In classical Galois theory, separability isn’t a real condition, since it holds for every extension of Q. But in the most general setting, it’s useful to treat it, especially since it does apply to function fields.
Let K be a field, and f be an irreducible polynomial over K. There’s a splitting field for f, where it by definition splits into linear factors. If the linear factors are all distinct, the polynomial is called separable; otherwise, it is called inseparable. For example, x^2 + 1 splits into (x + i)(x – i), so it’s separable since i != –i. A reducible polynomial f is called separable iff all of its irreducible factors are separable.
If char(K) = 0, then every polynomial over K is separable. This is because a repeated root of any polynomial divides its derivative. Further, if f(x) = a(n)x^n + … + a0 is irreducible, then f‘(x) = na(n)x^(n-1) + … + a1 is a nonzero polynomial of smaller degree, so it is not divisible by f. As f is irreducible, f‘ must then be coprime to f, which means it can’t have any root in common with f.
The above fails to hold in positive characteristic because the factor n might be 0; for instance, if f = x^p – 1, where p = char(K), then f‘ = 0. However, there’s still a large class of fields of positive characteristic over which every polynomial is separable, including all finite ones.
To see why, first note that if char(K) = p, then the function on K defined by f(a) = a^p is a ring homomorphism: it’s clearly multiplicative, and it’s also additive because the binomial expansion of (a + b)^p is such that all factors except a^p and b^p have coefficients divisible by p, making them equal to 0 in K. That function is called the Frobenius endomorphism on K; it’s injective because every nonzero homomorphism whose domain is a field is injective.
An inseparable irreducible polynomial must be a polynomial in x^p, because if x^k has a nonzero coefficient and p is not divisible by k, then f‘ includes the nonzero term ka(k)x^(k-1). But if it’s of the form (b(n)^p)x^(np) + … + (b(p)^p)x^p + b0^p then it’s equal to (a(n)x^n + … + a0)^p, contradicting its irreducibility.
So if every a in K can be written as b^p, then every irreducible polynomial over K is separable. This is clearly equivalent to the condition that the Frobenius endomorphism is surjective, i.e. an automorphism. The converse is also true, because if a in K can’t be written as b^p, then x^p – a is an inseparable irreducible polynomial.
If K is finite, then the Frobenius endomorphism has to be surjective by a counting argument. If K is merely algebraic over a finite field then it’s still surjective, because then K/F is algebraic, so for every a in K, F(a)/F is finite. The finiteness of F then turns F(a) into a finite field, on which the Frobenius endomorphism is surjective; that yields a suitable b with b^p = a. Therefore, the simplest field over which there exists an inseparable polynomial is F(p)(t), over which x^p – t is inseparable.
Now, let L/K be a field extension. An element a of L is said to be separable over K if its minimal polynomial is. That clearly holds iff in the normal closure of L, m(a) has precisely deg(m(a)) distinct roots. The extension is then said to be separable if every element in it is.
Clearly, if M/L/K is a tower of extensions and M/K is separable, then so is L/K. In fact so is M/L, since the minimal polynomial of each a in M over L is a factor of the minimal polynomial of a over K; therefore, if the minimal polynomial over K has distinct roots, then so does the minimal polynomial over L.
If L/K is a field extension, then a in L is inseparable iff m(a) only has nonzero coefficients in front of powers of x that are divisible by p; in that case, a^p has minimal polynomial m(a^p)(x) = a(n)x^n + … + a0 where a(i) is the coefficient of x^ip in m(a), so that [K(a):K(a^p)] = p. Conversely, if [K(a):K(a^p)] = p, then a^p has no pth root in K(a^p), so K(a)/K(a^p) is inseparable. But that implies K(a)/K is inseparable by taking the tower K(a)/K(a^p)/K. Therefore, a in L is separable iff K(a) = K(a^p).
The set of all separable elements in L forms an intermediate field. To see why, note that if K(a) = K(a^p) and K(b) = K(b^p), then K(a, b) = K(a^p, b^p), whence it’s readily seen that K(a + b) = K(a^p + b^p) and K(ab) = K((ab)^p). In addition, K(1/a) = K(a) = K(-a), and every element of K is separable over K with minimal polynomial x – a.
Therefore, if L is normal over K, it is separable iff the polynomials it is a splitting field for are all separable over K.
If L/K is any field extension, then let M be the intermediate field consisting of all separable elements over K. If M = K, the extension L/K is called purely inseparable.
An equivalent condition for L/K to be purely inseparable is for every element of L outside K to have a minimal polynomial of the form x^(p^k) – b.
To see why, note that if every element of L has such a minimal polynomial, then L/K is trivially purely inseparable. Conversely, if a has a different minimal polynomial, then let p^k be the p-power portion of the greatest common divisor of all exponents of x with nonzero coefficients in m(a). Then m(a^(p^k)) has an exponent not divisible by p, so that a^(p^k) is separable. Therefore a^(p^k) is in K, so a satisfies x^(p^k) – b = 0 for some b in K.
The characterization of pure inseparability in terms of only m(a) makes it sensible to call a purely inseparable over K iff its minimal polynomial is of the form x^(p^k) – b, or, equivalently, if a^(p^k) is in K for some integer k. It’s fairly clear that this holds iff the only conjugate of a in the normal closure is a itself, since the minimal polynomial splits as (x – a)^(p^k). Elements of K are purely inseparable, by letting k = 0.
The set of purely inseparable elements of L/K forms an intermediate field. To see why, note that if a^(p^k) and b^(p^l) is in K, say with k <= l, then (a + b)^(p^l) = a^(p^l) + b^(p^l) is in K by repeating the Frobenius endomorphism l times, and similarly for ab. Therefore, the splitting field of a family of purely inseparable polynomials is itself purely inseparable.