Norm homomorphism between ideal class groups

I have been working through Number Fields by D.A Marcus, and I'm stuck and need a hint, the question is in chapter 3 question 16 which goes as follows:

Let $K,L$ be number fields and $K \subset L$, where $R,S$ are the rings of integers of K and L resp. Then denote $G(R)$, $G(S)$ for the ideal class groups of $R,S$,

I need to show that there is a homomorphism $G(S) \rightarrow G(R)$ which sends any ideal I in a given class C to the class containing $N^{L}_{K}(I)$, the thing is I'm not quite sure how to show this homomorphism sends the identity to the identity.

Here for a prime ideal $Q$ of $S$, $N^{L}_{K}(Q)=P^{f(Q|P)}$, $P$ the prime ideal lying under $Q$ and $f(Q|P)$ the inertia degree.

Thank you

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I'm not quite sure what you're asking. By that point Marcus has surely shown that the group $I_L$ of non-zero fractional ideals of $L$ [maybe it would be better to write $S$, but everyone does this] is free on the set of prime ideals of $S$. So to give a homomorphism $I_L \to I_K$ you just say where the primes go, and that's what you've done. So I think your question is whether we can pass to a homomorphism $G(S) \to G(R)$, i.e., whether the image of a principal fractional ideal is principal, which is not obvious from this definition. Do I have that right? – Dylan Moreland Jun 22 '12 at 15:01
Well hes not defined fractional ideals yet, but has shown that the ideals of R factor uniquely into prime ideals, and in a previous exercise, we prove that principal ideals will stay principal under this map – Chris Birkbeck Jun 22 '12 at 17:34
I'll try to take a look at the book later. – Dylan Moreland Jun 23 '12 at 0:28

The trick for this exercise is that you need to look at exercises 14 and 15. Let us define a homomorphism $f : G(S) \to G(R)$ that sends $[I] \mapsto [N_{L/K}(I)]$. We need to check that this is well-defined and is a group homomorphism.

Suppose that there is another ideal $J$ such that $I \sim J$. Then there are $\alpha,\beta \in S$ such that $\alpha I = \beta J$. The trick now is we can say that $$(\alpha)I = (\beta)J.$$

Taking ideal norms (as defined in exercise 15) we get that $$N_{L/K}\left((\alpha) \right) N_{L/K}(I) = N_{L/K}\left( (\beta)\right) N_{L/K}(J).$$ Now by exercise 15(b) we get that $$\begin{array}{ccc} N_{L/K}\left((\alpha) \right) &=& \left( N_{L/K} (\alpha) \right) \\ N_{L/K}\left((\beta) \right) &=& \left( N_{L/K} (\beta) \right).\end{array}$$ Hence $$\left(N_{L/K}(\alpha)\right) N_{L/K}(I) = \left(N_{L/K}(\beta)\right) N_{L/K} (J)$$

which by the same trick we used in the beginning means that $$N_{L/K}(\alpha)N_{L/K}(I) = N_{L/K}(\beta) N_{L/K} (J),$$

hence $[N_{L/K}(I) ] = [ N_{L/K} (J)].$

By 15(c) again we see that $f$ takes the identity class to the identity class. By multiplicativity of the norm we conclude finally that $f$ is a well - defined group homomorphism from $G(S)$ to $G(R)$.

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