Vipul: New page: ==Statement== An fact about::irreducible element in an integral domain need not be a fact about::universal side divisor. In fact, an irreducible element need not be a universa...

2009-01-31T20:30:23Z

New page: ==Statement== An fact about::irreducible element in an integral domain need not be a fact about::universal side divisor. In fact, an irreducible element need not be a universa...

New page

==Statement==

An [[fact about::irreducible element]] in an [[integral domain]] need not be a [[fact about::universal side divisor]]. In fact, an irreducible element need not be a universal side divisor even in a [[fact about::Euclidean domain]].

==Related facts==

===Converse===

* [[Universal side divisor implies irreducible]]: In an integral domain, any universal side divisor is irreducible.
* [[Universal side divisor not implies prime]]: In an integral domain, a universal side divisor need not be prime.

===Caveats===

There are some Euclidean domains where every irreducible element is a universal side divisor. For instance:

* Suppose <math>k</math> is an [[algebraically closed field]]. Then, the [[polynomial ring over a field|polynomial ring]] <math>k[x]</math> has the property that every irreducible polynomial is a universal side divisor. This is because in a polynomial ring, the universal side divisors are precisely the nonconstant linear polynomials, and the field being algebraically closed is precisely equivalent to saying that these are the only irreducible polynomials.
* In a [[discrete valuation ring]], the uniformizing parameter, which is the unique irreducible (up to associates) is a universal side divisor. (Note that since both the property of being irreducible and the property of being a universal side divisor are preserved up to associates in an integral domain, this makes sense).

==Proof==

===Example of the ring of integers===

{{further|[[Particular example::ring of rational integers]]}}

In the ring of rational integers <math>\mathbb{Z}</math>, the only universal side divisors are the elements <math>\pm 2, \pm 3</math>. However, there are infinitely many primes. In particular, <math>\pm 5</math> are irreducibles that are not universal side divisors.

===Example of the polynomial ring over a field that is not algebraically closed===

{{further|[[Particular example::polynomial ring over a field]]}}

Let <math>k</math> be a field that is not algebraically closed. In the polynomial ring <math>k[x]</math>, the universal side divisors are precisely the nonconstant linear polynomials. However, since <math>k</math> is not algebraically closed, there exists an irreducible polynomial of degree greater than <math>1</math> in the polynomial ring <math>k[x]</math>. Thus, there is an irreducible element that is not a universal side divisor.

For instance, if <math>k = \R</math>, the polynomial <math>x^2 + 1 \in \R[x]</math> is irreducible but not a universal side divisor.

Irreducible not implies universal side divisor - Revision history

Vipul: New page: ==Statement== An fact about::irreducible element in an integral domain need not be a fact about::universal side divisor. In fact, an irreducible element need not be a universa...