[proofplan] We prove the equivalence between normality, being a splitting field of a set of polynomials, and being a directed union of finite normal subextensions. The forward direction from normality to splitting field uses the minimal polynomials of all elements of $K$: normality forces each to split completely in $K$. The directed union characterisation follows by taking splitting fields of finite subsets. The converse from splitting field to normality uses the fact that any $k$-irreducible polynomial with a root in $K$ must divide some polynomial in $S$, hence splits in $K$. [/proofplan] [step:Show that normality implies $K$ is a splitting field of a set of polynomials] Assume $K/k$ is a normal algebraic extension. For each $\alpha \in K$, let $f_\alpha = \operatorname{min}(\alpha, k) \in k[x]$ be the minimal polynomial of $\alpha$ over $k$. Define $S = \{f_\alpha : \alpha \in K\}$. We claim $K$ is the splitting field of $S$ over $k$. First, each $f_\alpha$ has at least one root in $K$ (namely $\alpha$ itself). By the definition of normality, every irreducible polynomial in $k[x]$ that has a root in $K$ splits completely in $K$. Since each $f_\alpha$ is irreducible in $k[x]$ and has the root $\alpha \in K$, $f_\alpha$ splits completely in $K$. It remains to check that $K$ is generated over $k$ by the roots of polynomials in $S$. But for every $\alpha \in K$, $\alpha$ is a root of $f_\alpha \in S$, so $K \subset k(\{\text{roots of } f : f \in S\}) \subset K$. Hence $K$ equals the splitting field of $S$. [/step] [step:Show that a splitting field of a set of polynomials is a directed union of finite normal subextensions] Assume $K$ is the splitting field of $S = \{f_i\}_{i \in I} \subset k[x]$ over $k$. For each finite subset $J \subset I$, let $K_J$ be the splitting field of $\{f_j : j \in J\}$ over $k$ inside $K$. By the [Normal Extensions Are Splitting Fields](/theorems/1269) characterisation (applied to each finite splitting field), $K_J/k$ is a finite normal extension. The family $\{K_J\}_{J \subset I, \, |J| < \infty}$ is directed under inclusion: for finite subsets $J_1, J_2 \subset I$, we have $K_{J_1}, K_{J_2} \subset K_{J_1 \cup J_2}$. We claim $K = \bigcup_{J} K_J$ where $J$ ranges over all finite subsets of $I$. Since $K$ is generated over $k$ by the roots of all $f_i$, every element of $K$ is a $k$-rational expression in finitely many roots of finitely many $f_i$. Hence every element lies in some $K_J$ for a finite $J$. This gives $K \subset \bigcup_J K_J$. The reverse inclusion $\bigcup_J K_J \subset K$ is immediate since each $K_J$ is a subfield of $K$. [guided] Assume $K$ is the splitting field of $S = \{f_i\}_{i \in I}$ over $k$. We want to express $K$ as a directed union of finite normal subextensions. The idea is natural: for each finite subset $J \subset I$, define $K_J$ to be the splitting field of $\prod_{j \in J} f_j$ over $k$ (constructed inside $K$). By the [Normal Extensions Are Splitting Fields](/theorems/1269) characterisation, $K_J/k$ is a finite normal extension (it is the splitting field of a single polynomial $\prod_{j \in J} f_j \in k[x]$, which has finite degree). Why is the family directed? Given finite subsets $J_1, J_2$, the field $K_{J_1 \cup J_2}$ contains all roots of $f_j$ for $j \in J_1 \cup J_2$, hence contains both $K_{J_1}$ and $K_{J_2}$. So $K_{J_1}, K_{J_2} \subset K_{J_1 \cup J_2}$, establishing the directed property. Why is $K = \bigcup_J K_J$? Every element of $K$ is algebraic over $k$ (since $K/k$ is algebraic) and can be expressed as a $k$-rational expression in finitely many roots of polynomials from $S$. Those roots all lie in $K_J$ for some finite $J$, so the element itself lies in $K_J$. Conversely, each $K_J \subset K$ since $K$ contains all roots of every $f_i \in S$. [/guided] [/step] [step:Show that a directed union of finite normal subextensions is normal] Assume $K = \bigcup_i K_i$ where each $K_i/k$ is a finite normal extension and the $K_i$ are directed under inclusion. We show $K/k$ is normal. Let $f \in k[x]$ be an irreducible polynomial with a root $\alpha \in K$. We must show $f$ splits completely in $K$. Since $\alpha \in K = \bigcup_i K_i$, there exists some $K_i$ containing $\alpha$. Since $K_i/k$ is normal and $f$ is irreducible in $k[x]$ with a root $\alpha \in K_i$, $f$ splits completely in $K_i$. In particular, all roots of $f$ lie in $K_i \subset K$, so $f$ splits completely in $K$. Since every irreducible polynomial in $k[x]$ with a root in $K$ splits completely in $K$, the extension $K/k$ is normal. [/step] [step:Show that being a splitting field of a set of polynomials implies normality] Assume $K$ is the splitting field of $S \subset k[x]$ over $k$. We show $K/k$ is normal directly. Let $g \in k[x]$ be irreducible with a root $\beta \in K$. We must show $g$ splits completely in $K$. Let $\beta' \in \overline{k}$ be any root of $g$. Since $g$ is irreducible over $k$ and both $\beta, \beta'$ are roots, there exists a $k$-isomorphism $\sigma_0: k(\beta) \xrightarrow{\sim} k(\beta')$ with $\sigma_0(\beta) = \beta'$. By the [Extension of Isomorphisms to Splitting Fields](/theorems/3313) applied to the isomorphism $\sigma_0: k(\beta) \xrightarrow{\sim} k(\beta')$, the polynomial set $S$ viewed over $k(\beta)$, and the splitting field $K$ of $S$ over $k$ (hence also over $k(\beta)$, since $k \subset k(\beta) \subset K$), there exists an isomorphism $\tilde{\sigma}: K \xrightarrow{\sim} K'$ extending $\sigma_0$, where $K'$ is a splitting field of $\sigma_0(S) = S$ over $k(\beta')$ (here $\sigma_0(S) = S$ because $S \subset k[x]$ and $\sigma_0$ fixes $k$). Since $K'$ is a splitting field of $S$ over $k(\beta') \supset k$, and $K$ is also a splitting field of $S$ over $k$, the uniqueness part of the [Existence and Uniqueness of Splitting Fields](/theorems/1258) (applied iteratively to the polynomials in $S$) shows we may take $K' = K$ (both are splitting fields of $S$ over $k$ inside $\overline{k}$). Thus $\tilde{\sigma}: K \xrightarrow{\sim} K$ with $\tilde{\sigma}(\beta) = \beta'$. In particular, $\beta' = \tilde{\sigma}(\beta) \in K$. Since every root $\beta'$ of $g$ lies in $K$, $g$ splits completely in $K$. This holds for every irreducible $g \in k[x]$ with a root in $K$, so $K/k$ is normal. [guided] Assume $K$ is the splitting field of $S \subset k[x]$ over $k$. We must show: if $g \in k[x]$ is irreducible and has a root $\beta \in K$, then $g$ splits completely in $K$. Let $\beta' \in \overline{k}$ be any other root of $g$. We need to show $\beta' \in K$. The idea is to build a $k$-automorphism of $K$ that sends $\beta$ to $\beta'$. Since $g$ is the minimal polynomial of $\beta$ over $k$ (being irreducible with $\beta$ as a root), there is a $k$-isomorphism $\sigma_0: k(\beta) \xrightarrow{\sim} k(\beta')$ mapping $\beta \mapsto \beta'$. We want to extend $\sigma_0$ to all of $K$. Now $K$ is the splitting field of $S$ over $k$. Since $S \subset k[x]$ and $\sigma_0$ fixes $k$ pointwise, $\sigma_0$ maps each polynomial in $S$ to itself. By the [Extension of Isomorphisms to Splitting Fields](/theorems/3313), $\sigma_0$ extends to an isomorphism $\tilde{\sigma}$ between the splitting field of $S$ over $k(\beta)$ and the splitting field of $S$ over $k(\beta')$. Both of these splitting fields equal $K$ (since $K$ is the splitting field of $S$ over $k$, and $k(\beta), k(\beta') \subset K$ in $\overline{k}$, the splitting field of $S$ over any intermediate field between $k$ and $K$ is still $K$). So $\tilde{\sigma}: K \xrightarrow{\sim} K$ with $\tilde{\sigma}(\beta) = \beta'$. In particular, $\beta' = \tilde{\sigma}(\beta) \in K$. Since $\beta'$ was an arbitrary root of $g$, all roots of $g$ lie in $K$, and $g$ splits completely in $K$. This confirms $K/k$ is normal. Note the key insight: the splitting field condition provides enough structure to extend any partial $k$-automorphism. This is exactly what fails for non-normal extensions -- there exist $k$-embeddings whose images escape the field. [/guided] [/step]