Consider the polynomial $x^2 + 1 \in \mathbb{Q}[x]$. Over $\mathbb{Q}$, this polynomial has no roots — there is no rational number whose square is $-1$. The same failure occurs in $\mathbb{R}$: the equation $x^2 + 1 = 0$ is simply unsolvable. This is not a minor inconvenience. It means that $\mathbb{R}[x]$ contains polynomials that refuse to factor into linear factors, and this algebraic incompleteness forces us to work with irreducible polynomials of arbitrarily high degree, making the structure of $\mathbb{R}[x]$ considerably more complicated than it could be. The passage to $\mathbb{C}$ fixes this particular failure. But it raises a more fundamental question: does $\mathbb{C}$ itself have the same problem? Could there be a polynomial with complex coefficients that has no complex root? The Fundamental Theorem of Algebra asserts that no such polynomial exists — every non-constant polynomial in $\mathbb{C}[x]$ splits completely into linear factors over $\mathbb{C}$. This property is not accidental; it is the defining feature of an algebraically closed field. The theory of algebraic closures asks: given any field $k$, can we always find a "big enough" extension that is algebraically closed? And if so, is this extension unique in any reasonable sense? The answers — yes and yes, up to isomorphism — are among the most beautiful results in abstract algebra. The construction requires tools from both field theory and set theory (specifically, Zorn's lemma), and the resulting object, the algebraic closure $\overline{k}$, plays the same role for $k$ that $\mathbb{C}$ plays for $\mathbb{R}$: it is the smallest algebraically closed field containing $k$, and every polynomial over $k$ splits completely within it. [example: Factoring a Quadratic over Various Fields] The polynomial $f(x) = x^2 + 1$ behaves differently depending on the base field. Over $\mathbb{Q}$, it is irreducible: there is no $a \in \mathbb{Q}$ with $a^2 = -1$, since $-1$ is not a perfect square in $\mathbb{Q}$. Over $\mathbb{R}$, it remains irreducible for the same reason. Over $\mathbb{C}$, it factors as $(x - i)(x + i)$ where $i^2 = -1$. Over the field $\mathbb{F}_5 = \mathbb{Z}/5\mathbb{Z}$, we check: $1^2 + 1 = 2$, $2^2 + 1 = 5 = 0$. So $x = 2$ is a root, and $f(x) = (x - 2)(x - 3)$ in $\mathbb{F}_5[x]$, since $3^2 + 1 = 10 = 0$ as well. Over $\mathbb{F}_3$, we check: $0^2 + 1 = 1$, $1^2 + 1 = 2$, $2^2 + 1 = 5 = 2$. None are zero, so $x^2 + 1$ is irreducible over $\mathbb{F}_3$. The pattern here is subtle: whether $-1$ is a square in $\mathbb{F}_p$ depends on whether $p \equiv 1 \pmod{4}$ (by quadratic reciprocity). This example illustrates that algebraic closure is not a property of a polynomial in isolation — it depends entirely on the ambient field. [/example] ## Definition The central notion is the idea of a field in which every polynomial has a root. This might seem like a modest requirement, but it turns out to be remarkably strong: if every polynomial has at least one root, then every polynomial splits completely. The field $\mathbb{Q}$ does not have this property. The polynomial $x^2 - 2$ has no root in $\mathbb{Q}$. But even $\mathbb{Q}(\sqrt{2})$ fails: the polynomial $x^2 + 1$ has no root there. The process of adjoining roots one by one never terminates for $\mathbb{Q}$ using finite extensions alone — there are always new polynomials without roots. The algebraically closed fields are precisely those for which this iterative process is unnecessary. [definition: Algebraically Closed Field] A field $k$ is **algebraically closed** if every non-constant polynomial $f \in k[x]$ has at least one root in $k$. [/definition] Equivalently, $k$ is algebraically closed if and only if every non-constant polynomial $f \in k[x]$ factors completely into linear factors: \begin{align*} f(x) = a(x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_n) \end{align*} where $a, \alpha_1, \ldots, \alpha_n \in k$ and $n = \deg f$. The equivalence between these two characterizations is itself a theorem: if every polynomial has a root, then by induction on degree, every polynomial splits into linear factors. If $f \in k[x]$ has degree $n \geq 1$, take a root $\alpha_1 \in k$; then $f(x) = (x - \alpha_1)g(x)$ for some $g \in k[x]$ of degree $n-1$, and the process continues. [remark: The Constant Polynomials] The restriction to non-constant polynomials is necessary. The polynomial $f(x) = c$ for a nonzero constant $c \in k$ has no roots by definition (the equation $c = 0$ has no solution), so including constant polynomials in the requirement would make the definition vacuously impossible to satisfy. [/remark] The notion of algebraic closure of a specific field $k$ is distinct from the notion of an algebraically closed field. We want not just any algebraically closed field, but the smallest one containing $k$. The intuition here is closure under algebraic operations. Every element of $\overline{k}$ should be reachable from $k$ by taking roots of polynomials. If we also added transcendental elements — elements with no polynomial relation to $k$ — we would produce a field larger than necessary. [definition: Algebraic Closure] Let $k$ be a field. An **algebraic closure** of $k$ is a field $\overline{k}$ satisfying: 1. $\overline{k}$ is an algebraic extension of $k$, meaning every element $\alpha \in \overline{k}$ is algebraic over $k$: there exists a non-zero polynomial $f \in k[x]$ with $f(\alpha) = 0$. 2. $\overline{k}$ is algebraically closed. [/definition] Condition (1) ensures $\overline{k}$ contains no "unnecessary" transcendental elements. Condition (2) ensures $\overline{k}$ is large enough that every polynomial over $k$ (or even over $\overline{k}$ itself) has roots. Together, they pin down $\overline{k}$ as tightly as possible. The notation $\overline{k}$ for the algebraic closure is standard (see §18.3 of the Notation Standards). In the important special case $k = \mathbb{Q}$, the algebraic closure $\overline{\mathbb{Q}}$ is the field of all algebraic numbers — complex numbers that satisfy a polynomial equation with rational coefficients. This is a countable field properly contained in $\mathbb{C}$. [explanation: Why Both Conditions Are Needed] Each condition alone is insufficient. A field satisfying only condition (1) — algebraic over $k$ — need not be algebraically closed. For example, $\mathbb{Q}(\sqrt{2})$ is algebraic over $\mathbb{Q}$, but $x^2 + 1$ has no root there. A field satisfying only condition (2) — algebraically closed — might be far larger than necessary. The field $\mathbb{C}(t)$ of rational functions over $\mathbb{C}$ is not algebraically closed (the polynomial $x^2 - t$ has no root in $\mathbb{C}(t)$), but its algebraic closure contains many transcendental elements over $\mathbb{Q}$. Even $\mathbb{C}$ itself, while algebraically closed, is not algebraic over $\mathbb{Q}$: it contains transcendental numbers like $\pi$ and $e$. The two conditions together force $\overline{k}$ to be "just right": large enough to contain all algebraic elements, small enough to contain nothing more. [/explanation] [example: The Algebraic Closure of the Rationals] The algebraic closure $\overline{\mathbb{Q}}$ consists of all complex numbers that are algebraic over $\mathbb{Q}$. This includes: - All rational numbers $\mathbb{Q}$ - Quadratic irrationalities like $\sqrt{2}$, $i = \sqrt{-1}$, $\sqrt{-3}$ - Higher algebraic numbers like $2^{1/3}$, primitive $n$-th roots of unity $\zeta_n = e^{2\pi i/n}$ - Numbers like $\sqrt{2 + \sqrt{3}}$: setting $\alpha = \sqrt{2 + \sqrt{3}}$, we have $\alpha^2 = 2 + \sqrt{3}$, so $\alpha^2 - 2 = \sqrt{3}$, and squaring gives $(\alpha^2 - 2)^2 = 3$, i.e., $\alpha^4 - 4\alpha^2 + 4 = 3$, so $\alpha^4 - 4\alpha^2 + 1 = 0$ It does not contain $\pi$ or $e$, which are transcendental. Note that $\overline{\mathbb{Q}}$ is a subfield of $\mathbb{C}$ — in fact $\overline{\mathbb{Q}} \subsetneq \mathbb{C}$ since $\mathbb{C}$ contains uncountably many elements while $\overline{\mathbb{Q}}$ is countable. [/example] ## Existence and Uniqueness The most significant question is whether every field has an algebraic closure. Unlike the passage from $\mathbb{R}$ to $\mathbb{C}$, which can be written down explicitly as pairs of real numbers with a specific multiplication rule, the general construction is unavoidably abstract and requires Zorn's lemma. There is an obvious obstruction to a naive approach. Given a field $k$, we might try to adjoin a root of each irreducible polynomial one by one. But there are infinitely many irreducible polynomials over $k$ (in fact, infinitely many of arbitrarily large degree), and adjoining roots one at a time produces only a countable chain of extensions — fine for countable fields like $\mathbb{Q}$, but the argument needs care even there: after adjoining roots of all polynomials over $\mathbb{Q}$, we must check that no new polynomials (with coefficients in the enlarged field) have been left without roots. The cleaner approach is to work with all irreducible polynomials simultaneously. We consider the polynomial ring $k[\{x_f\}_{f \text{ irred}}]$ in one variable $x_f$ for each irreducible $f \in k[x]$, and force $x_f$ to be a root of $f$ by passing to an appropriate quotient. A maximality argument via Zorn's lemma then produces the desired closure. [quotetheorem:1313] The proof runs as follows. Let $I$ be the ideal of the polynomial ring $k[\{x_f\}_{f \text{ irred}}]$ generated by all $f(x_f)$ for irreducible $f \in k[x]$. One shows that $I$ is a proper ideal (using the fact that any finite collection of irreducible polynomials can be simultaneously rooted in some finite extension of $k$). By Zorn's lemma, $I$ is contained in a maximal ideal $\mathfrak{m}$, and the quotient $k_1 = k[\{x_f\}]/\mathfrak{m}$ is a field extension of $k$ in which every irreducible polynomial over $k$ has a root. Iterating this construction transfinitely (or equivalently, using Zorn's lemma on the poset of algebraic extensions in which every polynomial over $k$ has a root) produces $\overline{k}$. The uniqueness statement requires a separate argument. Two algebraic closures of $k$ need not be equal as sets, but they are always isomorphic via an isomorphism that fixes $k$ pointwise. [quotetheorem:1260] This theorem is a consequence of a more general result: any embedding of $k$ into an algebraically closed field $\Omega$ extends to an embedding of any algebraic extension $K/k$ into $\Omega$. Applying this twice — first extending the identity embedding $k \hookrightarrow \overline{k}'$ to $\overline{k} \hookrightarrow \overline{k}'$, then noting that the image of $\overline{k}$ is both algebraic over $k$ and algebraically closed, hence equals $\overline{k}'$ — gives the isomorphism. [remark: The Isomorphism Is Not Canonical] The isomorphism $\overline{k} \xrightarrow{\sim} \overline{k}'$ is guaranteed to exist but is generally not unique. Complex conjugation $z \mapsto \bar{z}$ is a non-trivial automorphism of $\mathbb{C}$ over $\mathbb{R}$, and indeed $\operatorname{Gal}(\mathbb{C}/\mathbb{R}) \cong \mathbb{Z}/2\mathbb{Z}$. For $\overline{\mathbb{Q}}$, the automorphism group $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ — the absolute Galois group — is an extraordinarily complicated profinite group, central to modern number theory. The fact that we cannot choose a canonical isomorphism between algebraic closures reflects this richness. [/remark] [illustration:algebraic-extensions-lattice] ## Algebraically Closed Fields and Their Structure Having established existence and uniqueness, we turn to the