Question about Rings, Homomorphisms and Ideals.

Question

Let $\mathbb{Z}[i]= \{a+bi: a,b \in \mathbb{Z}\}$be a ring. The ideal $\langle 5 \rangle= \{a+bi: a,b \in \mathbb{Z}, 5\mid a$ and $5\mid b \}$ is a maximal ideal of $\mathbb{Z}[i]$??

$\textbf{My attempt:}$

I've tried to make a homomorphism between $\mathbb{Z}[i]/\langle 5 \rangle$ and $\mathbb{Z}_m$, for some $m \in \mathbb{Z}$. Then, if $m$ is a prime number we have a field $\mathbb{Z}[i]/\langle 5 \rangle$ and $\langle 5 \rangle$ is maximal ideal. For otherwise, if $m$ is not a prime number $\mathbb{Z}[i]/\langle 5 \rangle$ is not a field therefore $\langle 5 \rangle$ is not maximal.

However, it's hard to make a homomorphism between that rings (yes, I've tried m=5) and of course, I've tried to use the homomorphism theorem...

Can you help me??? What is the homomorphism?

Answer 1

Hint: the ring $\mathbb{Z}[i]$ is a PID. Thus, an ideal $0 \neq I = (x)$ is prime if and only if it is maximal, if and only if $x$ is prime (which for PIDs, coincides with the notion of irreducible).

Your question boils down to "Is $5$ prime in $\mathbb{Z}[i]$?

Note that $5 = (1+2i)(1-2i)$ and so $(5) \subsetneq (1+2i) \subsetneq \mathbb{Z}[i]$.

Answer 2

Hint:

This means that $5$ is irreducible in $\mathbf Z[i]$. However note that in this ring $$5=(2-i)(2+i),$$ and if $5$ were irreducible, one of .these factors would be a unit in the ring.