= An irreducible nonprime in the order Z sqrt minus seven
{title2=$2\in\mathbb Z[\sqrt{-7}]$}
The multiplicative <field norm> $N(m+n\sqrt{-7})=m^2+7n^2$ is an integer. <Units> have <field norm> one and are exactly $\pm1$. A nontrivial <factorization> of 2 would force both factors to have <field norm> two, which the displayed quadratic form cannot represent. Thus 2 is an <irreducible element>. It is not a <prime element>, since it divides $(1+\sqrt{-7})(1-\sqrt{-7})=8$ but neither factor. This is an example in a nonmaximal quadratic order, rather than the full <ring> of integers of $\mathbb Q(\sqrt{-7})$; the precise <ring> matters for <factorization>.
Back to article page
