Towers and linear disjointness: when adjoining information multiplies

Suppose F is contained in K and K is contained in L, with both successive dimensions finite. Choose an F-basis u₁ through uₘ of K and a K-basis v₁ through vₙ of L. Expand an element of L in the v-basis, then expand each coefficient in the u-basis. The products uᵢvⱼ span L over F. To prove independence, group a proposed F-coefficient linear relation by vⱼ, use independence over K, then use independence of the uᵢ over F.

This proof gives degree mn and explains why the middle field matters. If you know [K:F] and [M:F] for two fields neither containing the other, those values do not themselves form a tower. First pass to their compositum KM, the smallest field containing both, and compute how much M still adds after K is already present. A relation that was irreducible over F may have smaller irreducible factors over K.

For finite K/F and any M/F in a common overfield, an F-basis of K spans KM as an M-vector space. It may cease to be independent once coefficients from M are allowed. The extensions are linearly disjoint over F precisely when every F-linearly independent finite list from K stays independent over M. In the finite case this means the degree of KM over M equals the degree of K over F, so neither extension creates hidden linear relations in the other.

A common subfield larger than F prevents linear disjointness, but trivial intersection alone is not a universal substitute. For example, two distinct cubic root fields generated by different roots of X³ − 2 have intersection Q, yet together form its degree-six splitting field, not a degree-nine field. Their prime degrees force the intersection to be Q, while a relation involving a root of unity reduces the compositum degree. Stronger hypotheses can restore intersection criteria; the definition in terms of independence remains the safe starting point.

Consider Q(√2, √3) over Q(√2). If √3 equaled a + b√2 with rational a and b, squaring would give 3 = a² + 2b² + 2ab√2. Irrationality forces ab = 0. The possibilities then require either a² = 3 or b² = 3/2, neither a rational square. Consequently √3 is not in Q(√2), its quadratic remains irreducible there, and the combined field has degree four over Q.

The resulting basis 1, √2, √3, √6 makes the two independent quadratic choices explicit. By comparison, adjoining √8 after √2 has degree one because its generator is already present. These calculations reveal the operational meaning of degree: it measures new independent coordinates relative to what you already know. It does not count written square-root signs, and it does not promise that two numerical approximations that look different generate different fields.