Separable polynomial

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In mathematics, a polynomial P(X) is separable over a field K if its roots in an algebraic closure of K are distinct - that is P(X) has distinct linear factors in some large enough field extension. Equivalently, P is separable if and only if it is coprime to its formal derivative P′.

Irreducible polynomials over perfect fields are separable, which includes in particular all fields of characteristic 0, and all finite fields. This criterion is of technical importance in Galois theory. In this connection, the concept of separability is of lesser importance if P is not assumed irreducible, since repeated roots may then just reflect that P is not square-free.

The criterion above leads to the quick conclusion that if P is irreducible and not separable, then

P′(X) = 0.

This is only possible as a characteristic p phenomenon: we must have

P(X) = Q(X^p)

where the prime number p is the characteristic.

With this clue we can construct an example:

P(X) = X^p − T

with K the field of rational functions in the indeterminate T over the finite field with p elements. Here one can prove directly that P(X) is irreducible, and not separable. This is actually a typical example of why inseparability matters; in geometric terms P represents the mapping on the projective line over the finite field, taking co-ordinates to their pth power. Such mappings are fundamental to the algebraic geometry of finite fields. Put another way, there are coverings in that setting that cannot be 'seen' by Galois theory. (See radical morphism for a higher-level discussion.)

If L is the field extension

K(T^1/p),

in other words the splitting field of P, then L/K is an example of a purely inseparable field extension. It is of degree p, but has no automorphism fixing K, other than the identity, because T^1/p is the unique root of P. This shows directly that Galois theory must here break down. A field such that there are no such extensions is called perfect. That finite fields are perfect follows a posteriori from their known structure.

One can show that the tensor product of fields of L with itself over K for this example has nilpotent elements that are non-zero. This is another manifestation of inseparability: that is, the tensor product operation on fields need not produce a ring that is a product of fields (so, not a commutative semisimple ring).

If P(x) is separable, and its roots form a group (a subgroup of the field K), then P(x) is an additive polynomial.