Why the rationals are hard
Over ℂ rank classifies; over ℝ rank plus signature classifies. Over ℚ neither suffices, because the square classes ℚ*/(ℚ*)² form an infinite group: 1, −1, 2, −2, 3, … and all products are inequivalent. Is the rational form ⟨1, 1⟩ (i.e. x² + y²) equivalent to ⟨2, 2⟩? They have the same rank and the same real signature (2, 0), yet ⟨1, 1⟩ does not represent 2 rationally as a *single* square while the discriminant analysis differs — you need finer invariants. The genius idea is to study ℚ not directly but through all its completions at once.
The completions of ℚ are ℝ (the “place at infinity”) and the p-adic fields ℚ_p, one for each prime p. Over each ℚ_p there is a tidy local invariant, the [[hasse-invariant|Hasse invariant]] ε_p(q) ∈ {±1}, built from Hilbert symbols of the diagonal entries. Together with the rank, the discriminant in ℚ*/(ℚ*)², and the real signature, these local data form a complete list of invariants — once you know the rule that ties them together.
The Witt ring: forms as a ring
Fix a field K (char ≠ 2) and look at all nondegenerate quadratic forms at once. Orthogonal sum ⊥ adds them and tensor product ⊗ multiplies them, so isometry classes almost form a ring — except ⊥ has no inverses. Two fixes, mirroring how ℤ is built from ℕ. First take the Grothendieck-style group completion (a Grothendieck group) to force subtraction. Then declare the hyperbolic plane ℍ to be zero. The quotient is the [[witt-ring|Witt ring]] W(K): its elements are anisotropic forms (the q_an cores from Guide 4), since every hyperbolic summand has been killed.
Why is killing ℍ the right move? By Witt cancellation, two forms become equal after adding hyperbolic planes iff their anisotropic cores agree — so “equal in W(K)” means “same anisotropic part”, which is precisely the classification we want. The class of ⟨a⟩ in W(K) is denoted ⟪a⟫; addition is ⟪a⟫ + ⟪b⟫ = class of ⟨a, b⟩, and ⟪a⟫·⟪b⟫ = ⟪ab⟫. The identity is ⟨1⟩ and ⟪a⟫ + ⟪−a⟫ = 0 because ⟨a, −a⟩ ≅ ℍ.
Three Witt rings, computed.
W(C): over an algebraically closed field every form is <1,...,1>,
and <1,1> = H = 0. So the only anisotropic forms are 0 and <1>.
W(C) = Z/2Z. (rank mod 2 is the whole story.)
W(R): the anisotropic forms are <1,...,1> and <-1,...,-1>; the invariant
that survives is the SIGNATURE p - m, which is additive on (+)
and multiplicative on (x).
W(R) = Z, via q |-> signature(q).
W(F_q), q odd: square classes F_q*/(F_q*)^2 has order 2, pick a nonsquare s.
Anisotropic forms have dimension <= 2; one finds
W(F_q) = Z/2Z[s]/(s^2 - 1)-type ring of order 4:
= Z/4Z if q = 3 (mod 4),
= Z/2Z[t]/(t^2) if q = 1 (mod 4).
Sanity check in W(R): <1,1,1> + <-1,-1> has signature 3 + (-2) = 1 = <1>.
Indeed <1,1,1,-1,-1> = <1> (+) 2H, and 2H = 0. Consistent.What the Witt ring organizes
The invariants we collected are exactly ring homomorphisms (or filtration data) on W(K). Rank mod 2 is a homomorphism W(K) → ℤ/2ℤ; its kernel is the fundamental ideal I. The discriminant lives on I/I², the Hasse invariant on I²/I³, the real signature(s) give W(K) → ℤ at each ordering of K, and so on. The deep Milnor conjecture (now a theorem) identifies the graded pieces Iⁿ/Iⁿ⁺¹ with Galois cohomology — linking quadratic forms back to the Galois machinery of Volume I. The little arithmetic of squares from Guide 1 has grown into a bridge between algebra, number theory, and topology.