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Math Help - Irreducible, inseparable polynomials

  1. #1
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    Irreducible, inseparable polynomials

    Let K be a field of characteristic p which is not a perfect field (K neq K^{p})

    Prove there exist irreducible inseparable polynomials over K. Conclude that there exist inseparable finite extensions of K.

    I am a little confused, because I thought that fields of characteristic p were isomorphic to the finite field with p elements. In this finite field, every element is a p'th power. This is what I thought the definition of a perfect field was.
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  2. #2
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    Quote Originally Posted by robeuler View Post
    Let K be a field of characteristic p which is not a perfect field (K neq K^{p})

    Prove there exist irreducible inseparable polynomials over K. Conclude that there exist inseparable finite extensions of K.
    Since K is not perfect it means K^p \not = K and so there exists a\in K so that a\not \in K^p. Define the polynomial f(x) = x^p - a. Let L be an extension field having a zero \alpha of f(x). Notice that if \alpha \in L is a zero of this polynomial then \alpha^p - a = 0 \implies a = \alpha^p, but this means x^p - \alpha^p = (x-\alpha)^p therefore we have shown that f(x) splits over any extension field that has a zero of f(x). This means either f(x) is irreducible or it factors into linear factors (this is a result about polynomials of prime degree, if you do not know I can prove it for you) over F. It cannot factor into linear factors since it has no zeros in F and so f(x) is irreducible. Thus, L/F is an inseperable extension. In fact, what we have constructed is a purely inseperable extension, this is like the worst case scenario of seperability.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Since K is not perfect it means K^p \not = K and so there exists a\in K so that a\not \in K^p. Define the polynomial f(x) = x^p - a. Let L be an extension field having a zero \alpha of f(x). Notice that if \alpha \in L is a zero of this polynomial then \alpha^p - a = 0 \implies a = \alpha^p, but this means x^p - \alpha^p = (x-\alpha)^p therefore we have shown that f(x) splits over any extension field that has a zero of f(x). This means either f(x) is irreducible or it factors into linear factors (this is a result about polynomials of prime degree, if you do not know I can prove it for you) over F. It cannot factor into linear factors since it has no zeros in F and so f(x) is irreducible. Thus, L/F is an inseperable extension. In fact, what we have constructed is a purely inseperable extension, this is like the worst case scenario of seperability.
    thank you. I know the result about irreducible/linear factors.

    .
    Last edited by ThePerfectHacker; March 16th 2009 at 07:16 PM.
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