Let \(k\) be a field, let \(L/k\) be an algebraic extension, and let \(\bar{k}\) be an algebraic closure of \(k\) containing \(L\). Then \(L/k\) is normal if and only if every \(k\)-embedding \(\sigma\colon L\to \bar{k}\) satisfies \(\sigma(L) = L\). In that case every \(k\)-embedding \(\sigma\colon L\to\bar k\) restricts to a \(k\)-automorphism of \(L\):
Suppose \(L/k\) is normal and let \(\sigma\colon L\to\bar k\) be a \(k\)-embedding. For \(\alpha\in L\) with minimal polynomial \(m_\alpha\in k[x]\), \(\sigma(\alpha)\) is a root of \(m_\alpha\), and all roots of \(m_\alpha\) lie in \(L\), so \(\sigma(L)\subseteq L\). Moreover \(\sigma\) injects the finite set of roots of \(m_\alpha\) in \(L\) into itself, hence permutes it, so \(\alpha\) lies in \(\sigma(L)\); thus \(\sigma(L)=L\).
Conversely, suppose every \(k\)-embedding \(L\to\bar k\) has image \(L\). Let \(f\in k[x]\) be irreducible with a root \(\alpha\in L\), and let \(\beta\in\bar k\) be any root of \(f\). There is a \(k\)-embedding \(k(\alpha)\to\bar k\) with \(\alpha\mapsto\beta\), and since \(L/k(\alpha)\) is algebraic and \(\bar k\) is algebraically closed, it extends to a \(k\)-embedding \(\sigma\colon L\to\bar k\). Then \(\beta=\sigma(\alpha)\in\sigma(L)=L\), so \(f\) splits in \(L\) and \(L/k\) is normal.