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Math Help - Is Simple Extension Minimal?

  1. #1
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    Is Simple Extension Minimal?

    If you have fields F\leq E with \alpha \in E algebraic over F, show that,
    F(\alpha) is the minimal field containing F and \alpha.


    (Note: F(\alpha)=\phi_{\alpha}[F[x]]-a simple extension).
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    The map \phi_\alpha : F[X] \rightarrow E given by evaluation at \alpha is a ring homomorphism and hence its image, which you denote F(\alpha), is a subring of E. All you need to show is that it is closed under taking multiplicative inverses. Note for use in a moment that a subring of a field is an integral domain. The fact that \alpha is algebraic is equivalent to saying that F(\alpha) is a finite-dimensional F-vector space. Let \beta be a non-zero element of F(\alpha). Multiplication by \beta is an F-linear endomorphism and has trivial kernel (integral domain), hence is surjective by finiteness of the dimension and the rank--nullity formula. So 1 is in the image, hence a multiple of \beta, which is thus invertible.
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    I was thinking about this today, maybe you can do this?

    You have F\leq E with \alpha algebraic over F. Let there exist a field K with \alpha \in K such as,
    F\leq K\leq F(\alpha)\leq E,
    Any element \beta \in F(\alpha) can be expressed as,
    \beta =a_0+a_1\alpha +...+ a_{n-1}\alpha ^{n-1} with a_i \in K because every finite dimensional vector basis has a finite basis. Therefore \beta \in K thus, K\leq F and F\leq K thus, K=F.

    Am I missing something?
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    I realise that I didn't actually address the issue of being the minimal field, just that it actually was a field. But any field containing \alpha clearly contains any polynomial in \alpha. Hence the minimality.
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    Quote Originally Posted by rgep
    \alpha clearly contains any polynomial in \alpha. Hence the minimality.
    What do you mean by that?
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    Any field that contains F and contains \alpha must contain any polynomial in \alpha with coefficients in F. The previous part of my argument showed that the ring comprised of polynomials in \alpha with coefficients in F is actually a field. Hence it is the minimal subfield of E containing both F and \alpha.
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  7. #7
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    Quote Originally Posted by rgep
    Any field that contains F and contains \alpha must contain any polynomial in \alpha with coefficients in F. The previous part of my argument showed that the ring comprised of polynomials in \alpha with coefficients in F is actually a field. Hence it is the minimal subfield of E containing both F and \alpha.
    You mean a polynomial F[x] evaluated at \alpha.
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  8. #8
    DMT
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    Isn't this all circular? I thought  F(\alpha) was defined to be the smallest the field containing both F and \alpha.
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  9. #9
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    Quote Originally Posted by DMT
    Isn't this all circular? I thought  F(\alpha) was defined to be the smallest the field containing both F and \alpha.
    Maybe in some texts. The version I seen is that F(a) is defined as above.
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