# Prime element

The term prime element in commutative algebra is a generalization of the term prime number to commutative unitary rings .

## definition

An element of a commutative unitary ring is called a prime element if it is neither 0 nor a unit and applies to all : divides the product , then divides also or . ${\ displaystyle c}$${\ displaystyle (R, +, \ cdot, 0,1)}$${\ displaystyle c}$${\ displaystyle a, b \ in R}$ ${\ displaystyle c}$${\ displaystyle a \ cdot b}$${\ displaystyle c}$${\ displaystyle a}$${\ displaystyle b}$

In symbol notation: ${\ displaystyle c {\ mbox {is prime}} \ Leftrightarrow \ c \ neq 0 \ \ land \ c \ nmid 1 \ \ land \ \ forall a, b \ in R: \ c \ mid (a \ cdot b) \ Rightarrow (c \ mid a) \ lor (c \ mid b).}$

Prime elements are those elements, apart from 0 and units, which, if they occur in any product, also occur in at least one of the factors.

## Irreducible elements

Another generalization of the notion of prime numbers are irreducible elements, which are defined by the fact that they are not units and cannot be represented as the product of two non-units. In general, neither every prime element is irreducible nor every irreducible element prime (see examples ). But in an integrity ring every prime element is irreducible, and conversely in a factorial ring every irreducible element is also prime.

• If a prime element and a unit is also a prime element.${\ displaystyle c}$${\ displaystyle e}$${\ displaystyle c \ cdot e}$
• A non-unit is a prime element if and only if the main ideal is a prime ideal .${\ displaystyle c \ neq 0}$ ${\ displaystyle (c)}$
• A field consists only of the zero and units and therefore does not contain any prime elements.
• In a factorial ring, every element except 0, except for unit factors and order, can be clearly represented as a product of prime elements.

## Examples

• The prime elements in the ring of whole numbers are exactly the prime numbers (2, 3, 5, 7, 11, ...) and their opposite numbers ( −2, −3, −5, −7, −11, ...).
• The prime elements in the ring of Gaussian numbers are, apart from the unit factors, exactly the prime numbers of the form and the elements for which a prime number is, i.e. they are prime elements, but not , or (for proof see Fermat's two-squares theorem ).${\ displaystyle \ mathbb {Z} [i]}$${\ displaystyle \ pm 1, \ pm i}$${\ displaystyle 4k + 3, \ k \ in \ mathbb {Z}}$${\ displaystyle a + b \ cdot i, \ a, b \ in \ mathbb {Z}}$${\ displaystyle a ^ {2} + b ^ {2}}$${\ displaystyle 3, \, 7, \, 11, \, 1 + i, \, 2 + 3i}$${\ displaystyle 2 = (1 + i) \ cdot (1-i)}$${\ displaystyle 5 = (2 + i) \ cdot (2-i)}$${\ displaystyle 3 + i = (1 + i) \ cdot (2-i)}$
• In the integrity ring (contains all numbers of the form with ) the number 2 is irreducible, but not prime. This is because 6 is divided by 2, but it can be written as a product , with none of the factors being divisible by 2.${\ displaystyle \ mathbb {Z} [i {\ sqrt {5}}]}$${\ displaystyle a + b \ cdot i {\ sqrt {5}}}$${\ displaystyle a, b \ in \ mathbb {Z}}$${\ displaystyle (1 + i {\ sqrt {5}}) \ cdot (1-i {\ sqrt {5}})}$
• In the product ring there is a prime element that is not irreducible.${\ displaystyle \ mathbb {Z} \ times \ mathbb {Z}}$${\ displaystyle (1.0) = (1.0) \ cdot (1.0)}$

## Individual evidence

1. ^ Siegfried Bosch : Algebra. 7th edition 2009, Springer-Verlag, ISBN 3-540-40388-4 , p. 201.