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# Irreducible elements

A nonzero nonunit element $n$ of an integral domain $R$ is said to be irreducible if it is not a product of two nonunits. If $n$ is irreducible and $n=ab$ (where $a,b\in R$ ), then either $a$ is a unit or $b$ is (and then the other is either an associate of $n$ or equal to it).

For example, 7 is irreducible in $\mathbb {Z} [{\sqrt {10}}]$ . It can be expressed as $7=1\times 7=(-1)(-7)$ and there is no other such expression using only elements of $\mathbb {Z} [{\sqrt {10}}]$ . In the latter expression, –1 is a unit, and –7 is an associate of 7. But 10 is not irreducible in $\mathbb {Z} [{\sqrt {10}}]$ , which is to say that it is reducible, it can be expressed as $10=2\times 5=({\sqrt {10}})^{2}$ , but neither 2, 5 nor ${\sqrt {10}}$ are units, nor associates of 10.

## Relationship of irreducibles to prime elements

Irreducible elements should not be confused with prime elements. (A nonunit 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 unique factorization domains (UFDs, or, more generally, GCD domains).

### Prime and irreducible

If $R$ is a UFD, then all its irreducible elements are also prime elements. For example, 3 is both prime and irreducible in $\mathbb {Z}$ , since it's divisible only by the units 1 and –1, and by its associate, –3; and, in every case in which $3|ab$ , then either $3|a$ or $3|b$ (or maybe both).

### Irreducible but not prime

In all quadratic integer rings with class number greater than 1, the irreducible elements are not necessarily prime. This is a consequence of some elements having more than one factorization. Given some number $ab=pq$ , where $a$ , $b$ , $p$ , $q$ are all distinct, nonunit, nonzero numbers, it can happen that $p|ab$ yet $p\not |a$ and $p\not |b$ .

For example, $\mathbb {Z} [{\sqrt {-5}}]\,$ , it can be shown using field norm arguments that the number 3 is irreducible. However, it is not a prime in this ring since, for example, $3\,{\big |}\,\left(2+{\sqrt {-5}}\right)\,\left(2-{\sqrt {-5}}\right)\,=\,9,\,$ but 3 does not divide either of the two factors.

Now, just because a given ring is not a UFD one should not assume that every element has more than one factorization; this applies to some reducible elements. For example, in $\mathbb {Z} [{\sqrt {-6}}]$ , which is not a UFD, 2 and 5 are irreducible, but $7=(1-{\sqrt {-6}})(1+{\sqrt {-6}})$ , though that is its only factorization. The classic example to show we're not dealing with a UFD here is $10=2\times 5=(2-{\sqrt {-6}})(2+{\sqrt {-6}})$ . It therefore stands to reason that all positive multiples of 10 have more than one factorization, but we should not be led astray to think that all these numbers have more than two factorizations. For example, $70=2\times 5\times (1-{\sqrt {-6}})(1+{\sqrt {-6}})=(1-{\sqrt {-6}})(1+{\sqrt {-6}})(2-{\sqrt {-6}})(2+{\sqrt {-6}})=(8-{\sqrt {-6}})(8+{\sqrt {-6}})$ . But, as it turns out, that third factorization is not distinct because the two factors can factored further still, e.g., ${\frac {8+{\sqrt {-6}}}{2-{\sqrt {-6}}}}=(1+{\sqrt {-6}})$ . There are even more traps of this sort in real rings, where the multitude of units can disguise incomplete factorizations to an even greater extent than in imaginary rings.

### Prime but not irreducible

Whereas the concept of irreducibles may seem unnecessary in UFDs, it may seem downright nonsensical in reduced residue systems. Given a composite integer $n$ and $\mathbb {Z} _{n}$ consisting of the integers 0 to $n-1$ , the primes which divide $n$ in $\mathbb {Z}$ are still primes in $\mathbb {Z} _{n}$ , but they might be reducible to products of themselves.

For example, consider $\mathbb {Z} _{6}$ . There are only two elements, 1 and 3 itself, which divide 3 evenly. It happens that in this context that 3 is a zero-divisor, and there are only two nontrivial ways to obtain 0 as a product: $2\times 3=3\times 4=0$ , with 3 being one of the factors in both cases. But since $3^{2}=3$ , it follows that 3 can be reduced indefinitely.

## Relationship of irreducibles to ideals

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 $D\,$ .

## Irreducible factorizations

(...) (Irreducible factorizations)