Suppose you want to take the derivative of a polynomial $f \in k[x]$ and use it to detect repeated roots. Over $\mathbb{Q}$ or $\mathbb{R}$, this works perfectly: $f$ has a repeated root $\alpha$ if and only if $\alpha$ is also a root of the formal derivative $f'$. But over a field of characteristic $p > 0$, something surprising happens. Consider the polynomial $f(x) = x^p - t \in \mathbb{F}_p(t)[x]$, where $t$ is an indeterminate. Its formal derivative is $f'(x) = px^{p-1} = 0$ — identically zero, because the characteristic kills every coefficient. So $\gcd(f, f') = f$, suggesting all roots are repeated. But $f$ is irreducible! In the algebraic closure $\overline{\mathbb{F}_p(t)}$, the element $\alpha = t^{1/p}$ satisfies $\alpha^p = t$, and \begin{align*} x^p - t &= x^p - \alpha^p = (x - \alpha)^p, \end{align*} so $f$ has a single root $\alpha$ of multiplicity $p$. The extension $\mathbb{F}_p(t)(\alpha)/\mathbb{F}_p(t)$ is a degree-$p$ extension in which every element is algebraic — yet the minimal polynomial of $\alpha$ has only one root in $\overline{\mathbb{F}_p(t)}$, so there can be no nontrivial automorphism of the extension. The Galois group has order 1, but the degree is $p$. The Galois correspondence cannot work here: there are far fewer automorphisms than dimensions. This is the phenomenon of inseparability. An extension is separable when the pathology above cannot occur: when the minimal polynomial of every element has no repeated roots, and hence enough automorphisms exist to make the extension behave well. Separability, alongside normality, is one of the two conditions that must both hold for an extension to be Galois — for the Galois correspondence to be a perfect bijection rather than a partial one. The theory of separability is therefore not just about polynomial roots: it is about understanding which field extensions admit the full Galois machinery, and what replaces it when they do not. [example: An Inseparable Extension] Let $k = \mathbb{F}_2(t)$ and $\alpha = t^{1/2} \in \overline{\mathbb{F}_2(t)}$. The minimal polynomial of $\alpha$ over $k$ is $f(x) = x^2 - t$. In characteristic 2, we have $x^2 - t = (x - \alpha)^2$, since $\alpha^2 = t$. So $f$ has a single root $\alpha$ of multiplicity 2 in $\overline{k}$. Let $K = k(\alpha) = \mathbb{F}_2(t^{1/2})$. Then $[K : k] = 2$, since $f = x^2 - t$ is the minimal polynomial of $\alpha$ and is irreducible over $k$ (it has no root in $k$: if $g(t)/h(t) \in k$ satisfied $(g/h)^2 = t$, then $g^2 = th^2$, but comparing degrees modulo 2 gives a contradiction). Now suppose $\sigma \in \operatorname{Aut}(K/k)$ is an automorphism of $K$ fixing $k$. Then $\sigma(\alpha)$ must be a root of $x^2 - t$ in $K$. But $x^2 - t = (x - \alpha)^2$ over $K$, so $\alpha$ is the only root. Therefore $\sigma(\alpha) = \alpha$, which means $\sigma$ is the identity. We have $|\operatorname{Aut}(K/k)| = 1$ but $[K : k] = 2$: the extension is degree 2 but has only the trivial automorphism. This is the hallmark of an inseparable extension. [/example] ## Definition The failure in the example is that the minimal polynomial of $\alpha$ has repeated roots. The concept of separability isolates exactly when this cannot happen. Before defining separable extensions, we need the notion of a separable polynomial. What goes wrong with $x^p - t$? Its formal derivative is zero, so $\gcd(f, f') = f$, and $f$ divides its own formal derivative. The formal derivative is a purely algebraic operation — it does not require limits — and it detects repeated roots without reference to an algebraic closure. [definition: Separable Polynomial] Let $k$ be a field and $f \in k[x]$ a nonconstant polynomial. The polynomial $f$ is **separable** if it has no repeated