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?

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    $\begingroup$ By general theory the ring of algebraic integers is a rank four free abelian group, so you should get away with using just four integer parameters, say, $a,b,c,d$. A method for this job is to use the trace and norm to limit the denominators, and then check the remaining cases by brute force. $\endgroup$ Mar 7, 2017 at 22:34
  • $\begingroup$ In addition to my answer, I would also note: $\Bbb Q\cap\Bbb Z=\Bbb Z$ in case you were confused as to that part of my edit. $\endgroup$ Mar 7, 2017 at 22:59
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    $\begingroup$ @Adam That's what he wrote, but I don't think that's what he meant. I think he meant "rational numbers ($\mathbb Q$) that are not integers ($\mathbb Z$)", e.g., $$a = b = c = d = \frac{1}{4}$$ (of course that specific example is not an algebraic integer). $\endgroup$ Mar 8, 2017 at 19:22
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    $\begingroup$ And @David, by now you should have realized $e$ falls somewhere between redundant and unnecessary. If $e \neq 0$, we can adjust $b$ to $b + e$ and $d$ to $d + e$ and then discard $e$. $\endgroup$ Mar 8, 2017 at 19:24
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    $\begingroup$ @RobertSoupe I agree. I have changed things accordingly. $\endgroup$ Mar 8, 2017 at 19:53

3 Answers 3


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 $\Bbb Z$.


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 $e$.

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    $\begingroup$ My gut tells me that this is the right answer. But there are several things I had to look up and have yet to fully understand. First, $\zeta_8 = \sqrt i$, right? And $(\sqrt i)^8 = 1$? And $$\sqrt i = \frac{\sqrt 2 + \sqrt{-2}}{2},$$ so that $$\Re(\sqrt i) = \Im(\sqrt i) = \frac{1}{2}$$ ($\Im(a + bi) = b$ rather than $bi$). I admit, $e(\sqrt 2 + i)$ was a silly mistake on my part. Can we salvage $a, b, c, d,$ even though that might not be efficient? And then would those be drawn from $\textbf Z[\frac{1}{2}]$? $\endgroup$
    – David R.
    Mar 10, 2017 at 22:23
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    $\begingroup$ Hey @DavidR. Yes, $\zeta_8$ is something which squares to $\pm i$. You're a bit off though, first of all there are multiple $8^{th}$ roots of unity, you've chosen a particular one, and for that one $\Re(\zeta_8)=\Im(\zeta_8){1\over\sqrt 2}$ not ${1\over 2}$. In general you should try to avoid explicit embeddings into $\Bbb C$ to keep things purely algebraic. To see if another basis candidate is a valid one you need only see if the matrix $A\in GL_4(\Bbb Z)$ which takes $1, \zeta_8, \zeta_8^2,\zeta_8^3$ to your candidate is invertible, i.e. it is in $SL_4(\Bbb Z)$. I will edit to illustrate. $\endgroup$ Mar 10, 2017 at 22:34
  • $\begingroup$ @DavidR. there you go, an expanded answer. Even without $e$ your candidate basis fails. $\endgroup$ Mar 10, 2017 at 22:45

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?

  • $\begingroup$ This is impossible, the module has rank four. This is about algebraic integers too, you seem to potentially think this is about rational integers. $\endgroup$ Mar 7, 2017 at 22:57
  • $\begingroup$ @AdamHughes Yeah, spanned by $1, i, \sqrt{2}$ and $\frac{\sqrt{2}+\sqrt{-2}}{2}$. What's the problem? $\endgroup$
    – Bob Jones
    Mar 7, 2017 at 22:59
  • $\begingroup$ Oh I see, your pronouns were confusing me, but I see it now: I retract that objection. $\endgroup$ Mar 7, 2017 at 23:01
  • $\begingroup$ @AdamHughes Also, I'm confused by your other objection. When I say that this thing is an integer, how could I be talking about rational integers...? $\endgroup$
    – Bob Jones
    Mar 7, 2017 at 23:02
  • $\begingroup$ I meant the original written objection in its entirety in my retraction. $\endgroup$ Mar 7, 2017 at 23:03

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.

  • $\begingroup$ Hey Bill, a few clarifications: "The" correct one is a bit misleading, there are infinitely many. And since the set given is a spanning set, the op's specific choices aren't particularly special in terms of helping to find a basis--any spanning set would have the same effect by the general theory of vector spaces. One could pick 4-7 numbers from the field at random, and odds are pretty good they would be spanning. I'm also not sure what you mean with the literal coefficients: most number rings don't have power bases, and all separable field extensions do (primitive element theorem). $\endgroup$ Mar 28, 2017 at 1:01

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