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# Prime elements

A prime element of a domain is an element ${\displaystyle p}$ which is neither zero nor a unit divisible only by units and associates and which also satisfies the following condition: if ${\displaystyle p|ab}$, where ${\displaystyle a}$ and ${\displaystyle b}$ are also in the same domain, then either ${\displaystyle p|a}$ or ${\displaystyle p|b}$, maybe both; but if neither of those holds true, then ${\displaystyle p}$ may be irreducible but it is not prime. In fact, if the domain is not a unique factorization domain, it does not have prime elements though it may have irreducible elements.
For example, in ${\displaystyle \mathbb {Z} [i]}$ (see: Gaussian integers), we see that ${\displaystyle (1-i)|(2\times 5)}$ and that ${\displaystyle {\frac {2}{1-i}}=1+i}$. Though ${\displaystyle {\frac {5}{1-i}}={\frac {5+5i}{2}}\not \in \mathbb {Z} [i]}$, this does not detract from the fact that ${\displaystyle 1-i}$ is a prime element of ${\displaystyle \mathbb {Z} [i]}$.
Although 2 is irreducible in ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$, it is not prime. For example, ${\displaystyle 2|((1-{\sqrt {-5}})(1+{\sqrt {-5}}))}$,[1] but ${\displaystyle {\frac {1-{\sqrt {-5}}}{2}}={\frac {1}{2}}-{\frac {\sqrt {-5}}{2}}\not \in \mathbb {Z} [{\sqrt {-5}}]}$ and ${\displaystyle {\frac {1+{\sqrt {-5}}}{2}}={\frac {1}{2}}+{\frac {\sqrt {-5}}{2}}\not \in \mathbb {Z} [{\sqrt {-5}}]}$ either. Note that 6 has two factorizations into irreducibles: ${\displaystyle 6=2\times 3=(1-{\sqrt {-5}})(1+{\sqrt {-5}})}$.