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Consider the ring $\mathbb{Z}[i]$ of Gaußian integers. The principal ideal $(1+i)$ is maximal ideal in this ring. Since ideals are kernels of some homomorphisms, I would like to see a homomorphism from $\mathbb{Z}[i]$ to some ring whose kernel is $(1+i)$.

Of course, it is kernel of the natural homomorphism

$$f:\;\mathbb{Z}[i]\rightarrow \mathbb{Z}[i]\big/(1+i).$$

However, I tried to define some nice map $f$ whose kernel will be $(1+i)$. I tried this one: $f\colon \mathbb{Z}[i]\rightarrow \mathbb{Z}$, $m+ni\mapsto m-n$, but this $f$ is not a homomorphism. Can you help me to modify (or to define another homomorphism) whose kernel is the ideal $(1+i)$.

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    $\begingroup$ $(m+in)\mapsto (m+n)\mod 2\in\mathbb Z/2\mathbb Z$ $\endgroup$
    – user8268
    Jan 31, 2015 at 8:58
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    $\begingroup$ You should try to describe explicitly that quotient ring you mentioned (how many elements does it have, for example?) instead of using user8268's magic formula which is, really, of no help to you. $\endgroup$ Jan 31, 2015 at 9:02

2 Answers 2

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For completeness, I'll do everything from scratch in case future viewers are wondering how you know $(1+i)$ is maximal.

You can note that $(1+i)$ is maximal because it has a prime norm (hence it is irreducible) and $\Bbb Z[i]$ is a Euclidean ring under the norm (hence also a PID), so prime ideals are maximal and so the quotient is a field.

We note that $(1+i)|2$--indeed $(1+i)(1-i)=2$--hence $2\equiv 0\pmod{1+i}$, i.e. the field has characteristic $2$. Indeed, we can see even more, since $2\equiv 0\pmod{1+i}$ we may reduce $m,n\mod 2$ to one of four total choices (in the quotient)

$$\begin{cases} m\equiv n\equiv 0\mod 2 \\ m\equiv n+1\equiv 0\mod 2 \\ m\equiv n+1\equiv 1\mod 2 \\ m\equiv n\equiv 1\mod 2\end{cases}$$

So the field has at most $4$ elements. Noting that also (and even more trivially)

$$1+i\equiv 0\pmod{1+i}$$

we may identify the first and last case as well as the second and third, so the image is the field with two elements, namely $\Bbb Z/2\Bbb Z$. Any homomorphism which sends things of the same parity to $0\mod 2$ must agree with the canonical projection

$$\Bbb Z[i]\to\Bbb Z[i]/(1+i)\cong\Bbb Z/2\Bbb Z$$

so clearly

$$m+ni\mapsto m+n\mod 2$$

is a formula for the homomorphism onto the quotient, and since $1\not\equiv 0\pmod{1+i}$ we have that we may choose the image of $1, i\in \Bbb Z[i]$ to be $1$ in the quotient and the image of $0, 1+i$ to be $0$, and all other cases reduce to these by the earlier $\mod 2$ discussion.

Note: Your choice of $m+ni\mapsto m-n$ is actually not a bad one, the only thing you forgot is that you must take $m,n\mod 2$. We see that

$$1\equiv-1\mod 2\implies m+n\equiv m-n\mod 2$$

so this is just another way of expressing the same homomorphism.

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You can also think of this in the following way:

Look at the composition

$\mathbb Z[i] \to \mathbb Z[i]/(1+i) \cong \mathbb Z[X]/(X^2+1,X+1) = \mathbb Z[X]/(2,X+1) \cong (\mathbb Z/2\mathbb Z[X])/(X+1) \cong \mathbb Z/2\mathbb Z$

The first isomorphism is given by $a+bi \mapsto a+bX$ and the last isomorphism is given by $X \mapsto -1$. Hence the composition is given by $a+bi \mapsto a-b \mod 2$.

The same calculation works for every maximal ideal $(c+di) \subset \mathbb Z[i]$ with $c \neq 0 \neq d$ (this imples $c^2+d^2=p$ for some prime $p$ and we have the equality of ideals $(X^2+1,dX+c)=(p,dX+c)$ in $\mathbb Z[X]$). We get an explicit isomorphism

$\mathbb Z[i]/(c+di) \to \mathbb Z/p\mathbb Z, a+bi \mapsto a-bcd^{-1} \mod p$

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