# Transcendental extension fields

• Oct 18th 2012, 07:32 AM
Bingk
Transcendental extension fields
Hi, I'm having problems with the second part of this problem.

Let $\displaystyle F$ be an extension of $\displaystyle K$ and $\displaystyle a \in F$ transcendental over $\displaystyle K$.
Let $\displaystyle f \in K[x]$ with $\displaystyle \text{deg } f >0$. Show:
a) $\displaystyle f(a)$ is transcendental over $\displaystyle K$.
b) If $\displaystyle b \in F$ and $\displaystyle f(b)=a$, then $\displaystyle b$ is transcendental over $\displaystyle K$.

For part a, I assumed otherwise, so there exists a $\displaystyle g \in K[x]$ such that $\displaystyle g(f(a))=0$, but $\displaystyle g(f(x))=h(x) \in K[x]$ and $\displaystyle h(a)=0$ which contradicts $\displaystyle a$ being transcendental over $\displaystyle K$.

For part b, I'm not sure. I tried doing it by contradiction also, but I'm not getting anywhere. I feel like it's relatively simple, and I'm just forgetting something obvious. Any help would be appreciated! Thanks!
• Oct 19th 2012, 12:13 AM
Vlasev
Re: Transcendental extension fields
Tell me if you have covered such a result or not but finite sums and products of algebraic elements are algebraic. So we can prove the contrapositive. Let $\displaystyle b$ be algebraic. We need to show $\displaystyle f(b) \neq a$. To this end, $\displaystyle f(b)$ is just a finite sum of products of $\displaystyle b$ and the coefficients of $\displaystyle f(x)$, that is $\displaystyle f(b)$ is algebraic and cannot be transcendental, i.e. $\displaystyle f(b) \neq a$.
• Oct 19th 2012, 01:09 AM
FernandoRevilla
Re: Transcendental extension fields
Another way. If $\displaystyle b$ is algebraic over $\displaystyle K$, then $\displaystyle [K(b):K]$ is finite and as a consequence $\displaystyle K(b)$ is an algebraic extension of $\displaystyle K.$ As $\displaystyle a\in K(b)$, $\displaystyle a$ would be algebraic over $\displaystyle K$ (contradiction).
• Oct 19th 2012, 10:25 AM
Bingk
Re: Transcendental extension fields
I have not encountered that result, although it seems to make sense (it would be frightening to think of transcendentals being the result of finite sums and/or products of algebraic elements).

I like Fernando's proof though, and it's something I should have realized, it's just that the problem got my mind set on using functions.

This was my solution:
(Preliminary stuff)
Notation: $\displaystyle aK=\{ak:k \in K\}$ and $\displaystyle a+K=\{a+k:k \in K\}$
If $\displaystyle a \in K$ then $\displaystyle a( F \setminus K) = F \setminus K$
Proof: Since $\displaystyle a \in K$, $\displaystyle aK = K$. Now, $\displaystyle aF=F$ but $\displaystyle aF = a(K \cup (F \setminus K)) = aK \cup a(F \setminus K) = K \cup a(F \setminus K)$, thus $\displaystyle a( F \setminus K) = F \setminus K$.
Similarly, If $\displaystyle a \in K$ then $\displaystyle a + ( F \setminus K) = F \setminus K$.
Thus, if $\displaystyle b \in F \setminus K$, then $\displaystyle bK=F \setminus K$ and $\displaystyle b+K=F \setminus K$.

Let $\displaystyle f(x)=\sum_{i=0}^n f_i x^i \in K[x]$, so $\displaystyle f(b)=a$.
Now $\displaystyle f(b) - f_0 = b \sum_{i=1}^n f_i b^{i-1} = a - f_0$.

Note that: $\displaystyle a \in F \setminus K$, $\displaystyle b \in F \setminus K$ (because if it is in $\displaystyle K$ then $\displaystyle f(b) \in K$ which means $\displaystyle a \in K$ which contradicts $\displaystyle a$ being transcendental).
So, using the preliminary stuff, $\displaystyle a - f_0 \in F \setminus K$ implies that $\displaystyle b \sum_{i=1}^n f_i b^{i-1} \in F \setminus K$ and since $\displaystyle b \in F \setminus K$, we have that $\displaystyle \sum_{i=1}^n f_i b^{i-1} \in K$.
So, $\displaystyle h(x) = (x - f_0)(\sum_{i=1}^n f_i b^{i-1})^{-1} \in K[x]$ and $\displaystyle h(a)=b$. Thus, by part (a), $\displaystyle b$ is transcendental.

Is there anything wrong with what I've done?
What I think I basically did is:
If $\displaystyle \exists f \in K[x]$ such that $\displaystyle f(b)=a$, then $\displaystyle \exists g \in K[x]$ such that $\displaystyle g(a)=b$.