 Solution

Suppose otherwise and choose a relationwith the least possible number $m$ of nonzero coefficients. After relabelling and scaling, take every ${\lambda }_i\ne 0$ 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 givesThe $m$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.