I need help with the following exam exercise, my teacher didn’t post the answer and I can’t manage to solve it.
In $ A=\mathbb{Z}[i]=\{a+bi \ : \ a,b \in \mathbb{Z}\} $ we consider $a=7+56i; \ b=3+3i; \ c=1+8i$. We will write $(a)$ to refer to the ideal generated by $a$
1.- Prove that $\frac{A}{(a)}$ isn’t a field. Where $\frac{A}{(a)}$ is the quotient ring of $A$ by the ideal generated by $a$.
However, prove there exists an isomorphism between $\frac{A}{(a)}$ and $A \times B \times C$ where $A,B,C$ are three fields.
2.- Find out whether the elements $\overline{b}:= b + (a), \ \overline{c} \in \frac{A}{(a)}$ are invertible elements in $\frac{A}{(a)}$ or not. If they are, calculate their inverses.
EDIT: With the help of the answers below I’ve managed to do the following.
1.- A is an Euclidean domain, so it is a PID. Let $N$ be the norm application, so $N(a)=3185=5\cdot 637 = (7+56i)(7-56i)$ therefore $a$ isn’t prime in $A$, as a result $(a)$ isn’t a prime ideal in $A$. This implies $(a)$ isn’t a maximal ideal in the set of principal ideals of $A$, but $A$ is a PDI so $(a)$ isn’t maximal in $A$ so $\frac{A}{(a)}$ isn’t a field.
For the second part, we can use the Chinese Remainder Theorem, if we can express $(a)$ as the intersection of three coprime ideals (i.e I.J coprime ideals iff $I+J=A$) we can make use of it.
$a=7(1+8i)$ and $N(1+8i)=65=5 \cdot 13 = (2+i)(2-i)13$ and one can easily check that $(1+8i)/(2+i)=2+3i$. That gives us $a=7(2+i)(2+3i)$ where all of them are prime in $A$ therefore their ideals are prime too. (I haven’t managed to see if they are coprime)
We also know that $(n) \cap (m)=(lcm(m,n))$ and as $a=7(2+i)(2+3i)$ we have $(a)=(7) \cap(2+i) \cap(2+3i)$ but all of them are prime, so that intersection equals the product of all of them.
Now by the Chinese Remainder Theorem we get that there exists an isomorphism
$$ f: \frac{A}{(a)} \longrightarrow \frac{A}{(7)} \times \frac{A}{(2+i)} \times \frac{A}{(2+3i)} $$ where those quotient rings are fields since their ideals are prime which implies maximal.
2.- I’m still stuck on this one, but I’ve managed to do something: To see if $\overline{b}$ is invertirle we have to find $\overline{t}$ such that $\overline{b} \overline{t}= \overline{1}$ which translates into $\overline{1-bt}=0 \Rightarrow 1-bt \in (a) \Rightarrow \lambda a =1-bt \Rightarrow 1=\lambda a +bt $ and this last expression I know it’s a Bézout Identity but I don’t know how to work with it in $\mathbb{Z}[i]$.
Note that $a=7+56i=7(1+8i)$. Note also that $|1+8i|^2=65=5\cdot13$. Try $$ \frac{1+8i}{2+i} $$ and you'll have a full factorization of $a$. The rest should follow easily.