Given a polynomial $f \in k[x]$, the most natural question is: where do its roots live? Over $\mathbb{Q}$, the polynomial $x^2 - 2$ has no roots — but the moment we pass to $\mathbb{Q}(\sqrt{2})$, it splits completely into linear factors. Over $\mathbb{Q}$, the polynomial $x^2 + 1$ has no roots — but over $\mathbb{Q}(i)$ it splits as $(x - i)(x + i)$. In each case, the field we adjoin roots to is not arbitrary: it is the *smallest* field extension of $\mathbb{Q}$ over which the polynomial factors into linear factors. This smallest field — the **splitting field** — is one of the central objects of Galois theory. The splitting field is the arena in which a polynomial's symmetries become visible. Two polynomials may look structurally different over the base field yet have isomorphic splitting fields; the Galois group, which is the automorphism group of the splitting field over the base, encodes exactly these symmetries. Before we can study the Galois group, we must understand the splitting field itself: its existence, its uniqueness up to isomorphism, its degree over the base field, and how it relates to the roots of the polynomial. [example: A Splitting Field over the Rationals] Consider $f(x) = x^3 - 2 \in \mathbb{Q}[x]$. The three roots of $f$ are \begin{align*} \alpha_1 &= \sqrt[3]{2}, \quad \alpha_2 = \omega \sqrt[3]{2}, \quad \alpha_3 = \omega^2 \sqrt[3]{2}, \end{align*} where $\omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$ is a primitive cube root of unity. The field $\mathbb{Q}(\sqrt[3]{2})$ contains $\alpha_1$ but not $\alpha_2$ or $\alpha_3$, since $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$ and $\alpha_2, \alpha_3 \notin \mathbb{R}$. To split $f$ completely, we need both $\sqrt[3]{2}$ and $\omega$. The field $K = \mathbb{Q}(\sqrt[3]{2}, \omega)$ contains all three roots and is the smallest field extension of $\mathbb{Q}$ with this property. Over $K$, we have \begin{align*} x^3 - 2 &= (x - \sqrt[3]{2})(x - \omega\sqrt[3]{2})(x - \omega^2\sqrt[3]{2}). \end{align*} The degree of this extension is $[K : \mathbb{Q}] = 6$, computed via the tower $\mathbb{Q} \subset \mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{Q}(\sqrt[3]{2}, \omega)$: the first step has degree $3$ (the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$ is $x^3 - 2$, which is irreducible by Eisenstein's criterion at $p = 2$), and the second step has degree $2$ (the minimal polynomial of $\omega$ over $\mathbb{Q}(\sqrt[3]{2})$ is $x^2 + x + 1$, since $\omega \notin \mathbb{R}$ but $\omega$ satisfies $x^2 + x + 1 = 0$). Thus $[K : \mathbb{Q}] = 3 \cdot 2 = 6$. [/example] This example illustrates the key features: existence (we can always adjoin all the roots), uniqueness (the result is the smallest such field), and degree bounds (the degree is controlled by the degree of $f$). ## Definition Before giving the formal definition, we must be precise about what it means for a polynomial to "split" over a field. A polynomial $f \in k[x]$ of degree $n$ splits into linear factors over a field $K \supset k$ if there exist elements $\alpha_1, \ldots, \alpha_n \in K$ and a nonzero constant $a \in k$ such that \begin{align*} f(x) &= a(x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_n). \end{align*} If $K$ contains no proper subfield that also splits $f$, then $K$ is generated over $k$ by the roots $\alpha_1, \ldots, \alpha_n$ — that is, $K = k(\alpha_1, \ldots, \alpha_n)$. [definition: Splitting Field] Let $k$ be a field and let $f \in k[x]$ be a nonconstant polynomial. A **splitting field** of $f$ over $k$ is a field extension $K / k$ satisfying: 1. $f$ splits into linear factors over $K$: there exist $\alpha_1, \ldots, \alpha_n \in K$ (not necessarily distinct) and $a \in k^\times$ such that \begin{align*} f(x) &= a \prod_{i=1}^{n} (x - \alpha_i) \quad \text{in } K[x]; \end{align*} 2. $K$ is generated by the roots: $K = k(\alpha_1, \ldots, \alpha_n)$. [/definition] The second condition is essential — it ensures minimality. Without it, the algebraic closure $\overline{k}$ would satisfy condition (1) for any polynomial, but $\overline{k}$ is far too large; it contains roots of every polynomial, not just $f$. The splitting field is the *smallest* field containing all roots of $f$. [remark: Splitting Field of a Set of Polynomials] The definition extends naturally to a finite set $\{f_1, \ldots, f_m\} \subset k[x]$: a splitting field is a minimal extension $K / k$ over which each $f_i$ splits completely. Equivalently, it is the splitting field of the single polynomial $f_1 \cdots f_m$. This is useful when studying Galois closures of non-Galois extensions, which arise as splitting fields of minimal polynomials. [/remark] The notation $\operatorname{Split}(f, k)$ will denote the splitting field of $f$ over $k$ (well-defined up to $k$-isomorphism, as we shall prove). ## Existence and Uniqueness The two foundational questions about splitting fields are whether they exist and whether they are unique. Existence is constructive: we build the splitting field by repeatedly adjoining roots. Uniqueness is subtler and requires comparing two abstract extensions. **Existence** proceeds by induction on the degree of $f$. If $f$ splits over $k$ already (all roots lie in $k$), then $K = k$ is the splitting field. Otherwise, let $p \in k[x]$ be an irreducible factor of $f$ with $\deg p \ge 2$. The quotient ring $k[x]/(p)$ is a field (since $(p)$ is a maximal ideal in $k[x]$, as $p$ is irreducible), and it contains a root $\alpha = \bar{x}$ of $p$. The inclusion $k \hookrightarrow k[x]/(p)$ identifies $k$ with a subfield, and $[k(\alpha) : k] = \deg p \ge 2$. Over $k(\alpha)$, the polynomial $f$ has one fewer irreducible factor of degree $\ge 2$, since $p(x) = (x - \alpha) \cdot q(x)$ in $k(\alpha)[x]$. Repeating this process at most $n = \deg f$ times yields a tower of extensions $k = K_0 \subset K_1 \subset \cdots \subset K_m$ over which $f$ splits completely, and $K_m = k(\alpha_1, \ldots, \alpha_m)$ is the splitting field. [quotetheorem:1312] The bound $n!$ is sharp in general — it is achieved, for example, by a generic irreducible polynomial of degree $n$ whose Galois group is the full symmetric group $S_n$, in which case $[K : k] = n!$. But many polynomials have splitting fields of much smaller degree: $x^4 - 1$ splits over $\mathbb{Q}(i)$, which has degree $2$, and the splitting field of $x^n - 2$ over $\mathbb{Q}$ has degree $n \cdot \varphi(n)$ (where $\varphi$ is Euler's totient function), which equals $n!$ only for $n \le 3$. The uniqueness theorem says that any two splitting fields of $f$ over $k$ are isomorphic as extensions of $k$. The proof uses the following key lemma, which is the engine of the entire theory. [quotetheorem:3313] Taking $\sigma = \operatorname{id}_k$, $k = k'$, and $K, K'$ both splitting fields of the same $f$ over $k$, the theorem immediately gives: [quotetheorem:1258] This uniqueness justifies speaking of *the* splitting field $\operatorname{Split}(f, k)$, even though in practice we may construct it in different ways. The isomorphism between two constructions need not be unique — indeed, the number of such isomorphisms is related to the Galois group. ## The Degree of a Splitting Field How large can the splitting field of a degree-$n$ polynomial be? The bound $[K : k] \le n!