Prove $\{x(2+i)\}$ is a maximal ideal of a ring

Question

I am showing that $M=\{x(2+i)\mid x\in R\}$ is a maximal ideal of $R=\{a+bi\mid a,b\in \mathbb{Z}\}$.

$M\ne R$ because $1\notin M$.

I should do this in at least two ways:

first I should show that there is exactly one ideal $N$ (namely: $N=R$) such that $M\subsetneqq N$. I take a generic $a+bi\in R$ and set $a+bi \ne (c+di)(2+i)$ to see which element I can have in $N$.

The equality holds iff $a=5d$ and $b=5d$, so all elements in $M$ are multiples of $5$.

I am trying to find a way to get to $1\in N$; but I can't see how to factorize $(a+bi)(c+di)$ so that $5$ does not divide that. The combinations of $(a+bi)(a-bi)$ and alike I tried have a combination that is divisible by $5$.

Secondly, using a homomorphism theorem. I tried using $f:R\rightarrow R/M$ defined by $f(r)=r+M$ with $r\in R$. $\ker f=\{x\in R\mid f(x)=M\}=\{0\}$ so $f$ is injective. I think $f$ is surjective because $Im\space f=R$, because there is an element $r+M$ for every $r$.

The provided solution goes that in $R/M$, $2+M=-i+M$, I think because $1(2+i)+M$ is the identity and $(2+i)+M=M$ iff $2+M=-i+M$. "The image of any $a+bi$ in $R/M$ is of the form $a-2b+M$"; why? $R/M$ consists of $a+M$ where $a$ is an integer. $(2-i)(2+i)=5\in M$ so "if $a=5k+b$ with $0\leq b < 5$"; how can I say this previous sentence? It seems very arbitrary. Then $a+M=5k+b+M=b+M$, hence isomorphism to $\mathbb{Z}_5$, a field.

EDIT: only using elementary methods; I haven't studied anything other than basic ring theorems (homomorphisms, axioms, quotient) and I should work on this in basic terms before moving on

Answer 1

$\mathbb Z[i]$ is PID. So if the ideal $M=\langle i+2 \rangle$ is not maximal then there is an ideal $I=(a+bi)$ such that $M \subseteq I=(a+bi)$.

Let $m\in M$ then $m = (a+bi)(c+di) \implies 5=N(m) = N((a+bi))N((c+di))=(a^2+b^2)(c^2+d^2)$

So, possible candidates for $N((a+bi))$ is $1$ or $5$. It is not $1$ as it is not unit(if it is a unit then $I$ becomes whole ring.). It can not be also $5$ because that leads to fact that the other norm is $1$ and same reason is valid for it. Contradiction.

Answer 2

Here is a rewriting of what I understand from your "two ways".

• Let $N$ be an ideal such that $M\subsetneqq N$ and let $a+ib$ be an element of $N\setminus M$. Since $a+ib\notin M$, there is no $(c,d)\in\Bbb Z^2$ such that $a+ib=(c+id)(2+i)$ i.e. such that $a=2c-d$ and $b=c+2d$, or equivalently, such that $2a+b=5c$ and $2b-a=5d$. This means that $2a+b$ is not divisible by $5$ or $a-2b$ is not divisible by $5$. Actually, these two conditions are equivalent, hence (considering the first one) there exist integers $e,f$ such that $e(2a+b)=1+5f$, from which we derive: $$\begin{align}2e(a+ib)-1&=2ea-1+2i(1+5f-2ea)\\&=(2ea-1)(1-2i)+10if\\&=(2+i)[-i(2ea-1)+(2-i)2if]\in M\subset N\end{align}$$ and therefore, $1\in N$.
• Let us consider the morphism $f:R\to R/M,\;r\mapsto r+M$ (surjective by definition but clearly not injective). Since $2+i\in M=\ker f$, $$f(a+ib)=f(a)+f(i)f(b)=f(a)+f(-2)f(b)=f(a-2b),$$ which proves that every element of $R/M$ is of the form $f(c)$ for some integer $c$. Moreover, by Euclidean division in $\Bbb Z$, $c=5k+d$ for some integers $k,d$ such that $0\le d<5$, and since $5=(2-i)(2+i)\in M=\ker f$, $$f(c)=f(5)f(k)+f(d)=0f(k)+f(d)=f(d),$$ which proves that $$R/M=\{f(d)\mid d\in\{0,1,2,3,4\}\},$$ whence a surjective morphism $g:\Bbb Z_5\to R/M,(d\bmod 5)\mapsto f(d)$. Since $M\neq R$, the subgroup $\ker g$ cannot be the whole group $\Bbb Z_5$, hence $g$ is an isomorphism and $R/M$ is a field.