internal structure of algebraically closed fields. What do they look like? What distinguishes one algebraically closed field from another? The first surprise is that the property of being algebraically closed places very strong constraints on the polynomial ring $k[x]$. [quotetheorem:3319] This is a direct consequence of the definition. If $k$ is algebraically closed and $f \in k[x]$ is irreducible of degree $n \geq 2$, then $f$ has a root $\alpha \in k$, so $(x - \alpha) \mid f$. Since $f$ is irreducible, $f = c(x - \alpha)$ for some constant $c \in k^\times$, contradicting $\deg f = n \geq 2$. Conversely, if every irreducible is linear, then any $f \in k[x]$ factors completely into linear factors, and in particular has a root. This characterization has a clean consequence for field extensions of algebraically closed fields. [quotetheorem:3320] The proof is immediate: any $\alpha \in K$ satisfies a minimal polynomial $m_{\alpha,k} \in k[x]$. Since $k$ is algebraically closed, this polynomial splits completely over $k$, and in particular $\alpha \in k$ (as $m_{\alpha,k}$ is irreducible, hence linear). So $K \subset k$, meaning $K = k$. This theorem confirms the intuition that algebraic closures are "maximally algebraic": you cannot algebraically extend them further. The two algebraically closed fields that arise most naturally in practice are $\mathbb{C}$ (the algebraic closure of $\mathbb{R}$) and $\overline{\mathbb{F}_p}$ (the algebraic closure of the finite field $\mathbb{F}_p$). These are fundamentally different in character — one has characteristic zero, the other characteristic $p$ — and this difference is captured by the following structural result. [quotetheorem:3321] The prime field of an algebraically closed field of characteristic zero is $\mathbb{Q}$, and of characteristic $p$ is $\mathbb{F}_p$. The transcendence degree measures the "size" of the transcendental part of the field. For $\mathbb{C}$, the transcendence degree over $\mathbb{Q}$ is uncountable (equal to the cardinality of $\mathbb{R}$, or equivalently $2^{\aleph_0}$). For $\overline{\mathbb{F}_p}$, the transcendence degree over $\mathbb{F}_p$ is zero — the extension is purely algebraic. [example: The Algebraic Closure of a Finite Field] Let $p$ be a prime. The algebraic closure $\overline{\mathbb{F}_p}$ has a beautiful explicit description. Recall that for each $n \geq 1$, there is a unique field $\mathbb{F}_{p^n}$ of order $p^n$ (up to isomorphism), and $\mathbb{F}_{p^m} \subset \mathbb{F}_{p^n}$ if and only if $m \mid n$. The algebraic closure is the union: \begin{align*} \overline{\mathbb{F}_p} = \bigcup_{n=1}^{\infty} \mathbb{F}_{p^n}. \end{align*} This union is taken along the system of inclusions $\mathbb{F}_{p^n} \hookrightarrow \mathbb{F}_{p^{n!}}$ (or any compatible directed system). To verify this is algebraically closed: any polynomial $f \in \overline{\mathbb{F}_p}[x]$ of degree $d$ has coefficients lying in $\mathbb{F}_{p^n}$ for some $n$. The splitting field of $f$ over $\mathbb{F}_{p^n}$ is a finite extension of $\mathbb{F}_{p^n}$, hence has order $p^{nm}$ for some $m \geq 1$. By the theory of finite fields, $\mathbb{F}_{p^{nm}} \subset \mathbb{F}_{p^{n \cdot (nm)!}}$, and every finite field of order $p^r$ is contained in $\mathbb{F}_{p^s}$ whenever $r \mid s$. Since $nm \mid (nm)!$, all roots of $f$ lie in $\mathbb{F}_{p^{nm!}} \subset \mathbb{F}_{p^{(nd)!