Results 1 to 5 of 5

Math Help - Algebraicity of arbitrary extensions

  1. #1
    Junior Member
    Joined
    Aug 2012
    From
    Sweden
    Posts
    37
    Thanks
    1

    Algebraicity of arbitrary extensions

    Let F \subset K \subset E be field extensions. Assume that F \neq K and that there is x \in E s.t E=F(x). Show that E is algebraic over K.

    I have no idea how to even begin with this one. I'm sitting with my books trying to find some quality to use, but all are about finite extensions, something I suppose I can't assume here. As you know, if E=F(x) is finite over K then it is also algebraic over K, and I'm thinking about using the contrapositive, but that doesn't seem to solve the problem in its entirety.

    Thank you for any advice.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Aug 2012
    From
    Sweden
    Posts
    37
    Thanks
    1

    Re: Algebraicity of arbitrary extensions

    I made some progress on my own, which I suppose was the easy part.

    So, x is either algebraic or transcendental over F. If x is algebraic over F, then F(x)/F is finite => [F(x):F] = [F(x) : K][K : F] finite. The product if finite only if both factors are finite. In particular [F(x) : K] is finite, and hence algebraic. (I have one introductory algebra book stating the tower law (?) only for finite extensions, but Lang's Algebra states it for any extensions.)

    As for the case x transcendental, then F(x)/F is infinite => either F(x)/K or K/F is infinite, at which point things get circular...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2012
    From
    Sweden
    Posts
    37
    Thanks
    1

    Re: Algebraicity of arbitrary extensions

    Bump. Anyone?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member jakncoke's Avatar
    Joined
    May 2010
    Posts
    388
    Thanks
    80

    Re: Algebraicity of arbitrary extensions

    So Assume, F < K \leq M and M = F(r_1). Assume r_1 was transcendental.
    Now if there exist no intermediate field between F and M, then F(r_1) = K and so M is algebraically closed over K.
    If there does exist an intermediate field, then there exists an element r_2 \in K which is not in F.
    if r_2 was transcendental over F, then K(r_2) \cong F(x) \cong F(r_1) \cong M a contradiction.
    if r_2 was algebraic over F, kthen it implies, M is a transcendental extension of K, since it is a transcendental extension of F. There exists no polynomial in K[X], which makes r_1 0. so K(r_1) \cong K(x) < F(r_1) = M . Since K is strictly bigger then F, this cannot be, thus with both contradictions, we see that M = K thus again since no intermediate field exists M=K is algebraically closed.

    Now if r_1 can algebraic there is a min deg monic polynomial in F such that f(r_1) = 0. Then [M:F] = n this implies, [M:K][K:F] = [M:F] so [M:K] is a finite extension, thus an algebraic extension.
    if so.
    Last edited by jakncoke; March 8th 2013 at 11:50 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2012
    From
    Sweden
    Posts
    37
    Thanks
    1

    Re: Algebraicity of arbitrary extensions

    jakncoke, thank you! I haven't been able to look through your explanation, but at least now I know where to go when I return to the subject
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Normal extensions
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 29th 2013, 09:47 AM
  2. Algebraic Extensions
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 6th 2011, 01:10 AM
  3. Replies: 0
    Last Post: March 15th 2010, 08:49 AM
  4. Extensions
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 1st 2009, 09:05 AM

Search Tags


/mathhelpforum @mathhelpforum