[proofplan] We prove each direction separately. For the backward direction ($\Leftarrow$), we assume $L$ is the splitting field of some $f \in K[t]$ and show that for every $\alpha \in L$, the minimal polynomial $P_\alpha$ splits over $L$. The argument lifts the isomorphism $K(\alpha) \cong K(\beta)$ (given by any root $\beta$ of $P_\alpha$ in $\bar{K}$) to a $K$-homomorphism $\tilde{\sigma}: L \to \bar{K}$ using the Roots-Homomorphisms Correspondence, then shows $\tilde{\sigma}(L) = L$ because $\tilde{\sigma}$ permutes the roots of $f$ and $L$ is generated by those roots. For the forward direction ($\Rightarrow$), we write $L = K(\alpha_1, \ldots, \alpha_r)$, take $f = P_{\alpha_1} \cdots P_{\alpha_r}$, and observe that normality forces each factor to split over $L$. [/proofplan]

[step:($\Leftarrow$) Assume $L$ is a splitting field and take any $\alpha \in L$ with minimal polynomial $P_\alpha$] Suppose $L$ is the splitting field of some polynomial $f \in K[t]$ over $K$. Let $\alpha_1, \ldots, \alpha_m \in L$ be the roots of $f$ in $L$, so that $f$ splits as a product of linear factors over $L$ and $L = K(\alpha_1, \ldots, \alpha_m)$. To show $L/K$ is normal, we must prove that for every $\alpha \in L$, the minimal polynomial $P_\alpha \in K[t]$ splits completely over $L$. Fix such an $\alpha$ and let $\beta \in \bar{K}$ be an arbitrary root of $P_\alpha$. It suffices to show $\beta \in L$. [/step]

[step:Lift the root $\beta$ to a $K$-homomorphism $\tilde{\sigma}: L \to \bar{K}$ via the Roots-Homomorphisms Correspondence]Since $P_\alpha$ is the minimal polynomial of $\alpha$ over $K$, it is irreducible in $K[t]$ by the [Properties of the Minimal Polynomial](/theorems/1250). Both $\alpha$ and $\beta$ are roots of $P_\alpha$ in $\bar{K}$. By the [Roots-Homomorphisms Correspondence](/theorems/1256) applied to the simple extension $K(\alpha)/K$ with target $\bar{K}$, there is a bijection between $\operatorname{Hom}_K(K(\alpha), \bar{K})$ and $\operatorname{Root}_{P_\alpha}(\bar{K})$ given by $\sigma \mapsto \sigma(\alpha)$. Since $\beta$ is a root of $P_\alpha$ in $\bar{K}$, there exists a $K$-homomorphism \begin{align*} \sigma: K(\alpha) &\to \bar{K} \end{align*} with $\sigma(\alpha) = \beta$. We extend $\sigma$ to a $K$-homomorphism $\tilde{\sigma}: L \to \bar{K}$. Since $L = K(\alpha_1, \ldots, \alpha_m)$ is a finite extension of $K$ and $K(\alpha) \subset L$, we apply the [Roots-Homomorphisms Correspondence](/theorems/1256) iteratively. Starting from $\sigma: K(\alpha) \to \bar{K}$, we extend one generator at a time: at each stage, $\alpha_j$ is algebraic over the current intermediate field, and the Roots-Homomorphisms Correspondence guarantees that any $K$-homomorphism from that intermediate field into $\bar{K}$ extends to one that includes $\alpha_j$, by choosing an image for $\alpha_j$ among the roots of its minimal polynomial (over the intermediate field) in $\bar{K}$. After finitely many extensions, we obtain \begin{align*} \tilde{\sigma}: L \to \bar{K} \end{align*} a $K$-homomorphism extending $\sigma$, with $\tilde{\sigma}(\alpha) = \sigma(\alpha) = \beta$.[/step]

