I am here not for the proof of this theorem, so this is not a duplicate.
Here are the theorem and the proof:
I can follow the first few steps of the proof (prove that the quotient is a field, the image of $F$ is isomorphic to $F$), but I couldn't understand how the author gets the natural projection "out of the polynomial". In my understanding,
$p(\pi(x))=\sum a_n(\pi(x))^n$ and $\pi(p(x))=\sum \pi(a_n)\left(\pi(x)\right)^n$. Why is $\pi(a_n)=a_n$ true?
My guess is that the author says he "identifies $F$ with its image in $K$". Any help will be appreciated. Thanks in advance.
Best Answer
Your guess is completely correct.
The thing is that the restriction of $\pi$ to $F$ is an injection (so an embedding of fields), therefore we identify elements of $K$ with their image in $F$ under $\pi$.
That's what "have a root in a bigger field" means in any case if you allow field embeddings instead of strict inclusion.
Another way to see it is that the map $\pi_{\mid F} : F\to K$ makes $K$ into an $F$-algebra, and the definition of this $F$-algebra structure is $\lambda \cdot v := \pi(\lambda)v, \lambda \in F, v\in K$; in particular by definition $a_n(\pi(x))^n = \pi(a_n)(\pi(x))^n$ (in fact you can see that this is the only sensible way of giving the expression $a_n(\pi(x))^n$ any meaning in the first place ! or more generally of giving the expression $p(\pi(x))$ any meaning)