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Let $K$ be a number field of degree $n$, $\sigma_1 , \ldots , \sigma_n$ be the distinct embeddings of $K$ into $\mathbb C$, and define $\Delta(x_1, \ldots , x_n) = \det(\sigma_i (x_j)^2) = \det(\mathrm{Tr}_{K/\mathbb Q} (x_i x_j) )$.

Apparently, if $x_i' = \sum_{j=1}^n a_{ij} x_j$, where $a_{ij} \in \mathbb Q$ and $A = a_{ij}$, then $\Delta(x_1', \ldots , x_n ') = (\det A)^2 \Delta(x_1, \ldots ,x_n)$

I can't see why this is true (though I think it should be fairly easy). I'd appreciate if someone could explain.

Thanks

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2 Answers 2

up vote 3 down vote accepted

This exercise boils down to plug-and-chug and careful inspection. You need to use the linearity of the number field trace as well as recognize matrix multiplication in coordinate form. For ease we can set the matrix $\Lambda= \big(\operatorname{Tr}(x_ix_j) \,\big)$. Now observe

$$\Delta(x_1',\cdots,x_n')=\det\big(\operatorname{Tr}(x_i'x_j')\big)=\det\left(\operatorname{Tr}\left[\left(\sum_{l=1}^n a_{il}x_l\right)\left(\sum_{k=1}^n a_{jk}x_k\right)\right]\right)$$

$$=\det\left(\operatorname{Tr}\left(\sum_{l=1}^n\sum_{k=1}^n a_{il}a_{jk}x_lx_k\right)\right)=\det\left(\sum_{l=1}^n\sum_{k=1}^n a_{il}a_{jk}\operatorname{Tr}(x_lx_k)\right)=\det\left(A\Lambda A^T\,\right)$$

$$=(\det A)^2\Delta(x_1,\cdots,x_n).$$

$$ $$

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Here's an alternate method to avoid excessive indices :

Let $S,S',A$ be the matrices $(S)_{ij}=\sigma_i(x_j)$, $(S')_{ij}= \sigma_i(x_j')$ and $A _{ij} = a_{ij}$. Then:

$$(S')_{ij} = \sigma_i(x_j') = \sigma_i(\sum a_{jk}x_k)=\sum a_{jk}\sigma_i(x_k)=(SA^T)_{ij}$$

And the result follows.

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