roots in $\overline{k}$, where $\overline{k}$ is an algebraic closure of $k$. [/definition] Equivalently, $f$ is separable if and only if $\gcd(f, f') = 1$ in $k[x]$, where $f'$ is the formal derivative of $f$. The equivalence between the two conditions is a theorem, not part of the definition — we state it separately below. The formal-derivative condition is more computational; the no-repeated-roots condition is more geometric. [remark: Formal Derivative] For $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \in k[x]$, the formal derivative is \begin{align*} f'(x) &= n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \cdots + a_1. \end{align*} This is defined purely algebraically, with no notion of limit. In characteristic zero, $f' = 0$ only if $f$ is a nonzero constant. In characteristic $p > 0$, $f'$ can vanish even when $f$ is nonconstant: for example, $(x^p)' = p x^{p-1} = 0$. [/remark] The formal derivative gives a clean computational test for separability: rather than factoring over an algebraic closure, one can check whether a polynomial shares a common factor with its own derivative. [quotetheorem:3329] The key observation behind this theorem is that $\alpha$ is a repeated root of $f$ if and only if $(x - \alpha)^2 \mid f(x)$ in $\overline{k}[x]$, which occurs if and only if $(x - \alpha) \mid f(x)$ and $(x - \alpha) \mid f'(x)$ — that is, $\alpha$ is a common root of $f$ and $f'$. But common roots in $\overline{k}$ are detected by the $\gcd$ in $k[x]$ (via the Euclidean algorithm, which is valid over any field). Irreducible polynomials behave particularly well with respect to separability. An irreducible polynomial over a field of characteristic zero is always separable; the same holds over finite fields. But over imperfect fields of characteristic $p$, inseparable irreducible polynomials can exist — and they are always of a special form. [quotetheorem:3330] In characteristic zero, $f' = 0$ forces $f$ to be a constant, contradicting irreducibility. So every irreducible polynomial over a field of characteristic zero is separable. Over a field of characteristic $p$, the inseparable irreducibles are exactly those of the form $g(x^p)$: polynomials in $x^p$ alone. The simplest examples are $x^p - a$ for $a \in k$ that is not a $p$-th power in $k$. With the polynomial notion in hand, we can lift it to elements and extensions. The point of isolating a separable *element* is that separability is a local property: it depends only on the minimal polynomial of one element, not on the whole extension. An extension can contain a mix of separable and inseparable elements, and it is worth being precise about which elements cause trouble before asking about the whole extension. [definition: Separable Element] Let $K/k$ be an algebraic extension and $\alpha \in K$. The element $\alpha$ is **separable over $k$** if its minimal polynomial $\operatorname{min}(\alpha, k) \in k[x]$ is a separable polynomial — that is, $\operatorname{min}(\alpha, k)$ has no repeated roots in $\overline{k}$. [/definition] Once we can classify individual elements, the extension-level notion is immediate: an extension is separable when every element in it is. The requirement is not vacuous even for small extensions — an extension of degree $p$ can be entirely inseparable if it is generated by a single inseparable element. [definition: Separable Extension] An algebraic extension $K/k$ is a **separable extension** if every element $\alpha \in K$ is separable over $k$. [/definition] The most important special cases: all algebraic extensions of fields of characteristic zero are separable, and all algebraic extensions of finite fields are separable. Inseparability is a purely characteristic-$p$ phenomenon, and