I'm having a little trouble with this one. Help would be appreciated!Quote:

Let be an extension field of . If is transcendental over , then show that every element of that is not in is also transcendental over .

- April 13th 2010, 07:02 AMhatsoff[SOLVED] Prove elements of K(u) are transcendental if u is transcendentalQuote:

Let be an extension field of . If is transcendental over , then show that every element of that is not in is also transcendental over .

- April 13th 2010, 07:24 AMLaurent
I've not been working on this stuff for a while, so my answer is to be taken with caution...

If , then for some non-constant polynomial . Let be such that . We have hence , which belongs to , must be 0 by transcendance of ; however, if the degree of is at least 1, then so is that of (P is non-constant), hence . I think this proves that is transcendental over . - April 13th 2010, 07:32 AMhatsoff
Thanks, but I do not believe this is correct. The definition I have for is . So if , then for some nonzero .

I have tried assuming towards a contradiction that is algebraic, so that we have for some . In that case, is a root of . However, I am unable to show that , or anything else which might yield a contradiction.

But thanks for the effort! Keep the suggestions coming! - April 13th 2010, 07:46 AMLaurent
Oh, right... :o I guessed it wouldn't be that simple...

Quote:

I have tried assuming towards a contradiction that is algebraic, so that we have for some . In that case, is a root of . However, I am unable to show that , or anything else which might yield a contradiction.

Consider however the following: if , then , which tells that is algebraic over . If was algebraic, then so would by "transitivity" (cf. multiplicativity of degrees). Hence is transcendental. - April 13th 2010, 12:24 PMFancyMouse
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. # - April 13th 2010, 05:00 PMhatsoff
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 with , and assume towards a contradiction that is algebraic over . Then there are with , and . Now, has the form

,

where , and each . Consider the polynomial

. [EDIT: Assuming this polynomial is nonzero,] we have

,

such that is algebraic over . However, this contradicts the hypothesis that is transcendental over . So our assumption must be false, and the conclusion follows. - April 14th 2010, 04:00 AMLaurent
- April 14th 2010, 07:09 AMhatsoff
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.