Help with proof that $I = \langle 2 + 2i \rangle$ is not a prime ideal of $Z[i]$ (Note: $Z[i] = \{a + bi\ |\ a,b\in Z \}$)
This is what I have so far. 
Proof:
If $I$ is a prime ideal of $Z[i]$ then $Z[i]/I$ must also be an integral domain.
Now (I think this next step is right, I'm not sure though), 
$$
Z[i]/I = \{a+bi + \langle 2 + 2i \rangle\ | a,b,\in Z \}.
$$ 
So, let $a=b=2$, then we can see that 
$$
(2+2i) \cdot \langle 2 + 2i \rangle = \langle 4 - 4 \rangle = 0
$$
Thus, $Z[i]/I$ has a zero-divisor. Thus, $Z[i]/I$ is not an integral domain which means that $I$ is not a prime ideal of $Z[i]$.
$\square$
Now if my proof is right, am I right to think that $(2+2i) \cdot \langle 2 + 2i \rangle $ represents the following:
$$
(2+2i) \cdot \langle 2 + 2i \rangle = (2+2i) \cdot \{2 +2i^2, 4 + 4i^2, 6 +6i^2, 8 + 8i^2, \ldots \} = \{0, 0, 0, 0, \ldots\} = 0?
$$

A much simpler argument:
Note that $2,1+i\notin \langle 2+2i\rangle$, yet $2\cdot (1+i)=2+2i\in \langle 2+2i\rangle$.