$ from the existence theorem is crude but useful. Understanding when this bound is achieved — and when it can be improved — is intimately connected to the irreducibility of $f$ and the relationships among its roots. The degree computation proceeds via the tower law: if $K = k(\alpha_1, \ldots, \alpha_m)$ is built by adjoining the roots one at a time, then \begin{align*} [K : k] &= [k(\alpha_1, \ldots, \alpha_m) : k(\alpha_1, \ldots, \alpha_{m-1})] \cdots [k(\alpha_1) : k]. \end{align*} Each factor $[k(\alpha_1, \ldots, \alpha_i) : k(\alpha_1, \ldots, \alpha_{i-1})]$ equals the degree of the minimal polynomial of $\alpha_i$ over $k(\alpha_1, \ldots, \alpha_{i-1})$. This degree divides the degree of $f$ over $k$, and it decreases as we adjoin more roots (since each new root satisfies a polynomial of lower degree over the current field). [example: Computing the Degree of a Splitting Field] Let $f(x) = x^4 - 5x^2 + 6 = (x^2 - 2)(x^2 - 3) \in \mathbb{Q}[x]$. The roots of $f$ are $\pm\sqrt{2}$ and $\pm\sqrt{3}$. The splitting field is $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$. To compute $[K : \mathbb{Q}]$, use the tower $\mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt{2}, \sqrt{3})$. We have $[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = 2$ since $x^2 - 2$ is irreducible over $\mathbb{Q}$ (it has no rational roots). Now, $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}(\sqrt{2})]$ equals the degree of the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})$. We claim this is $2$. If $\sqrt{3} \in \mathbb{Q}(\sqrt{2})$, then $\sqrt{3} = a + b\sqrt{2}$ for some $a, b \in \mathbb{Q}$. Squaring: $3 = a^2 + 2b^2 + 2ab\sqrt{2}$. Since $\sqrt{2} \notin \mathbb{Q}$, we need $2ab = 0$, so either $a = 0$ or $b = 0$. If $a = 0$: $3 = 2b^2$, giving $b = \sqrt{3/2} \notin \mathbb{Q}$. If $b = 0$: $3 = a^2$, giving $a = \sqrt{3} \notin \mathbb{Q}$. Both cases lead to a contradiction, so $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$, and the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})$ is $x^2 - 3$, of degree $2$. Therefore $[K : \mathbb{Q}] = 2 \cdot 2 = 4$. Note that $\deg f = 4$ and $4 \le 4! = 24$: the splitting field is much smaller than the bound. Note also that $-\sqrt{2} = -1 \cdot \sqrt{2} \in \mathbb{Q}(\sqrt{2})$ and $-\sqrt{3} \in \mathbb{Q}(\sqrt{3}) \subset K$, so all four roots of $f$ lie in $K$ automatically once we adjoin $\sqrt{2}$ and $\sqrt{3}$. [/example] [example: A Splitting Field of Degree Strictly Less Than $n!$] For the polynomial $f(x) = x^3 - 3x - 1 \in \mathbb{Q}[x]$, the splitting field has degree $3$ over $\mathbb{Q}$, far below the upper bound $3! = 6$. This happens because once we adjoin one root $\alpha$, the other two roots can be expressed as rational functions of $\alpha$. To see this concretely: $f$ is irreducible over $\mathbb{Q}$ (no rational roots, since the only candidates $\pm 1$ satisfy $f(1) = -3 \neq 0$ and $f(-1) = 1 \neq 0$). Let $\alpha$ be a root. The three roots are related by the identity \begin{align*} \alpha_1 + \alpha_2 + \alpha_3 &= 0, \quad \alpha_1 \alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3 = -3, \quad \alpha_1\alpha_2\alpha_3 = 1, \end{align*} coming from Vieta's formulas. Direct computation shows that $\alpha_2 = \alpha^2 - 2$ and $\alpha_3 = -\alpha^2 - \alpha + 2$ are also roots of $f$ (a direct computation: substituting $\alpha^2 - 2$ into $f$ and using $\alpha^3 = 3\alpha + 1$). Thus $\mathbb{Q}(\alpha)$ contains all three roots, and the splitting field is $\mathbb{Q}(\alpha)$ itself, of degree $3$. By contrast, a generic irreducible cubic $f \in \mathbb{Q}[x]$ has splitting field of degree $6 = 3!$ over $\mathbb{Q}$, since the three roots are typically algebraically independent over $\mathbb{Q}$. [/example] The general principle is that $[\operatorname{Split}(f, k) : k]$ divides $n!