Perfect field

In algebra, a field k is perfect if any one of the following equivalent conditions holds:

Otherwise, k is called imperfect.

In particular, all fields of characteristic zero and all finite fields are perfect.

Perfect fields are significant because Galois theory over these fields becomes simpler, since the general Galois assumption of field extensions being separable is automatically satisfied over these fields (see third condition above).

Another important property of perfect fields is that they admit Witt vectors.

More generally, a ring of characteristic p (p a prime) is called perfect if the Frobenius endomorphism is an automorphism.[1] (This is equivalent to the above condition "every element of k is a pth power" for integral domains.)

Examples

Examples of perfect fields are:

In fact, most fields that appear in practice are perfect. The imperfect case arises mainly in algebraic geometry in characteristic p>0. Every imperfect field is necessarily transcendental over its prime subfield (the minimal subfield), because the latter is perfect. An example of an imperfect field is

the field ${\displaystyle \mathbb {F} _{q}(x)}$

It embeds into the perfect field

${\displaystyle \mathbb {F} _{q}(x,x^{1/p},x^{1/p^{2}},\ldots )}$

Imperfect fields cause technical difficulties because irreducible polynomials can become reducible in the algebraic closure of the base field. For example,[2] consider ${\displaystyle f(x,y)=x^{p}+ay^{p}\in k[x,y]}$ for ${\displaystyle k}$ an imperfect field of characteristic ${\displaystyle p}$. Then in its algebraic closure ${\displaystyle k^{\operatorname {alg} }[x,y]}$, the following equality holds:

${\displaystyle f(x,y)=(x+a^{1/p}y)^{p}}$.

Geometrically, this means that ${\displaystyle f}$ does not define an affine plane curve in ${\displaystyle k[x,y]}$.

Field extension over a perfect field

Any finitely generated field extension K over a perfect field k is separably generated, i.e. admits a separating transcendence base, that is, a transcendence base Γ such that K is separably algebraic over k(Γ).[3]

Perfect closure and perfection

One of the equivalent conditions says that, in characteristic p, a field adjoined with all pr-th roots (r≥1) is perfect; it is called the perfect closure of k and usually denoted by ${\displaystyle k^{p^{-\infty }}}$.

The perfect closure can be used in a test for separability. More precisely, a commutative k-algebra A is separable if and only if ${\displaystyle A\otimes _{k}k^{p^{-\infty }}}$ is reduced.[4]

In terms of universal properties, the perfect closure of a ring A of characteristic p is a perfect ring Ap of characteristic p together with a ring homomorphism u : AAp such that for any other perfect ring B of characteristic p with a homomorphism v : AB there is a unique homomorphism f : ApB such that v factors through u (i.e. v = fu). The perfect closure always exists; the proof involves "adjoining p-th roots of elements of A", similar to the case of fields.[5]

The perfection of a ring A of characteristic p is the dual notion (though this term is sometimes used for the perfect closure). In other words, the perfection R(A) of A is a perfect ring of characteristic p together with a map θ : R(A) → A such that for any perfect ring B of characteristic p equipped with a map φ : BA, there is a unique map f : BR(A) such that φ factors through θ (i.e. φ = θf). The perfection of A may be constructed as follows. Consider the projective system

${\displaystyle \cdots \rightarrow A\rightarrow A\rightarrow A\rightarrow \cdots }$

where the transition maps are the Frobenius endomorphism. The inverse limit of this system is R(A) and consists of sequences (x0, x1, ... ) of elements of A such that ${\displaystyle x_{i+1}^{p}=x_{i}}$ for all i. The map θ : R(A) → A sends (xi) to x0.[6]