# Field Extensions: Algebraic and Transcendental Elements

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• Mar 1st 2010, 09:12 AM
roninpro
Field Extensions: Algebraic and Transcendental Elements
Hello there. I have the following problem:

Suppose that $v$ is algebraic over $K(u)$ for some $u$ in an extension field of the field $K$. If $v$ is transcendental over $K$, show that $u$ is algebraic over $K(v)$.

I've tried writing down the minimal polynomial for $v$ over $K(u)$ and playing around with that, but I haven't had any luck. I am also somewhat baffled by the transcendental assumption - where does it even come in to play, and how can I write it down? (I usually see it defined as a negative.)

Just a hint in the correct direction would be appreciated.
• Mar 1st 2010, 11:22 AM
clic-clac
Hi

I just write what the hypotheses mean:

$\mu_{P,K(u)}\in K(u)[X]-K[X]$ . I think it is sufficient to conclude ;)
• Mar 1st 2010, 02:28 PM
Krahl
Not sure if this is right but have a look at it see what you think;

Let v be algebraic so that there is $f(x) \in K(u)[x]$
with $f(v)=\sum_{i=0}^ma_iv^i=0$ and $a_i \in K(u)$.
let v be transcendental over K so that $a_i=u^n$ for some n>0.

so $f(x)=\sum_{j=0}^na_{i_j}x^{i_j}+\sum_{j=n+1}^mu^{i _j}x^{i_j}$
$a_{i_j} \in K$

then
$f(v)=\sum_{j=0}^na_{i_j}v^{i_j}+\sum_{j=n+1}^mu^{i _j}v^{i_j}=\sum_{j=0}^m(a_{i_j}v^{i_j}+v^{i_j})u^{ i_j}=0$

and $a_{i_j},v^{i_j} \in K(v)$ so $a_{i_j}v^{i_j} \in K(v)$

so letting

$g(x)=\sum_{j=0}^mc_jx^{i_j}$

where

$c_j=a_{i_j}v^{i_j}+v^{i_j} \in K(v)[x]$

and g(u)=0 so v is algebraic over K(u).
• Mar 1st 2010, 03:04 PM
clic-clac
Quote:

let v be transcendental over K so that http://www.mathhelpforum.com/math-he...6c1d069a-1.gif for some n>0.
Need a bit more than that. What is important is that you can write your $f(x)$ under the form $q(x)+up(x)$ with $p(v)\neq 0$ and $q(x)\in K[x].$