Perfect field

In mathematics, particularly in the field of algebra, a perfect field is defined as a field K in which every algebraic extension is separable, meaning that the minimal polynomial of every element in the extension over K has distinct roots.^[1] Equivalently, K is perfect if every irreducible polynomial over K is separable, or if K has characteristic zero, or if K has positive characteristic p and the Frobenius endomorphism (raising elements to the p-th power) is surjective, i.e., every element of K is a p-th power.^[2] This property ensures that there are no purely inseparable extensions of K, simplifying the structure of field extensions and algebraic geometry over such fields.^[3] Fields of characteristic zero, such as the rational numbers \mathbb{Q}, the real numbers \mathbb{R}, and the complex numbers \mathbb{C}, are always perfect, as separability holds automatically in characteristic zero due to the nonzero derivative of polynomials.^[4] In characteristic p > 0, finite fields like \mathbb{F}_p (the prime field of p elements) are perfect because the Frobenius map is bijective on finite sets.^[2] Algebraically closed fields and algebraic extensions of perfect fields are also perfect.^[3] However, not all fields of positive characteristic are perfect; for instance, the field of rational functions \mathbb{F}_p(t) over a finite field is imperfect, as it admits inseparable extensions like the extension adjoining a p-th root of t.^[4] The concept of perfect fields plays a crucial role in algebraic geometry and number theory, where imperfect fields lead to complications such as non-reduced schemes or inseparable morphisms; over perfect fields, reduced algebras remain geometrically reduced after base change.^[1] For any field K, there exists a unique minimal perfect extension called the perfect closure K^{perf}, obtained by iteratively adjoining p-th roots in characteristic p, which is purely inseparable over K.^[1] This closure is essential for studying properties invariant under perfection, such as étale cohomology or the behavior of algebraic varieties.^[3]

Definition

Formal Definition

A field k is perfect if every algebraic extension L/k is separable.^[1] An algebraic extension L/k is separable if every element \alpha \in L is separable over k, meaning that the minimal polynomial of \alpha over k has distinct roots in an algebraic closure of k.^[5] In a perfect field, every finite extension is separable and thus Galois if normal.^[2]

Equivalent Characterizations

A field k is perfect if and only if every irreducible polynomial over k has distinct roots, or equivalently, is separable.^[2] This condition ensures that the derivative of any such polynomial does not share roots with it, as separability requires \gcd(f, f') = 1 for irreducible f.^[6] This characterization is equivalent to the statement that every algebraic extension of k is separable.^[2] To see this, suppose every irreducible polynomial over k is separable; then for any algebraic extension L/k, the minimal polynomial of any \alpha \in L over k is separable, implying \alpha is separable over k, and thus L/k is separable (since every element is separable over k). Conversely, if an irreducible polynomial f over k were inseparable, adjoining a root \alpha would yield a separable extension only if f splits into distinct linear factors, but inseparability implies multiple roots, contradicting the assumption that all algebraic extensions are separable.^[2]^[6] In characteristic p > 0, k is perfect if and only if the Frobenius endomorphism F: k \to k, defined by F(x) = x^p, is surjective, meaning every element of k is a p-th power.^[2] Equivalently, for every a \in k, there exists b \in k such that b^p = a.^[6] This surjectivity holds if and only if the image of the Frobenius map equals k, i.e., k = k^p.^[2] If k^p \neq k, then for a \notin k^p, the polynomial T^p - a is irreducible (by Eisenstein or direct verification) but inseparable, as its derivative vanishes and it factors as (T - \alpha)^p in a splitting field, violating the separability condition.^[2]^[6]

