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I am trying to prove that $\mathbb Z[i]/ \langle 1+2i \rangle$ is isomorphic to $\mathbb Z_5$.

The only thing that came to my mind was trying to apply the first isomorphism theorem using an appropiate function. If I consider the euclidean function $N: Z[i] \setminus \{0 \} \to \mathbb N$ defined as $N(a+bi)=a^2+b^2$, then I can express any element $z$ in $\mathbb Z[i]$ as $z=(1+2i)q+r$ with $N(r)=1,2,3,4$ or $r=0$.

If I define $f:\mathbb Z[i] \to \mathbb Z_5$ as $f(x=(1+2i)q_x+r_x)=\overline{r_x}$, then it is clear that $f(x)=0$ if and only if $x \in \langle 1+2i \rangle$. The problem with this function is that it doesn't satisfy $f(x+y)=f(x)+f(y)$ and it is not surjective for if $r_x=f(x)=3 (5)$, then if $r_x=a+bi$, we have $a^2+b^2=3$, which is absurd.

I don't know what else to do, any suggestions would be appreciated.

user156441
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4 Answers4

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One more proof. It turns out that in this case, it is easier to obtain the isomorphism by going the other direction.

Let $f: \Bbb{Z}\to \Bbb{Z}[i]/(1+2i)$ be the natural mapping, i.e. let $n$ map to $\overline{n}$ where $\overline{n}$ is the residue of $n$ as a gaussian integer mod $1+2i$. Now what is the image of this mapping? Note that $(1+2i)=(-2+i)$ since $i(1+2i)=-2+i$. Therefore $a+bi-b(-2+i)=a+2b\in \Bbb{Z}$ so $a+bi \equiv a+2b \pmod{-2+i}$. In other words, every element of $\Bbb{Z}[i]/(1+2i)$ is the image of some integer under $f$, since $f(a+2b)= a+bi + (1+2i)$.

Now we just need to show $f$ has kernel $(5)$ and we will be done since then by the first isomorphism theorem, $\Bbb{Z}/(5)\cong \Bbb{Z}[i]/(1+2i)$. But of course if $n\in \ker f$, then $1+2i \mid n$, so $1-2i$ also divides $n$, and since $1-2i$ and $1+2i$ are relatively prime, $(1+2i)(1-2i)=5 \mid n$ so $n\in (5)$. Since $1+2i \mid 5$, it is clear that $5\in\ker f$. Therefore $\ker f=(5)$ and we are done.

In general the same proof goes through for any Gaussian prime that is not an integer or $1+i$ since the argument that shows $1+i \mid n$ implies $2 \mid n$ fails here. Although the proof still goes through if you note that if $1+i \mid n$, then $n=(1+i)(a+bi)=a-b+(a+b)i$, and $b=-a$ so that $n=(1+i)(a-ai)=(1+i)(1-i)a=2a$.

jgon
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Since $$ \Bbb{Z}[i] \cong \Bbb{Z}[x] / \langle x^2 + 1 \rangle, $$ you have the isomorphism $$ R = \Bbb{Z}[i] / \langle 1 + 2i \rangle \cong \Bbb{Z}[x] / \langle 1 + x^2, 1 + 2x \rangle. $$

In order to construct a ring homomorphism $\varphi:R \to \Bbb{Z}_5$, it suffices to describe what $x$ maps to. Say $\varphi(x) = y \in \Bbb{Z}_5$. In order for $\varphi$ to be a ring homomorphism, $y$ must satisfy $$ 1 + y^2 = 0 \quad\text{and}\quad 1 + 2y = 0. $$

Then, you need to check that this map is both injective and surjective to verify that it's an isomorphism.

Sammy Black
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Sometimes it is better to give elementary proofs to elementary problems, and a constructive, explicit proof gives us the feeling that we are in control. Therefore, if we denote by $\widehat {x + y \Bbb i}$ elements of $\Bbb Z [\Bbb i] / (1 + 2 \Bbb i)$ and by $\bar n$ elements of $\Bbb Z _5$, it is really easy to check that the map sending $\widehat {x + y \Bbb i}$ to $\overline {x + 2y}$ is your desired isomorphism. This definition is suggested by the fact that in $\Bbb Z [\Bbb i] / (1 + 2 \Bbb i)$ you have $\hat 1 = \widehat {- 2 \Bbb i}$, so $\widehat {a + b \Bbb i} = \widehat {a + b \Bbb i (- 2 \Bbb i)} = \widehat {a + 2b}$.

We can be even more explicit: let us see what element is mapped where. In your notations, if $N(x) \ge 5$ then $\widehat x = \widehat {r_x}$, so we need only study the elements with $N(x) < 5$ which are easily shown to be $0, 1, \Bbb i, 1 + \Bbb i, -1 + \Bbb i, 2, 2 \Bbb i$ and their additive inverses. The images in $\Bbb Z [\Bbb i] / (1 + 2 \Bbb i)$ of these elements are $0, 1, 2, 3, 1, 2, 4$ (and their additive inverses modulo $5$, respectively), which are immediately mapped to the corresponding elements of $\Bbb Z _5$.

Alex M.
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Another approach

$R=\mathbb Z[i]/ \langle 1+2i \rangle$= $\{a+bi+\langle 1+2i \rangle |\ a,b \in \Bbb{Z}\}$.

Now we have $\overline{1+2i}=\bar0$ so we can say $1+2i=0$ and reduce our number of elements using this relation.

Now we can eliminate $i $ term from $a+bi$ by using $2i=-1$ (if $b$ is even)
OR
we can eliminate $i$ from $a+bi$ by using $8i=1$ and thus $bi=bi.1=bi.8i=-8b$ (if $b$ is odd)

So now all elements of $R=\{c+\langle 1+2i \rangle |\ c \in \Bbb{Z}\}$.

But $\langle 1+2i \rangle$ is the zero element in $R$, so $1+2i=0 \implies 2i=-1 \implies -4=1 \implies 5=0.$

So $c=5q+r$ where $r \in \Bbb{Z_5}$ makes $R=\{r+\langle 1+2i \rangle |\ r \in \Bbb{Z_5}\} \cong \Bbb{Z_5}$

Bhaskar Vashishth
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