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In [[abstract algebra]], a non-zero non-[[unit (ring theory)|unit]] element in an [[integral domain]] is said to be '''irreducible''' if it is not a product of two non-units. | |||
Irreducible elements should not be confused with [[prime element]]s. (A non-zero non-unit element <math>a</math> in a [[commutative ring]] <math>R</math> is called prime if whenever <math>a | bc</math> for some <math>b</math> and <math>c</math> in <math>R</math>, then <math>a|b</math> or <math>a|c</math>.) In an [[integral domain]], every prime element is irreducible,<ref>Consider p a prime that is reducible: p=ab. Then p | ab => p | a or p | b. Say p | a => a = pc, then we have: p=ab=pcb => p(1-cb)=0. Because R is a integral domain we have: cb=1. So b is a unit and p is irreducible.</ref> but the converse is not true in general. The converse ''is'' true for [[unique factorization domain|UFD]]s (or, more generally, [[GCD domain]]s.) | |||
Moreover, while an ideal generated by a prime element is a [[prime ideal]], it is not true in general that an ideal generated by an irreducible element is an [[irreducible ideal]]. However, if <math>D</math> is a GCD domain, and <math>x</math> is an irreducible element of <math>D</math>, then the ideal generated by <math>x</math> ''is'' an irreducible ideal of <math>D</math>.<ref>http://planetmath.org/encyclopedia/IrreducibleIdeal.html</ref> | |||
== Example == | |||
In the [[quadratic integer ring]] <math>\mathbf{Z}[\sqrt{-5}]</math>, it can be shown using [[Field norm|norm]] arguments that the number 3 is irreducible. However, it is not a prime in this ring since, for example, | |||
:<math>3 | \left(2 + \sqrt{-5}\right)\left(2 - \sqrt{-5}\right)=9</math> | |||
but <math>3</math> does not divide either of the two factors.<ref>William W. Adams and Larry Joel Goldstein (1976), ''Introduction to Number Theory'', p. 250, Prentice-Hall, Inc., ISBN 0-13-491282-9</ref> | |||
== References == | |||
{{reflist}} | |||
{{DEFAULTSORT:Irreducible Element}} | |||
[[Category:Ring theory]] | |||
{{Abstract-algebra-stub}} | |||
Revision as of 12:53, 19 December 2013
In abstract algebra, a non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units.
Irreducible elements should not be confused with prime elements. (A non-zero non-unit element in a commutative ring is called prime if whenever for some and in , then or .) In an integral domain, every prime element is irreducible,[1] but the converse is not true in general. The converse is true for UFDs (or, more generally, GCD domains.)
Moreover, while an ideal generated by a prime element is a prime ideal, it is not true in general that an ideal generated by an irreducible element is an irreducible ideal. However, if is a GCD domain, and is an irreducible element of , then the ideal generated by is an irreducible ideal of .[2]
Example
In the quadratic integer ring , it can be shown using norm arguments that the number 3 is irreducible. However, it is not a prime in this ring since, for example,
but does not divide either of the two factors.[3]
References
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Template:Abstract-algebra-stub
- ↑ Consider p a prime that is reducible: p=ab. Then p | ab => p | a or p | b. Say p | a => a = pc, then we have: p=ab=pcb => p(1-cb)=0. Because R is a integral domain we have: cb=1. So b is a unit and p is irreducible.
- ↑ http://planetmath.org/encyclopedia/IrreducibleIdeal.html
- ↑ William W. Adams and Larry Joel Goldstein (1976), Introduction to Number Theory, p. 250, Prentice-Hall, Inc., ISBN 0-13-491282-9