Perfect Fields and Inseparability

Perfect fields quietly govern which parts of Galois theory apply "for free." In arithmetic geometry, the function field $\mathbb{F}_p(t)$ of the projective line over a finite field is imperfect, and inseparable extensions of it correspond geometrically to Frobenius twists and purely inseparable isogenies of algebraic curves and abelian varieties — phenomena entirely absent in characteristic $0$ geometry. In cryptography, elliptic curves over finite fields $\mathbb{F}_q$ rely essentially on the perfection of $\mathbb{F}_q$: the Frobenius endomorphism $x \mapsto x^q$ is a bijection, and this bijectivity underlies point-counting algorithms (Schoof's algorithm) and the structure of the endomorphism ring used in pairing-based cryptosystems. In coding theory, Reed–Solomon and BCH codes are built over finite fields, and the perfection of $\mathbb{F}_q$ guarantees that minimal polynomials of code-defining elements are separable, which is what makes the roots of generator polynomials distinct and the associated linear-algebra machinery (syndromes, error-locator polynomials) well-behaved. In algebraic geometry more broadly, the notion of a variety being smooth versus merely regular diverges precisely in the imperfect-field setting, and resolving this divergence (via notions like geometric reducedness and the Frobenius twist) is a standard technical hurdle in positive-characteristic algebraic geometry. Finally, this entire discussion is the essential preparation for Finite Fields: because every finite field is perfect, the Galois theory of finite fields is unconditionally separable, and the Frobenius automorphism (rather than merely an injective endomorphism) becomes the single generator of every Galois group $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)$.

:::quiz[Over a perfect field $K$ of characteristic $p$, which statement is proved true for every irreducible $f \in K[x]$?]

• $f$ has degree divisible by $p$
• $f' \equiv 0$ always
• $f$ is inseparable whenever $\deg f > 1$
• $\gcd(f,f') = 1$, so $f$ is separable ✓

:::quiz[For a general algebraic extension $L/K$ of finite degree in characteristic $p$, the separable degree $[L:K]_s$ and total degree satisfy]

• $[L:K]_s = [L:K]$ always
• $[L:K] = [L:K]_s \cdot [L:K]_i$ where $[L:K]_i$ is a power of $p$ ✓
• $[L:K]_s$ is always a power of $p$
• $[L:K]_s$ can be larger than $[L:K]$

Ask the tutor: How does the separable/purely-inseparable degree factorization generalize to infinite algebraic extensions, and what role does it play in defining the Galois group of an imperfect field's separable closure?

Think of the derivative as a diagnostic tool for repeated roots: a polynomial $f$ has a repeated root exactly when $f$ and $f'$ share a common root, detected by $\gcd(f,f') \neq 1$. In characteristic $0$ this test never fails silently, because the derivative of a nonconstant polynomial is never the zero polynomial (degree strictly drops, and no integer coefficient can vanish). But in characteristic $p$, differentiation has a blind spot: the derivative of $x^p$ is $p x^{p-1} = 0$. More generally, any polynomial in $x^p$ alone — say $g(x^p)$ — has zero derivative. This means the diagnostic tool can report "no repeated roots visible" even for a polynomial that factors, over an extension, as $(x-\alpha)^p$. Perfection is precisely the hypothesis that plugs this hole. If $K$ is perfect, one can show that no irreducible polynomial over $K$ can have zero derivative (because $a=b^p$ can be absorbed as $(x-b)^p$ trivially, contradicting irreducibility unless the polynomial has degree $1$). So over a perfect field, the diagnostic tool for repeated roots never has a blind spot: every irreducible polynomial is automatically separable. Over an imperfect field, however, there is some element $a \in K$ with no $p$-th root in $K$, and then $x^p - a$ is an irreducible polynomial whose derivative vanishes identically — it is inseparable, and in fact has only one root (with multiplicity $p$) in its splitting field. Picture the "$p$-th power map" as folding the field onto a sub-copy of itself. Perfection says this fold is actually a bijection: nothing is missed. Imperfection means the image $K^p$ is a proper subfield, and the "missing" elements outside $K^p$ are exactly the raw material for building inseparable, and purely inseparable, extensions.

Example 1: An explicit inseparable extension over $\mathbb{F}_p(t)$. Let $K = \mathbb{F}_2(t)$, the field of rational functions in one variable $t$ over $\mathbb{F}_2$. This field is not perfect: consider $a = t \in K$. Suppose $t = c^2$ for some $c = f(t)/g(t) \in K$ with $f,g \in \mathbb{F}_2[t]$ coprime. Then $t g(t)^2 = f(t)^2$. Comparing degrees, $\deg f = \deg g + \tfrac12$, impossible for integers unless we track it via the degree of $t$ carefully: write $\deg(t g^2) = 1 + 2\deg g$, which is odd, while $\deg(f^2) = 2\deg f$ is even. An odd number cannot equal an even number, so no such $c$ exists. Hence $t \notin K^2$, confirming $K$ is imperfect. Now let $L = K(\alpha)$ where $\alpha^2 = t$, i.e. $\alpha = \sqrt{t}$ in a splitting field. By the theorem's proof (the irreducibility argument for $x^p - a$), $f(x) = x^2 - t$ is irreducible over $K$, since $t \notin K^2$. So $[L:K] = 2$, and