[guided]The goal of this step is to connect the abstract root $\beta$ back to $L$ through a field homomorphism. Why do we need this? The root $\beta$ lives in $\bar{K}$, and we have no direct reason to believe it lies in $L$. The strategy is to build a $K$-homomorphism $\tilde{\sigma}: L \to \bar{K}$ that sends $\alpha \mapsto \beta$, and then show that this homomorphism maps $L$ into itself — which will force $\beta = \tilde{\sigma}(\alpha) \in L$. The first ingredient is the [Roots-Homomorphisms Correspondence](/theorems/1256): for a simple algebraic extension $K(\alpha)/K$ and any extension $E/K$, the $K$-homomorphisms $K(\alpha) \to E$ are in bijection with the roots of $P_\alpha$ in $E$, via $\sigma \mapsto \sigma(\alpha)$. The theorem requires that $\alpha$ is algebraic over $K$ (which holds since $L/K$ is finite, hence algebraic) and that $E$ is a field extension of $K$ (which holds for $\bar{K}$). Since $P_\alpha$ is irreducible and $\beta$ is a root of $P_\alpha$ in $\bar{K}$, the correspondence produces a $K$-homomorphism $\sigma: K(\alpha) \to \bar{K}$ sending $\alpha \mapsto \beta$. To extend $\sigma$ from $K(\alpha)$ to all of $L = K(\alpha_1, \ldots, \alpha_m)$, we adjoin the generators of $L$ one at a time. Write $L_0 := K(\alpha)$ and $L_j := L_{j-1}(\alpha_j)$ for $j = 1, \ldots, m$ (reindexing and possibly passing through generators already in $K(\alpha)$, which does not change the field). At each stage, $\alpha_j$ is algebraic over $L_{j-1}$, and the Roots-Homomorphisms Correspondence applied to the simple extension $L_j / L_{j-1}$ with target $\bar{K}$ (viewing $\bar{K}$ as an extension of $\tilde{\sigma}_{j-1}(L_{j-1})$) ensures that each extension step has at least one valid lift. After $m$ steps, we have $\tilde{\sigma}: L \to \bar{K}$ with $\tilde{\sigma}|_{K(\alpha)} = \sigma$. This iterative extension is sometimes called the "extension lemma" or "lifting lemma" for field homomorphisms. The point is that since $\bar{K}$ is algebraically closed, the minimal polynomial of each $\alpha_j$ over the current intermediate field always has a root in $\bar{K}$, so the extension never gets stuck.[/guided]

[step:Show $\tilde{\sigma}(L) = L$ by observing that $\tilde{\sigma}$ permutes the roots of $f$]We now prove that $\tilde{\sigma}(L) = L$, which will complete the backward direction. Since $\tilde{\sigma}$ is a $K$-homomorphism, it fixes every element of $K$ and preserves polynomial relations over $K$. For each root $\alpha_i$ of $f$ in $L$, we have $f(\alpha_i) = 0$. Applying $\tilde{\sigma}$: \begin{align*} f(\tilde{\sigma}(\alpha_i)) = \tilde{\sigma}(f(\alpha_i)) = \tilde{\sigma}(0) = 0, \end{align*} where the first equality uses the fact that $\tilde{\sigma}$ is a $K$-homomorphism (so it commutes with evaluation of polynomials with coefficients in $K$). Therefore $\tilde{\sigma}(\alpha_i)$ is a root of $f$ in $\bar{K}$. Since $L$ is the splitting field of $f$ over $K$, the polynomial $f$ splits completely over $L$, so all roots of $f$ in $\bar{K}$ lie in $L$. (If $f$ had a root $\gamma \in \bar{K} \setminus L$, then $f$ would not split completely over $L$, contradicting the definition of splitting field; alternatively, the factorisation of $f$ into linear factors over $L$ accounts for all $\deg f$ roots counting multiplicity, and a polynomial of degree $n$ has at most $n$ roots in any field.) Therefore $\tilde{\sigma}(\alpha_i) \in L$ for each $i = 1, \ldots, m$. Since $L = K(\alpha_1, \ldots, \alpha_m)$ and $\tilde{\sigma}$ is a ring homomorphism fixing $K$, every element of $L$ is a polynomial expression in $\alpha_1, \ldots, \alpha_m$ with coefficients in $K$, and \begin{align*} \tilde{\sigma}\bigl(g(\alpha_1, \ldots, \alpha_m)\bigr) = g\bigl(\tilde{\sigma}(\alpha_1), \ldots, \tilde{\sigma}(\alpha_m)\bigr) \end{align*} for any $g \in K[t_1, \ldots, t_m]$. Since each $\tilde{\sigma}(\alpha_i) \in L$, the right-hand side lies in $L$. Therefore $\tilde{\sigma}(L) \subset L$. To conclude equality, we use the [Finite $K$-Homomorphisms Are Isomorphisms](/theorems/1254) theorem. The field $\tilde{\sigma}(L)$ is a subfield of $L$ containing $K$, and $\tilde{\sigma}: L \to \tilde{\sigma}(L)$ is a $K$-isomorphism (field homomorphisms are injective). Therefore $[\tilde{\sigma}(L) : K] = [L : K]$. Since $\tilde{\sigma}(L) \subset L$ and both are finite-dimensional $K$-vector spaces of the same dimension, we conclude $\tilde{\sigma}(L) = L$. In particular, $\beta = \tilde{\sigma}(\alpha) \in \tilde{\sigma}(L) = L$. Since $\beta$ was an arbitrary root of $P_\alpha$ in $\bar{K}$, the polynomial $P_\alpha$ splits completely over $L$. Since $\alpha \in L$ was arbitrary, $L/K$ is normal.[/step]

