symmetric monoidal (∞,1)-category of spectra
The notion of a Dedekind-Hasse norm is a generalization of the notion of the degree function in a Euclidean domain. It makes clear the links between a principal ideal domain and an Euclidean domain: a pid is like an Euclidean domain but with a weaker notion of Euclidean division.
Let $R$ be an integral domain. A Dededekind-Hasse norm is a function $v \colon R \rightarrow \mathbb{N}$ such that:
$v(s)=0 \Leftrightarrow s=0$
$\forall a \in R$, $\forall (b \neq 0) \in R$:
$b|a$, or
$\exists p,q,r \in R$ such that $pa = bq + r$ and $0 \lt v(r) \lt v(b)$
Let $R$ be an integral domain. Then it is a principal ideal domain iff it possesses a Dedekind-Hasse norm.
Suppose that $R$ possesses a Dedekind-Hasse norm. Let $I$ be a non-zero ideal. Let $b$ be a non-zero element of $I$ of minimal norm. We know that $(b) \subseteq I$. Let $a$ be an element of $I$. Suppose that $b$ doesn’t divide $a$. Then, there exists $p,q,r \in R$ such that $pa = bq+r$ and $0 \lt v(r) \lt v(b)$. Thus, $r=pa-bq \in I$, $r \neq 0$ and $v(r) \lt v(b)$, absurd! Therefore $b|a$ and $a \in (b)$. Thus, $I \subseteq (b)$ and $I = (b)$. We have proved that $R$ is a pid.
Suppose that $R$ is a pid. Thus, it is a UFD. Put $v(s)=0$ if $s=0$, and $v(s)$ equal to $2^{n}$ where $n$ is the number of irreducible elements in the factorization of $n$, if $s \neq 0$. Let $a \in R$ and $(b \neq 0) \in R$. Suppose that $b$ doesn’t divide $a$. We know that $(a,b) = (r)$ and thus there exists $p,q,r \in R$ such that $pa+bq = r$. $r$ divides $b$ but $b$ doesn’t divide $r$ because it would imply that $b$ divides $a$. Thus, there is strictly less irreducible elements in the factorization of $r$ than in the one of $b$ and $v(r) \lt v(b)$. Moreover $r \neq 0$ because $(a,b) = (r)$ and $b \neq 0$. Thus $0 \lt v(r) \lt v(b)$. We have proved that $v$ is a Dedekind-Hasse norm.
Named after Helmut Hasse.
Last revised on August 13, 2022 at 15:49:08. See the history of this page for a list of all contributions to it.