}}$, which is already in the union. Every element of the union is algebraic over $\mathbb{F}_p$ since every $\mathbb{F}_{p^n}$ is a finite (hence algebraic) extension of $\mathbb{F}_p$. Note that $\overline{\mathbb{F}_p}$ is a countably infinite field, in contrast to $\mathbb{C}$, which is uncountable. [/example] ## The Extension Theorem and Embeddings The uniqueness of algebraic closure is a special case of a far more useful result: the ability to extend field embeddings to larger fields. This extension theorem is the engine behind much of Galois theory. It tells us that algebraic closures are, in a precise sense, "universal" among algebraic extensions. Why does this matter? In Galois theory, we study automorphisms of field extensions — field isomorphisms that fix the base field. These automorphisms arise by asking: given a root $\alpha$ of an irreducible polynomial $f \in k[x]$, can we send $\alpha$ to another root $\beta$ of the same polynomial? The extension theorem tells us exactly when and how such partial maps extend to full automorphisms of the closure. [quotetheorem:3322] The proof uses Zorn's lemma on the poset of pairs $(E, \tau_E)$ where $k \subset E \subset K$ is an intermediate field and $\tau_E: E \to \Omega$ extends $\sigma$. Any chain has an upper bound (take the union), so a maximal element $(E_{\max}, \tau_{\max})$ exists. If $E_{\max} \subsetneq K$, pick $\alpha \in K \setminus E_{\max}$; then $\alpha$ is algebraic over $E_{\max}$, and its minimal polynomial over $E_{\max}$ has a root in $\Omega$ (since $\Omega$ is algebraically closed), allowing us to extend $\tau_{\max}$ to $E_{\max}(\alpha)$, contradicting maximality. [example: Counting Embeddings of a Cube Root Extension] Let $k = \mathbb{Q}$ and $K = \mathbb{Q}(2^{1/3})$. The minimal polynomial of $2^{1/3}$ over $\mathbb{Q}$ is $f(x) = x^3 - 2$. The extension theorem says any embedding $\sigma: \mathbb{Q} \to \mathbb{C}$ (there is only one, the inclusion) extends to an embedding $\tau: K \to \mathbb{C}$. To specify $\tau$, we must choose where $2^{1/3}$ goes — it must go to a root of $f$ in $\mathbb{C}$. The three roots are: \begin{align*} \alpha_1 &= 2^{1/3}, \quad \alpha_2 = 2^{1/3} \cdot \zeta_3, \quad \alpha_3 = 2^{1/3} \cdot \zeta_3^2, \end{align*} where $\zeta_3 = e^{2\pi i/3}$ is a primitive cube root of unity. Each choice gives a distinct embedding $\tau_j: K \to \mathbb{C}$ defined by $\tau_j(2^{1/3}) = \alpha_j$ and $\tau_j|_{\mathbb{Q}} = \operatorname{id}_{\mathbb{Q}}$. Only $\tau_1$ maps $K$ into the real subfield $\mathbb{R}$; the others map into the complex plane. This count — three embeddings, matching $[K : \mathbb{Q}] = 3$ — is no coincidence and reflects the separability of $x^3 - 2$ over $\mathbb{Q}$. [/example] A key corollary of the extension theorem is that every algebraic extension embeds into the algebraic closure. [quotetheorem:3323] This follows immediately from the extension theorem applied to $\sigma = \operatorname{id}_k: k \hookrightarrow \overline{k}$ and the algebraically closed field $\Omega = \overline{k}$. The practical consequence: when studying algebraic extensions of $k$, we may always assume our fields live inside a fixed algebraic closure $\overline{k}$. This is the working assumption throughout Galois theory — all field extensions are subfields of $\overline{k}$, and Galois groups are groups of automorphisms of these subfields. ## Galois Groups and the Absolute Galois Group The most profound structure associated to the algebraic closure is the absolute Galois group. This object encodes the symmetries of all algebraic extensions of $k$ simultaneously. An automorphism of $\overline{k}$ over $k$ is a field isomorphism $\sigma: \overline{k} \to \overline{k}$ that fixes every element of $k$. The collection of all such automorphisms forms a group under composition. [definition: Absolute Galois Group] Let $k$ be a field with algebraic closure $\overline{k}$. The **absolute Galois group** of $k$ is \begin{align*} G_k := \operatorname{Gal}(\overline{k}/k) = \{\sigma: \overline{k} \xrightarrow{\sim} \overline{k} : \sigma|_k = \operatorname{id}_k\}. \end{align*} [/definition] For finite Galois extensions $K/k$ (that is, normal and separable extensions of finite degree), the Galois group $\operatorname{Gal}(K/k)$ is a finite group. The absolute