Properties

In Characteristic Zero

In fields of characteristic zero, every such field is perfect.^[2] This follows from the fact that the formal derivative of any non-constant polynomial over such a field is itself a non-zero polynomial.^[6] Specifically, for a polynomial f(x) = \sum_{i=0}^n a_i x^i with a_n \neq 0 and n \geq 1, the derivative is f'(x) = \sum_{i=1}^n i a_i x^{i-1}; since the characteristic is zero, the coefficients i a_i are non-zero for at least the leading term (where i = n), ensuring \deg f' = n-1 \geq 0.^[2] A key theorem states that every irreducible polynomial over a field of characteristic zero is separable.^[1] To see this, suppose f(x) is irreducible of degree n \geq 1. If f(x) had a multiple root in some extension, then f(x) and f'(x) would share a common root, implying \gcd(f, f') \neq 1.^[6] However, since f'(x) \neq 0 and \deg f' < \deg f, the only possible non-constant common divisor would be f(x) itself, which would require f(x) to divide f'(x); this is impossible because \deg f' < \deg f.^[2] Thus, \gcd(f, f') = 1, so f(x) has distinct roots in its splitting field and is separable.^[1] As a consequence, there are no inseparable extensions of fields of characteristic zero, meaning every algebraic extension is separable.^[6] This absence of inseparability simplifies the structure of Galois theory over such fields, as finite normal extensions coincide with Galois extensions without additional separability conditions.^[2]

In Characteristic p

In fields of characteristic p > 0, a field k is perfect if and only if the Frobenius endomorphism F: k \to k, defined by F(x) = x^p, is an automorphism.^[7] This map is always a ring homomorphism in characteristic p, and for fields, it is injective because the kernel of a nonzero field homomorphism must be trivial.^[7] Surjectivity of F is equivalent to k = k^p, the condition that every element of k is a p-th power in k.^[7] A key consequence is that perfect fields of characteristic p admit no nontrivial purely inseparable extensions. In such fields, every algebraic extension is separable, as the separability of irreducible polynomials over k follows from the surjectivity of the Frobenius map, ensuring that derivatives and multiple roots do not arise in the expected manner.^[8] Purely inseparable extensions, which rely on elements whose minimal polynomials have repeated roots, thus reduce to the trivial case over perfect k.^[8] Moreover, the automorphism property extends iteratively: the n-th iterate F^n: k \to k, given by x \mapsto x^{p^n}, is also an automorphism for every positive integer n, implying k^{p^n} = k.^[9] This allows the extraction of p^n-th roots within algebraic closures of k. For imperfect fields, the perfect closure is obtained by iteratively adjoining all p^n-th roots over increasing n, yielding the minimal perfect extension.^[9]

Examples

Perfect Fields

Finite fields provide fundamental examples of perfect fields in positive characteristic. Every finite field \mathbb{F}_q, where q = p^n for a prime p and positive integer n, is perfect because the Frobenius endomorphism x \mapsto x^p is an injective field automorphism, and injectivity on a finite set implies surjectivity.^[10] This surjectivity ensures that every element has a p-th root in the field, satisfying the criterion for perfection in characteristic p.^[1] Fields of characteristic zero are perfect by definition, as the Frobenius map is irrelevant and all algebraic extensions are separable due to the absence of inseparable polynomials.^[11] The rational numbers \mathbb{Q}, real numbers \mathbb{R}, and complex numbers \mathbb{C} exemplify this, with every irreducible polynomial over them having distinct roots in any extension.^[12] Similarly, p-adic fields such as the field of p-adic numbers \mathbb{Q}_p and its finite extensions are perfect, inheriting the characteristic zero property from \mathbb{Q}.^[13] Number fields, which are finite extensions of \mathbb{Q}, are also perfect for the same reason.^[14] For instance, quadratic extensions like \mathbb{Q}(\sqrt{2}) or \mathbb{Q}(\sqrt{-3}) exhibit this perfection, as their characteristic zero nature guarantees separability of all extensions.^[11] Algebraically closed fields offer another class of perfect fields, independent of characteristic. Any algebraically closed field, such as the algebraic closure \overline{\mathbb{Q}} of the rationals, is perfect because every irreducible polynomial over it is linear, hence separable, ensuring all algebraic extensions are separable.^[15] In characteristic p > 0, examples include the algebraic closure of \mathbb{F}_p, where the surjectivity of the Frobenius map holds due to the field's properties.^[1]

