In mathematics, an irreducible polynomial P(X) is separable if its roots in an algebraic closure are distinct - that is P(X) has distinct linear factors in some large enough field extension. This is the 'usual' case, since it turns out to be true if P is defined over a field K that is either (a) of characteristic 0, or (b) a finite field. This criterion is of technical importance in Galois theory.
If P is not assumed irreducible the concept is of lesser importance, since repeated roots may then just reflect that P is not square-free . We can test for common factors of P(X) and the derivative P′(X) over any field, using the calculus formula: any repeated root will divide the highest common factor of P and P′. This 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(Xp) where the prime number p is the characteristic.
With this clue we can construct an example:
- P(X) = Xp − 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.
If L is the field extension K(T1/p) (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 T1/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.