Dedekind-Hasse norm in nLab

Proof

Suppose that RR possesses a Dedekind-Hasse norm. Let II be a non-zero ideal. Let bb be a non-zero element of II of minimal norm. We know that (b)⊆I(b) \subseteq I. Let aa be an element of II. Suppose that bb doesn’t divide aa. Then, there exists p,q,r∈Rp,q,r \in R such that pa=bq+rpa = bq+r and 0<v(r)<v(b)0 \lt v(r) \lt v(b). Thus, r=pa−bq∈Ir=pa-bq \in I, r≠0r \neq 0 and v(r)<v(b)v(r) \lt v(b), absurd! Therefore b|ab|a and a∈(b)a \in (b). Thus, I⊆(b)I \subseteq (b) and I=(b)I = (b). We have proved that RR is a pid.

Suppose that RR is a pid. Thus, it is a UFD. Put v(s)=0v(s)=0 if s=0s=0, and v(s)v(s) equal to 2 n2^{n} where nn is the number of irreducible elements in the factorization of nn, if s≠0s \neq 0. Let a∈Ra \in R and (b≠0)∈R(b \neq 0) \in R. Suppose that bb doesn’t divide aa. We know that (a,b)=(r)(a,b) = (r) and thus there exists p,q,r∈Rp,q,r \in R such that pa+bq=rpa+bq = r. rr divides bb but bb doesn’t divide rr because it would imply that bb divides aa. Thus, there is strictly less irreducible elements in the factorization of rr than in the one of bb and v(r)<v(b)v(r) \lt v(b). Moreover r≠0r \neq 0 because (a,b)=(r)(a,b) = (r) and b≠0b \neq 0. Thus 0<v(r)<v(b)0 \lt v(r) \lt v(b). We have proved that vv is a Dedekind-Hasse norm.