**Question:** "Extending the result: $Z[i]/(a−ib)≅Z/(a^2+b^2)Z$, if $a,b$ are relatively prime.

**Answer:** Let $a-ib:=1-(-3)i=1+3i$ and let $B:=\mathbb{Z}/(1+3^2)\cong\mathbb{Z}/(10)$ and let $A:=\mathbb{Z}[i]/(1+3i)$. This approach uses the chinese remainder lemma and it illustrates the "unique factorization of ideals" into products of powers of maximal ideals in Dedekind domains: 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 $K:=\mathbb{Q}(i)$.

**Question:** "A very basic ring theory question, which I am not able to solve. How does one show that $\mathbb{Z}[i]/(3−i)≅\mathbb{Z}/10\mathbb{Z}$?"

**Example:** We may use the "same method": There a factorization into products of ideals

$$(3-i)=(1-i)(2+i)=\mathfrak{m}\mathfrak{n}$$

and the ideals $\mathfrak{m},\mathfrak{n}$ are coprime maximal ideals with finite residue field. There is by the CRL isomorphisms

$$\mathbb{Z}[i]/(3-i)\cong \mathbb{Z}[i]/\mathfrak{m}\oplus \mathbb{Z}[i]/\mathfrak{n} \cong \mathbb{Z}/(2)\oplus \mathbb{Z}/(5) \cong \mathbb{Z}/(10).$$

As above you may write down explicit maps and verify they are isomorphisms.