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Math Help - ring+conditon=field

  1. #1
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    ring+conditon=field

    Let A be a ring with p^n elements with characteristic p , p prime, n\geq 1, and f\in\mathbb{Z}[X] a polynom of degree n ireductibile over \mathbb{Z}_p.If  f have a solution a \in A then A is field .
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  2. #2
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    Quote Originally Posted by petter View Post

    Let A be a ring with p^n elements with characteristic p , p prime, n\geq 1, and f\in\mathbb{Z}[X] a polynom of degree n ireductibile over \mathbb{Z}_p.If  f have a solution a \in A then A is field .
    i'll suppose that the ring is unitary. since \text{char} A = p, the map [k] \hookrightarrow k.1_A is an embedding of \mathbb{Z}_p into A. let \overline{f(x)} be the image of f(x) under the natural homomorphism \mathbb{Z}[x] \longrightarrow \mathbb{Z}_p[x].

    now let \overline{f(a)}=0, where a \in A. define the map \nu: \mathbb{F}_{p^n}= \frac{\mathbb{Z}_p[x]}{<\overline{f(x)}>} \longrightarrow A by: \nu(\overline{g(x)})=g(a), \ \forall g(x) \in \mathbb{Z}_p[x]. it's easily seen that \nu is a ring monomorphism. thus \nu is an isomorphism

    because |\mathbb{F}_{p^n}|=|A|=p^n. \ \ \Box
    Last edited by NonCommAlg; January 3rd 2009 at 12:06 AM.
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  3. #3
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    Can u give me a elementary solution please ?
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  4. #4
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    Quote Originally Posted by petter View Post

    Can u give me a elementary solution please ?
    well, there's nothing advanced about your problem and my solution. so ...
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