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I would like to show directly that $C(K)$ is trivial, where $K = \mathbb{Q}(\sqrt[4]{-2})$. Write $\delta = \sqrt[4]{-2}$. It is pretty easy to see that $\mathcal{O}_K = \mathbb{Z}[\delta] = R$. Then $\Delta_K = 2048$, and the Minkowski bound is $$ \frac{4!}{4^4}\left(\frac{4}{\pi}\right)^2\sqrt{2048}\approx 6.88,$$ so we need only check primes lying over $2$, $3$, $5$.

Over $2$, since $\delta R = \mathfrak{B}_2$ has norm $2$, it is prime; since $\mathfrak{B}_2^4 = (2)$, this is the only prime lying over $2$, so $[\mathfrak{B}_2]$ is trivial in $C(K)$.

Over $5$, since $x^4+2$ remains irreducible mod $5$, it follows that $5R = \mathfrak{B}_5$ is prime, and is the only prime lying over $5$; it too is trivial in $C(K)$.

Over $3$, $x^4+2 = (x+1)(x+2)(x^2+1)$, so that $3R = \mathfrak{B}_3\mathfrak{B}'_3\mathfrak{B}''_3$, where $N(\mathfrak{B}_3) = N(\mathfrak{B}'_3) = 3$ and $N(\mathfrak{B}''_3)=9$.

This is where I'm not clear on how to proceed. It doesn't seem obvious that each of these is principal. Am I going about this in a reasonable way? What other tools are available to compute such groups "from first principles"?

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  • $\begingroup$ So we now have an explicit description, say $(3, \sqrt[4]{2}+1)$. It's not clear to me how that helps. $\endgroup$
    – rogerl
    Jul 24, 2014 at 21:43
  • $\begingroup$ @AdamHughes, there’s something wrong with your analysis, since $\sqrt2+1$ is a unit of $\mathbb Z[\sqrt2\,]$ $\endgroup$
    – Lubin
    Jul 24, 2014 at 21:50
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    $\begingroup$ You use $\mathbb{Q}(\sqrt[4]{2})$ in your title and question, but then talk about $x^4+2$ in the bulk of your question; shouldn't you be looking at $x^4-2$? $\endgroup$ Jul 24, 2014 at 21:53
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    $\begingroup$ Whoops, perhaps someone should point out that the appropriate polynomial is $X^4-2$, not $X^4+2$ $\endgroup$
    – Lubin
    Jul 24, 2014 at 21:53
  • $\begingroup$ @Lubin yes, I was fooled by the wrong polynomial showing up, I'm typing up a fixed version now, thanks for the note. $\endgroup$ Jul 24, 2014 at 21:55

1 Answer 1

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Let's just do it all the way. Modulo $3$ we get a factorization $x^4-1=(x^2+1)(x-1)(x+1)$ which gives you three ideals, one of norm $9$ and two of norm $3$:

$$\begin{cases}\mathfrak{B}_3 = (3,\sqrt[4]{-2}+1) \\ \mathfrak{B}_3' = (3, \sqrt[4]{-2}-1) \\ \mathfrak{B}_3''= (3, 1+\sqrt{-2})\end{cases}$$

which we know works because the ring $\Bbb Z[\sqrt[4]{-2}]$ has trivial conductor in $\mathcal{O}_K$ as you already note.

After computing norms in your favorite fashion of the elements which are not $3$, we see that $N(1+i\sqrt{2})=9$, hence it generates that ideal because clearly $(3,\alpha)|(\alpha)$ is always true, and if the norms agree then the two ideals are equal. Similarly for the other two, we get their minimal polynomial easily from that of $\sqrt[4]{-2}$ by shifting and it is easy to see that the constant term in both $(x-1)^4+2$ and $(x+1)^4+2$ is $3$, hence the norms, $N(\sqrt[4]{-2}+1),N(\sqrt[4]{-2}-1)$ are both $3$ (norms down to $\Bbb Q$ of an algebraic number are always the constant term in the minimal polynomial), hence those elements generate the ideals $\mathfrak{B}_3, \mathfrak{B}_3'$.

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