Vipul: New page: ==Statement== A commutative unital ring that admits a fact about::Dedekind-Hasse norm must be a fact about::principal ideal ring. In particular, an [[fact about::integral dom...

2009-01-23T18:23:02Z

New page: ==Statement== A commutative unital ring that admits a fact about::Dedekind-Hasse norm must be a fact about::principal ideal ring. In particular, an [[fact about::integral dom...

New page

==Statement==

A [[commutative unital ring]] that admits a [[fact about::Dedekind-Hasse norm]] must be a [[fact about::principal ideal ring]].

In particular, an [[fact about::integral domain]] that admits a Dedekind-Hasse norm must be a [[fact about::principal ideal domain]].

==Definitions used==

===Dedekind-Hasse norm===

{{further|[[Dedekind-Hasse norm]]}}

A Dedekind-Hasse norm on a [[commutative unital ring]] <math>R</math> is a function <math>N</math> from the nonzero elements of <math>R</math> to the nonnegative integers with the property that for any elements <math>a,b \in R</math> with <math>b \ne 0</math>, it is true that either <math>a \in (b)</math> or there exists an element <math>r</math> in the ideal <math>(a,b)</math> such that <math>N(r) < N(b)</math>.

===Principal ideal ring===

{{further|[[Principal ideal ring]]}}

A commutative unital ring <math>R</math> is termed a [[principal ideal ring]] if every ideal of <math>R</math> is principal.

==Proof==

'''Given''': A commutative unital ring <math>R</math> with a Dedekind-Hasse norm <math>N</math>. An ideal <math>I</math> of <math>R</math>.

'''To prove''': There exists <math>b \in I</math> such that <math>I = (b)</math>.

'''Proof''': If <math>I = 0</math>, we can take <math>b = 0</math>, and we are done.

So, suppose <math>I \ne 0</math>. Consider the function <math>N</math> on the nonzero elements of <math>I</math>. Since <math>N</math> maps to a well-ordered set, there is an element <math>b \in I \setminus \{ 0 \}</math> such that <math>N(b) \le N(r)</math> for all <math>r \in I \setminus \{ 0 \}</math>.

Pick any <math>a \in I</math>. If <math>a = 0</math>, then <math>a \in (b)</math> or there exists <math>r \in (a,b)</math> with <math>N(r) < N(b)</math>.

Note that in the latter case, we have <math>r \in (a,b) \subseteq I</math>, so <math>r \in I</math>, and <math>N(r) < N(b)</math>, contradicting the assumption that <math>b</math> has minimum norm among the nonzero elements of <math>I</math>. Hence, we have the case <math>a \in (b)</math>. Thus, <math>I \subseteq (b)</math>. Conversely, we clearly have <math>(b) \subseteq I</math>, so <math>I = (b)</math>.

Dedekind-Hasse norm implies principal ideal ring - Revision history

Vipul: New page: ==Statement== A commutative unital ring that admits a fact about::Dedekind-Hasse norm must be a fact about::principal ideal ring. In particular, an [[fact about::integral dom...