I am currently reading about binary cubic forms and cubic number fields (mainly about using binary cubic forms with integer coefficients to parametrize orders in the cubic field) and I thought it might be good to do some computations to understand how this theory works by taking a concrete example.

Given a cubic field I want to find the corresponding binary cubic form associated with it or rather with the ring of integer of $K$ should I say?

Suppose we take $$K=\mathbb{Q}(\sqrt[3]{2})\supset\mathbb{Q}$$

Since $$x^{3}-2\in\mathbb{Q}[x]$$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion with $p=2$ we get $$[K:\mathbb{Q}]=3$$

The ring of integers of $K$ is $$\mathcal{O}_{k}=\mathbb{Z}[\alpha]=\mathbb{Z}[\sqrt[3]{2}]$$ with integral basis $\{1,\alpha,\alpha^2\}$ hence $$\mathbb{Z}[\sqrt[3]{2}]=\{a+b\sqrt[3]{2}+c\sqrt[3]{2^2}|a,b,c\in\mathbb{Z}\}$$

In the next step I would like to find (if possible) the corresponding binary cubic form. Will this form correspond to the field of to the ring?

I follow Belabas and Cohen ''Binary cubic forms and cubic number fields'': Let $K$ be number field defined by a root $\theta$ of the polynomial $x^3+px^2+qx+r$ with $p,q,r \in\mathbb{Z}$ such that there exists an integral basis of the form $(1,\theta,(\theta^2+t\theta+u)/f)$ with $t,u,f\in \mathbb{Z}$ and $f=[\mathbb{Z}_{k}:\mathbb{Z}[\theta]]$ Then we choose $\alpha=\theta$ and $\beta=(\theta^2+t\theta+u)/f)$ we have explicitly: $$\begin{split} F_{B}(x,y)&=((t^3-2t^2p+t(q+p^2)+r-pq)/f^2)x^3\\&+((-3t^2+4tp-(p^2+q))/f))x^2y\\ &+(3t-2p)xy^2-fy^3\end{split}$$

In case of $x^{3}-2$ we have $\begin{cases}q=0\\r=-2\\p=0\end{cases}$

And $f=[\mathbb{Z}_{k}:\mathbb{Z}[\theta]]=1$

Plugging everything in I get something like this $F_{B}(x,y)=-2x^3-y^3$

Does this work? Or is it completely incorrect? $f=[\mathbb{Z}_{k}:\mathbb{Z}[\theta]]$ shouldn't this expression give index form? (i.e binary cubic form?)

Thank you.

  • $\begingroup$ This is not an appropriate site for homeworks. $\endgroup$ – Alex Degtyarev Feb 14 '14 at 8:28
  • 2
    $\begingroup$ This is not a homework. Did you actually read what I wrote? I am studying it on my own and I posted this question on mathstackexchange but nobody there seems to have any suggestions to this problem $\endgroup$ – Heidi Feb 14 '14 at 8:32
  • $\begingroup$ Perhaps wait a little longer, like a few days. At the very least say you posted on the other site and link to it. $\endgroup$ – David Roberts Feb 14 '14 at 8:42
  • $\begingroup$ Here is the link: math.stackexchange.com/questions/674159/…. $\endgroup$ – Dietrich Burde Feb 15 '14 at 16:31

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