Question:
In $\mathbb{Z} \left[ \sqrt{-5} \ \right]$ show that $1+3\sqrt{-5}$ is irreducible but not prime.
Attempt:
Suppose $1+3\sqrt{-5} =\left ( a+b\sqrt{-5} \right )\cdot \left ( c+d\sqrt{-5} \right )$ Where $\left ( a+b\sqrt{-5} \right ), \left ( c+d\sqrt{-5} \right )$ are not units. $\cdot \cdot \cdot$ I attempted without assuming that $\left ( a+b\sqrt{-5} \right ),\left ( c+d\sqrt{-5} \right )$ are not units. Par for the course, the attempt did not span very far.
What is peculiar is that when I look at the definition of irreducible, there was no such condition requiring $\left ( a+b\sqrt{-5} \right ),\left ( c+d\sqrt{-5} \right )$ to be non-units.
Recall: a non-zero element a in the integral domain D is called irreducible if a is not a unit and whenever $\exists b,c \in D$: $a=b \cdot c$ then either b or c is a unit.
Any help is appreciated.
The definition says: If $a=bc$ then at least one of $a,b$ is a unit. Equivalently: It is not possible to write $a=bc$ with $a,b$ non-units. So as proof strategy: Assume $a=bc$ with $b,c$ non-units and try to arrive at a contradiction.