Can this be salvaged to give a proof that $\mathbb{Z}[\sqrt[3]{2}]$ is the ring of integers of $\mathbb{Q}(\sqrt[3]{2})$?

Question

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.

Answer 1

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 $k$.

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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.