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Math Help - algebraic elements and evaluation homomorphism

  1. #1
    ynj
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    algebraic elements and evaluation homomorphism

    Let F be a subfield of E, \alpha\in Eand \alphais an algebraic over F. Let \phi_\alphabe the evaluation homomorphism. F(\alpha)is the minimal field containing F,\alpha
    True or false: \phi_\alpha(F[x])\cong F(\alpha)?
    One main problem is: \phi_\alpha(F[x])is a field????
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  2. #2
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    Quote Originally Posted by ynj View Post
    Let F be a subfield of E, \alpha\in Eand \alphais an algebraic over F. Let \phi_\alphabe the evaluation homomorphism. F(\alpha)is the minimal field containing F,\alpha
    True or false: \phi_\alpha(F[x])\cong F(\alpha)?
    One main problem is: \phi_\alpha(F[x])is a field????
    yes! let f(x) \in F[x] be the minimal polynomial of \alpha, i.e. f is monic and it has the smallest degree among all g(x) \in F[x] with this property that g(\alpha)=0. then \ker \phi_{\alpha} = <f(x)> and, since

    \phi_{\alpha}:F[x] \longrightarrow F[\alpha] is surjective, we have \frac{F[x]}{<f(x)>} \cong \phi_{\alpha}(F[x]) = F[\alpha]. finally, since f(x) is irreducible and F[x] is a PID, the ideal <f(x)> is maximal and thus F[\alpha] is a field. as a result

    F[\alpha]=F(\alpha). the converse is also true: if F[\alpha] is a field, then \alpha is algebraic over F. the reason is that if \alpha is transcendental over F, then F[\alpha] \cong F[x] and obviously F[x] is never a field.
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  3. #3
    ynj
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    emm...I ignored the fact that f(x)is irreducible..
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