$ and is divisible by $[k(\alpha) : k]$ for any root $\alpha$ of $f$. If $f$ is irreducible, then $[k(\alpha) : k] = n$, so $n$ divides $[\operatorname{Split}(f, k) : k]$. ## Splitting Fields and Normal Extensions The connection between splitting fields and the theory of normal extensions is one of the most important structural results in Galois theory. It explains why splitting fields are precisely the right class of extensions to study. Recall that a field extension $K / k$ is *normal* if every irreducible polynomial in $k[x]$ that has at least one root in $K$ splits completely over $K$. Normality is not automatic: the extension $\mathbb{Q}(\sqrt[3]{2}) / \mathbb{Q}$ is not normal, because $x^3 - 2$ is irreducible over $\mathbb{Q}$ with one root $\sqrt[3]{2} \in \mathbb{Q}(\sqrt[3]{2})$, yet the other roots $\omega\sqrt[3]{2}$ and $\omega^2\sqrt[3]{2}$ do not lie in $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$. The relationship between splitting fields and normal extensions is exact: [quotetheorem:1316] This theorem clarifies the geometric meaning of normality: a normal extension is one that is "closed under all symmetries" in the sense that every way of embedding it into the algebraic closure lands back in itself. Non-normal extensions lack this closure property. [explanation: Why Non-Normal Extensions Fail to Be Splitting Fields] The failure of $\mathbb{Q}(\sqrt[3]{2}) / \mathbb{Q}$ to be normal has a precise geometric interpretation. Inside $\overline{\mathbb{Q}} \subset \mathbb{C}$, there are three $\mathbb{Q}$-embeddings of $\mathbb{Q}(\sqrt[3]{2})$: \begin{align*} \sigma_1: \sqrt[3]{2} &\mapsto \sqrt[3]{2}, \\ \sigma_2: \sqrt[3]{2} &\mapsto \omega\sqrt[3]{2}, \\ \sigma_3: \sqrt[3]{2} &\mapsto \omega^2\sqrt[3]{2}. \end{align*} The images $\sigma_1(\mathbb{Q}(\sqrt[3]{2})) = \mathbb{Q}(\sqrt[3]{2})$ (the real cube root), $\sigma_2(\mathbb{Q}(\sqrt[3]{2})) = \mathbb{Q}(\omega\sqrt[3]{2})$, and $\sigma_3(\mathbb{Q}(\sqrt[3]{2})) = \mathbb{Q}(\omega^2\sqrt[3]{2})$ are three *different* subfields of $\mathbb{C}$. This is the hallmark of a non-normal extension: the different embeddings give genuinely different subfields. The splitting field $K = \mathbb{Q}(\sqrt[3]{2}, \omega)$ is normal: any $\mathbb{Q}$-embedding of $K$ must send $\sqrt[3]{2}$ to one of the three cube roots of $2$ and must send $\omega$ to one of the two primitive cube roots of unity, both of which lie in $K$. So every $\mathbb{Q}$-embedding of $K$ into $\overline{\mathbb{Q}}$ has image contained in (and hence equal to, by a dimension argument) $K$ itself. [/explanation] [illustration:embeddings-normality] The theorem implies that the splitting field of any polynomial is automatically a normal extension. Combined with the concept of separability, normality is half of the definition of a Galois extension. [definition: Galois Extension] A finite field extension $K / k$ is a **Galois extension** if it is both normal and separable. The **Galois group** of $K / k$ is the group of all field automorphisms of $K$ that fix $k$ pointwise: \begin{align*} \operatorname{Gal}(K / k) &:= \{ \sigma \in \operatorname{Aut}(K) : \sigma(a) = a \text{ for all } a \in k \}. \end{align*} [/definition] When $K = \operatorname{Split}(f, k)$ and $f$ is separable over $k$ (meaning $f$ has no repeated roots in $\overline{k}$, equivalently $\gcd(f, f') = 1$ in $k[x]$), the extension $K / k$ is Galois. In characteristic zero — and over finite fields — every irreducible polynomial is separable, so in the most common settings, splitting fields of irreducible polynomials are automatically Galois. ## The Galois Correspondence The power of working with splitting fields becomes fully apparent through the Galois correspondence: a bijection between the subgroups of $\operatorname{Gal}(K / k)$ and the intermediate fields $k \subset F \subset K$. This correspondence is one of the deepest results in algebra, and it applies precisely to Galois extensions — that is, to splitting fields of separable polynomials. To understand why the Galois group governs the intermediate fields, we