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

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April 13th 2010, 07:02 AM
hatsoff

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

Quote:

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!
April 13th 2010, 07:24 AM
Laurent

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 $K$ .
April 13th 2010, 07:32 AM
hatsoff

Quote:

Originally Posted by Laurent

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!
April 13th 2010, 07:46 AM
Laurent

Quote:

Originally Posted by hatsoff

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... :o I guessed it wouldn't be that simple...

Quote:

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.
April 13th 2010, 12:24 PM
FancyMouse

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 PM
hatsoff

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. $\blacksquare$
April 14th 2010, 04:00 AM
Laurent

Quote:

Originally Posted by hatsoff

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 $p(x)=0$ ...
April 14th 2010, 07:09 AM
hatsoff

Quote:

Originally Posted by Laurent

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.