If I and J are coprime ideals of a Dedekind domain R, that is I + J = R, then by unique factorization into prime ideals, I and J have no prime ideal factors in common. The intersection of I and J is just IJ, since it’s defined by the maximum powers of the ideal factors, and IJ is defined by the sum of powers. More importantly, R/IJ and (R/I)*(R/J) have the same structure, where R1*R2 is defined like Z^2 = Z*Z.
To see why, it’s enough to construct an invertible ring homomorphism from one to the other. We can send a + IJ to (a + I, a + J); it’ll clearly be a ring homomorphism. The homomorphism is 1-to-1, since (a + I, a + J) = (b + I, b + J) implies that a–b is in I and J, so that it’s in IJ and a + IJ = b + IJ. It’s onto, since given a + I and b + J, we can find c with c + I = a + I and c + J = b + J.
The last statement is true because otherwise, a + I and b + J have empty intersection. That would mean that I and (b–a) + J have empty intersection. But that’s a contradiction: R = I + J implies that (b–a) = r1 + r2 where r1 is in I and r2 is in J, so I and (b–a) + J intersect in r1 = (b–a) – r2.
In particular, rings with the same structure have the same number of elements. So if R is a ring of integers of a number field, we have N(IJ) = N(I)N(J).
To show it holds even when I and J are not coprime, it’s enough to show it holds when I is prime and J is a power of I, I^n. Then it’ll hold for all ideals by decomposing them into primes. In fact, it’s enough to show N(I^n) = N(I)N(I^(n-1)). But it’s true that R/I^(n-1) has the same structure as (R/I^n)/(I^(n-1)/I^n); just think about it for a little while, and remember that I^(n-1) is a bigger ideal than I^n since it contains more elements.
Now, that sameness means that N(I^(n-1)) times the number of elements of I^(n-1)/I^n is equal to N(I^n). So it’s enough to show that I^(n-1)/I^n has N(I) elements. To see that, fix an element c in I^(n-1) that isn’t in I^n, and consider the function that sends a + I to ac + I^n. This is not a ring homomorphism – it’s only a homomorphism of the additive groups – but it’s good enough.
This function is 1-to-1, since if ac + I^n = bc + I^n, then (a–b)c is in I^n. The set of all elements a–b of R that satisfy that condition is an ideal containing I and not equal to R, since 1 is not in it by choice of c. I is prime, so it’s a maximal ideal, which means that ideal is just I; then a + I = b + I.
This function is also onto. Otherwise, there’s some b + I^n, where b is in I^(n-1), that doesn’t intersect (c) + I^n. But the ideal factorization of c includes I to the power of n-1 by the choice of c, so (c) + I^n = I^(n-1), and we can write b in I^(n-1) as cr + d where r is in R and d is in I^n. Then b + I^n and (c) + I^n both contain cr + d.
This shows that for all ideals I and J, N(I)N(J) = N(IJ). This is useful for, among other things, showing that there are only finitely many ideals of a given norm.
That is true since if P is a prime ideal, then the set of all rational integers in P is a prime ideal of Z, generated by p. Since p is in I, N(p) = p^n (since all conjugates of the field K fix Q) is divisible by N(P), so that N(P) = p^k for some integer k. In particular, P contains its norm, p^k. So if I = P1P2…P(r) then I contains N(P1)N(P2)…N(P(r)) = N(I).
So every ideal of norm m contains m. Writing out (m) as P1P2…P(r), we get that every ideal of norm m is some combination of the P(i)’s, since in divides (m). There are only finitely many combinations; hence, only finitely many ideals of norm m. More generally, there are only finitely many ideals of norm at most m.