Normal closure

Remarks

A normal closure exists and is unique up to an isomorphism of fields that is the identity on $L$. Let $L = K(\alpha_1, \ldots, \alpha_n)$ be finite over $K$, and let $f_i$ be the minimal polynomial of $\alpha_i$ over $K$. Then a normal closure is the splitting field of $f_1 \cdots f_n$ over $L$, and it is again finite over $K$. Let $N$ be a normal extension of $K$ that contains $L$. Inside $N$, the normal closure is the compositum of the fields $\tau(L)$, for all embeddings $\tau \colon L \to N$ over $K$.

Example: for $L = \Q(\sqrt[3]{2})$ over $\Q$, let $M = \Q(\sqrt[3]{2}, \omega)$ with $\omega = e^{2\pi i/3}$. Check: $M$ is the splitting field of $x^3 - 2$ over $\Q$, so it is normal over $\Q$. A normal subfield containing $\sqrt[3]{2}$ contains all roots $\sqrt[3]{2}, \omega\sqrt[3]{2}, \omega^2\sqrt[3]{2}$ of its minimal polynomial, hence contains $M$.

Non-example: an algebraic closure $\overline{\Q}$ of $\Q$ contains $L$ and is normal over $\Q$, but it fails clause 2, since $M$ is a proper normal subfield containing $L$.