Separable polynomial

If P is not assumed irreducible the concept is of lesser importance, since repeated roots may then just reflect that P isn't 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(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 p-th 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(T^1/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 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).

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