Perfect field

In algebra, a field ${\displaystyle K}$ is perfect if any one of the following equivalent conditions holds:

• Every irreducible polynomial over ${\displaystyle K}$ has no multiple roots in any field extension ${\displaystyle L/K}$.
• Every irreducible polynomial over ${\displaystyle K}$ has non-zero formal derivative.
• Every irreducible polynomial over ${\displaystyle K}$ is separable.
• Every finite extension of ${\displaystyle K}$ is separable.
• Every algebraic extension of ${\displaystyle K}$ is separable.
• Either ${\displaystyle K}$ has characteristic 0, or, when ${\displaystyle K}$ has characteristic ${\displaystyle p>0}$, every element of ${\displaystyle K}$ is a ${\displaystyle p}$-th power.
• Either ${\displaystyle K}$ has characteristic 0, or, when ${\displaystyle K}$ has characteristic ${\displaystyle p>0}$, the Frobenius endomorphism ${\displaystyle x\mapsto x^{p}}$ is an automorphism.
• The separable closure of ${\displaystyle K}$ is algebraically closed.
• Every reduced commutative ${\displaystyle K}$-algebra ${\displaystyle A}$ is a separable algebra; i.e., ${\displaystyle A\otimes _{k}L}$ is reduced for every field extension ${\displaystyle L/K}$.

Otherwise, ${\displaystyle K}$ 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 ${\displaystyle p}$ (${\displaystyle p}$ a prime) is called perfect if the Frobenius endomorphism is an automorphism. When restricted to integral domains, this is equivalent to the above condition "every element of ${\displaystyle K}$ is a ${\displaystyle p}$-th power".