Units in $\mathbb Z[\sqrt{-5}]$

Detailed Solution

An element $u=a+b\sqrt{-5}$ is a unit if and only if its norm is $1$. Thus we need to solve $$ a^2+5b^2=1, \qquad a,b\in\mathbb Z. $$ Since $a^2\geq0$ and $5b^2\geq0$, we have $$ 0\leq 5b^2\leq1. $$ Because $b$ is an integer, this forces $b=0$. Substituting $b=0$ into the norm equation gives $$ a^2=1, $$ so $$ a=\pm1. $$ Therefore, the only elements of norm $1$ are $$ u=1\quad\text{and}\quad u=-1. $$ Both are units because $$ 1^{-1}=1,\qquad (-1)^{-1}=-1. $$ Hence the complete set of units is $$ \boxed{\mathbb Z[\sqrt{-5}]^\times=\{1,-1\}}. $$ Thus option A is correct. Option B is incorrect because $$ N(\sqrt{-5})=0^2+5(1)^2=5\neq1. $$ Therefore, neither $\sqrt{-5}$ nor $-\sqrt{-5}$ is a unit. Option C is incorrect. For example, the element $1+\sqrt{-5}$ has norm $$ N(1+\sqrt{-5})=1^2+5(1)^2=6. $$ Since $6$ divides $6$ but the norm is not $1$, this element is not a unit. Option D is incorrect because $1$ and $-1$ are units. Thus the correct option is $\boxed{\mathrm{A}}$.