[guided]This is the heart of the backward direction. We must show that the image $\tilde{\sigma}(L)$ lands back inside $L$, even though $\tilde{\sigma}$ maps into the much larger field $\bar{K}$. The key observation is that $\tilde{\sigma}$ acts on the roots of $f$. Why? Because $\tilde{\sigma}$ is a $K$-homomorphism: it fixes the coefficients of $f$ (which lie in $K$) and preserves the field operations. So if $f(\alpha_i) = 0$, then \begin{align*} f(\tilde{\sigma}(\alpha_i)) = \tilde{\sigma}(f(\alpha_i)) = \tilde{\sigma}(0) = 0. \end{align*} This means $\tilde{\sigma}$ sends each root of $f$ to another root of $f$ — it permutes (not necessarily bijectively at this stage, but we will see it is) the roots of $f$ in $\bar{K}$. Now, where do the roots of $f$ in $\bar{K}$ lie? Since $L$ is the splitting field of $f$, $f$ factors completely into linear factors over $L$. A polynomial of degree $n$ over a field has at most $n$ roots (counting multiplicity), and all $n$ roots are accounted for by the linear factors in $L[t]$. So every root of $f$ in $\bar{K}$ already belongs to $L$. This means $\tilde{\sigma}(\alpha_i) \in L$ for all $i$. Since $L = K(\alpha_1, \ldots, \alpha_m)$ — that is, $L$ is generated over $K$ by the roots $\alpha_1, \ldots, \alpha_m$ — and since $\tilde{\sigma}$ fixes $K$ and maps each $\alpha_i$ into $L$, every element of $L$ (being a rational expression in $\alpha_1, \ldots, \alpha_m$ over $K$) maps into $L$. Therefore $\tilde{\sigma}(L) \subset L$. Could the inclusion be strict? No. A field homomorphism is always injective (the kernel is an ideal of a field, hence $\{0\}$ or the whole field; since $\tilde{\sigma}(1) = 1 \neq 0$, the kernel is $\{0\}$). So $\tilde{\sigma}: L \to \tilde{\sigma}(L)$ is a $K$-isomorphism, giving $[\tilde{\sigma}(L) : K] = [L : K]$. Since $\tilde{\sigma}(L) \subset L$ and both are finite-dimensional $K$-vector spaces of the same dimension, the inclusion is an equality: $\tilde{\sigma}(L) = L$. This uses the elementary linear algebra fact that an injective linear map between vector spaces of the same finite dimension is surjective. Therefore $\beta = \tilde{\sigma}(\alpha) \in L$, as required.[/guided]

