Irreducible element

In abstract algebra, a non-zero non-unit element in an integral domain $D$ is said to be irreducible or an atom if it is not a product of two non-units. In the litterature there appear many different equivalent definitions for irreducible element, such as: an element $a$ in an integral domain $D$ is irredusible,

1. If for every factorization $a=bc$, $b$ or $c$ is a unit (an element that has an inverse element with respect to multiplication), also $a$ is not a product of two non-units 2. If for every factorization $a=bc$, $b$ or $c$ is an associate of $a$ (elements $x$,$y$ are associates if $x|y$ and $y|x$) 3. If all factors of $a$ are units or associates of $a$ 4. If all factors of $a$ are units or of the form $ua$ where $u$ is a unit, also $a$ has only so-called trivial factors 5. If every factorization of $a$ to two factors is of the form $a=u^{-1}*(ua)$, where $u$ is a unit, also $a$ has only so-called trivial factorizations 6. The ideal generated by $a$, $(a)=aD$ is maximal element in the set of proper principal ideals of $D$

The concept of irreducible element or atom can be generalized into commutative rings (possibly without but usually with identity). In generalizing this concept one must however be VERY CAREFUL since in a ring that is not an integral domain the above definitions are no longer equivalent. This means that the concept of irreducible element on a ring that is not an integral domain is highly dependent on which of the above definitions is taken as the definition in rings that are not integral domains. Furthermore, there are also other definitions than those listed above for irreducible element in rings that are not integral domains. The terminology is not as of yet quite standard, but the following very wide article <href>http://projecteuclid.org/euclid.rmjm/1181072068</href> attempts to survey the subject and describes 5 different concepts of irreducible element (the authors call these concepts irreducible element, strongly irreducible element. very strongly irreducible element, irreducible element in the sense of Flecther and m-irreducible element) in a commutative unital ring that is not an integral domain, and proves different kind of relations between these concepts (for example, irreducible elements are exactly the irreducible elements in the sense of Fletcher).

Irreducible elements should not be confused with prime elements. (A non-zero non-unit element $a$ in a commutative ring $R$ is called prime if whenever $a|bc$ for some $b$ and $c$ in $R$, then $a|b$ or $a|c$.) In an integral domain, every prime element is irreducible, but the converse is not true in general. The converse is true for UFDs (or, more generally, GCD domains.) In a ring that is not an integral domain, prime elements are not necessary irreducible (according to some definitions of irreducible element; see above for the discussion of the difficulty of defining irreducible element in rings that are not integral 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 $D$ is a GCD domain, and $x$ is an irreducible element of $D$, then the ideal generated by $x$ is an irreducible ideal of $D$.

Example

In the quadratic integer ring $\mathbf {Z} [{\sqrt {-5}}]$, 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,

$3|\left(2+{\sqrt {-5}}\right)\left(2-{\sqrt {-5}}\right)=9$

but $3$ does not divide either of the two factors.

In the ring of integers modulo 6 $\mathbf {Z} /6\mathbf {Z}$, which is a commutative unital ring that is not an integral domain ($[2]*[3]=[0]$ in math>\mathbf{Z}/6\mathbf{Z}</math>, $[2]\neq [0]$, $[3]\neq [0]$), the element $[2]$ is a prime element, but it is not irreducible if we defineirreducible element in a ring that is not an integral domain so that an irreducible element is not a product of two non-units, as in this case we have $[2]=[2]*[4]$ but neither $[2]$ nor $[4]$ is a unit. Also, if we define the irreducible element in a ring that is not an integral domain as follows (this is the definition for irreducible element in the article <href>http://projecteuclid.org/euclid.rmjm/1181072068</href> above): an element $a$ is irredusible, if for every factorization $a=bc$, $b$ or $c$ is an associate of $a$ (elements $x$,$y$ are associates if $x|y$ and $y|x$), then we see that by the fact that $[2]$ is prime, it is also irreducible (Note that in the factorization $[2]=[2]*[4]$ both $[2]$ and $[4]$ are associates of $[2]$). Indeed, it is clear that with this definition of irreducible element, all prime elements on any commutative unital rings are always irreducible elements. Finally one can check that in the ring $\mathbf {Z} /6\mathbf {Z}$, $[2]$ has only trivial factors, meaning that the factors are units or $[2]$ multiplied by unit; we note that $[2]=[1]*[2]$, $[4]=[5]*[2]$ in $\mathbf {Z} /6\mathbf {Z}$, and $[1]$ and $[5]$ are units in $\mathbf {Z} /6\mathbf {Z}$ ($[1]=[5]*[5]$ in $\mathbf {Z} /6\mathbf {Z}$). However one sees from the factorization $[2]=[2]*[4]$ that not all the factorizations of $[2]$ are trivial factorization, id est, of the form $[2]=u^{-1}*(u*[2])$ where $u$ is a unit element of $\mathbf {Z} /6\mathbf {Z}$ ($[5]^{-1}=[5]$ in $\mathbf {Z} /6\mathbf {Z}$). Thus it is clear that the different definitions of irreducible element that are equivalent in integral domains are not equivalent in rings that are not integral domains, and thus the concept of irreducible element in a ring that is not an integral domain varies according to which definition of irreducible element in a ring that is not an integral domain is adopted.