TITLE: How can I prove that something is an irreducible element?
QUESTION [8 upvotes]: I want to show that $4 + i$ is an irreducible element of $\mathbb{Z}[i]$.  
My current approach is to let $4+i$ = $AB$, where $A$ and $B$ are elements of $\mathbb{Z}[i]$, where $A=a+ia'$ and $B=b+ib'$.
$AB = 4+i$ so:
$$4+i = (ab-a'b') + i(ab' + a'b)$$
therefore:
$$ab-a'b' = 4 $$
and: $$ab' + a'b = 1  $$
I don't know where to go from here though.  Is this a good start? What are some general strategies for showing something is an irreducible element of a ring?

REPLY [7 votes]: Here's a different approach.  Since prime elements are irreducible, it suffices to show that $4+i$ is prime.  (In fact, since $\mathbb{Z}[i]$ is a UFD, then an element is irreducible iff it is prime.)
Recall an ideal $I$ of a commutative ring $R$ is prime iff $R/I$ is a domain.  Since $\mathbb{Z}[i] = \mathbb{Z}[x]/\langle x^2 + 1 \rangle$, then
\begin{align*}
\frac{\mathbb{Z}[i]}{\langle 4 + i \rangle} \cong \frac{\mathbb{Z}[x]}{\langle x^2 + 1, x + 4 \rangle} \cong \frac{\mathbb{Z}[x]/\langle x+4 \rangle}{\langle x^2 + 1, x + 4 \rangle/\langle x+4 \rangle} \cong \frac{\mathbb{Z}}{\langle (-4)^2 + 1 \rangle} = \frac{\mathbb{Z}}{\langle 17 \rangle}
\end{align*}
by the Third Isomorphism Theorem, where the last isomorphism is induced by the evaluation map at $-4$.  Since $\mathbb{Z}/\langle 17 \rangle$ is a domain (a field, even), then $\langle 4 + i \rangle$ is prime (actually, maximal).  Thus $4+i$ is prime.