Let
- $K$ be a field of characteristic $0$;
- $L$ be an extension of $K$ of degree $n$; and
- $N$ be a normal closure of $L$.
There's a theorem that guarantees the existence of exactly $n$ distinct $K$-monomorphisms $\tau_1, \tau_2, \dots, \tau_n$ from $L$ into $N$.
With these definitions in place, my textbook1 goes on as follows:
For each element $x$ of $L$, we define the norm $\mathrm{N}_{L/K}(x)$ and the trace $\mathrm{Tr}_{L/K}(x)$ by $$\mathrm{N}_{L/K}(x) = \prod_{i=1}^n \tau_i(x),$$ $$\mathrm{Tr}_{L/K}(x) = \sum_{i=1}^n \tau_i(x).$$
Then the book proceeds to state and prove the following theorem:
Theorem 8.16
The mapping $\mathrm{N}_{L/K}$ is a group homomorphism from $(L^*, \cdot)$ into $(K^*, \cdot)$. The mapping $\mathrm{Tr}_{L/K}$ is a non-zero group homomorphism from $(L, +)$ into $(K, +)$.
The proof begins by showing that $\mathrm{N}_{L/K}(xy) = \mathrm{N}_{L/K}(x) \mathrm{N}_{L/K}(y)$ and $\mathrm{Tr}_{L/K}(x + y) = \mathrm{Tr}_{L/K}(x) + \mathrm{Tr}_{L/K}(y)$, which is straightforward, but then, out of the blue, it flatly asserts:
thus $\mathrm{N}_{L/K}$ and $\mathrm{Tr}_{L/K}$ are (respectively) homomorphisms into $(L^*, \cdot)$ and $(L, +)$.
Huh??? The $\tau_i$ are maps $L \to N$, so I don't see how one can assert that $\prod_{i=1}^n \tau_i(x) \in L^*$ (when $x \in L^*$), and $\sum_{i=1}^n \tau_i(x) \in L$ (when $x \in L$).
Could someone please fill the missing steps in the argument?
1 John M. Howie, Fields and Galois theory, p. 142.