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 .
Need some clarifications: "Show that the ideal (2(1 + i)) of Z(i) has for Z-basis 4,2(1 + i)"
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blamocur
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I am guessing that [imath]u,v \in I[/imath] are Z-basis of ideal [imath]I[/imath] iff any element of [imath]I[/imath] can be expressed as [imath]au+bv[/imath] for some [imath]a,b\in \mathbb Z[/imath].
The next step is to show that any element of the ideal can be represented as [imath]4a+2(1+i)b[/imath] where [imath]a,b \in \mathbb Z[/imath]. I.e., that the set [imath]4a+2(1+i)b[/imath] is not only a subset of the ideal, but covers all of it.
You proved that 4 belongs to the ideal. Thus any combination [imath]4a + 2(1+i)b[/imath] (where [imath]a,b\in \mathbb Z[/imath]) belongs to the ideal.
The next step is to show that any element of the ideal can be represented as [imath]4a+2(1+i)b[/imath] where [imath]a,b \in \mathbb Z[/imath]. I.e., that the set [imath]4a+2(1+i)b[/imath] is not only a subset of the ideal, but covers all of it.