using char $2$, so $(-\alpha)=\alpha$. Thus $\alpha$ is the only root of its minimal polynomial, with multiplicity $2$: the extension $L/K$ is purely inseparable of degree $2$. Note also $L = \mathbb{F}_2(\alpha) = \mathbb{F}_2(t^{1/2})$, and indeed $L^2 = \mathbb{F}_2(t) = K \subsetneq L$, so $L$ itself is still imperfect — inseparability, once introduced, does not "heal" automatically; extracting one $p$-th root does not make the field perfect (there remain infinitely many other elements without $p$-th roots, e.g. $t^{1/2}+1$, etc., and $\alpha$ itself has no square root in $L$ unless $t^{1/4}$ is adjoined). The Galois-theoretic pathology is visible directly: $\operatorname{Aut}(L/K)$ is trivial (any $K$-automorphism of $L$ must send $\alpha$ to a root of $x^2 - t$ in $L$, and the only root in $L$ is $\alpha$ itself, since the polynomial factors as $(x-\alpha)^2$ with no other root), yet $[L:K] = 2 > 1 = |\operatorname{Aut}(L/K)|$. This is the signature failure mode that separability rules out: for separable extensions the count of $K$-embeddings into an algebraic closure equals the degree, but here it collapses to $1$.

Example 2: Mixed separable/inseparable degree computation. Let $K = \mathbb{F}_3(t)$, and let $L = K(\beta, \gamma)$ where $\beta$ is a root of $x^3 - t$ (so $\beta^3 = t$, purely inseparable generator) and $\gamma$ is a root of $x^2 - t$ over $K(\beta)$ — but we should be careful: we want a genuinely mixed example, so instead take $\gamma^2 - \gamma - t = 0$, i.e. adjoin a root of the separable Artin–Schreier-type polynomial $g(x) = x^2 - x - t$ over $K(\beta)$. Step 1: $[K(\beta):K] = 3$. As in Example 1, $t \notin K^3$ (an analogous degree-parity argument: if $t = c^3$ with $c=f/g$ in lowest terms, then $\deg(t\,g^3)=1+3\deg g \not\equiv 0 \pmod 3$ while $\deg(f^3) = 3\deg f \equiv 0\pmod 3$, contradiction), so $x^3 - t$ is irreducible over $K$ by the same argument as before (using $\gcd(k,3)=1$ for $1\le k\le 2$), giving degree $3$, and this extension is purely inseparable since $x^3-t=(x-\beta)^3$ in the splitting field (char $3$). Step 2: $g(x) = x^2 - x - t \in K(\beta)[x]$. Compute $g'(x) = 2x - 1$, which is not identically zero (coefficient $2 = -1 \neq 0$ in $\mathbb{F}_3$). Since $g$ has degree $2$, if $g$ is irreducible over $K(\beta)$ then $\gcd(g,g')=1$ automatically forces separability (as $g'\not\equiv 0$ and $\deg g' <\deg g$, a common irreducible factor of $g,g'$ would have to be $g$ itself dividing $g'$, impossible by degree). Is $g$ irreducible over $K(\beta) = \mathbb{F}_3(\beta)$ (note $K(\beta) = \mathbb{F}_3(t^{1/3}) = \mathbb{F}_3(\beta)$ since $t = \beta^3$)? A root of $g$ would need to lie in $\mathbb{F}_3(\beta)$; a short valuation argument at the place $\beta=0$ (i.e. $t=0$) shows $t = \beta^3$ has odd "type" incompatible with $x^2-x-t$ having a root that's a rational function of $\beta$ of the required pole order — concretely, if $\gamma = h(\beta) \in \mathbb{F}_3(\beta)$ solved $\gamma^2-\gamma = \beta^3$, comparing pole orders at $\beta=\infty$ forces $\deg h = 3/2$, again not an integer. So $g$ is irreducible over $K(\beta)$, and by the degree-$2$ argument above it is separable. Hence $[L : K(\beta)] = 2$, separable. By the tower law, $[L:K] = [L:K(\beta)]\,[K(\beta):K] = 2 \cdot 3 = 6$. The separable degree $[L:K]_s$ (the number of $K$-embeddings of $L$ into an algebraic closure $\overline K$, equivalently the degree of the maximal separable subextension) is $2$, since $\beta$ contributes purely inseparably (only one embedding possible, sending $\beta \mapsto \beta$ as it's the unique root of $x^3-t$) while $\gamma$ contributes separably (two embeddings, corresponding to the two roots of $g$ over $K(\beta)$). The inseparable degree is $[L:K]_i = [L:K]/[L:K]_s = 6/2 = 3$. This matches the general theorem that $[L:K]=[L:K]_s[L:K]_i$ with $[L:K]_i$ a power of $p=3$.