Perfect field

This article defines a field property: a property that can be evaluated to true/false for any field.
View a complete list of field properties|View a complete list of field extension properties

Definition

A field is termed perfect if it satisfies the following equivalent conditions:

1. It is either a field of characteristic zero or it has characteristic $p$ for some prime $p$, and the Frobenius map is surjective.
2. Every finite extension of the field is a separable extension.
3. Every irreducible polynomial over the field is a separable polynomial.

Equivalence of definitions

Further information: Equivalence of definitions of perfect field

Relation with other properties

Stronger properties

• Algebraically closed field
• Field of characteristic zero: For proof of the implication, refer Characteristic zero implies perfect and for proof of its strictness (i.e. the reverse implication being false) refer Perfect not implies characteristic zero.
• Finite field: For proof of the implication, refer Finite implies perfect and for proof of its strictness (i.e. the reverse implication being false) refer Perfect of prime characteristic not implies finite.

Metaproperties

Subfield-realizability

This field property is subfield-realizable: every field can be realized as a subfield of a field with this property.
View other subfield-realizable field properties

Every field can be realized as a subfield of a perfect field. For a field of characteristic zero, the field itself is perfect. For a field of characteristic $p$, the perfect closure is the field obtained by adjoining $(p^n{)}^{th}$ roots of all elements of the field for all $n$.

Finite-extension-closedness

This field property is closed under taking finite extensions. In other words, any finite extension of a field with this property also has this property.
View other such properties

A finite extension of a perfect field is perfect. For full proof, refer: Perfectness is finite-extension-closed

Suppose $K_1$ and $K_2$ are two perfect subfields of a field $K$. Then, the composite of $K_1$ and $K_2$, i.e., the subfield they generate, is also perfect. For full proof, refer: Perfectness is composite-closed