Galois group $G_k$ is a profinite group — a topological group that is the inverse limit of the finite Galois groups $\operatorname{Gal}(K/k)$ over all finite Galois extensions $K/k$ inside $\overline{k}$: \begin{align*} G_k = \varprojlim_{K/k \text{ finite Galois}} \operatorname{Gal}(K/k). \end{align*} The topology on $G_k$ is the inverse limit topology (equivalently, the topology of pointwise convergence on $\overline{k}$, treating $\overline{k}$ as a discrete set). [example: The Absolute Galois Group of R] The algebraic closure of $\mathbb{R}$ is $\mathbb{C}$. The extension $\mathbb{C}/\mathbb{R}$ has degree $2$ and is Galois (it is the splitting field of $x^2 + 1$). Its Galois group has exactly two elements: the identity $\operatorname{id}_\mathbb{C}$ and complex conjugation $\sigma: z \mapsto \bar{z}$. Therefore: \begin{align*} G_\mathbb{R} = \operatorname{Gal}(\mathbb{C}/\mathbb{R}) \cong \mathbb{Z}/2\mathbb{Z}. \end{align*} This is one of the simplest possible absolute Galois groups — a consequence of the Artin–Schreier theorem, which characterizes all fields with finite absolute Galois group: they are either algebraically closed, or real closed fields (like $\mathbb{R}$) with absolute Galois group $\mathbb{Z}/2\mathbb{Z}$. [/example] [example: The Absolute Galois Group of a Finite Prime Field] The absolute Galois group $G_{\mathbb{F}_p} = \operatorname{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p)$ is a profinite group topologically generated by a single element: the Frobenius automorphism $\operatorname{Frob}_p$, defined by \begin{align*} \operatorname{Frob}_p: \overline{\mathbb{F}_p} &\to \overline{\mathbb{F}_p} \\ x &\mapsto x^p. \end{align*} For each $n$, the restriction $\operatorname{Frob}_p|_{\mathbb{F}_{p^n}}$ generates $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) \cong \mathbb{Z}/n\mathbb{Z}$. Taking the inverse limit: \begin{align*} G_{\mathbb{F}_p} \cong \hat{\mathbb{Z}} = \varprojlim_{n \geq 1} \mathbb{Z}/n\mathbb{Z}, \end{align*} the profinite completion of $\mathbb{Z}$. The element $1 \in \hat{\mathbb{Z}}$ corresponds to the Frobenius. This is topologically a compact abelian group containing $\mathbb{Z}$ as a dense subgroup. [/example] The absolute Galois group $G_{\mathbb{Q}} = \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ is far more complex than either of these. Its structure is one of the deepest open problems in mathematics. The known information about $G_{\mathbb{Q}}$ comes through its representations: for each prime $\ell$, there is a continuous homomorphism $G_{\mathbb{Q}} \to GL_2(\mathbb{Z}_\ell)$ attached to each elliptic curve $E/\mathbb{Q}$, encoding the action of $G_{\mathbb{Q}}$ on the $\ell$-power torsion points $E[\ell^n]$. The modularity theorem (formerly the Taniyama–Shimura conjecture) identifies these representations with those arising from modular forms — weight-$2$ cusp forms for congruence subgroups of $SL_2(\mathbb{Z})$. The precise dictionary between $G_{\mathbb{Q}}$ and the world of automorphic forms is the subject of the Langlands program, which predicts that all motivic Galois representations — those arising from the $\ell$-adic cohomology of algebraic varieties over $\mathbb{Q}$ — are automorphic. ## Splitting Fields and Normal Extensions The algebraic closure is too large for most purposes. It contains roots of every polynomial in $k[x]$ simultaneously — including polynomials we may have no interest in — and it is in general an infinite, even uncountably infinite, extension of $k$. When studying a specific polynomial $f \in k[x]$, adjoining all of $\overline{k}$ is like bringing in the entire library to read one chapter: most of it is irrelevant, and the excess structure obscures what is specific to $f$. In practice, we want the smallest extension in which $f$ itself splits — no more, no less. This is the splitting