= Solution
Any extending <field homomorphism> $j$ is determined by $j(\alpha)$, since every element of $K(\alpha)$ is represented by a <polynomial> in $\alpha$ with coefficients in $K$. Applying $j$ to the minimal-<polynomial> equation shows $i(m_\alpha)(j(\alpha))=0$. Thus there are at most $m$ choices.
For each listed root $\beta_r$, define $K[x]\to L$ by $\sum a_kx^k\mapsto\sum i(a_k)\beta_r^k$. Its kernel contains $(m_\alpha)$, so it induces a unital map from the <field> $K[x]/(m_\alpha)\cong K(\alpha)$ into $L$. A unital map from a <field> is injective. This gives an extending homomorphism $j_r$ with $j_r(\alpha)=\beta_r$. Distinct roots give distinct maps, proving \b[exactly $m$ extensions].
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