Euclidean implies principal ideal - Revision history

Vipul at 16:06, 12 November 2023

2023-11-12T16:06:37Z

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	{{curing property implication}}		{{curing property implication\|stronger = Euclidean ring\|weaker = principal ideal ring}}

	==Statement==		==Statement==

Vipul: 3 revisions

2008-05-12T16:20:10Z

3 revisions

← Older revision	Revision as of 16:20, 12 May 2008
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Vipul at 18:39, 5 January 2008

2008-01-05T18:39:04Z

← Older revision		Revision as of 18:39, 5 January 2008
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	===Verbal statement===		===Verbal statement===

	Any [[Euclidean ring]] is a [[principal ideal ring]].		Any [[Euclidean ring]] is a [[principal ideal ring]]. In particular, any [[Euclidean domain]] is a [[principal ideal domain]].

	==Proof==		==Proof==

Vipul: Euclidean ring implies principal ideal ring moved to Euclidean implies principal ideal

2008-01-05T18:13:56Z

Euclidean ring implies principal ideal ring moved to Euclidean implies principal ideal

← Older revision	Revision as of 18:13, 5 January 2008
(No difference)

Vipul at 18:07, 5 January 2008

2008-01-05T18:07:37Z

New page

{{curing property implication}}

==Statement==

===Property-theoretic statement===

The [[property of commutative unital rings]] of being a [[Euclidean ring]] is stronger than the property of being a [[principal ideal ring]].

===Verbal statement===

Any [[Euclidean ring]] is a [[principal ideal ring]].

==Proof==

The key idea behind the proof is to, given an ideal, exhibit an element that generates the ideal. The claim is as follows:

''Consider the [[Euclidean norm]] as a function on the ideal. Pick any element with minimum norm (such an element exists because the set of possible norms is a well-ordered set). Then, the ideal is generated by that element''.

(Note that when the ideal is the zero ideal, the element of smallest possible norm is zero, whose norm is assumed to be <math>\infty</math>).

Let's prove this claim. Suppose <math>R</math> is a [[Euclidean ring]] with [[Euclidean norm]] <math>N</math>, and <math>I</math> is an [[ideal]] inside <math>R</math>. Let <math>d</math> be the smallest possible norm among elements of <math>I</math>, and let <math>a \in I</math> be an element such that <math>N(a) = d</math>. Now pick and <math>c \in I</math>. We want to show that <math>c</math> is a multiple of <math>a</math>.

By the division algorithm over <math>R</math>, there exist <math>q,r \in R</math> such that:

<math>c = aq + r, \qquad N(r) < N(a)</math> or <math>r = 0</math>

Now note that <math>r = c - aq</math>, and since <math>c,a \in I</math> and <math>I</math> is an [[ideal]], <math>r \in I</math>. Since <math>a</math> was chosen with minimum norm, <math>N(r) < N(a)</math> is impossible, so <math>r = 0</math>, and thus <math>c</math> is a multiple of <math>a</math>.