# Integral ideals of norm less than the Minkowski Bound

Consider a number field $K$ and suppose I want to find the class group and class number of $K$. One of the first steps is to compute the the Minkowski bound. Suppose our bound is $B$. In all the expositions I've read about this, we only consider the ideals with norm a prime $< B$. Why is this? Why don't we consider the ideals with norm 1 or those which are not prime?

-
Think about what it means for an ideal to have norm 1. –  Gerry Myerson Nov 9 '11 at 23:23

## 1 Answer

Every ideal is equivalent (modulo principal ideals) to an ideal satisfying the Minkowski bound; that's why it's enough to consider such ideals to find the ideal class group. Since every ideal $I$ splits to a product of prime ideal, and their norms can't be larger than the norm of $I$, the class group is generated by prime ideals satisfying the Minkowski bound. Their norms are powers of primes. I don't believe it's enough to consider ideals with prime norms $<B$. On the other hand, the prime ideals with norm $<B$ are usually found by factorizing primes $<B$ to prime ideals.

-