= Solution
Suppose otherwise and choose a relation
$$
\lambda_1\sigma_1(x)+\cdots+\lambda_m\sigma_m(x)=0
\qquad(x\in K)
$$
with the least possible number $m$ of nonzero coefficients. After relabelling and scaling, take every $\lambda_i\ne0$ and $\lambda_m=1$. Since the <field embeddings> $\sigma_1$ and $\sigma_m$ are distinct, there is a $y\in K$ such that $\sigma_1(y)\ne\sigma_m(y)$.
Apply the relation to $xy$ and subtract $\sigma_m(y)$ times the relation at $x$. Because each <field embedding> is multiplicative, this gives
$$
\sum_{i=1}^{m-1}
\lambda_i\bigl(\sigma_i(y)-\sigma_m(y)\bigr)\sigma_i(x)=0
\qquad(x\in K).
$$
The $m$th term has disappeared, while the coefficient of $\sigma_1$ is nonzero. This is a shorter nontrivial relation, contradicting minimality. Hence distinct embeddings are linearly independent as functions. This is the <Artin independence theorem>.
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