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*) = *r*1 + *r*2 where *r*1 is in I and *r*2 is in J, so I and (*b*–*a*) + J intersect in *r*1 = (*b*–*a*) – *r*2.

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*.