Prove that $\mathbb Z[i]/(1+3i)\simeq \mathbb Z_{10}$ I am confused how to do the proof that $\mathbb Z[i]/(1+3i)\simeq \mathbb Z/10\mathbb Z$.
It is clear to me intuitively.
$1+3i=0$ in the quotient ring, so $3i=-1\implies 9i^2=1\implies 10=0$ in quotient ring.
But when trying to show rigorously, I am stuck as I cannot get any natural epimorphism from $\mathbb Z[i]\to \mathbb Z/10\mathbb Z$ with kernel $(1+3i)$.
I asked our instructor and he gave a proof which is beyond my understanding:
He wrote
\begin{equation}
\mathbb Z[i]/(1+3i)\simeq \frac{\frac{\mathbb Z[x]}{(1+x^2)}}{\frac{(1+3x,1+x^2)}{(1+x^2)}}\simeq \frac{\mathbb Z[x]}{(1+3x,1+x^2)}\simeq\frac{\frac{\mathbb Z[x]}{(1+3x)}}{\frac{(1+3x,1+x^2)}{(1+3x)}}\simeq\mathbb Z/10\mathbb Z.
\end{equation}
I don't understand the proof and I want a more natural proof of this statement.
Can someone help me?
 A: Question: "I am stuck as I cannot get any natural epimorphism from Z[i]→Z/10Z with kernel (1+3i). I asked our instructor and he gave a proof which is beyond my understanding."
Answer: In $B:=\mathbb{Z}/(10)$ and let $A:=\mathbb{Z}[i]/(1+3i)$. It follows $-1 \cong 10-1 \cong 9$ hence you get a well defined map
$$\phi: \mathbb{Z}[i] \rightarrow B$$
by definining $\phi(a+bi):=a+3b$. It follows
$$-1=\phi(-1)=\phi(i^2)=\phi(i)^2=3^2=9=10-1 =-1.$$
hence the map is well defined. By definition $\phi(1+3i)=1+3^2=10=0$, hence you get a well defined surjective map $\phi: A \rightarrow B$.
In $A$ there are two ideals $I:=(1+i),J:=(2+i)$ and
$$(1+i)(2+i)=2+i+2i+i^2=1+3i=0.$$
The ideals $I,J$ are coprime, hence
$$A \cong A/IJ \cong A/I \oplus A/J \cong \mathbb{Z}/(2) \oplus \mathbb{Z}/(5)$$
and
$$\mathbb{Z}/(10) \cong \mathbb{Z}/(2) \oplus \mathbb{Z}/(5)$$
by the chinese remainder lemma. There are explicit isomorphisms
$$f:A/I\cong \mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/(2)$$
defined by sending $i$ to $1$.
There is an isomorphism
$$g:A/J \cong \mathbb{Z}[i]/(2+i) \cong \mathbb{Z}/(5)$$
defined by sending $i$ to $3$.  It follows
$$g(1+i^2)=1+3^2=10=0$$
and
$$g(2+i)=2+3=5=0$$
hence $g$ is well defined. You may check that $f,g$ are isomorphisms and you should do this as an exercise in "commutative ring theory/abstract algebra".
Note: This is a general fact: In the ring of integers $\mathcal{O}_K$ in a number field $K$, any ideal $\mathfrak{a}$ may be written (uniquely up to the order) as a product of powers of distinct maximal ideals
$$(*) \mathfrak{a}=\mathfrak{m}_1^{p_1} \cdots \mathfrak{m}_l^{p_l}.$$
The ideals above $I,J$ are maximal since the quotients $A/I,A/J$ are (finite) fields. The equality $(*)$  is proved in Theorem I.3.3 in Neukirch "Algebraic number theory" (this property was in fact one of the reason for the introduction of ideals in algebraic number theory and commutative algebra).
In the above case the multiplicities $l_I,l_J=1$ and you get the decomposition
$$(1+3i)=IJ$$
in $\mathcal{O}_K\cong \mathbb{Z}[i]$ where $K:=\mathbb{Q}(i)$.
A: Clearly we have to find a homomorphism $f:\mathbb Z[i]\to \mathbb Z_{10}$ such that it kills $1+3i$,that is $1+3i=0\implies i=-1/3$ in $\mathbb Z_{10}$.So,$i$ should be mapped to the additive inverse of the multiplicative inverse of $3$ in $\mathbb Z_{10}$ which means $i=3$.So define $f$ by $f(a+ib)=\overline{a+3b}$ which is an epimorphism with kernel $(1+3i)$.So,by first isomorphism theorem ,we are done.