even there it requires the base field to be imperfect. [definition: Perfect Field] A field $k$ is **perfect** if every algebraic extension of $k$ is separable. [/definition] The condition can be made explicit: a field $k$ is perfect if and only if either $\operatorname{char}(k) = 0$, or $\operatorname{char}(k) = p > 0$ and every element of $k$ is a $p$-th power (i.e., the Frobenius endomorphism $x \mapsto x^p$ is surjective on $k$). These are not additional axioms — they are equivalent characterizations, and the equivalence is a theorem. Over a perfect field, inseparability is impossible, and every algebraic extension is automatically separable. Finite fields $\mathbb{F}_{p^n}$ are perfect: the Frobenius $x \mapsto x^p$ is a bijection on a finite field (injective implies bijective for finite sets). The field $\mathbb{F}_p(t)$ is imperfect: $t$ is not a $p$-th power in $\mathbb{F}_p(t)$, since any $p$-th power has numerator and denominator degrees divisible by $p$. ## Separability Degree and the Primitive Element Theorem Having defined separable extensions, the next question is how to measure "how separable" an extension is, and whether every separable extension is generated by a single element — which would make the structure far more tractable. The number of embeddings of $K$ into $\overline{k}$ over $k$ controls the Galois group size. For a simple extension $k(\alpha)/k$ with minimal polynomial $f = \operatorname{min}(\alpha, k)$ of degree $n$, the number of $k$-embeddings $k(\alpha) \hookrightarrow \overline{k}$ equals the number of roots of $f$ in $\overline{k}$. If $f$ is separable, this is exactly $n = [k(\alpha) : k]$; if $f$ is inseparable, it is strictly less. [definition: Separable Degree] Let $K/k$ be a finite algebraic extension. The **separable degree** of $K/k$, denoted $[K : k]_s$, is the number of distinct $k$-embeddings $\sigma: K \hookrightarrow \overline{k}$. The **inseparable degree** (or **purely inseparable degree**) is \begin{align*} [K : k]_i &:= \frac{[K : k]}{[K : k]_s}. \end{align*} [/definition] The inseparable degree is always a power of $p = \operatorname{char}(k)$ when $\operatorname{char}(k) = p > 0$, and equals 1 in characteristic zero. A finite extension is separable if and only if $[K : k]_i = 1$, equivalently $[K : k]_s = [K : k]$. [quotetheorem:3331] This multiplicativity is the analogue of the tower law for degrees. It tells us that separability is a transitive property: we can check separability one step at a time in a tower. The primitive element theorem is one of the most useful structural results in the theory. It says that a finite separable extension is always simple — generated by a single element over the base field. Why might we need a condition? Consider a purely inseparable extension: $K = \mathbb{F}_p(t^{1/p}, s^{1/p})$ over $k = \mathbb{F}_p(t, s)$ (two algebraically independent variables). This extension has degree $p^2$ but cannot be generated by a single element — any single element $\alpha = a \cdot t^{1/p} + b \cdot s^{1/p}$ (with $a, b \in k$) generates a degree-$p$ extension, not degree $p^2$. The failure is entirely due to inseparability. [quotetheorem:1267] The proof is elegant in the infinite case: if $k$ is infinite, take any $\alpha \in K$ with minimal polynomial of maximum degree, or more directly, for any two elements $\beta, \gamma$ with $K = k(\beta, \gamma)$, show that $k(\beta + c\gamma) = K$ for all but finitely many $c \in k$ — and an infinite field supplies such $c$. For finite $k$, the argument uses the fact that the multiplicative group of a finite extension of a finite field is cyclic. [example: Finding a Primitive Element] Consider $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $k = \mathbb{Q}$. This is a degree-4 extension (since $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$, as $[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = 2$ and $\sqrt{3}$ satisfies the irreducible polynomial $x^2 - 3$ over $\mathbb{Q}(\sqrt{2})$). The primitive element theorem guarantees a single generator. We claim $\alpha = \sqrt{2} + \sqrt{3}$ is a primitive element, i.e., $K = \mathbb{Q}(\alpha)$. First, $\alpha \in K$ and $\mathbb{Q}(\alpha) \subset K$. We show the reverse inclusion by expressing $\sqrt{2}$ and $\sqrt{3}$ in terms of $\alpha$. Compute $\alpha^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}$, so $\alpha^2 - 5 = 2\sqrt{6}$, giving $\sqrt{6} = \frac{\alpha^2 - 5}{2} \in \mathbb{Q}(\alpha)$. Now $\alpha \cdot \sqrt{6} = (\sqrt{2} + \sqrt{3})\sqrt{6} = \sqrt{12} + \sqrt{18} = 2\sqrt{3} + 3\sqrt{2}$. So: \begin{align*} \alpha \cdot \sqrt{6} &= 2\sqrt{3} + 3\sqrt{2}. \end{align*} We also have $\alpha = \sqrt{2} + \sqrt{3}$. Solving the system: \begin{align*} \sqrt{3} - \sqrt{2} &= \frac{\alpha \cdot \sqrt{6} - 3\alpha}{2} = \frac{\alpha(\sqrt{6} - 3)}{2}. \end{align*} But more directly: from $\alpha = \sqrt{2} + \sqrt{3}$ and $\alpha\sqrt{6} = 2\sqrt{3} + 3\sqrt{2}$, subtract $3\alpha = 3\sqrt{2} + 3\sqrt{3}$: \begin{align*} \alpha\sqrt{6} - 3\alpha &= 2\sqrt{3} + 3\sqrt{2} - 3\sqrt{2} - 3\sqrt{3} = -\sqrt{3}, \end{align*} so $\sqrt{3} = \alpha(3 - \sqrt{6}) = 3\alpha - \alpha\sqrt{6}$. Since $\alpha, \sqrt{6} \in \mathbb{Q}(\alpha)$, we have $\sqrt{3} \in \mathbb{Q}(\alpha)$, and then $\sqrt{2} = \alpha - \sqrt{3} \in \mathbb{Q}(\alpha)$. Thus $K = \mathbb{Q}(\sqrt{2}, \sqrt{3}) \subset \mathbb{Q}(\alpha)$, so $K = \mathbb{Q}(\alpha)$. The minimal polynomial of $\alpha = \sqrt{2} + \sqrt{3}$ over $\mathbb{Q}$ is: \begin{align*} (\alpha^2 - 5)^2 &= (2\sqrt{6})^2 = 24, \end{align*} so $\alpha^4 - 10\alpha^2 + 25 = 24$, giving $\alpha^4 - 10\alpha^2 + 1 = 0$. Thus $\operatorname{min}(\alpha, \mathbb{Q}) = x^4 - 10x^2 + 1$, a degree-4 polynomial — consistent with $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 4$. [/example] ## Separability and Galois Theory The true importance of separability emerges from its role in the Galois correspondence. A Galois extension is one that is both normal and separable, and separability is what ensures there are enough automorphisms — that $|\operatorname{Gal}(K/k)| = [K : k]$. To see why separability is necessary for the Galois correspondence, consider what happens when it fails. For the inseparable extension $k(\alpha)/k$ where $\operatorname{min}(\alpha, k) = x^p - a$ (with $a$ not a $p$-th power in $k$), the minimal polynomial has only one root $\alpha$ in $\overline{k}$ (with multiplicity $p$). So any $k$-embedding $\sigma: k(\alpha) \hookrightarrow \overline{k}$ must send $\alpha$ to a root of $x^p - a$, and the only such root is $\alpha$ itself. Every $k$-embedding fixes $\alpha$, hence is the identity. The automorphism group has order 1, but $[k(\alpha) : k] = p$. The fixed field of the trivial group is all of $k(\alpha)$, not $k$: the Galois correspondence would send $\{e\}$ to the fixed field of $\{e\}$, which is $K$, and it sends $G = \{e\}$ to the fixed field of all of $G$, which should be $k$. But $\{e\} = G$, so both fields would be simultaneously $k$ and $K$. The correspondence collapses. [quotetheorem:1265] Here $\operatorname{Hom}_k(K, \overline{k})$ denotes the set of $k$-algebra homomorphisms from $K$ to $\overline{k}$ — these are the $k$-embeddings of $K$ into $\overline{k}$. When $K/k$ is also normal, every such embedding maps $K$ to itself (see the [Normal Extension](/page/Normal%20Extension) page for details), so $\operatorname{Hom}_k(K, \overline{k}) = \operatorname{Aut}(K/k) = \operatorname{Gal}(K/k)$. [explanation: