How to prove $\Bbb Z[i]/(1+2i)\cong \Bbb Z_5$?
My method is
$$(1+2i)=\big\{a+bi丨a+2b≡0\pmod 5\big\},$$
So any $a+bi$ in $\Bbb Z(i)$,we got
$$a+bi=(b-2a)i+a(1+2i).$$
So $\Bbb Z[i]/(1+2i)=\big\{0,[i],[2i],[3i],[4i]\big\}$.
I know how to prove this ring is isomorphic to $\Bbb Z_5$, but how can I prove that $\Bbb Z[i]/(1+2i)$ equals to $\Bbb Z_5$ directly? Any suggestion ia appreciated.
On
If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $\mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.
Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.
On
Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)\supset (5)$. Hence $\mathbb Z[i]/(1+2i)\subset \mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)\cong \mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.
To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) $\{0,1,2,3,4\}$. It's easy to see none of them are the same.
On
A ring homomorphism $\varphi\colon\mathbb{Z}[i]\to\mathbb{Z}_5$ is uniquely determined by assigning $q=\varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.
With the second choice, you have $\varphi(a+bi)=[a-2b]$. Can you determine $\ker\varphi$?
I think it is easier to do it this way. Let $\varphi:\Bbb Z[x]\to \Bbb Z[i]$ sending $x\mapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $\Bbb Z[i]$ under $\varphi$ is the ideal $J=(x^2+1,x-2)$ of $\Bbb Z[x]$. Thus $\varphi$ induces the isomorphism $$\Bbb{Z}[i]/I\cong \Bbb{Z}[x]/J.$$
Now observe that $$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$ This shows that$$\frac{\Bbb Z[i]}{(1+2i)}=\frac{\Bbb{Z}[i]}{(i-2)}\cong \frac{\Bbb Z[x]}{(x^2+1,x-2)}=\frac{\Bbb{Z}[x]}{(5,x-2)}\cong\frac{\Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$ Because under the isomorphism $\Bbb{Z}[x]/(x-2)\cong \Bbb Z$ (which sends $p(x)+{\color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5\Bbb{Z}$ of $\Bbb Z$, we get $$\frac{\Bbb Z[i]}{(1+2i)}\cong\frac{\Bbb Z}{5\Bbb Z},$$ as required.