I'm trying to prove the following but it causes me a lot of trouble:
Let $L$ be a finite extension of degree $n$ of a field $K$ with characteristic $0$. Let $\sigma_1,\dots,\sigma_n$ be different $K$ linear field morphisms of $L$ into the algebraic closure of $K$. Define the norm $N(f)$ of $f\in L$ as the product of all $\sigma_i (f)$.
Then I have to show that the norm $N(f)$ belongs to $K$, i.e. $N(f)\in K$.
Let $L=K(\alpha)$. Now I have to express the norm of the element $\alpha$ in terms of the coefficients of its defining polynomial.
I would be so glad if anyone is able to help me with these. Thank you!
For the first part of your question, note that the elements of $K$ are exactly those elements which are fixed by each of the automorphisms $\sigma_1, \ldots, \sigma_n$. So show that the norm of an element has this property.
For the second part, note that each automorphism $\sigma_i$ must send a root of the defining polynomial to another root. Now, determine how the coefficients of a polynomial are related to the roots of the polynomial. (Hint: Write a polynomial as $p(x)=c_n x^n + \cdots + c_1 x + c_0 = (x-\alpha_1)\cdots(x-\alpha_n)$, so the $\alpha_i$ are the roots of the polynomial. Multiply out the right hand side, and compare coefficients.)