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Math Help - [SOLVED] Transitivity of Algebraic extension Fields

  1. #1
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    [SOLVED] Transitivity of Algebraic extension Fields

    so if we're given that L extends K, w/ L algebraic of K, and M extends L, w/ M algebraic over L, do we get that M is algebraic of K? (we may not assume finite extensions)

    that is:
    For all l in L, there exists p(t) in K[t] such that p(l)=0
    For all m in M, there exists q(t) in L[t] such that q(m)=0

    For all m in M, does there exist r(t) in K[t] such that r(m)=0?

    thanks for any help
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  2. #2
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    Quote Originally Posted by verdi View Post
    so if we're given that L extends K, w/ L algebraic of K, and M extends L, w/ M algebraic over L, do we get that M is algebraic of K? (we may not assume finite extensions)

    that is:
    For all l in L, there exists p(t) in K[t] such that p(l)=0
    For all m in M, there exists q(t) in L[t] such that q(m)=0

    For all m in M, does there exist r(t) in K[t] such that r(m)=0?

    thanks for any help

    Let m\in M\Longrightarrow \,\,\exists \,0\neq p(x)=x^n + a_{n-1}x^{n-1}+\ldots a_1x+a_0\in L[x] s.t. p(m)=0.
    Now look at F:=K(a_0,\ldots,a_{n-1}) . This is a finite extension of K ( why?) and m is algebraic over F , so it is algebraic over K (why?) .

    Tonio
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  3. #3
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    Thanks!

    I guess F:=K(a_0,\ldots,a_{n-1}) is a finite extension because, we have finitely many algebraic roots, each of which are the root of some minimal polynomial in K, so using the Tower Theorem, the degree is a product of a bunch of finite terms and thus finite. Then we simply use that finite \rightarrow algebraic (though the converse is false). So each m \in M is algebraic over K and M is algebraic over K.

    Does that sound right?
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  4. #4
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    Quote Originally Posted by verdi View Post
    Thanks!

    I guess F:=K(a_0,\ldots,a_{n-1}) is a finite extension because, we have finitely many algebraic roots, each of which are the root of some minimal polynomial in K, so using the Tower Theorem, the degree is a product of a bunch of finite terms and thus finite. Then we simply use that finite \rightarrow algebraic (though the converse is false). So each m \in M is algebraic over K and M is algebraic over K.

    Does that sound right?

    Sounds right and dandy. Good!

    Tonio
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