Transcendental extension fields

**Bingk** · October 18th 2012, 07:32 AM

Hi, I'm having problems with the second part of this problem.

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

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

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!

**Vlasev** · October 19th 2012, 12:13 AM

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 $b$ be algebraic. We need to show $f(b) \neq a$ . To this end, $f(b)$ is just a finite sum of products of $b$ and the coefficients of $f(x)$ , that is $f(b)$ is algebraic and cannot be transcendental, i.e. $f(b) \neq a$ .

**FernandoRevilla** · October 19th 2012, 01:09 AM

Another way. If $b$ is algebraic over $K$ , then $[K(b):K]$ is finite and as a consequence $K(b)$ is an algebraic extension of $K.$ As $a\in K(b)$ , $a$ would be algebraic over $K$ (contradiction).

**Bingk** · October 19th 2012, 10:25 AM

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

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

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

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

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