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Math Help - Polynomial in Z[X]

  1. #1
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    Polynomial in Z[X]

    Hi! I just have a short question: I want to proof the following claim: Let K be a field with characteristic p>0 and f be a polynomial in Z[X] (with Z being the set of integers). Then f(y^p)=f(y)^p for any y in K.

    My consideration is this: f=a_0 + a_1 x + a_2 x^2 + ... + a_n x^n Using the binomial theorem it easily follows that f(y)^p=(a_0)^p + (a_1 y)^p + (a_2 y^2)^p + ... + (a_n y^n)^p = a_0 + a_1 y^p + ... + a_n (y^p)^n = f(y^p), as (a_i)^p=a_i. But this last implication only holds if the coefficients are in Z/pZ which is not demanded.

    So my questions: Is there a mistake in the description of the exercise? Or are the coefficients to be treated modulo p; if yes, why? Thankful for any tips, Banach!
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  2. #2
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    Quote Originally Posted by Banach View Post
    Hi! I just have a short question: I want to proof the following claim: Let K be a field with characteristic p>0 and f be a polynomial in Z[X] (with Z being the set of integers). Then f(y^p)=f(y)^p for any y in K.

    My consideration is this: f=a_0 + a_1 x + a_2 x^2 + ... + a_n x^n Using the binomial theorem it easily follows that f(y)^p=(a_0)^p + (a_1 y)^p + (a_2 y^2)^p + ... + (a_n y^n)^p = a_0 + a_1 y^p + ... + a_n (y^p)^n = f(y^p), as (a_i)^p=a_i. But this last implication only holds if the coefficients are in Z/pZ which is not demanded.

    So my questions: Is there a mistake in the description of the exercise? Or are the coefficients to be treated modulo p; if yes, why? Thankful for any tips, Banach!
    Every field of characteristic p>0 contains the base field \mathbb{Z}/p\mathbb{Z}, so the equation {a_i}^p\equiv a_i is perfectly all right.
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  3. #3
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    The binomial theorem still holds in over a field.
    Thus, (a+b)^p = a^p + b^p + \sum_{k=1}^{p-1}{p\choose k}a^pb^{p-k}.
    But the coefficients in the summand is each divisible by p so each one dies off and gives you zero.
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,
    Assuming that p is prime...
    Or did I miss a point in the original text ?
    Yes you did. Characteristic of a finite field is always prime.
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  5. #5
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    Quote Originally Posted by Isomorphism View Post
    Yes you did. Characteristic of a finite field is always prime.
    Ok, I have to wake up, I missed this sentence in the wiki..
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  6. #6
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    Quote Originally Posted by Moo View Post
    Ok, I have to wake up, I missed this sentence in the wiki..
    you can see this by writing out \underbrace{1+1+1+\ldots+1}_\text{m times}=0 and use the fact that a field has no zero divisor.
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  7. #7
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    Hi! Thanks everyone for your answers. My only problem was the structure that is given between elements of the field and integers, but as i remembered the definition everything cleared up. I got it now. Have a nice day! Banach
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