Imperfect Fields

A prototypical example of an imperfect field is the rational function field \mathbb{F}_p(t) over the prime field \mathbb{F}_p of characteristic p > 0. In this field, the transcendental element t is not a p-th power of any element, and the polynomial x^p - t \in \mathbb{F}_p(t) is irreducible but inseparable, as it has a multiple root in its splitting field.^[2] This inseparability demonstrates that \mathbb{F}_p(t) admits purely inseparable algebraic extensions, violating the condition for perfection. The imperfection of \mathbb{F}_p(t) can also be seen through the Frobenius endomorphism \phi: a \mapsto a^p, which fails to be surjective, as t does not lie in the image \mathbb{F}_p(t)^p.^[2] More generally, this non-surjectivity aligns with the characterization that fields of characteristic p are imperfect precisely when the Frobenius map is not onto. Function fields of curves over finite fields provide further examples of imperfect fields in positive characteristic. For a smooth projective curve C over a finite field k = \mathbb{F}_q with q = p^n, the function field k(C) is imperfect, as it contains transcendental elements over k that are not p-th powers, leading to inseparable polynomials analogous to x^p - t. Unless the base field is algebraically closed, such global fields typically exhibit this behavior due to their transcendental nature over the perfect finite base. Certain local fields constructed as fields of formal Laurent series in characteristic p over an imperfect coefficient field, such as \mathbb{F}_p(t)((u)), are also imperfect. Here, the residue field \mathbb{F}_p(t) being imperfect ensures that the Frobenius map on the series field remains non-surjective, propagating inseparability.^[2] A key implication of imperfection in these fields is the existence of non-trivial purely inseparable extensions. For instance, adjoining a p-th root of t to \mathbb{F}_p(t) yields a degree-p extension that is purely inseparable, as the minimal polynomial x^p - t has no roots in the base but splits with multiplicity p in the extension.^[2]

Extensions

Separability of Extensions

A fundamental consequence of a field k being perfect is that every algebraic extension L/k is separable.^[1] For finite extensions, this follows because every irreducible polynomial over a perfect field has distinct roots in its splitting field, ensuring that the extension is generated by separable elements.^[2] Infinite algebraic extensions are then separable as unions (or direct limits) of finite separable subextensions.^[2] A key corollary is that every finite extension of a perfect field is separable, and thus every finite normal extension is Galois.^[2] In particular, the separable closure of a perfect field k coincides with its algebraic closure, as all algebraic elements are separable.^[16] This guaranteed separability simplifies the study of Galois groups over perfect fields in positive characteristic, eliminating the need to handle inseparable extensions separately.^[2]

Perfect Closure

For an imperfect field k of characteristic p > 0, the perfect closure k^{\mathrm{perf}} is defined as the minimal perfect field extension of k.^[1] It can be characterized as the direct limit of the directed system obtained by iteratively applying the Frobenius endomorphism F: k \to k^p, a \mapsto a^p, yielding k^{\mathrm{perf}} = \varinjlim (k \xrightarrow{F} k^p \xrightarrow{F} k^{p^2} \xrightarrow{F} \cdots), where elements are equivalence classes of sequences (a_n)_{n \geq 0} satisfying a_{n+1}^p = a_n for all n.^[17] Equivalently, k^{\mathrm{perf}} is obtained by adjoining all p^n-th roots of elements of k iteratively for n \geq 0.^[1] The explicit construction of k^{\mathrm{perf}} proceeds as the ascending union k^{\mathrm{perf}} = \bigcup_{n=0}^\infty k^{1/p^n}, where k^{1/p^0} = k and, for each n \geq 1, k^{1/p^n} is the unique algebraic extension of k^{1/p^{n-1}} obtained by adjoining a p^n-th root for every element of k^{1/p^{n-1}}, ensuring that every element of k^{1/p^n} raised to the p^n-th power lies in k.^[1] This tower of extensions is purely inseparable. The natural map k \to k^{\mathrm{perf}} is injective, as the Frobenius endomorphism on a field is injective.^[1] The perfect closure k^{\mathrm{perf}} enjoys several key properties: it is perfect by construction, as every element admits a p-th root within it; it is flat as a k-algebra; and it is the maximal purely inseparable extension of k, containing every algebraic purely inseparable extension of k.^[1]^[18] Moreover, k^{\mathrm{perf}} is unique up to unique isomorphism over k, and any perfect extension of k contains k^{\mathrm{perf}} as an intermediate field.^[1] If k is imperfect, such as the field \mathbb{F}_p(t) of rational functions over the prime field, then [k^{\mathrm{perf}} : k] = \infty.^[19]