should think carefully about what an element $\sigma \in \operatorname{Gal}(K / k)$ does. Since $K = k(\alpha_1, \ldots, \alpha_n)$ where $\alpha_1, \ldots, \alpha_n$ are the roots of $f$, any $k$-automorphism of $K$ must permute the roots: if $f(\alpha) = 0$ then $f(\sigma(\alpha)) = \sigma(f(\alpha)) = \sigma(0) = 0$. This gives an injective group homomorphism \begin{align*} \operatorname{Gal}(K / k) &\hookrightarrow S_n, \end{align*} where $S_n$ is the symmetric group on the $n$ roots of $f$. The Galois group is thus a *subgroup* of $S_n$, and understanding which subgroup it is — its order, its transitivity properties — tells us about the arithmetic of $f$. [quotetheorem:1274] The fixed field $K^H$ is the subfield of elements of $K$ that are fixed by every automorphism in $H$. The correspondence says: larger subgroups correspond to smaller intermediate fields (inclusion is reversed), and the degree of an intermediate field over $k$ equals the index of the corresponding subgroup. [example: Galois Correspondence for the Splitting Field of $x^3 - 2$] We return to $K = \mathbb{Q}(\sqrt[3]{2}, \omega)$, the splitting field of $x^3 - 2$ over $\mathbb{Q}$. We computed $[K : \mathbb{Q}] = 6$, so $G = \operatorname{Gal}(K / \mathbb{Q})$ has order $6$. Since $G$ embeds into $S_3$ (the permutations of the three roots $\alpha_1 = \sqrt[3]{2}$, $\alpha_2 = \omega\sqrt[3]{2}$, $\alpha_3 = \omega^2\sqrt[3]{2}$) and $|S_3| = 6$, we have $G \cong S_3$. The six elements of $G$ are determined by where they send $\sqrt[3]{2}$ and $\omega$: | Element | $\sqrt[3]{2} \mapsto$ | $\omega \mapsto$ | Order | |---|---|---|---| | $e$ | $\sqrt[3]{2}$ | $\omega$ | $1$ | | $\sigma$ | $\omega\sqrt[3]{2}$ | $\omega$ | $3$ | | $\sigma^2$ | $\omega^2\sqrt[3]{2}$ | $\omega$ | $3$ | | $\tau$ | $\sqrt[3]{2}$ | $\omega^2$ | $2$ | | $\sigma\tau$ | $\omega^2\sqrt[3]{2}$ | $\omega^2$ | $2$ | | $\sigma^2\tau$ | $\omega\sqrt[3]{2}$ | $\omega^2$ | $2$ | The subgroups of $S_3$ are: $\{e\}$, $\langle\sigma\rangle = \{e, \sigma, \sigma^2\} \cong C_3$ (the unique subgroup of order $3$, which is normal), three subgroups of order $2$ ($\langle\tau\rangle$, $\langle\sigma\tau\rangle$, $\langle\sigma^2\tau\rangle$), and $S_3$ itself. The Galois correspondence gives: - $\{e\} \leftrightarrow K = \mathbb{Q}(\sqrt[3]{2}, \omega)$, with $[K : \mathbb{Q}] = 6$. - $\langle\sigma\rangle \leftrightarrow K^{\langle\sigma\rangle} = \mathbb{Q}(\omega)$: the elements fixed by $\sigma$ are those in $\mathbb{Q}(\omega)$, since $\sigma$ permutes the cube roots but fixes $\omega$. We have $[K^{\langle\sigma\rangle} : \mathbb{Q}] = [S_3 : \langle\sigma\rangle] = 2$. - $\langle\tau\rangle \leftrightarrow K^{\langle\tau\rangle} = \mathbb{Q}(\sqrt[3]{2})$: $\tau$ fixes $\sqrt[3]{2}$ and sends $\omega \mapsto \omega^2$, so the fixed field is $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$. We have $[K^{\langle\tau\rangle} : \mathbb{Q}] = 3$. - $S_3 \leftrightarrow K^{S_3} = \mathbb{Q}$, the base field. The normal subgroup $\langle\sigma\rangle \trianglelefteq S_3$ corresponds to the normal extension $\mathbb{Q}(\omega) / \mathbb{Q}$, with $\operatorname{Gal}(\mathbb{Q}(\omega)/\mathbb{Q}) \cong S_3 / \langle\sigma\rangle \cong C_2$. The non-normal subgroups $\langle\tau\rangle, \langle\sigma\tau\rangle, \langle\sigma^2\tau\rangle$ correspond to the extensions $\mathbb{Q}(\sqrt[3]{2})$, $\mathbb{Q}(\omega\sqrt[3]{2})$, $\mathbb{Q}(\omega^2\sqrt[3]{2})$ over $\mathbb{Q}$, none of which is normal (since $x^3 - 2$ does not split completely over any of them). [/example] [illustration:galois-correspondence-s3] ## Splitting Fields in Characteristic $p$ Everything above applies in characteristic zero. In characteristic $p > 0$, splitting fields still exist and are unique, but new phenomena appear: the Frobenius endomorphism