Decide if $\mathbb Z[i]/\langle i\rangle$ and $\mathbb Z$ are isomorphic, if $\mathbb Z[i]/\langle i+1\rangle$ and $\mathbb Z_2$ are isomorphic

Question

Decide if $\mathbb Z[i]/\langle i\rangle$ and $\mathbb Z$ are isomorphic, if $\mathbb Z[i]/\langle i+1\rangle$ and $\mathbb Z_2$ are isomorphic

I know that in the first case if there exist such homomorphism then $f(i)=0$ (and in the second case $f(i+1)=0$), but I don't know exactly how to prove it.

Answer 1

To show the first isomorphism you can use one of the isomorphism theorems:

$$(A+I)/I \cong A/(A\cap I) $$

In this case $A=\mathbb{Z}\subseteq \mathbb{Z}[i]$ and $I=(i)$.

In the second case, note that $(i+1)^{2}=(2)$, so we have $\mathbb{Z}\cap (i+1)^{2}=(2)$. Then apply again the same theorem.

Answer 2

Here is an answer for the second question:

Since $2=(1+i)(1-i)$, we have $2 \in (1+i)$.

Thus, $a+bi \equiv (a\bmod 2)+(b\bmod 2)\,i \equiv (1+i)$.

Therefore, the classes of $\mathbb Z[i]$ mod $(1+i)$ are reduced to the classes of $0,1,i,1+i$.

Now $0 \equiv 1+i$ and $1 \equiv i$ and so $\mathbb Z[i]$ mod $(1+i)$ has only two classes and must be $\mathbb Z_2$.

Answer 3

It seems you are in the context of rings.

Note that $i$ is invertible in $\mathbb{Z}[i]$, so the ideal it generates is the whole ring. Hence the quotient $\mathbb{Z}[i]/\langle i\rangle$ is the trivial ring.

For the second part, consider $a+bi=a-b+b(1+i)$ and the ring homomorphism $$ \varphi\colon\mathbb{Z}[i]\to\mathbb{Z}/2\mathbb{Z} $$ defined by $\varphi(a+bi)=[a-b]$ where $[x]$ denotes the residue class of $x$ modulo $2$ (prove it is a ring homomorphism and that its kernel is $\langle 1+i\rangle$).