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Math Help - extension fields question

  1. #1
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    extension fields question

    Problem: If F is algebraic over K and D is an integral domain such that $K\subset D\subset F$, then D is a field.

    Sketch of Proof: We want to show every element of D is a unit. Let u in D. Since F is a field, u^{-1} exists at least in F. Since F is algebraic over K, there exists f in K[x]\{0} such that f(u^{-1})=b_0+b_1u^{-1}+...+b_m u^{-m}=0. Now using a nonzero element of D (such as u^{m-1}) I want to be able to factor the last expression and divide by the nonzero element to obtain a polynomial expression for u^{-1} in K[u], which will complete the proof since K[u]\subset D.

    Arlington
    Last edited by arlingtonbassett; January 20th 2013 at 09:19 AM.
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  2. #2
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    Re: extension fields question

    why bother with taking f(u-1)? F is algebraic over K, and u is in F (since D is contained in F), so there is some polynomial f(x) in K[x]\{0} such that f(u) = 0.

    say f(x) = b0 + b1x + b2x2 +...+ bnxn. we may without loss of generality assume f is irreducible over K

    (since D is a domain, if f(x) = g(x)k(x), then 0 = f(u) = g(u)k(u), so either g(u) = 0, or k(u) = 0).

    then since f(u) = 0:

    b0 = -b1u - b2u2 - .... - bnun.

    if b0 = 0, then this contradicts the minimality of f, so b0 is non-zero, thus:

    1 = [-(b1/b0) - (b2/b0)u -...- (bn/b0)un-1](u)

    which shows u is a unit in D (since the left factor is in K[u], contained in D).
    Thanks from arlingtonbassett
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