Why the Count Equals the Separable Degree] For a simple extension $k(\alpha)/k$ with minimal polynomial $f$ of degree $n$, the $k$-embeddings $k(\alpha) \hookrightarrow \overline{k}$ are in bijection with the roots of $f$ in $\overline{k}$: each embedding is determined by where it sends $\alpha$, and that image must be a root of $f$. If $f$ has $s$ distinct roots in $\overline{k}$ (with $s \le n$), there are exactly $s$ embeddings. For separable $f$, $s = n$, so there are $n = [k(\alpha) : k]$ embeddings. For inseparable $f$, $s < n$. For a tower $k \subset F \subset K$, the count multiplies: each embedding $K \hookrightarrow \overline{k}$ over $k$ restricts to an embedding $F \hookrightarrow \overline{k}$ over $k$ (there are $[F:k]_s$ of these), and for each such restriction, there are $[K:F]_s$ extensions to $K \hookrightarrow \overline{k}$ over $F$. The total is $[K:k]_s = [K:F]_s \cdot [F:k]_s$. [/explanation] The interplay between separability and the Galois correspondence is perfectly summarized by the following criterion. [quotetheorem:3310] This is the intrinsic characterization of Galois extensions: $K/k$ is Galois when the fixed field of all $k$-automorphisms of $K$ is exactly the base field $k$ — no larger, no smaller. ## Purely Inseparable Extensions Separability is one extreme; the opposite extreme is equally important and has its own theory. An element is purely inseparable when it is as far from separable as possible: when its minimal polynomial has a single root of high multiplicity. What forces an element to be purely inseparable? If $\operatorname{min}(\alpha, k) = x^p - a$ has a single root $\alpha$ of multiplicity $p$, then $\alpha$ is "purely inseparable" over $k$. More generally, the condition is that the minimal polynomial has only one root in $\overline{k}$. [definition: Purely Inseparable Element] Let $K/k$ be an algebraic extension and $\alpha \in K$. The element $\alpha$ is **purely inseparable over $k$** if $\operatorname{char}(k) = p > 0$ and $\alpha^{p^n} \in k$ for some $n \ge 0$. [/definition] Lifting from elements to extensions is the right move: just as we asked whether every element of an extension is separable, we now ask whether every element is purely inseparable. The resulting class of extensions — purely inseparable extensions — is the orthogonal complement of separable extensions: the two notions together account for all finite extensions, via the separable-then-purely-inseparable decomposition that comes next. [definition: Purely Inseparable Extension] An algebraic extension $K/k$ is **purely inseparable** if every element $\alpha \in K$ is purely inseparable over $k$. [/definition] A purely inseparable extension is the opposite extreme from a separable extension: the only element common to both categories (separable and purely inseparable) over $k$ is the elements of $k$ itself. [quotetheorem:3332] The field $F$ consists of all elements of $K$ that are separable over $k$. The structure is layered: first a separable step (from $k$ to $F$), then a purely inseparable step (from $F$ to $K$). The degrees are related by \begin{align*} [K : k] &= [K : F] \cdot [F : k] = [K : k]_i \cdot [K : k]_s, \end{align*} where $[F : k] = [K : k]_s$ is the separable degree and $[K : F] = [K : k]_i$ is the inseparable degree. [example: Separable Closure Inside a Purely Inseparable Extension] Let $k = \mathbb{F}_p(s, t)$ (with $s, t$ algebraically independent over $\mathbb{F}_p$) and $K = k(s^{1/p}, t^{1/p})$. We have $[K : k] = p^2$, since $s^{1/p}$ satisfies $x^p - s$ (irreducible over $k$ by Eisenstein-type arguments), and then $t^{1/p}$ satisfies $x^p - t$ which remains irreducible over $k(s^{1/p})$. Every element of $K$ has the form $\sum_{i,j=0}^{p-1} c_{ij} s^{i/p} t^{j/p}$ with $c_{ij} \in k$. For such an element $\alpha$ to be separable over $k$, we need $\operatorname{min}(\alpha, k)$ to have no repeated roots. But every element of $K$ satisfies $\alpha^{p^2} \in k$ (since $(s^{i/p} t^{j/p})^{p^2} = s^{ip} t^{jp} \in k$), so $\alpha$ is purely inseparable over $k$. The only elements of $K$ that are separable over $k$ are the elements of $k$ itself. Thus the separable closure of $k$ in $K$ is $F = k$, and $K/k$ is entirely purely inseparable. The separable degree is $[K : k]_s = 1$ and the inseparable degree is $[K : k]_i = p^2$. [/example] [example: A Mixed Extension with a Nontrivial Separable Part] Let $k = \mathbb{F}_2(t)$ and $K = k(\alpha, \beta)$ where $\alpha^2 - \alpha - t = 0$ and $\beta^2 = t$ (so $\beta = t^{1/2}$). The element $\alpha$ satisfies $x^2 - x - t \in k[x]$. The formal derivative is $2x - 1 = -1 \ne 0$ (in characteristic 2, $2x = 0$ so the derivative is $-1 = 1$), and $\gcd(x^2 - x - t, 1) = 1$, so $\alpha$ is separable over $k$. The extension $k(\alpha)/k$ has degree 2 and is separable: $K_s = k(\alpha)$ is the maximal separable subextension. The element $\beta = t^{1/2}$ satisfies $x^2 - t \in k[x]$, which has derivative $2x = 0$ identically. So $\beta$ is purely inseparable over $k$, and over $k(\alpha)$: since $\beta^2 = t \in k \subset k(\alpha)$, the element $\beta$ is purely inseparable over $k(\alpha)$. The extension $k(\alpha, \beta)/k(\alpha)$ is purely inseparable of degree 2. The full degree is $[K : k] = 4$, decomposing as $[K : k]_s = 2$ (the separable part $k(\alpha)/k$) and $[K : k]_i = 2$ (the inseparable part $k(\alpha, \beta)/k(\alpha)$). [/example] ## Separability in Characteristic $p$: The Frobenius In characteristic $p$, the Frobenius endomorphism is the key tool for understanding separability. It is both the source of inseparability (it "collapses" $p$-th powers) and the tool for detecting it. The Frobenius $\varphi_p: k \to k$ given by $\varphi_p(x) = x^p$ is always a ring homomorphism (this is the famous identity $(a + b)^p = a^p + b^p$ in characteristic $p$, which follows from $\binom{p}{i} \equiv 0 \pmod{p}$ for $0 < i < p$). For fields, it is injective. The question is whether it is surjective — that is, whether every element of $k$ has a $p$-th root in $k$, which is exactly the condition that $k$ is perfect. [quotetheorem:3333] The trace characterization in (iv) is particularly useful: the trace is nonzero if and only if the extension is separable. For inseparable extensions, the trace vanishes identically — there are so few embeddings that their sum is always zero. [illustration:separable-inseparable-lattice] [example: Trace of a Separable Extension] Let $K = \mathbb{F}_{p^2}$ over $k = \mathbb{F}_p$. This is a degree-2 separable extension (finite fields are perfect). The $k$-embeddings are $\operatorname{id}$ and the Frobenius $\varphi_p: x \mapsto x^p$. For $\alpha \in K$: \begin{align*} \operatorname{Tr}_{K/k}(\alpha) &= \alpha + \alpha^p. \end{align*} This is not identically zero: take $\alpha = 1$ (if $\mathbb{F}_p$ has $p \ne 2$), giving $\operatorname{Tr}(1) = 1 + 1^p = 2 \ne 0$ for $p \ne 2$; or take any element $\alpha$ with $\alpha \notin \mathbb{F}_p$, for which $\alpha^p \ne \alpha$ and $\alpha + \alpha^p \ne 0$ (the trace form is a nondegenerate bilinear form on separable extensions). The trace is surjective onto $\mathbb{F}_p$. Now compare with the inseparable extension $K = \mathbb{F}_p(t^{1/p})$ over $k = \mathbb{F}_p(t)$. Here $[K:k] = p$ but there is only one $k$-embedding $\sigma: K \hookrightarrow \overline{k}$ over $k$, namely the identity (as established in the opening example: the only root of $x^p - t$ in $\overline{k}$ is $t^{1/p}$, so $\sigma(t^{1/p}) = t^{1/p}$). The trace defined via the sum over all $k$-embeddings gives \begin{align*} \operatorname{Tr}_{K/k}(\alpha) &= \sigma(\alpha) = \alpha \end{align*} for any $\alpha \in K$. But this formula returns elements of $K$, not of $k$: for instance $\operatorname{Tr}_{K/k}(t^{1/p}) = t^{1/p} \notin k$. The trace map is identically zero as a $k$-valued function — that is, $\operatorname{Tr}_{K/k}$ does not take values in $k$ for inseparable extensions, which forces it to be the zero map as a functional $K \to k$. This is exactly the failure predicted by criterion (iv) of the theorem above. [/example] ## The Separable Closure Just as every field has an algebraic closure, it also has a separable closure — the largest separable algebraic extension. The separable closure is the natural domain for the absolute Galois group. Why do we need the separable closure when we already have the algebraic closure? Because over imperfect fields, the algebraic closure is "too big": it contains purely inseparable elements, and the Galois theory of the algebraic closure is ill-behaved. The absolute Galois group $\operatorname{Gal}(\overline{k}/k)$ is not well-defined in the usual sense for imperfect $k$ — the algebraic closure is not a Galois extension of $k$. The separable closure is the right object. [definition: Separable Closure] Let $k$ be a field with algebraic closure $\overline{k}$. The **separable closure** of $k$, denoted $k^{\mathrm{sep}}$, is the compositum inside $\overline{k}$ of all finite separable extensions of $k$: \begin{align*} k^{\mathrm{sep}} &:= \bigcup \{ F : k \subset F \subset \overline{k}, \, F/k \text{ finite and separable} \}. \end{align*} [/definition] The separable closure $k^{\mathrm{sep}}$ is the unique maximal separable extension of $k$ inside $\overline{k}$. It is a Galois extension of $k$ (it is both normal and separable over $k$), and its Galois group $\operatorname{Gal}(k^{\mathrm{sep}}/k)$ is the **absolute Galois group** of $k$. [quotetheorem:3334] For a perfect field, $k^{\mathrm{sep}} = \overline{k}$ and the absolute Galois group is the Galois group of the algebraic closure. For a finite field $\mathbb{F}_p$, the separable closure is $\overline{\mathbb{F}}_p = \bigcup_{n \ge 1} \mathbb{F}_{p^n}$, and the absolute Galois group is the profinite completion $\hat{\mathbb{Z}} = \varprojlim_{n} \mathbb{Z}/n\mathbb{Z}$, topologically generated by the Frobenius. For an imperfect field like $k = \mathbb{F}_p(t)$, the separable closure $k^{\mathrm{sep}}$ is strictly smaller than $\overline{k}$. The element $t^{1/p} \in \overline{k}$ is purely inseparable over $k$ and does not lie in $k^{\mathrm{sep}}$. The absolute Galois group $G_k = \operatorname{Gal}(k^{\mathrm{sep}}/k)$ in this case is a much more complicated profinite group, still not fully understood. [example: The Absolute Galois Group of a Finite Field] Let $k = \mathbb{F}_p$. Since $\mathbb{F}_p$ is perfect, $k^{\mathrm{sep}} = \overline{\mathbb{F}}_p$. The finite Galois extensions of $\mathbb{F}_p$ are exactly the finite fields $\mathbb{F}_{p^n}/\mathbb{F}_p$, with Galois groups: \begin{align*} \operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) &\cong \mathbb{Z}/n\mathbb{Z}, \end{align*} generated by the Frobenius $\varphi_p: x \mapsto x^p$. The absolute Galois group is the inverse limit: \begin{align*} G_{\mathbb{F}_p} &= \operatorname{Gal}(\overline{\mathbb{F}}_p / \mathbb{F}_p) \cong \varprojlim_{n} \mathbb{Z}/n\mathbb{Z} = \hat{\mathbb{Z}}. \end{align*} Here $\hat{\mathbb{Z}} = \prod_{\ell \text{ prime}} \mathbb{Z}_\ell$ is the profinite completion of $\mathbb{Z}$, where $\mathbb{Z}_\ell$ is the ring of $\ell$-adic integers. The profinite