becomes central, and the notions of separability and inseparability acquire real teeth. Over a finite field $\mathbb{F}_q$ where $q = p^n$, the Frobenius $\varphi_p: x \mapsto x^p$ is a field automorphism. Every finite extension of $\mathbb{F}_q$ is a splitting field of a cyclotomic polynomial $x^{q^m} - x$ for some $m$, and these extensions are always Galois with cyclic Galois group generated by the Frobenius. [quotetheorem:1275] In characteristic $p$, inseparable polynomials arise. A polynomial $f \in k[x]$ is *inseparable* if it has a repeated root in $\overline{k}$. The canonical example is $f(x) = x^p - a \in k[x]$ when $a$ is not a $p$-th power in $k$: over $\overline{k}$, $f(x) = (x - \alpha)^p$ where $\alpha^p = a$. The splitting field of $f$ over $k$ is $k(\alpha)$, a purely inseparable extension. The extension $k(\alpha)/k$ has a peculiar structure: the single repeated root $\alpha$ generates a degree-$p$ extension, yet there are no non-trivial $k$-automorphisms of $k(\alpha)$ — the Galois group is trivial, and the Galois correspondence sees nothing. What makes this extreme case worth isolating? The key observation is that iterating the Frobenius eventually returns every element to the base field: $\alpha^{p^1} = a \in k$. A general algebraic extension in characteristic $p$ can always be factored into a separable part (which Galois theory governs) followed by a purely inseparable part (which the Galois group cannot detect). Identifying the purely inseparable part precisely — those elements whose $p$-power lands in $k$ — is therefore essential for understanding the full structure of algebraic extensions in characteristic $p$. [definition: Purely Inseparable Extension] A field extension $K / k$ of characteristic $p > 0$ is **purely inseparable** if for every $\alpha \in K$, there exists $n \ge 0$ such that $\alpha^{p^n} \in k$. [/definition] Purely inseparable extensions are the "opposite" of Galois extensions: a purely inseparable extension $K / k$ satisfies $\operatorname{Gal}(K / k) = \{e\}$ (only the identity automorphism fixes $k$ pointwise, since any $k$-automorphism $\sigma$ satisfies $\sigma(\alpha)^{p^n} = \sigma(\alpha^{p^n}) = \alpha^{p^n} \in k$, forcing $\sigma(\alpha) = \alpha$). [remark: Separable Closures and the Full Closure] For a general field $k$ of characteristic $p$, the algebraic closure $\overline{k}$ decomposes in a precise way. The separable closure $k^{\text{sep}} \subset \overline{k}$ is the compositum of all finite Galois extensions of $k$; it is the splitting field of all separable polynomials over $k$. The extension $\overline{k} / k^{\text{sep}}$ is then purely inseparable. This decomposition $k \subset k^{\text{sep}} \subset \overline{k}$ is the algebraic analogue of the factorization of a polynomial into its separable and purely inseparable parts. [/remark] ## Solvability by Radicals The deepest application of splitting fields is to the classical problem of solving polynomial equations by radicals. A polynomial $f \in k[x]$ is *solvable by radicals* if its roots can be expressed by a formula involving the coefficients of $f$ and the operations $+, -, \times, \div$, and $n$-th root extraction. The quadratic, cubic, and quartic formulas show that all polynomials of degree $\le 4$ are solvable by radicals. Galois's theorem settles the general case. The key insight is that adjoining a radical $\sqrt[n]{a}$ to a field $k$ (where $k$ contains the $n$-th roots of unity) produces an extension with cyclic Galois group $\mathbb{Z}/n\mathbb{Z}$. A tower of such extensions — a radical tower — has a Galois group that is assembled from abelian pieces: at each stage the Galois group of one step over the previous one is cyclic, so the full group is built from abelian quotients via repeated extensions. What abstract group-theoretic