Need some clarifications: "Show that the ideal (2(1 + i)) of Z(i) has for Z-basis 4,2(1 + i)"

[Thanos_10]

The question is as follows : "Show that the ideal (2(1 + i)) of Z(i) has for Z-basis 4,2(1 + i)". It is clear to me that i can obtain 4 by multiplicating 2(1+i) by (1-i), but if 4 can be obtained that way, what is the purpose of that specific base. I clearly am missing something, since i do not know,exactly, what a Z-basis is. Any help would be appreciated .

 

[blamocur]

I am guessing that $u,v \in I$ are Z-basis of ideal $I$ iff any element of $I$ can be expressed as $au+bv$ for some $a,b\in \mathbb Z$.

It is clear to me that i can obtain 4 by multiplicating 2(1+i) by (1-i),

You proved that 4 belongs to the ideal. Thus any combination $4a + 2(1+i)b$ (where $a,b\in \mathbb Z$) belongs to the ideal.
The next step is to show that any element of the ideal can be represented as $4a+2(1+i)b$ where $a,b \in \mathbb Z$. I.e., that the set $4a+2(1+i)b$ is not only a subset of the ideal, but covers all of it.