Universal side divisor implies irreducible: Difference between revisions

Latest revision as of 16:01, 5 February 2009

Statement

In an commutative unital ring, any universal side divisor is an irreducible element.

Related facts

Converse

The converse is not true, even in a Euclidean domain. Further information: Irreducible not implies universal side divisor

Other related facts

Proof

Given: A commutative unital ring $R$ , a universal side divisor $x\in R$ such that $x=ab$ .

To prove: $x$ is neither zero nor a unit, and either $a$ is a unit or $b$ is a unit.

Proof: The fact that $x$ is neither zero nor a unit follows form the definition of universal side divisor, so it remains to show that if $x=ab$ , then either $a$ is a unit or $b$ is a unit.

Since $x$ is a universal side divisor, we obtain that either $x|a$ or there exists a unit $u$ such that $x|a-u$ .

Case $x|a$ : In this case we have $x|a$ and $a|x$ , so $x$ and $a$ are associates, and we are done.
Case there exists a unit $u$ such that $x|a-u$ : We have $a-u=xy$ for some $y$ , yielding $u=a-xy=a-aby=a(1-by)$ . Since $u$ is a unit, there exists $v$ such that $uv=1$ , yielding $a(1-by)v=1$ , so $a$ is a unit with inverse $(1-by)v$ . Thus, $x$ is associate with $b$ , and we are done.

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 ==Statement==
-In an [[integral domain]], any [[fact about::universal side divisor]] is an [[fact about::irreducible element]].
+In an [[commutative unital ring]], any [[fact about::universal side divisor]] is an [[fact about::irreducible element]].
 ==Related facts==
@@ Line 14: / Line 14: @@
 * [[Associate implies same orbit under multiplication by group of units in integral domain]]
 * [[Associate not implies same orbit under multiplication by group of units]]
-* [[Element of smallest norm among non-units in Euclidean ring is a universal side divisor]]
+* [[Element of minimum norm among non-units in Euclidean ring is a universal side divisor]]
 * [[Euclidean ring that is not a field has a universal side divisor]]
 ==Proof==
-'''Given''': An integral domain <math>R</math>, a universal side divisor <math>x \in R</math> such that <math>x = ab</math>.
+'''Given''': A commutative unital ring <math>R</math>, a universal side divisor <math>x \in R</math> such that <math>x = ab</math>.
 '''To prove''': <math>x</math> is neither zero nor a unit, and either <math>a</math> is a unit or <math>b</math> is a unit.
@@ Line 26: / Line 26: @@
 Since <math>x</math> is a universal side divisor, we obtain that either <math>x | a</math> or there exists a unit <math>u</math> such that <math>x | a - u</math>.
-# Case <math>x | a</math>: We have <math>a = xy</math> for some <math>y</math>, and we get <math>x = xyb</math>, yielding <math>x(1 - yb) = 0</math>. Since <math>x \ne 0</math> and <math>R</math> is an integral domain, we obtain <math>by = 1</math>, so <math>b</math> is a unit.
+# Case <math>x | a</math>: In this case we have <math>x | a</math> and <math>a | x</math>, so <math>x</math> and <math>a</math> are associates, and we are done.
-# Case there exists a unit <math>u</math> such that <math>x | a - u</math>: We have <math>a - u = xy</math> for some <math>y</math>, yielding <math>u = a - xy = a - aby = a(1 - by)</math>. Since <math>u</math> is a unit, there exists <math>v</math> such that <math>uv = 1</math>, yielding <math>a(1-by)v = 1</math>, so <math>a</math> is a unit with inverse <math>(1-by)v</math>.
+# Case there exists a unit <math>u</math> such that <math>x | a - u</math>: We have <math>a - u = xy</math> for some <math>y</math>, yielding <math>u = a - xy = a - aby = a(1 - by)</math>. Since <math>u</math> is a unit, there exists <math>v</math> such that <math>uv = 1</math>, yielding <math>a(1-by)v = 1</math>, so <math>a</math> is a unit with inverse <math>(1-by)v</math>. Thus, <math>x</math> is associate with <math>b</math>, and we are done.