$\mathbb Z[i]/ \langle 1+2i \rangle \cong \mathbb Z_5$

I am trying to prove that $\mathbb Z[i]/ \langle 1+2i \rangle$ is isomorphic to $\mathbb Z_5$.

The only thing that came to my mind was trying to apply the first isomorphism theorem using an appropiate function. If I consider the euclidean function $N: Z[i] \setminus \{0 \} \to \mathbb N$ defined as $N(a+bi)=a^2+b^2$, then I can express any element $z$ in $\mathbb Z[i]$ as $z=(1+2i)q+r$ with $N(r)=1,2,3,4$ or $r=0$.

If I define $f:\mathbb Z[i] \to \mathbb Z_5$ as $f(x=(1+2i)q_x+r_x)=\overline{r_x}$, then it is clear that $f(x)=0$ if and only if $x \in \langle 1+2i \rangle$. The problem with this function is that it doesn't satisfy $f(x+y)=f(x)+f(y)$ and it is not surjective for if $r_x=f(x)=3 (5)$, then if $r_x=a+bi$, we have $a^2+b^2=3$, which is absurd.

I don't know what else to do, any suggestions would be appreciated.

One more proof. It turns out that in this case, it is easier to obtain the isomorphism by going the other direction.

Let $f: \Bbb{Z}\to \Bbb{Z}[i]/(1+2i)$ be the natural mapping, i.e. let $n$ map to $\overline{n}$ where $\overline{n}$ is the residue of $n$ as a gaussian integer mod $1+2i$. Now what is the image of this mapping? Note that $(1+2i)=(-2+i)$ since $i(1+2i)=-2+i$. Therefore $a+bi-b(-2+i)=a+2b\in \Bbb{Z}$ so $a+bi \equiv a+2b \pmod{-2+i}$. In other words, every element of $\Bbb{Z}[i]/(1+2i)$ is the image of some integer under $f$, since $f(a+2b)= a+bi + (1+2i)$.

Now we just need to show $f$ has kernel $(5)$ and we will be done since then by the first isomorphism theorem, $\Bbb{Z}/(5)\cong \Bbb{Z}[i]/(1+2i)$. But of course if $n\in \ker f$, then $1+2i \mid n$, so $1-2i$ also divides $n$, and since $1-2i$ and $1+2i$ are relatively prime, $(1+2i)(1-2i)=5 \mid n$ so $n\in (5)$. Since $1+2i \mid 5$, it is clear that $5\in\ker f$. Therefore $\ker f=(5)$ and we are done.

In general the same proof goes through for any Gaussian prime that is not an integer or $1+i$ since the argument that shows $1+i \mid n$ implies $2 \mid n$ fails here. Although the proof still goes through if you note that if $1+i \mid n$, then $n=(1+i)(a+bi)=a-b+(a+b)i$, and $b=-a$ so that $n=(1+i)(a-ai)=(1+i)(1-i)a=2a$.

Since $$ \Bbb{Z}[i] \cong \Bbb{Z}[x] / \langle x^2 + 1 \rangle, $$ you have the isomorphism $$ R = \Bbb{Z}[i] / \langle 1 + 2i \rangle \cong \Bbb{Z}[x] / \langle 1 + x^2, 1 + 2x \rangle. $$

In order to construct a ring homomorphism $\varphi:R \to \Bbb{Z}_5$, it suffices to describe what $x$ maps to. Say $\varphi(x) = y \in \Bbb{Z}_5$. In order for $\varphi$ to be a ring homomorphism, $y$ must satisfy $$ 1 + y^2 = 0 \quad\text{and}\quad 1 + 2y = 0. $$

Then, you need to check that this map is both injective and surjective to verify that it's an isomorphism.