# What is the symmetry of $\mathbb{Z}[\sqrt{1 + i}]$?

If we make a plot of the units and primes from among the Gaussian integers, $$\mathbb{Z}[i]$$, we see a fourfold symmetry. For example, $$2 + i$$ and $$1 + 2i$$. I'm tempted to say there is also eightfold symmetry, but I'm not completely sure, mostly because there are four units, not eight.

The field $$\mathbb{Q}(\sqrt{1 + i})$$ is of degree $$4$$ but it has only one intermediate field, $$\mathbb{Q}(i)$$, according to LMFDB. This suggests $$\mathbb{Z}[\sqrt{1 + i}]$$ has at least fourfold symmetry.

The fundamental unit is $$\sqrt{1 + i} - i - (\sqrt{1 + i})^3$$. I've found it very awkward and error-prone to do arithmetic with this number. My calculations suggests that the powers of this unit escape to infinity, like in a real quadratic ring, but my calculations could very easily be wrong. And even if my calculations are correct, I could have misunderstood them.

If we plot the units and primes in the ring of integers of $$\mathbb{Q}(\sqrt{1 + i})$$, will we find it has at least fourfold symmetry like $$\mathbb{Q}(i)$$?

• I think I know what you mean by awkward, error-prone arithmetic. But that's why we've got computers. For example, even in Wolfram Alpha you can do Table[N[(Sqrt[1 + I] - I - Sqrt[1 + I]^3)^n], {n, 0, 19}] and you will readily see that it indeed gets farther and farther away from 0. – Mr. Brooks Sep 12 '19 at 22:50

It seems clear that you haven’t seen any Galois Theory. You have to check whether your extension is normal. For this you find the minimal polynomial of $$\alpha=\sqrt{1+i}$$, and see whether all roots are expressible as polynomial expressions (or rational, but that shouldn’t be necessary here) in $$\alpha$$. Once you do this, you should be able to find the symmetries easily.