generator is the Frobenius $\varphi_p$, which acts on $\overline{\mathbb{F}}_p$ by $\varphi_p(x) = x^p$. The Galois correspondence for $\overline{\mathbb{F}}_p / \mathbb{F}_p$ (via the infinite Galois correspondence with the Krull topology on $\hat{\mathbb{Z}}$) gives a bijection between closed subgroups of $\hat{\mathbb{Z}}$ and intermediate fields. The closed subgroups of $\hat{\mathbb{Z}}$ are exactly $n\hat{\mathbb{Z}}$ for $n \ge 0$ (including $0 \cdot \hat{\mathbb{Z}} = \hat{\mathbb{Z}}$ for the closed subgroup corresponding to $\mathbb{F}_p$, and $\{0\}$ corresponding to $\overline{\mathbb{F}}_p$). The intermediate fields are the $\mathbb{F}_{p^n}$ for $n \ge 1$. The subgroup $n\hat{\mathbb{Z}}$ (of index $n$) corresponds to $\mathbb{F}_{p^n}$, consistent with $[\mathbb{F}_{p^n} : \mathbb{F}_p] = n$. [/example] ## Separability in Function Fields and Transcendence In algebraic geometry and function field arithmetic, separability appears in a different guise: a morphism of varieties (or a map of function fields) is separable when the corresponding field extension is separable. This connects the algebraic notion to geometric and differential conditions. Let $K/k$ be a finitely generated field extension of transcendence degree 1 (a function field of a curve). Such an extension is not algebraic, but there is still a meaningful notion of separability. The notion of separability extends to transcendental extensions through the concept of a separating transcendence basis. The key question is whether we can choose a transcendence basis $\{t_1, \ldots, t_n\}$ inside $K$ such that the residual algebraic step $K/k(t_1, \ldots, t_n)$ is separable. If such a basis exists, the transcendental part contributes no inseparability — all the inseparability, if any, would be hiding in the algebraic step. Over imperfect fields this can fail: the algebraic step may be forced to be inseparable no matter which transcendence basis we pick. [definition: Separably Generated Extension] A finitely generated field extension $K/k$ of transcendence degree $n$ is **separably generated** over $k$ if there exist elements $t_1, \ldots, t_n \in K$ such that $K/k(t_1, \ldots, t_n)$ is a finite separable extension. The set $\{t_1, \ldots, t_n\}$ is called a **separating transcendence basis**. [/definition] Over a perfect field, every finitely generated extension is separably generated (this is Mac Lane's theorem). Over imperfect fields, this can fail. [quotetheorem:3335] The geometric content is this: a morphism $f: X \to Y$ of varieties over a perfect field $k$ is separable (in the sense of algebraic geometry) if and only if the function field extension $k(X)/k(Y)$ is separable. Over imperfect fields, the distinction between separably generated and non-separably-generated extensions corresponds precisely to the distinction between smooth and non-smooth morphisms in characteristic $p$. ## References Emil Artin, *Galois Theory* (1944). The foundational source for separability and the relationship between automorphisms and field degrees. Serge Lang, *Algebra*, 3rd edition (2002). Chapters V and VI develop separability, purely inseparable extensions, and the separable closure systematically. Nathan Jacobson, *Basic Algebra II* (1980). Chapter 8 gives an exhaustive treatment of separability including the norm and trace, the characteristic polynomial, and separably generated extensions. David Cox, *Galois Theory*, 2nd edition (2012). Chapter 6 gives an accessible treatment of separability with many examples over function fields. Alexandre Grothendieck and Jean Dieudonné, *Éléments de Géométrie Algébrique IV* (1964). The definitive source for separability in the context of algebraic geometry and morphisms of schemes.