1 Fields , and the Degree of a Field Extension

version. In the abstract theory, if V is a K vector space of dimension n, there there exists a basis B = {e1, . . . , en} for V . That is, every v ∈ V is represented uniquely as a linear combination v = c1e1 + · · ·+ cnen for some coefficients cj ∈ K. Then the coordinate vector for v (relative to that basis) is Cv = (c1, . . . , cn) T ∈ Kn. If B′ = {e1, . . . , en} is another basis for V , the vector v is represented by a different coordinate vector C ′ v = (c ′ 1, . . . , c ′ n) T. There is an invertible n × n matrix P that expresses the new basis in terms of the old. Then C ′ v = PCv for every V . Definition. A bilinear form is a map b : V × V → K that is K-linear in each slot. That form is symmetric if b(v, w) = b(w, v) for every v, w ∈ V . The form b is nonsingular if for every nonzero v ∈ V , there exists w ∈ V such that b(v, w) 6= 0. Equivalently, if b(v, w) = 0 for every w, then v = 0. The quadratic form associated to b is q : V → K defined by: q(v) = b(v, v). For a basis B = {e1, . . . , en} for V , let aij = b(ei, ej), and define M = (aij) to be the Gram matrix of b, relative to B. Then M is a symmetric n× n matrix over K. Exercise. Continue the notations above. (1) If v, w ∈ V have coordinate (column) vectors Cv, Cw for basis B, then: b(v, w) = CT vMCw, and q(v) = C T vMCv. Then if v is represented as the column vector (x1, . . . , xn) T, then q(v) = ∑ ij aijxixj . (2) The form b is nonsingular ⇐⇒ M is nonsingular (i.e invertible) as a matrix. (3) If B′ is another basis, the new Gram matrix M ′ is related to the previous M by: M ′ = PTMP for an invertible n× n matrix P . (4) A map q : V → K is a quadratic form if and only if the map b(v, w) = 12 ( q(v + w)− q(v)− q(w) ) is bilinear. Exercise. (1) Quadratic form Q2 = 2xy has Gram matrix [ 0 1 1 0 ] . The form Q′(x, y) = x2− y2 has Gram matrix [ 1 0 0 −1 ] . Explain why those polynomials represent the same abstract quadratic form, using two choices of basis for the underlying 2-dimensional space. (That is, Q2 and Q ′ are isometric.) (2) Show that Qa(x, y) = axy is also isometric to Q ′, for every a ∈ K∗. That is: aQ′ ∼= Q′. Quadratic forms are types of polynomials in n variables, and were initially studied purely in that aspect. The standard dot product on Rn motivates a link between those quadratic forms as polynomials and generalized dot products on vector spaces. For column vectors u, v in Rn, their dot product is u · v = uTv = ∑ i uivi. As mentioned in freshman calculus (or physics) classes, dot products are related to angles. In fact, u · v = |u||v| cos(θ) where θ is the angle between u and v. Assumptions. Today assume that char(K) 6= 2 and every vector space is finite dimensional.