= Solution
Put $L_i=K(\alpha_1,\ldots,\alpha_i)$, with $L_0=K$, and let
$$
d_i=[L_i:L_{i-1}].
$$
Fix a $K$-embedding $\sigma:L_{i-1}\to\overline K$. If $m_i(t)$ is the minimal polynomial of $\alpha_i$ over $L_{i-1}$, an extension of $\sigma$ to $L_i$ is determined by the image of $\alpha_i$, which must be a root of the polynomial obtained by applying $\sigma$ to the coefficients of $m_i$. Conversely, each distinct root gives one extension. Since $\overline K$ is algebraically closed, there is at least one such root, and there are at most $d_i$ distinct roots.
Starting from the unique embedding of $K$, induction therefore gives
$$
1\leq |\operatorname{Hom}_K(L_i,\overline K)|
\leq d_1\cdots d_i=[L_i:K].
$$
If $L/K$ is separable, each $m_i$ is separable, so every transformed polynomial has exactly $d_i$ distinct roots. Every inequality is then an equality. Taking $i=n$ proves
$$
1\leq |\operatorname{Hom}_K(L,\overline K)|\leq[L:K],
$$
with equality in the separable case.
Back to article page
