if necessary).
    This doesn't make sense, we need more context. What would he even mean by component wise addition? Perhaps thinking of \mathbb{F}_{p^n} in its usual form as \mathbb{Z}[x]/(p(x)) for an appropriate p(x) from where the 'addition componentwise' would make sense for the addition in \mathbb{F}_{p^n}.
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    Re: Question on proof regarding structure of finite fields

    True...the componentwise part doesn't make sense without the rest. Here's the whole proof. There's another part I don't understand so I'll have that part ready too."

    "To see that the multiplicative group \mbox{GF}(p^n)^\times of nonzero elements of \mbox{GF}(p^n) is cyclic, we first note...that it is isomorphic to a direct product of the form \mathbb{Z}_{n_1} \otimes \mathbb{Z}_{n_2} \otimes \cdots \otimes \mathbb{Z}_{n_k}, where each n_{i+1} divides n_i. So, for any element a=(a_1, a_2, ..., a_k) in this product, we have a^{n_1} = (n_1a_1, n_1a_2, ..., n_1a_k). (Remember, the operation is componentwise addition.)
    Thus, the polynomial x^{n_1}-1 has p^n-1 zeros in \mbox{GF}(p^n). Since the number of zeros of a polynomial over a field cannot exceed the degree of the polynomial, we know that p^n-1 \le n_1. On the other hand, since \mbox{GF}(p^n)^\times has a subgroup isomorphic to \mathbb{Z}_{n_1}, we also have n_1 \le p^n-1. It follows, then, that \mbox{GF}(p^n)^\times is isomorphic to \mathbb{Z}_{p^n-1}."
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    Re: Question on proof regarding structure of finite fields

    Quote Originally Posted by AlexP View Post
    True...the componentwise part doesn't make sense without the rest. Here's the whole proof. There's another part I don't understand so I'll have that part ready too."

    "To see that the multiplicative group \mbox{GF}(p^n)^\times of nonzero elements of \mbox{GF}(p^n) is cyclic, we first note...that it is isomorphic to a direct product of the form \mathbb{Z}_{n_1} \otimes \mathbb{Z}_{n_2} \otimes \cdots \otimes \mathbb{Z}_{n_k}, where each n_{i+1} divides n_i. So, for any element a=(a_1, a_2, ..., a_k) in this product, we have a^{n_1} = (n_1a_1, n_1a_2, ..., n_1a_k). (Remember, the operation is componentwise addition.)
    Thus, the polynomial x^{n_1}-1 has p^n-1 zeros in \mbox{GF}(p^n). Since the number of zeros of a polynomial over a field cannot exceed the degree of the polynomial, we know that p^n-1 \le n_1. On the other hand, since \mbox{GF}(p^n)^\times has a subgroup isomorphic to \mathbb{Z}_{n_1}, we also have n_1 \le p^n-1. It follows, then, that \mbox{GF}(p^n)^\times is isomorphic to \mathbb{Z}_{p^n-1}."
    First of all, don't use tensor for direct product since that has a separate meaning. What he is really saying is this, if f:\mathbb{F}_{p^n}\to\mathbb{Z}_{n_1}\times\cdots \times\mathbb{Z}_{n_k} is an isomorphism we have that x^{n_1}-1=f^{-1}(f(x^{n_1}-1))=f^{-1}(n_1f(x)-1)=f^{-1}(0)=1 where I made use of the fact that n_1f(x)-1=0 since, if f(x)=(a_1,\cdots,a_k) we have that n_1(a_1,\cdots,a_k)=(na_1,\cdots,na_k)=(0,\cdots,0  ) since n_1\mid |\mathbb{Z}_{n_j}| for each j\in[k] and so n_1a_j=0 by Lagrange's theorem. Make sense?


    Also if you are having trouble understanding this I, at the risk of shamelessly self-promoting, suggest you see this post I wrote on my blog about this topic.
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    Re: Question on proof regarding structure of finite fields

    I understand that n_1(a_1, ..., a_k)=(n_1a_1, ..., n_1a_k) = (0,...,0), but with x^{n_1}-1=\cdots=0 I'm getting x^{n_1}=2 which does not make sense.

    That notation was a mistake, no idea what I was thinking. I meant to use what the book uses, which is \oplus.
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    Re: Question on proof regarding structure of finite fields

    more generally, any finite subgroup of the multiplicative group of a field is cyclic.
    this result, under some conditions, can be extended nicely to division rings. see here.
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    Re: Question on proof regarding structure of finite fields

    Quote Originally Posted by AlexP View Post
    I understand that n_1(a_1, ..., a_k)=(n_1a_1, ..., n_1a_k) = (0,...,0), but with x^{n_1}-1=\cdots=0 I'm getting x^{n_1}=2 which does not make sense.

    That notation was a mistake, no idea what I was thinking. I meant to use what the book uses, which is \oplus.
    How exactly do you manage that? Perhaps if you told us how you are getting that result we can help find your mistake.
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    Re: Question on proof regarding structure of finite fields

    Wow...I have no idea where my brain has been lately. That's scary.

    What you did say was x^{n_1}-1= \cdots =1, and that's where I got that. I'm clearly not following something.
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    Re: Question on proof regarding structure of finite fields

    Quote Originally Posted by AlexP View Post
    Wow...I have no idea where my brain has been lately. That's scary.

    What you did say was x^{n_1}-1= \cdots =1, and that's where I got that. I'm clearly not following something.
    I think that it's perhaps because I made an EGREGIOUS typo/mistake in my earlier post (what time of night was that)? Ok, so you're trying to show that x^{n_1}=1 but since f is an isomorphism this is EQUIVALENT to showing that f(x^{n_1})=f(1)=0 (we know that f(1)=0 since 1 is the identity of \mathbb{F}_{p^n} and so cooresponds to the identity  0=(0,\cdots,0) of \mathbb{Z}_{n_1}\times\cdots\times\cdots\times \mathbb{Z}_{n_m}). But, since f is a morphism we know that f(x^{n_1})=n_1f(x) from there I think you got the reason why this is zero.

    I'm sorry for the error, does that make more sense?
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    Re: Question on proof regarding structure of finite fields

    It's fine, everyone's entitled to some mistakes (I certainly made some as well). I do see it clearly now. Just one thing... You wrote our map as f: \mathbb{F}_{p^n} \to \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_k}, but should it really be f: \mathbb{F}_{p^n}^\times \to \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_k} (specifically from the multiplicative group of our field to the direct product)?
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    Re: Question on proof regarding structure of finite fields

    Quote Originally Posted by AlexP View Post
    It's fine, everyone's entitled to some mistakes (I certainly made some as well). I do see it clearly now. Just one thing... You wrote our map as f: \mathbb{F}_{p^n} \to \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_k}, but should it really be f: \mathbb{F}_{p^n}^\times \to \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_k} (specifically from the multiplicative group of our field to the direct product)?
    Yes.
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    Re: Question on proof regarding structure of finite fields

    ok. Thank you very much for the help (and patience).
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    Re: Question on proof regarding structure of finite fields

    Quote Originally Posted by AlexP View Post
    ok. Thank you very much for the help (and patience).
    Anytime
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