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Apologies. I was remembering more than I was thinking: you are right in that the relation doesn't hold at the level of elements[0]. The correct statement is

  (2) = (1+i)^2
where (2) and (1+i) represent the ideals generated by 2 and 1+i respectively. Officially, an ideal of a ring R is a subset I of R which is closed under addition and under and multiplication by elements of R:

  i, j ∈ I implies  i + j ∈ I
  i    ∈ I implies     ri ∈ I for all r ∈ R
For our purposes, a simpler case -- that of principal ideals -- suffices, and that's what I'll discuss in the rest of this comment.

The ideal generated by an element a of a ring R, usually denoted (a) or aR, is essentially the subset of all multiples of a in R. For instance, over the integers, we have ideals like

  (2) = {...,  -4, -2, 0, 2, 4,  ...} 
  (3) = {...,  -6, -3, 0, 3, 6,  ...}
  (4) = {...,  -8, -4, 0, 4, 8,  ...}
  (6) = {..., -12, -6, 0, 6, 12, ...}
One can also define the ideal generated by the elements a,b,c,... of R:

  (a,b,c,...) = {aa' + bb' + cc' + ... : a',b',c' ∈ R}
i.e. the "linear combinations" of a, b, c, and so on.

Here are a few things you might like to verify (i.e. just nod along) about some ideals of Z, to get familiar with the notation:

  (1) = Z (i.e. all the integers)
  (2) = (-2)
  (6) = (2) ∩ (3) 
  15Z =  3Z ∩ 5Z
where ∩ is the intersection of the two sets, or the set of common elements. ("Fact": The intersection of two ideals is also an ideal.)

In general, if you have an ideal of the form (a), then we always have

  (a) = (ua) for all units u
Write G (for Gauss?) for Z[i]. In G, the previous statement means that (2) = (2i). To check this, note that we have

      x ∈ (2) 
  iff x =  2y            for some y
  iff x = (2i)(-yi)
  iff x ∈ (2i)
so all elements of (2) are in (2i), and vice versa.

Now one defines the product of ideals: given ideals I and J in some ring R,

  IJ = ideal generated by the elements {ij : i ∈ I, j ∈ J}
In our case, if I = (a) and J = (b), IJ is just (ab). Question: what's the relation between (ab) and (a) ∩ (b)?

For instance, you can verify that

  (2)(3) = (6)
and, to bring us back to our original point,

  (1+i)(1-i) = (2i) = (2) = (1+i)^2
where all the products are ideal products.

--

There is a very incomplete treatment at [1] (with a bit of the LaTeX broken) that is meant to provide background for a sequence of posts on algebraic geometry -- it's been suspended for a while, but I'm going to look over a couple of drafts this week!

[1]: https://mrkgnao.github.io/prime-ideals/

[0]: As always in math: if you can't prove it, just add a few adjectives or simply change the language. Facts arrived at by alternative definitions, if you will. :)




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