Perfect field

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

• Every irreducible polynomial over k has distinct roots.
• Every irreducible polynomial over k is separable.
• Every finite extension of k is separable.
• Every algebraic extension of k is separable.
• Either k has characteristic 0, or, when k has characteristic p > 0, every element of k is a pth power.
• Either k has characteristic 0, or, when k has characteristic p > 0, the Frobenius endomorphism x→x^p is an automorphism of k
• The separable closure of k is algebraically closed.
• Every reduced commutative k-algebra A is a separable algebra; i.e., $A \otimes_k F$ is reduced for every field extension F/k. (see below)

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).

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

Examples

Examples of perfect fields are:

• every field of characteristic zero, e.g. the field of rational numbers or the field of complex numbers;
• every finite field, e.g. the field F_p = Z/pZ where p is a prime number;
• every algebraically closed field;
• the union of a set of perfect fields totally ordered by extension;
• fields algebraic over a perfect field.

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 $k(X)$ of all rational functions in an indeterminate $X$, where k has characteristic p>0 (because X has no p-th root in k(X)).

Field extension over a perfect field

Any finitely generated field extension over a perfect field is separably generated.

Perfect closure and perfection

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

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

In terms of universal properties, the perfect closure of a ring A of characteristic p is a perfect ring A_p of characteristic p together with a ring homomorphism u : A → A_p such that for any other perfect ring B of characteristic p with a homomorphism v : A → B there is a unique homomorphism f : A_p → B 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.

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 : B → A, there is a unique map f : B → R(A) such that φ factors through θ (i.e. φ = θf). The perfection of A may be constructed as follows. Consider the projective system

$\cdots\rightarrow A\rightarrow A\rightarrow A\rightarrow\cdots$

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 $x_{i+1}^p=x_i$ for all i. The map θ : R(A) → A sends (x_i) to x₀.

See also

• p-ring
• Quasi-finite field