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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 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.2012-01-25

1 Answers 1

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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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    I see, thank you. By the way, would you mind if I could see your Mathematica code? If not, I understand.2012-01-26