**Proof.**
Choose an algebraic closure $\overline{L}$ of $L$ (Theorem 9.10.4).

Let $\tau $ be as in (1). Then $\tau (M) = M$ as subfields of $\overline{L}$ by Lemma 9.15.5 and hence $\tau |_ M : M \to M$ is an automorphism.

Let $\sigma : M \to L$ be as in (2). By Lemma 9.10.5 we can extend $\sigma $ to a map $\tau : L \to \overline{L}$, i.e., such that

\[ \xymatrix{ L \ar[r]_\tau & \overline{L} \\ M \ar[u] \ar[ru]_\sigma & K \ar[l] \ar[u] } \]

is commutative. By Lemma 9.15.5 we see that $\tau (L) = L$. Hence $\tau : L \to L$ is an automorphism which extends $\sigma $.
$\square$

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