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My lecturer of Algebraic Number Theory was talking about the relative trace, but I have no idea what this means. For example he was talking about $\mathbb{Q}(\sqrt(6),i)$ with the subfield $\mathbb{Q}(i)$ and says that for an integral element $\alpha=p+qi+-r\sqrt(6)-s\sqrt(-6)$ the relative trace of $\alpha$ in $\mathbb{Q}(i)$ is $2p+2qi \in \mathbb{Z}[i]$. How do I calculate this trace and how do I know it is in $\mathbb{Z}[i]$?

TheBeiram
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  • Try this http://math.stackexchange.com/questions/297549/relative-trace-and-algebraic-integers – BLAZE Sep 25 '15 at 09:41

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If $L/K$ is an extension of number fields, and $\alpha \in L$, then there is a $K$-linear map given by $$m_\alpha: x \mapsto \alpha x$$

The relative trace, $\mathrm{Tr}_{L/K}(\alpha)$ is defined to be the regular linear algebraic trace of the linear map $m_\alpha$. Since traces are scalar valued, the trace will lie in $K$.

We can show using linear algebra that $$\mathrm{Tr}_{L/K}(\alpha)=\sum_{\sigma:L\hookrightarrow \mathbb C}\sigma(\alpha)$$ where the sum is over $K$-embeddings into $\mathbb C$. (Think of the values $\sigma(\alpha)$ as other roots of the minimal polynomial of $\alpha$.)

In particular, if $\alpha \in \mathcal O_L$ is an algebraic integer, then $\mathrm{Tr}_{L/K}(\alpha)\in\mathcal O_L\cap K = \mathcal O_K$.

Mathmo123
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  • Ok, I think I understand, but I still don't see how I get to $2p+2qi$ in this specific case. – TheBeiram Sep 25 '15 at 10:20
  • in this case you can work it out manually. You have a basis of $L/K$ given by $1,\sqrt 6$, so you should be able to write down the form of the matrix of $m_\alpha$ with respect to this basis, and then calculate the trace. – Mathmo123 Sep 25 '15 at 10:33
  • Alternatively you can use the second formula and work it out via the embeddings $L\hookrightarrow \mathbb C$ – Mathmo123 Sep 25 '15 at 10:34
  • But if I use the second formula to calculate the relative trace of $\alpha$ in $\mathbb{Q}(i)$, I have two embeddings, the first one sends i to i and the second one i to -i. Then I get that $\sigma_1(\alpha)=\alpha$ and $\sigma_2(\alpha)=p-qi-r\sqrt(6)+is\sqrt(6)$ so the sum gives me $2p-2r\sqrt(6)$ and not $2p+2qi$? – TheBeiram Sep 25 '15 at 12:16
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    Ah I already see, the embeddings are of course not from $i$ but from $\sqrt(6)$, since we extend from $\mathbb{Q}(i)$ to $\mathbb{Q}(i,\sqrt(6)$. Thank you for the help! – TheBeiram Sep 25 '15 at 13:03