property captures exactly this structure? We need a property that holds for abelian groups, is preserved under extensions, and fails for groups like $A_5$ that cannot arise from radical towers. The answer is *solvability*: a group is solvable if it can be broken into abelian layers via a normal series with abelian quotients. Abelian groups are solvable in one step; extensions of solvable groups by solvable groups remain solvable; and $A_5$, being simple and non-abelian, is not solvable. This is precisely the dividing line between polynomials admitting a radical formula and those that do not. [definition: Solvable Group] A group $G$ is **solvable** if there exists a normal series \begin{align*} \{e\} &= G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_r = G \end{align*} such that each quotient $G_i / G_{i-1}$ is abelian. [/definition] [quotetheorem:1321] The alternating group $A_5$ is the smallest non-abelian simple group; it is not solvable. For $n \ge 5$, the symmetric group $S_n$ contains $A_n$ as a normal subgroup, and $A_n$ is simple and non-abelian, so $S_n$ is not solvable. A general polynomial of degree $n \ge 5$ over $\mathbb{Q}$ has Galois group $S_n$ (this requires a separate argument, but it is true for "most" polynomials in a precise sense). Therefore: [quotetheorem:3312] [example: An Unsolvable Quintic] The polynomial $f(x) = x^5 - 4x + 2 \in \mathbb{Q}[x]$ is not solvable by radicals. We verify: **Irreducibility:** By Eisenstein's criterion with $p = 2$: $2 \mid 4$ and $2 \mid 2$, but $4 \nmid 2$. So $f$ is irreducible over $\mathbb{Q}$. **Galois group is $S_5$:** Since $f$ is irreducible of degree $5$, the Galois group $G = \operatorname{Gal}(\operatorname{Split}(f, \mathbb{Q}) / \mathbb{Q})$ embeds in $S_5$ and has order divisible by $5$ (because $5 = [{\mathbb{Q}(\alpha) : \mathbb{Q}}]$ divides $|G|$ for any root $\alpha$). By Cauchy's theorem, $G$ contains an element of order $5$, hence a $5$-cycle in $S_5$. By calculus, $f'(x) = 5x^4 - 4$, which has two real roots $\pm (4/5)^{1/4}$. Evaluating $f$ at these critical points (using a calculator or a careful sign analysis) shows that $f$ has exactly $3$ real roots and $2$ non-real complex roots. Complex conjugation restricts to a transposition in $G \le S_5$. A transposition and a $5$-cycle together generate $S_5$ (this is a standard group theory lemma: a transposition and a $p$-cycle in $S_p$ generate $S_p$ when $p$ is prime). Therefore $G = S_5$. **Not solvable:** $S_5$ is not solvable (since $A_5$ is simple non-abelian, the derived series of $S_5$ is $S_5 \supset A_5 \supset A_5 \supset \cdots$, never reaching $\{e\}$). By Galois's theorem, $f$ is not solvable by radicals. [/example] This example crystallizes the beauty of the theory: a question about the existence of a formula — purely algebraic and centuries old — is resolved by understanding the symmetry group of a splitting field. ## References Emil Artin, *Galois Theory* (1944). The canonical lecture notes that established the modern Galois-theoretic framework; the proof of the main theorem follows Artin's approach. Serge Lang, *Algebra*, 3rd ed. (2002). Chapters V–VI cover splitting fields, normal extensions, and the Galois correspondence in full generality, including the inseparable case. David Dummit and Richard Foote, *Abstract Algebra*, 3rd ed. (2004). Chapters 13–14 give a thorough, example-driven treatment of field theory and Galois theory including solvability by radicals. Ian Stewart, *Galois Theory*, 4th ed. (2015). An accessible introduction with emphasis on the historical problem of solvability by radicals and explicit Galois group computations. Jean-Pierre Serre, *Local Fields* (1979). Chapter I treats the characteristic-$p$ theory, including Frobenius and inseparable extensions, in the context of local fields.