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Math Help - Algebraic Extensions

  1. #1
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    Algebraic Extensions

    1.Let F=K(u) where u is transcedental over the field K. If E is a field such that K contained in E contained in F, then Show that u is algebraic over E.

    Let a
    be any element of E that is not in K. Then a = f(u)/g(u)

    for some polynomials
    f(x), g(x) inK[x]

    2.Let F be a finite extension of K such that [F:K]=p, a prime number. If u in F but u not in K, show that F=K(u)
    I'm not quite sure how to get started on this one
    3.Let K contained in E contained in F be fields. Prove that if F is algebraic over K, then F is algebraic over E and E is algebraic over K

    F is algebraic so F(u)=0
    We want to show E(u)=0
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  2. #2
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    Quote Originally Posted by mathematic View Post
    2.Let F be a finite extension of K such that [F:K]=p, a prime number. If u in F but u not in K, show that F=K(u)
    I'm not quite sure how to get started on this one

    u\in F, K \subset F \Rightarrow K(u) \subseteq F

    K \subset K(u)\Rightarrow \left[K(u):K \right]\neq 1

    \left[F:K(u) \right]\left[K(u):K \right]=\left[F:K \right] = p = prime \Rightarrow \left[F:K(u) \right]=1 \Rightarrow F=K(u)
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  3. #3
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    1.Let F=K(u) where u is transcedental over the field K. If E is a field such that K contained in E contained in F, then Show that u is algebraic over E.

    Let a
    be any element of E that is not in K. Then a = f(u)/g(u)

    for some polynomials f(x), g(x) inK[x]

    I get stuck after this point
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  4. #4
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    The hint I was given was:
    Let a be any element of E that is not in K. Then a = f(u)/g(u) for some polynomials f(x), g(x) in K[x]. Then I could use this to find a nonzero polynomial in E[x] with u as a root.
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  5. #5
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    3. elements of F have minimal polynomials of K
    If E/K is a subfield of F/K, then we have a minimal polynomial of E and thus a minimal polynomial of E of K. So algebraic?
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  6. #6
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    Quote Originally Posted by mathematic View Post
    3.Let K contained in E contained in F be fields. Prove that if F is algebraic over K, then F is algebraic over E and E is algebraic over K
    F algebraic over K means that every element of F is root of some polynomial with coefficients in K.

    Since K is a subfield of E we can see the coefficients as elements in E.
    So, F is algebraic over E.

    Since E is a subfield of F we have that E is algebraic over K.
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