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Recently, I was intrigued by the question asking for an easy way to show $\mathbb{Z}[\sqrt[3]{2}]$ is the ring of integers of $\mathbb{Q}(\sqrt[3]{2})$.

I was playing with the approach, trying to avoid a lot of field theory I don't really know. I take $\alpha=a+b\sqrt[3]{2}+c\sqrt[3]{2}$ be to an integral element of $\mathbb{Q}(\sqrt[3]{2})$. Viewing $\mathbb{Q}(\sqrt[3]{2})$ as a $\mathbb{Q}$-vector space with basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$, the action of left-multiplication by $\alpha$ can be represented as the matrix $$\begin{bmatrix} a & 2c & 2b \\ b & a & 2c \\ c & b & a \end{bmatrix}. $$ Now the trace and determinant must then be integers, so $3a\in\mathbb{Z}$ and $a^3+2b^3+4c^3-6abc\in\mathbb{Z}$.

Also, multiplying $\alpha$ by $\sqrt[3]{2}$ or $\sqrt[3]{4}$ is still an integral elements, and the matrices corresponding to multiplication by $\sqrt[3]{2}\alpha$ and $\sqrt[3]{4}$ are $$ \begin{bmatrix} 2c & 2b & 2a \\ a & 2c & 2b \\ b & a & 2c \end{bmatrix}, \qquad \begin{bmatrix} 2b & 2a & 4c \\ 2c & 2b & 2a \\ a & 2c & 2b \end{bmatrix}. $$ So by taking the trace I find $6b,6c\in\mathbb{Z}$ are also integers.

This gives a handful of relations about $a,b,c$. I've been trying to use them to conclude $a,b,c\in\mathbb{Z}$ actually, to prove the claim.

Perhaps my elementary number theory is not very sharp, because I've been struggling to conclude this. Is there some clever way to observe that $a,b,c$ are integers, and thus give a somewhat simple, low-level proof of the claim? Thanks, I would be most grateful to see if this works.

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The determinants that you've given for $\sqrt[3]2\alpha$ and $\sqrt[3]4\alpha$ are, not surprisingly, just $2$ and $4$ times the determinant you gave for $\alpha$, so they don't add much to the equation. –  Thomas Andrews Jan 25 '12 at 2:17
    
If you use traces instead of determinants, you get that $3a$, $6b$ and $6c$ are integers. The hard part is checking that $a$, $b$ and $c$ themselves have to be integers. This is why most of the answers to that question began by observing that all of the difficulty was with the primes $2$ and $3$. –  David Speyer Jan 25 '12 at 3:03
    
@DavidSpeyer I used both, and did realize that $3a,6b,6c$ are integers. Maybe I'm being too optimistic that using the determinant of one matrix will allow one to conclude $a,b,c$ are integers. –  Vika Jan 25 '12 at 3:22
2  
The determinant and the trace are two out of the three nontrivial coefficients of the minimal polynomial of your matrix. For finding the ring of integers, you need all three coefficients. –  franz lemmermeyer Jan 25 '12 at 8:49

1 Answer 1

up vote 5 down vote accepted

Once one adds in using traces, as the OP mentions in comments he has thought of, this happens to work. The matrices for multiplication by $\alpha$, $\sqrt[3]{2} \alpha$ and $\sqrt[3]{4} \alpha$ must all have integer trace, which gives $3a$, $6b$ and $6c \in \mathbb{Z}$. So we can write $(a,b,c) = (i/3, j/6, k/6)$. Plugging into Vika's determinant, we get $$f(i,j,k) := \frac{4 i^3 + j^3 + 2 k^3 - 6 i j k}{108}.$$ Notice that, if $i$, $j$ or $k$ all change by multiples of $18$, the numerator changes by an integer. So we can find out when $f$ is an integer by just running $i$, $j$ and $k$ through the integers $0$ through $17$.

Mathematica does this basically instantly. It turns out that $f$ is only an integer when $3$ divides $i$ and $6$ divides $j$ and $k$. So this does give a complete proof.

There was a request for code. There are surely better ways, but I just did

f[a_,b_,c_]:=a^3+2b^3+4c^3-6a*b*c ;
foo =Flatten[Table[{i,j,k,f[i/3,j/6,k/6]}, {i,0,17}, {j,0,17}, {k,0,17}], 2] ;
bar = Select[foo, IntegerQ[Last[#]]&] ; 

and then looked at bar by hand to see that it consisted of the 54 cases where $3$ divided $i$ and $6$ divided $j$ and $k$.


The reason that this works, from a higher perspective, is that the primes $2$ and $3$ are completely ramified in $\mathbb{Z}[\sqrt[3]{2}]$. Vika's determinant is the norm map. So what we are checking is that, if $\alpha$ is integral away from $2$ and $3$, and $N(\alpha)$ is an integer, then $\alpha$ is an algebraic integer.

Higher level proof: Let $p$ be $2$ or $3$ and let $\mathfrak{p}$ be the unique prime of $\mathbb{Q}(\sqrt[3]{2})$. Let $v_{\mathfrak{p}}$ be the valuation at $\mathfrak{p}$ and $v_p$ the valuation at $p$. By hypothesis, $N(\alpha)$ is an integer, so $v_p(N(\alpha)) \geq 0$. Since $\mathfrak{p}$ is a totally ramified prime, $v_p(N(\alpha)) = v_{\mathfrak{p}}(\alpha)$. We already assumed that $\alpha$ was integral away from $2$ and $3$, so $v_{\mathfrak{q}}(\alpha) \geq 0$ for primes $\mathfrak{q}$ other than the ones over $2$ and $3$. So all the valuations of $\alpha$ are nonnegative and $\alpha$ is an algebraic integer.

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Thank you Professor Speyer. If I can ask one small clarification, how did you notice that $i$, $j$, $k$ changing by multiples of $18$ changes the numerator by an integer? –  Vika Jan 25 '12 at 18:17
    
I thought that multiples of $6$ worked, tried to write down a false proof, and spotted the error :). If I were being methodical, I suppose I would have noticed that it is obvious that multiples of $108$ are fine, and then thought about which divisors of $108$ would still be good enough. –  David Speyer Jan 25 '12 at 18:26
    
Sorry, I guess what I meant is, what do you mean changing $i$, $j$, and $k$ by multiples of $18$? Do you mean adding multiples of 18 to each one? It's just that if I were to plug in say $i+5$, $j+2$ and $k+6$, then this also differs from the numerator by an integer, since I'll still get a product $4i^3+j^3+2k^3-6ijk$ when expanding, plus some other combinations of integers. What exactly does the word "changes" mean here, and how is 18 special to it. –  Vika Jan 26 '12 at 4:13
    
Right, if you replace $(i,j,k)$ by $(i',j',k')$ where $i-i'$, $j-j'$ and $k-k'$ are divisible by $18$, then the numerator changes by an integer. As far as your comment about changing $(i,j,k)$ to $(i+5, j+2, k+6)$, indeed, there are many other changes that also change the numerator by an integer. But I just needed get myself down to a reasonable number of cases to check. Once I got to $18^3$ cases, there was no reason find a way to get fewer. –  David Speyer Jan 26 '12 at 4:39
    
I see, thank you. By the way, would you mind if I could see your Mathematica code? If not, I understand. –  Vika Jan 26 '12 at 4:49

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