What are the integers of $\textbf Q(\sqrt 2 + i)$?

Question

Given $\{a, b, c, d, e\} \in \textbf Z$, it seems to me that any number of the form $$a + b i + c \sqrt{-2} + d \sqrt 2 + e (\sqrt 2 + i)$$ should be an algebraic integer in $\textbf Q(\sqrt 2 + i)$. But I doubt this is a complete characterization, there might be cases $\{a, b, c, d, e\}$ can be drawn from $\textbf Q\setminus\textbf Z$, or maybe I have overlooked algebraic integers by which to multiply the integers, maybe both.

The first thing I did was look at some positive integer powers of $\sqrt 2 + i$, like $(\sqrt 2 + i)^5$ (I also looked at negative integer exponents but am not sure what, if anything, I should make of them).

Then, figuring that $1 + \sqrt 2$ must be a unit in this domain, I verified that $$\frac{\sqrt 2 + i}{1 + \sqrt 2} = 2 - i - \sqrt 2 + \sqrt{-2},$$ $$\frac{\sqrt 2}{1 + \sqrt 2} = 2 - \sqrt 2,$$ $$\frac{i}{1 + \sqrt 2} = -i + \sqrt{-2}$$ and $$\frac{\sqrt{-2}}{1 + \sqrt 2} = 2i - \sqrt{-2}.$$

I acknowledge that what I have tried may be insufficient, or altogether on the wrong track. What have I overlooked, or what should I have been doing, to find the algebraic integers of this domain?

Answer 1

The answer has already been given, but if it doesn't seem clear, it's understandable.

For one thing, it has been hinted that your incorrect integral basis can actually be used to find the correct one. Set $$c = d = \frac{1}{2},$$ zero out $a, b, e$, to obtain $$\frac{\sqrt{-2} + \sqrt{2}}{2}.$$ This number is special because it is a root of $x^8 - 1$, meaning it's an eighth root of $1$ (and for that reason frequently denoted as $\zeta_8$). But that's not its minimal polynomial $x^4 + 1$, which means the field is of degree $4$. Furthermore, as you've already discovered, $$\frac{\sqrt{-2} + \sqrt{2}}{2} = \sqrt{i}.$$

A less obvious hint that your basis was incorrect is that the literal coefficients (remember that $a$ has a tacit literal coefficient of $1$) can't be arranged as a sequence of powers. From the LFMDB, you can derive $a + b \sqrt{i} + ci + d (\sqrt{i})^3$ (obviously $(\sqrt{i})^0 = 1$).

I suppose that on an intellectual level, it is satisfying to keep things purely algebraic. But if you're interested in computing norms of numbers and verifying that this is a unique factorization domain, it helps to know the correct integral basis. Though I imagine there might be a still more efficient way to express it for the purpose of computing norms.

Answer 2

Note that $F=\Bbb Q(\sqrt 2+i)\subseteq \Bbb Q(\sqrt 2, i)=\Bbb Q(\zeta_8)=K$ and that indeed it is equal to $K$ because all four non-trivial Galois elements act non-trivially on the field generator, $\sqrt 2+i$, so $F$ cannot be a proper subfield of $K$ by the Fundamental Theorem of Galois Theory. But then since the field is cyclotomic, we know the integer ring is just $\mathcal{O}_K=\Bbb Z[\zeta_8]$.

Note that even from the start you can see your answer cannot be correct (or at least not minimally correct) because you have $5$ elements in a basis, but even without knowing $F=K$: since $F\subseteq K$ we have $[F:\Bbb Q]\le [K:\Bbb Q]=4$ so the integer ring is at most a rank-$4\;$ module over $\Bbb Z$.

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Addendum

Per the op's follow-up question as to whether a different version of his candidates would work, we note that two bases for a finitely-generated, free $\Bbb Z$-module differ by an element of $SL_n(\Bbb Z)$. In our case $n=4$ and his basis options candidate is $\{1, i,\sqrt 2,i\sqrt 2\}$. Let us make a choice of $i$ so that $i=\zeta_8^2$ for simplicity--this is a formality just to avoid specific embeddings into $\Bbb C$, if you want to say $i=e^{i\pi /2}$ and $\zeta_8=e^{i\pi/4}$ it changes nothing.

Then let me reorder the new basis candidate as $\{1, \sqrt 2, i, i\sqrt 2\}$ for ease of the computation. Then using $\zeta_8^{-1}=\zeta_8^7=-\zeta_8^3$ the matrix is determined by

$$\begin{cases} 1\mapsto 1 \\ \zeta_8\mapsto \zeta_8+\zeta_8^{-1}=\zeta_8-\zeta_8^3 \\ \zeta_8^2\mapsto \zeta_8^2 \\ \zeta_8^3\mapsto i\sqrt 2 = \zeta_8^2(\zeta_8+\zeta_8^{-1})=\zeta_8^3+\zeta_8 \end{cases}$$

Then the matrix is given by

$$M=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{pmatrix}$$

which has determinant $2$, so your basis is inadmissible, even removing $e$.

Answer 3

One question you should ask yourself immediately is what the field $\mathbb{Q}(i+\sqrt{2})$ is; did you prove that it is equal to $\mathbb{Q}(i, \sqrt{2})$? If so, you can ignore $i+\sqrt{2}$ and focus on $1, i, \sqrt{2},$ and $\sqrt{-2}$. And yes, you should know that $\sqrt{2}+1$ is a unit; its inverse is $\sqrt{2}-1$.

So now the question is what the integers are in $\mathbb{Q}(i, \sqrt{2})$. Can you show that $\frac{\sqrt{2}+\sqrt{-2}}{2}$ is an integer? If you do that, then I claim that this and $1, i, \sqrt{2}$ span the integers. Do you know any methods to prove that some numbers span the set of integers?