Results 1 to 8 of 8

Math Help - [SOLVED] Prove elements of K(u) are transcendental if u is transcendental

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    410

    [SOLVED] Prove elements of K(u) are transcendental if u is transcendental

    Let F be an extension field of K. If u\in F is transcendental over K, then show that every element of K(u) that is not in K is also transcendental over K.
    I'm having a little trouble with this one. Help would be appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    I've not been working on this stuff for a while, so my answer is to be taken with caution...

    If v\in K(u)\setminus K, then v=P(u) for some non-constant polynomial P\in K[X]. Let Q\in K[X] be such that Q(v)=0. We have 0=Q(v)=Q(P(u)) hence Q\circ P, which belongs to K[X], must be 0 by transcendance of u; however, if the degree of Q is at least 1, then so is that of Q\circ P (P is non-constant), hence Q=0. I think this proves that v is transcendental over K.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by Laurent View Post
    If v\in K(u)\setminus K, then v=P(u) for some non-constant polynomial P\in K[X].
    Thanks, but I do not believe this is correct. The definition I have for K(u) is \{a(u)/b(u):a(x),b(x)\in K[x],b(u)\neq 0\}. So if v\in K(u)\setminus K, then v=a(u)/b(u) for some nonzero a(x),b(x)\in K[x].

    I have tried assuming towards a contradiction that v is algebraic, so that we have c(v)=0 for some c(x)\in K[x]. In that case, u is a root of c(a(x)/b(x)). However, I am unable to show that c(a(x)/b(x))\in K[x], or anything else which might yield a contradiction.

    But thanks for the effort! Keep the suggestions coming!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by hatsoff View Post
    Thanks, but I do not believe this is correct. The definition I have for K(u) is \{a(u)/b(u):a(x),b(x)\in K[x],b(u)\neq 0\}. So if v\in K(u)\setminus K, then v=a(u)/b(u) for some nonzero a(x),b(x)\in K[x].
    Oh, right... I guessed it wouldn't be that simple...

    I have tried assuming towards a contradiction that v is algebraic, so that we have c(v)=0 for some c(x)\in K[x]. In that case, u is a root of c(a(x)/b(x)). However, I am unable to show that c(a(x)/b(x))\in K[x], or anything else which might yield a contradiction.
    Maybe you can write c\left(\frac{a(x)}{b(x)}\right)=\frac{p(x)}{q(x)}, where p,q are mutually prime (by "reducing to the same denominator"), hence c(v)=0 implies \frac{p(u)}{q(u)}=0 and thus p(u)=0, hence p=0 identically. This proves c\left(\frac{a(x)}{b(x)}\right)=0, hence... c=0? This conclusion amounts to saying that non-constant rational fractions over K are transcendantal over K; must be a theorem? ** Actually, this is (up to isomorphism) what we want to prove: any y\in K(x)\setminus K is transcendental over K. So, not much progress.

    Consider however the following: if v=\frac{a(u)}{b(u)}, then b(u)v-a(u)=0, which tells that u is algebraic over K(v). If v was algebraic, then so would u by "transitivity" (cf. multiplicativity of degrees). Hence v is transcendental.
    Last edited by Laurent; April 13th 2010 at 10:29 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2010
    Posts
    78
    Let L=K(v). Then K(u)=L(u). It's easy to see that K(u)/L is algebraic, since v is a non-constant rational function on u by the fact that v is in K(u) but not in K.
    But then if L/K were algebraic, then K(u)/K would be algebraic since algebraic extensions behave well in towers. #
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Thanks to everyone for the help! However, I am not familiar with any theorems which show that algebraic extensions "behave well in towers." After a little bit more thinking, I have a solution which I think fits better with the methods of my text. I reproduce it here just in case anyone feels like checking it for errors:

    Proof: Let v\in K(u)=\{a(u)/b(u):a(x),b(x)\in K[x],b(u)\neq 0\} with v\notin K, and assume towards a contradiction that v is algebraic over K. Then there are a(x),b(x),c(x)\in K[x] with v=a(u)/b(u), and c(v)=0. Now, c(x) has the form

    \sum_{i=0}^n c_ix^i,

    where n=\deg( c(x)), and each c_i\in K. Consider the polynomial

    \sum_{i=0}^n c_i a(x)^i b(x)^{n-i}\in K[x]. [EDIT: Assuming this polynomial is nonzero,] we have

     \sum_{i=0}^n c_i a(u)^i b(u)^{n-i}

     =b(u)^n\sum_{i=0}^n c_i (a(u)/b(u))^i

     =b(u)^n c(a(u)/b(u))

     =b(u)^nc(v)=0,

    such that u is algebraic over K. However, this contradicts the hypothesis that u is transcendental over K. So our assumption must be false, and the conclusion follows. \blacksquare
    Last edited by hatsoff; April 14th 2010 at 08:12 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by hatsoff View Post
    Consider the polynomial

    \sum_{i=1}^n c_i a(x)^i b(x)^{n-i}\in K[x].
    You should prove that this polynomial is not 0, and this is not obvious. This is the analog of p(x) in my last post, and I couldn't find a contradiction once I proved p(x)=0...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by Laurent View Post
    You should prove that this polynomial is not 0, and this is not obvious. This is the analog of p(x) in my last post, and I couldn't find a contradiction once I proved p(x)=0...
    Whew, good catch! Unfortunately, I cannot prove that right now, so the larger "proof" is useless. Drat!

    I suppose I'll have to use transitivity after all. The only hitch is that the textbook has degree arguments in later sections, suggesting that I'm expected to use some other method here.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Transcendental
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 15th 2011, 09:37 AM
  2. Field Extensions: Algebraic and Transcendental Elements
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: March 1st 2010, 04:04 PM
  3. Transcendental numbers
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 24th 2009, 02:35 AM
  4. Transcendental
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 24th 2009, 12:26 PM
  5. Replies: 3
    Last Post: February 3rd 2009, 03:33 PM

Search Tags


/mathhelpforum @mathhelpforum