Separable Extensions, Linearly Disjoint

Linearly Disjoint

Two subgroups within a larger group are disjoint if they have only the identity element in common. This is a bit odd, since disjoint sets have nothing in common - but you'll get use to it.

Two ring extensions of R in a larger ring S are disjoint if they have only R in common. To field extensions L/K and M/K are disjoint if they have only K in common.

Ok, that was standard terminology; here comes a new adverb. If L/K is a finite field extension and M/K is an arbitrary field extension, L and M are linearly disjoint if (L+M)/M has the same dimension as L/K.

Let u be an element in L-K and adjoin it to M. The irreducible polynomial that establishes the dimension of K(u)/K may factor over M. In other words, M(u)/M could have a smaller dimension than K(u)/K. Its dimension can't be any larger. Continue adjoining the generators of L/K, and the dimension of L+M over M is bounded by the dimension of L over K.

If L and M are not disjoint, select the first u from M intersect L. Now K(u) is a proper field extension of K, but M(u) is trivial. From this point on, the dimension of the extension of M is less than the dimension of the corresponding extension of K. In other words, L and M cannot be linearly disjoint if they are not disjoint.

Coprime

Assume the dimensions of L/K and M/K are relatively prime. The dimension of L+M over K is divisible by both these dimensions, and is at least their product. We saw before that the dimension of L+M over M is at most the dimension of L over K, hence the dimension of L+M over K is at most the product of the two dimensions. We have dimensional equality, and L and M are linearly disjoint.

Galois

Suppose L is galois and apply natural irrationality. The dimension of L+M over M equals the dimension of L over L∩M. The extensions are disjoint iff they are linearly disjoint.

Subfields of L

Continue the above, where L and M are linearly disjoint, and L is finite and galois. Consider a subfield of L. For convenience, let's write it as a simple extension K(u). A subgroup of the galois group of L over K fixes K(u), and a subgroup of the galois group of L+M over M fixes M(u). These galois groups are isomorphic. Look at the isomorphism described under natural irrationality, and note that the subgroups fixing M(u) and L(u) correspond. They are the same size, hence M(u)/M has the same dimension as K(u)/K. Every subfield of L is linearly disjoint with M.

Let's rephrase the above. If L/K and M/K are field extensions, and L is a finite separable extension, and M is disjoint from the normal closure of L, then L and M are linearly disjoint.

If we're careful, we can remove the word "separable" from the above. Let L be purely inseparable over K, with characteristic p. Let L and M be disjoint extensions. Let x be an element in L, with order p^s. Thus x to the p^s equals v for some v in K. Let x to the p^r equal w, for some w in M. Clearly w lies in L, and since L and M are disjoint, w lies in K, and equals v. Therefore M(x)/M has the same dimension as K(x)/K, namely p^s.

Now let L/K be a finite field extension whose normal closure is disjoint from M. Start with M and adjoin the purely inseparable elements of L/K, then the separable elements. At each step the dimensions are the same. Therefore L and M are linearly disjoint.

Here is a further generalization - let M be a finite extension that is disjoint from the normal closure of L. This time L is arbitrary - it could even contain transcendental elements. First, adjoint the purely inseparable elements of M to L. Since M and L are disjoint, the dimensions correspond. (This was shown above.) That leaves a finite separable extension. For convenience, invoke the primitive element theorem, and call it a simple extension. Adjoin the element u, having irreducible polynomial p(x). If p remains irreducible over L then the dimensions are preserved, and L and M are linearly disjoint.

Suppose p(x) factors into g(x)*h(x) over L. Write g as a product of monomials x-u_i, for various conjugates of u. The conjugates of u are algebraic over K, and the same holds for the coefficients of g, and of h. The coefficients form a finite set of algebraic elements over K, contained in L. Let these coefficients generate the intermediate extension L′. Now p factors over the field L′. However, L′ is finite, and its normal closure is disjoint from M, hence the extensions are linearly disjoint. Adjoining u to K or to L′ yields the same dimension, hence p(u) does not factor over L′, or L.

In summary, L and M are linearly disjoint if M is disjoint from the normal closure of L, and either extension is finite.

Direct Product of Galois Groups

Let L and M be disjoint, finite, and galois. The galois group of L+M over M is that of L over K. These automorphisms fix M, and run independently of, and commute with, the automorphisms of M that fix K. Together they form a group whose order equals the dimension of L+M over K. Thus the galois group of L+M over K is the direct product of the galois groups of L and of M.

Conversely, if the galois group of F is a direct product, take the two fixed fields L and M, fixed by the (normal) summands of G. These galois extensions are disjoint, since their intersection is fixed by G. Thus they are linearly disjoint. The field L+M lives inside F, and has the proper dimension, hence it is equal to F.

This extends to a finite direct product of groups. If G is abelian, F is the compositum of the linearly disjoint subfields fixed by the cyclic subgroups of G.