 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 $\sum a_k{x}^k\mapsto \sum i(a_k){\beta }_r^k$. 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 exactly $m$ extensions.