Q:

Discriminant of n algebraic numbers equals $0$ iff the algebraic numbers linearly dependent

Let $K \subset L$ be two number fields with $[L:K] = n$. Let $\{\alpha_i:1 \leq i \leq n\} \subset L$. Then $\operatorname{disc}(\alpha_1 \dots \alpha_n) = 0 \iff \alpha_i$ are linearly dependent over $K$. 

$\operatorname{disc}(\alpha_1 \dots \alpha_n) = \det (\sigma_j(\alpha_i)), 1 \leq i,j \leq n$ where $\sigma_j$ are the $n$ embeddings into $\Bbb C$. 
All the proofs I have seen convert $A=(\sigma_j(\alpha_i))$ into its trace definition and go through that path.  That makes me wonder if there is something wrong with the following proof which seems much simpler:
$\operatorname{Disc}(\alpha_1 \dots \alpha_n) = 0$, therefore the columns of $A$ are linearly dependent and looking along each co ordinate, we get that $\{\sigma_j(\alpha_i), 0 \leq i \leq n\}$ is linearly dependent for all $j$. Since automorphisms commute with addition and scalar multiplication, doesen't this directly imply that the $\alpha_i$ are linearly dependent? 

A:

The only immediate doubt I have is the following: how do we know that the linear dependence relation has scalars that live inside $K$?  For instance, the matrix
$$ \begin{pmatrix} 1 & \pi\\ 1 & \pi \end{pmatrix} $$
has linearly dependent columns over $\mathbb{R}$, but not over any number field $K$.
If you could show that the scalars are actually in $K$, then you would be done.  This is one way of understanding why someone else might use traces, since the trace will always live in the base field $K$.

