Irreducible element not implies prime - Revision history

Vipul: /* Proof */

2009-02-01T16:08:26Z

Proof

← Older revision		Revision as of 16:08, 1 February 2009
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	But the only possibilities for this are <math>a = \pm 2, b = 0</math> or <math>a = \pm 1, b = 0</math>.		But the only possibilities for this are <math>a = \pm 2, b = 0</math> or <math>a = \pm 1, b = 0</math>.

			Note that this ring is a [[Dedekind domain]].

	===Example of a ring of integer-valued polynomials===		===Example of a ring of integer-valued polynomials===
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	{{proofat\|[[Every binomial polynomial is irreducible but not prime in the ring of integer-valued polynomials over rational integers]]}}		{{proofat\|[[Every binomial polynomial is irreducible but not prime in the ring of integer-valued polynomials over rational integers]]}}

			Unlike the previous example, the ring is ''not'' a [[Noetherian ring]].

Vipul: /* Example of a ring of integer-valued polynomials */

2009-02-01T16:07:26Z

Example of a ring of integer-valued polynomials

← Older revision		Revision as of 16:07, 1 February 2009
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	is irreducible but not prime in <math>R</math>.		is irreducible but not prime in <math>R</math>.

	{{proofat\|[[Every binomial polynomial is irreducible but not prime in ring of integer-valued polynomials over rational integers]]}}		{{proofat\|[[Every binomial polynomial is irreducible but not prime in the ring of integer-valued polynomials over rational integers]]}}

Vipul: /* Example of a ring of integer-valued polynomials */

2009-02-01T16:07:01Z

Example of a ring of integer-valued polynomials

← Older revision		Revision as of 16:07, 1 February 2009
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	is irreducible but not prime in <math>R</math>.		is irreducible but not prime in <math>R</math>.

	{{proofat\|[[~~Binomial~~ polynomial is irreducible but not prime in ring of integer-valued polynomials over rational integers]]}}		{{proofat\|[[Every binomial polynomial is irreducible but not prime in ring of integer-valued polynomials over rational integers]]}}

Vipul: New page: ==Statement== An fact about::irreducible element in an integral domain need not be a fact about::prime element. ==Related facts== An [[integral domain in which every irreduc...

2009-02-01T16:05:18Z

New page: ==Statement== An fact about::irreducible element in an integral domain need not be a fact about::prime element. ==Related facts== An [[integral domain in which every irreduc...

New page

==Statement==

An [[fact about::irreducible element]] in an [[integral domain]] need not be a [[fact about::prime element]].

==Related facts==

An [[integral domain in which every irreducible is prime]] is an integral domain where irreducible elements are all prime. Such integral domains are very common. In fact:

* [[Bezout implies every irreducible is prime]]: In a [[Bezout domain]], i.e., an integral domain where every finitely generated ideal is principal, every irreducible element is prime.
* [[Unique factorization implies every irreducible is prime]]: In a [[unique factorization domain]], every irreducible element is prime.

==Proof==

===Example of a quadratic integer ring===

Consider the ring <math>\mathbb{Z}[\sqrt{-5}]</math>. In this ring, we have:

<math>2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})</math>.

Thus, <math>2 | (1 + \sqrt{-5})(1 - \sqrt{-5})</math>, but <math>2</math> does not divide either factor, so <math>2</math> is not prime.

On the other hand, <math>2</math> is irreducible, as can be verified using the algebraic norm. If <math>(a + b\sqrt{-5}) | 2</math>, then <math>N(a + b \sqrt{-5}) | N(2)</math>, yielding:

<math>a^2 + 5b^2 | 4</math>.

But the only possibilities for this are <math>a = \pm 2, b = 0</math> or <math>a = \pm 1, b = 0</math>.

===Example of a ring of integer-valued polynomials===

{{further|[[Particular example::ring of integer-valued polynomials over rational integers]]}}

Let <math>R</math> be the ring of integer-valued polynomials over rational integers: this is the ring of those polynomials in <math>\mathbb{Q}[x]</math> that send integers to integers. Then, any [[binomial polynomial]], i.e., any polynomial of the form:

<math>\binom{x}{r} = \frac{x(x-1) \dots (x-r+1)}{r!}</math>

is irreducible but not prime in <math>R</math>.

{{proofat|[[Binomial polynomial is irreducible but not prime in ring of integer-valued polynomials over rational integers]]}}