[step:($\Rightarrow$) Assume $L/K$ is normal and construct a polynomial $f$ whose splitting field is $L$]Suppose $L/K$ is a finite normal extension. By the [Characterisation of Finite Extensions](/theorems/1252), since $L/K$ is finite, we can write $L = K(\alpha_1, \ldots, \alpha_r)$ for finitely many algebraic elements $\alpha_1, \ldots, \alpha_r \in L$. For each $i = 1, \ldots, r$, let $P_{\alpha_i} \in K[t]$ denote the minimal polynomial of $\alpha_i$ over $K$. Define \begin{align*} f := P_{\alpha_1} \cdot P_{\alpha_2} \cdots P_{\alpha_r} \in K[t]. \end{align*} We verify that $L$ is the splitting field of $f$ over $K$ by checking two conditions: **$f$ splits completely over $L$.** Since $L/K$ is normal, for each $i$ the minimal polynomial $P_{\alpha_i}$ splits completely over $L$ (by the definition of normality: every irreducible polynomial in $K[t]$ that has a root in $L$ splits over $L$, and $P_{\alpha_i}$ has the root $\alpha_i \in L$). Therefore $f = P_{\alpha_1} \cdots P_{\alpha_r}$ splits completely over $L$. **No proper subfield of $L$ over $K$ splits $f$.** Since $L = K(\alpha_1, \ldots, \alpha_r)$ and each $\alpha_i$ is a root of $f$, any subfield of $L$ over $K$ that contains all roots of $f$ must contain $K(\alpha_1, \ldots, \alpha_r) = L$. Therefore $L$ is the smallest field extension of $K$ over which $f$ splits.[/step]

[guided]The forward direction is considerably shorter. We need to produce a polynomial $f \in K[t]$ such that $L$ is its splitting field over $K$. The natural candidate is the product of the minimal polynomials of a generating set. Since $L/K$ is finite, the [Characterisation of Finite Extensions](/theorems/1252) provides finitely many algebraic elements $\alpha_1, \ldots, \alpha_r \in L$ with $L = K(\alpha_1, \ldots, \alpha_r)$. (This uses the fact that a finite extension is algebraic, so each generator is algebraic over $K$ and has a well-defined minimal polynomial.) Setting $f = P_{\alpha_1} \cdots P_{\alpha_r}$, we need to verify two things: First, does $f$ split over $L$? This is where normality is used. Each $P_{\alpha_i}$ is an irreducible polynomial in $K[t]$ that has a root in $L$ (namely $\alpha_i$ itself). The definition of normality states that every such polynomial splits completely over $L$. So each factor $P_{\alpha_i}$ splits over $L$, and therefore the product $f$ splits over $L$. Second, is $L$ the smallest such field? Yes: the roots of $f$ include $\alpha_1, \ldots, \alpha_r$ (each $\alpha_i$ is a root of $P_{\alpha_i}$, hence of $f$). Any field over $K$ containing all roots of $f$ must contain $\alpha_1, \ldots, \alpha_r$ and therefore contain $K(\alpha_1, \ldots, \alpha_r) = L$. So $L$ is the splitting field of $f$ over $K$. Note that $f$ is generally not irreducible — it is a product of irreducible polynomials. The choice of generators $\alpha_1, \ldots, \alpha_r$ is not unique, but any choice produces a valid polynomial $f$ (though different choices may yield different polynomials with the same splitting field).[/guided]

[step:Combine both directions to conclude the equivalence] The backward direction ($\Leftarrow$) established that if $L$ is the splitting field of some $f \in K[t]$, then $L/K$ is normal. The forward direction ($\Rightarrow$) established that if $L/K$ is a finite normal extension, then $L$ is the splitting field of the polynomial $f = P_{\alpha_1} \cdots P_{\alpha_r} \in K[t]$, where $\alpha_1, \ldots, \alpha_r$ generate $L$ over $K$. Together, the two directions prove that a finite extension $L/K$ is normal if and only if $L$ is the splitting field of some polynomial $f \in K[t]$. [/step]