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Let $L/K/\mathbb{Q}$ be a tower of number fields.

If $e_1,\dots,e_n$ and $m_1,\dots,m_k$ are $\mathbb{Z}$-bases of the ring of integers of $L$ and $K$, then $$d_L=\det(\text{tr}(e_ie_j)), \ d_K=\det(\text{tr}(m_im_j)).$$ We can write $m_i=\sum a_{ik} e_k$ for some $a_{ij}\in\mathbb{Z}$, so $$d_K=\det(\text{tr}(m_im_j))=\det\left(\sum a_{ik}a_{jl} \text{tr}(e_ke_l)\right)=\det(a_{ij})^2 d_L.$$ But this must be wrong: consider a cyclotomic field, whose discriminant has a square factor. It contains (by Galois theory) a quadratic field, whose discriminant is squarefree when it is $\mathbb{Q}(\sqrt{d})$ with $d\equiv 1\pmod{4}$)

Where is the mistake ?

Klaus
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1 Answers1

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(Reposted as an answer as suggested by @QiL'8) (EDIT: This is wrong, see below)

The reason your argument is a little fishy is that the trace map for the extension $K / \mathbf{Q}$ is not the same as the restriction to $K$ of the trace for $L / \mathbf{Q}$; in fact the two differ by a factor of $[L : K]$. The formula that actually comes out is

$$ [L : K]^2 d_K = \det(a_{ij})^2 d_L $$

which is no contradiction.

EDIT. Revisiting this 2 years later prompted by A.P.'s comments, it seems that this is wrong as well. On closer inspection, the manipulation in the question is meaningless, because the matrix $A = (a_{ij})$ is not a square matrix, so $\det(A)$ does not exist.

David Loeffler
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  • Are you sure about this? We have $\mathfrak{d}{L/\Bbb{Q}} = \mathfrak{d}{K/\Bbb{Q}}^{[L:K]} N_{K/\Bbb{Q}}(\mathfrak{d}{L/K})$ where $\mathfrak{d}{A/B}$ is the discriminant ideal of the extension $A/B$. Since $\mathfrak{d}{L/\Bbb{Q}}=(d_L)$ and $\mathfrak{d}{K/\Bbb{Q}}=(d_K)$, doesn't this imply that $d_K^{[L:K]} \mid d_L$ (which doesn't seem to follow from your formula)? – A.P. May 15 '15 at 11:20
  • I'm not convinced. By definition $\mathfrak{d}{K/\Bbb{Q}}$ is an ideal of $\Bbb{Z}$, thus principal, say generated by $g$. Now, $N{K/\Bbb{Q}}$ sends ideals of $\mathcal{O}K$ in ideals of $\Bbb{Z}$, so $N{K/\Bbb{Q}}(\mathfrak{d}{K/\Bbb{Q}})$ looks like an abuse of notation. Even ignoring this, $N{K/\Bbb{Q}}(g\Bbb{Z}) = \left( N_{K/\Bbb{Q}}(g\mathcal{O}K) = \right) N{K/\Bbb{Q}}(g) \Bbb{Z} = g^{[K:\Bbb{Q}]} \Bbb{Z}$. Then your formula implies that $d_K = | g^{[K:\Bbb{Q}]} |$. What am I missing here? – A.P. May 15 '15 at 17:41
  • Ah, that's the reason of the misunderstanding! I'm using the notation of Neukirch's Algebraic Number Theory, where $\mathfrak{d}{K/\Bbb{Q}}$ denotes the discriminant ideal of $\mathcal{O}_K$ over $\Bbb{Z}$, not the different (which Neukirch denotes $\mathfrak{D}{K\Bbb{Q}}$). – A.P. May 16 '15 at 12:39
  • In particular, for a finite separable field extension $F/G$, $\mathfrak{d}{F/G}$ is defined as the ideal of $\mathcal{O}_G$ generated by the discriminants $d(\alpha_1,\dotsc,\alpha_n) = \det(Tr(\alpha_i\alpha_j))$ for ${\alpha_1,\dotsc,\alpha_n}$ ranging over all the bases of $F$ over $G$ with elements in $\mathcal{O}_F$. As you suggest, it can then be proved (Neukirch's theorem III.2.9) that $\mathfrak{d}{F/G} = N_{F/G}(\mathfrak{D}_{F/G})$. – A.P. May 16 '15 at 12:48
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    You are right, this was wrong. I've edited my answer. – David Loeffler May 16 '15 at 15:22