Prove that $2$ and $2i$ are irreducible in $R = \{a+ 2b i \mid a,\,b\in \mathbb{Z}\}$ I am trying to show that $R= \{a+2b i \mid a,\,b\in \mathbb{Z}\}$ is not a UFD. To do that I need to show that $2$ and $2i$ are irreducible elements in $R$ and that they are not associated in order to show that $4$ has two different factorizations. That they are not associated is clear to me because $i$ is not an unit in $R$. But I have difficulties showing that they are irreducible elements. 
Thanks for your suggestions.
I know that an element $a$ which is not an unit is irreducible if whenever $a = x y$, with $x, y\in R$ then at least $x$ or $y$ is a unit in $R$. So in  this case have $$2 = (a+ 2b i )(c + 2di) $$ and I have to show that $(a+2bi)$ or $(c + 2di)$  are units, but not both.
 A: Let $\ 2= (a+2bi)(c+2di)$.
Then $\ a,b,c,d\in \Bbb{Z}$ 
We define $\ N(\alpha)=\alpha.\bar\alpha=(a^2+4b^2),\; \alpha(=a+2bi)\in R$
So obviously $\ N(\alpha)$ is the norm function.  
So  \begin{align} &N(2) =N(a+2bi)\cdot N(c+2di)\\ 
\implies &(a^2+4b^2)(c^2+4d^2)=4\\
\end{align}
This is possible if  $$\ \begin{align} \text{either}\; &i) (a^2+4b^2)=4\, \text{and} \,(c^2+4d^2)=1\\
\text{or}\quad & ii) (a^2+4b^2)=1\, \text{and} \,(c^2+4d^2)=4\\
\text{or}\quad & iii) (a^2+4b^2)=2\, \text{and} \,(c^2+4d^2)=2 \end{align}$$
In $\ i)\; (c^2+4d^2)=1\implies (c+2di)$ is a unit.
In $\ ii)\; (a^2+4b^2)=1\implies (a+2bi)$ is a unit.
$\ iii)$ is not possible. 
Therefore if $\ 2= (a+2bi)(c+2di)$, then either of the factors is a unit, proving that $\ 2$ is irreducible in $\ R$.
Now let Let $\ 2i= (a+2bi)(c+2di)$.
Then $\ a,b,c,d\in \Bbb{Z}$ 
We define $\ N(\alpha)=\alpha.\bar\alpha=(a^2+4b^2),\; \alpha(=a+2bi)\in R$
So obviously $\ N(\alpha)$ is the norm function.  
So  \begin{align} &N(2i) =N(a+2bi)\cdot N(c+2di)\\ 
\implies &(a^2+4b^2)(c^2+4d^2)=4\\
\end{align}
This is possible if  $$\ \begin{align} \text{either}\; &i) (a^2+4b^2)=4\, \text{and} \,(c^2+4d^2)=1\\
\text{or}\quad & ii) (a^2+4b^2)=1\, \text{and} \,(c^2+4d^2)=4\\
\text{or}\quad & iii) (a^2+4b^2)=2\, \text{and} \,(c^2+4d^2)=2 \end{align}$$
In $\ i)\; (c^2+4d^2)=1\implies (c+2di)$ is a unit.
In $\ ii)\; (a^2+4b^2)=1\implies (a+2bi)$ is a unit.
$\ iii)$ is not possible. 
Therefore if $\ 2i= (a+2bi)(c+2di)$, then either of the factors is a unit, proving that $\ 2i$ is irreducible in $\ R$.
