Let -span , whence -spans . Now

for some , so by Subgroup of a free abelian group we have

Now for the discriminant we have

by basic properties of the determinant and the definition of the discriminant.

Let . Invoking UFI, we have and . By the Chinese remainder theorem for rings,

Thus it suffices to show that for any prime ideal we have

Since we have the chain of ideals

it is enough to show that for we have , since

by the Third isomorphism theorem. We prove the stronger result

for each . First, we can construct a -module homomorphism

where is arbitrary. This induces a map

which will turn out to be a -module isomorphism.

To prove surjectivity of and thus , we can show that . Since (after all we are in a Containment-division ring) we also have . But is a proper divisor, so by unique factorization as required.

To prove injectivity, we can show , since hence follows if then so . Since we have . Conversely, let and write with and , where we have the Prime order of an ideal

Since by construction, it follows whence . Therefore and ^P2 is proven.

For ^P3, note that if , then . Since is finite by ^C1, there can only be finitely many such ideals by the Fourth isomorphism theorem, proving ^P3.