field. Its degree over $k$ is finite, it is uniquely determined by $f$ up to isomorphism, and it carries exactly the symmetry group — the Galois group of $f$ — that governs the roots of $f$. A polynomial $f \in k[x]$ of degree $n$ splits over $k$ if it factors as \begin{align*} f(x) = a(x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_n) \end{align*} with $a, \alpha_1, \ldots, \alpha_n \in k$. [definition: Splitting Field] Let $f \in k[x]$ be a non-constant polynomial. A **splitting field** of $f$ over $k$ is a field extension $K/k$ such that: 1. $f$ splits completely in $K[x]$: $f(x) = a(x - \alpha_1) \cdots (x - \alpha_n)$ with each $\alpha_i \in K$. 2. $K = k(\alpha_1, \ldots, \alpha_n)$: $K$ is generated over $k$ by the roots of $f$. [/definition] Every splitting field is a finite algebraic extension of $k$, hence a subfield of $\overline{k}$ (after choosing an embedding). Any two splitting fields of $f$ over $k$ are isomorphic over $k$. The splitting field of $f$ is the subfield of $\overline{k}$ generated by all roots of $f$. In this sense, $\overline{k}$ is the "splitting field of everything at once" — it is generated over $k$ by the roots of all polynomials in $k[x]$. The Galois group of a polynomial $f$ acts by permuting the roots of $f$. For this action to be well-defined on the splitting field $K$ — meaning every automorphism of $\overline{k}$ fixing $k$ sends $K$ to itself — we need $K$ to contain, along with each root $\alpha$ of an irreducible polynomial $p \in k[x]$, all other roots of $p$ as well. If $K$ contained only some roots of $p$, an automorphism could move $\alpha$ to a root outside $K$, breaking the action. The condition that isolates exactly these well-behaved extensions is normality. [definition: Normal Extension] An algebraic extension $K/k$ is **normal** if every irreducible polynomial in $k[x]$ that has at least one root in $K$ splits completely in $K[x]$. [/definition] [quotetheorem:1316] For finite extensions, the family can be taken to be a single polynomial. Normal extensions are exactly the fields that are "closed" under the action of $G_k$: if $K/k$ is normal and $\sigma \in G_k$, then $\sigma(K) = K$. [example: Normality and Its Failure] The extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ is normal: the minimal polynomial $x^2 - 2$ splits as $(x - \sqrt{2})(x + \sqrt{2})$ over $\mathbb{Q}(\sqrt{2})$, since $-\sqrt{2} \in \mathbb{Q}(\sqrt{2})$. The extension $\mathbb{Q}(2^{1/3})/\mathbb{Q}$ is not normal. The minimal polynomial $x^3 - 2$ has one root $2^{1/3} \in \mathbb{Q}(2^{1/3})$, but the other two roots $2^{1/3}\zeta_3$ and $2^{1/3}\zeta_3^2$ are not in $\mathbb{Q}(2^{1/3}) \subset \mathbb{R}$, since they are non-real. The splitting field of $x^3 - 2$ over $\mathbb{Q}$ is $\mathbb{Q}(2^{1/3}, \zeta_3)$, which has degree $6$ over $\mathbb{Q}$ and is a normal extension with Galois group $S_3$. [/example] ## Separability and Perfect Fields The theory described above is cleanest when every irreducible polynomial is separable — has no repeated roots in the algebraic closure. In characteristic zero, this is automatic. In characteristic $p$, it can fail. What goes wrong in characteristic $p$? Consider the polynomial $f(x) = x^p - t \in \mathbb{F}_p(t)[x]$, where $t$ is a transcendental variable. In the algebraic closure $\overline{\mathbb{F}_p(t)}$, there exists an element $\alpha$ with $\alpha^p = t$. Then: \begin{align*} x^p - t = x^p - \alpha^p = (x - \alpha)^p \end{align*} in $\overline{\mathbb{F}_p(t)}[x]$, using the Frobenius identity $(x - \alpha)^p = x^p - \alpha^p$ in characteristic $p$. So $f$ has only one distinct root $\alpha$, with multiplicity $p$. The polynomial $f$ is irreducible over $\mathbb{F}_p(t)$ (it has no roots there, and it is irreducible over $\mathbb{F}_p(t)$: by Eisenstein's criterion applied to the prime element $t \in \mathbb{F}_p[t]$, the polynomial $x^p - t$ has no roots in $\mathbb{F}_p(t)$ and does not factor into polynomials of smaller degree), yet it is not separable. [definition: Separable Polynomial] A polynomial $f \in k[x]$ is **separable** if it has no repeated roots in $\overline{k}$, i.e., if $f$ and its formal derivative $f'$ have no common factor of positive degree in $k[x]$. [/definition] The example above — $x^p - t = (x - \alpha)^p$ — is the prototype of an inseparable polynomial. The failure of separability for $x^p - t$ is visible in its derivative: the formal derivative of $x^p - t$ in characteristic $p$ is $px^{p-1} - 0 = 0$, which is the zero polynomial. So $\gcd(f, f') = f$, a factor of positive degree, confirming the repeated root. This observation motivates extending separability from polynomials to field extensions: we want to know whether an entire extension $K/k$ exhibits this failure. [definition: Separable Extension] An algebraic extension $K/k$ is **separable** if every element $\alpha \in K$ has a separable minimal polynomial over $k$. [/definition] [quotetheorem:1262] The proof is concise: if $f \in k[x]$ is irreducible and $f' = 0$, then $f' = 0$ as a polynomial, which forces $f$ to be a constant in characteristic zero (since the derivative of any non-constant polynomial is non-zero in characteristic zero). But $f$ is non-constant, so $f' \neq 0$. Since $f$ is irreducible and $\deg f' < \deg f$, we have $\gcd(f, f') = 1$, so $f$ is separable. The theorem above shows that separability of extensions is an intrinsic property of the base field: in characteristic zero, the problem never arises; in characteristic $p$, it arises precisely when the Frobenius fails to be surjective. Fields for which separability is guaranteed — regardless of which algebraic extension you take — are singled out not merely as a naming convenience, but because they are precisely the fields over which Galois theory works in its cleanest form: the fundamental theorem of Galois theory, the correspondence between subgroups and intermediate fields, holds without any inseparability correction. Over imperfect fields, the theory requires additional machinery to handle the purely inseparable part of the extension. [definition: Perfect Field] A field $k$ is **perfect** if every algebraic extension of $k$ is separable. [/definition] The following theorem classifies perfect fields. [quotetheorem:3324] All finite fields $\mathbb{F}_{p^n}$ are perfect: the Frobenius $x \mapsto x^p$ is an automorphism of $\mathbb{F}_{p^n}$ (it is injective on a finite set, hence surjective). The field $\mathbb{F}_p(t)$ of rational functions is not perfect: $t$ is not a $p$-th power in $\mathbb{F}_p(t)$. For perfect fields, the algebraic closure $\overline{k}$ is also the separable closure: every algebraic extension of $k$ is already separable, so there is no distinction. For imperfect fields, one must distinguish between the separable closure $k^{\mathrm{sep}}$ (the maximal separable extension, consisting of all separably algebraic elements) and the full algebraic closure $\overline{k}$, with the extension $\overline{k}/k^{\mathrm{sep}}$ being purely inseparable. [illustration:separable-closure-tower] ## References Emil Artin, *Galois Theory* (1944). Serge Lang, *Algebra* (3rd edition, 2002). Chapter V covers algebraic closure, splitting fields, and separability with full proofs. Michael Atiyah and Ian MacDonald, *Introduction to Commutative Algebra* (1969). Chapter 5 covers integral extensions and algebraic closures from the commutative algebra perspective. Patrick Morandi, *Field and Galois Theory* (1996). Chapter 1–3 develop the theory of algebraic extensions and algebraic closure at a careful pace. David Cox, *Galois Theory* (2nd edition, 2012). Chapter 3 covers splitting fields and algebraic closures with extensive examples.