Show that $a^2 \equiv a \mod (1+i)$ for all $a \in \mathbb{Z}[i]$. This is a problem statement from a section on quotient rings in Abstract Algebra, so I assume it requires the use of the FIT for rings. When looking around for similar problems, I was only able to find examples with number theory, which isn't really what I'm looking for.
My main question here is: what does $\mathbb{Z}[i]$ mean? I'm familiar with the notation $\mathbb{Z}(i) = \{a+bi : a,b\in\mathbb{Z}\}$ and $\mathbb{Z}[X]$ (a polynomial ring), but never have I seen $\mathbb{Z}[i]$.
I think that I'll be able to figure this out knowing that, but I still would very much appreciate a hint on how to start. I assume you can just show that $a^2 - a \equiv a(a - 1) \equiv 0 \mod (1 + i)$, but I'm not really sure where to go from there.
 A: Regarding the notation, $\Bbb Z [\Bbb i]$ is $\{a + b\Bbb i \mid a, b \in \Bbb Z\}$, i.e. exactly what you incorrectly call $\Bbb Z (\Bbb i)$.
Regarding the problem itself, since you want to use the fundamental isomorphism theorem, I suggest the following approach: find the quotient ring $\Bbb Z [\Bbb i] / (1 + \Bbb i)$ explicitly. To do this notice that taking the quotient by the ideal $(1 + \Bbb i)$ means, among others (and using hats to denote elements in the quotient), that in the quotient ring we shall have $\widehat {1 + \Bbb i} = \hat 0$, i.e. $\widehat {a + b\Bbb i} = \widehat {a-b}$. This suggest us to consider the map $f : \Bbb Z [\Bbb i] \to \Bbb Z_2$ given by $f (a + b \Bbb i) = (a - b) \pmod 2 = \overline {a-b}$, where the overline denotes classes in $\Bbb Z_2$.
Notice that $f (0 + 0 \Bbb i) = \bar 0$ and $f (1 + 0 \Bbb i) = \bar 1$, so $f$ is surjective.
To see that $f$ is a ring morphism, notice that
$$f \big( (a + b \Bbb i) + (c + d \Bbb i) \big) = \overline {(a + c) - (b + d)} = \overline {a-b} + \overline {c-d} = f \big( (a + b \Bbb i) \big) + f \big( (c + d \Bbb i) \big)$$
and that (remembering that $x-y = x+y$ in $\Bbb Z_2$)
$$f \big( (a + b \Bbb i)  (c + d \Bbb i) \big) = f \big( (ac - bd) + (ad + bc) \Bbb i \big) = \overline {(ac - bd) - (ad + bc)} = \\
\overline {(ac + bd) - (- ad - bc)} = \overline {(a-b) (c-d)} = f \big( (a + b \Bbb i) \big) f \big( (c + d \Bbb i) \big) ,$$
which, corroborated with $f(0) = \bar 0$ and $f(1) = \bar 1$, show that $f$ is indeed a surjective ring morphism.
It remains to find out its kernel. Notice that
$$f( a + b \Bbb i) = \bar 0 \iff \overline{a-b} = \bar 0 \iff \exists k \in \Bbb Z \text{ such that } a = b + 2k = b + (1 + \Bbb i) (1 - \Bbb i) k \\
\iff \exists k \in \Bbb Z \text{ such that } a + b \Bbb i = b + (1 + \Bbb i) (1 - \Bbb i) k + b \Bbb i = (1 + \Bbb i) [b + (1 - \Bbb i) k] \iff \\
a + b \Bbb i \in (1+i) ,$$
which shows that $\ker f = (1+\Bbb i)$. Applying the fundamental isomorphism theorem gives then $\Bbb Z[\Bbb i] / (1 + \Bbb i) \simeq \Bbb Z_2$ and since in $\Bbb Z_2$ we obviously have $x^2 = x$, the conclusion follows.
A: From the question What does $\mathbb{Z}[i]$ means? the answer can be They're the Gaussian integers, that are the elements of the form $a+bi$ with $a,b$ integers. They're interesting for a lot of things, for example in geometry and if you notice, all the prime numbers of structure $4k+1$ in $\mathbb{Z}$ become not prime.
From what you have, you need to see that if $a=m+in$, 
$$a^2-a=m^2-n^2+2imn-m-in= m^2-n^2+2imn--m-in$$
Notice that $2imn=(1+i)(1-i)imn$ is congruent to zero, so you only need to see if $m^2-n^2-in\equiv 0$. 
Then your congruence becomes 
$$a^2\equiv a\ \bmod 1+i\iff m^2-n^2-(i)n \equiv m^2-n^2-(-1)n-m \bmod 1+i $$
$$m^2-n^2-(-1)n-m\equiv (m+n)(m-n)-(m-n)$$$$\equiv (m+n-1)(m-n) \bmod 1+i$$
So you only need to see which between $m-n$ and $m+n-1$ is even (in the integers sense) to use the factorization
$$x=\frac{x}{2}(i+1)(1-i)$$
To finally see that 
$$(m-n)(m+n-1)\equiv 0 \bmod 1+i $$
A: Note that $ 2 = (1+i)(1-i) \equiv 0 \pmod{1+i} $ and any $ a + bi \in \mathbf Z[i] $ with $ a, b \in \mathbf Z $ is congruent to an integer modulo $ 1 + i $, thus any element of $ \mathbf Z[i] $ is congruent to either $ 0 $ or $ 1 $ modulo $ 1+i $.
A: I will be using the following result.
RESULT: Let $z\in\mathbb{Z}[i]$. Then $N(z)$ (read as norm of $z$) is even $\iff z\equiv 0\mod(1+i)$.
My answer to the question is as follows:
Let $a=m+in$. Then $$a^2-a=(m^2-n^2-m)+(2m-n)i.$$ We get
$$N(a^2-a)=(m^2-n^2-m)^2+(2m-1)^2.$$
Enumerating all possible cases for the integers $m,n$, we can show that $N(a^2-a)$ is even. Using the above result, the posted question holds.
A: By the Third Isomorphism Theorem,
$$
\frac{\mathbb{Z}[i]}{(1+i)} \cong \frac{\mathbb{Z}[x]}{(1+x, x^2 +1)} \cong \frac{\mathbb{Z}}{((-1)^2 +1)} = \frac{\mathbb{Z}}{(2)} \, .
$$
By Fermat's Little Theorem, every element $a \in \mathbb{Z}/2\mathbb{Z}$ satisfies $a^2 = a$.
