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 $\mathbb{Q}(i)$?

  • 2
    $\begingroup$ 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. $\endgroup$
    – Mr. Brooks
    Sep 12, 2019 at 22:50

1 Answer 1


You have to be careful with something like this. It’s a quadratic extension of a quadratic extension, where the situation can be various. There easily could be just two symmetries, for instance.

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.


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