The Witt Group and the Grothendieck–Witt Ring

Introduction

The isometry classes of non-degenerate quadratic forms over a field carry two operations: the orthogonal sum, which adds dimensions, and the tensor product, which multiplies them. Witt cancellation makes the first operation invertible, and the second distributes over it, so the isometry classes form a ring. Quotienting by the forms that are sums of hyperbolic planes removes the part of a form that carries no anisotropic information and produces the Witt ring $W(F)$, the home of the classical invariants of a quadratic form. This article constructs the ring, the Grothendieck–Witt ring $GW(F)$ from which it descends, the invariants — dimension, discriminant, signature, Hasse invariant — the ideal filtration whose successive quotients are the Milnor K-theory groups, and the Albert form of a biquaternion algebra, whose anisotropy is the division criterion for that algebra.

The base is a field $F$ of characteristic not $2$ throughout, since every construction uses the cancellation theorem of Witt's Theorems. The isometry class, the discriminant and the signed discriminant are from Bilinear Forms and The Rotation Group and Orientation; the Witt index and the decomposition $V \cong V_0 \perp m\,H$ are from Witt's Theorems; the tensor product of quadratic forms uses the tensor product of vector spaces from category 04. The characteristic classes are stated as standard results and the final identification with Milnor K-theory is cited.

The Grothendieck–Witt Ring

The Monoid of Isometry Classes

Let $\mathcal{M}(F)$ be the set of isometry classes of non-degenerate quadratic forms over $F$. The orthogonal sum induces a well-defined operation

$$ [q] + [q'] = [q \perp q'], $$

which is associative and commutative with identity the class of the zero form on the zero-dimensional space, the empty sum of squares; this makes $\mathcal{M}(F)$ a commutative monoid. By the cancellation theorem of Witt's Theorems, the operation is cancellative: $q \perp q_1 \cong q \perp q_2$ implies $q_1 \cong q_2$. Hence the natural map from $\mathcal{M}(F)$ to its Grothendieck group is injective.

Definition

Definition. The Grothendieck–Witt group $GW(F)$ is the Grothendieck group of the monoid $\mathcal{M}(F)$ under orthogonal sum; its elements are formal differences $[q] - [q']$ of isometry classes. The dimension gives a homomorphism

$$ \dim : GW(F) \longrightarrow \mathbb{Z}, \qquad \dim([q] - [q']) = \dim q - \dim q'. $$

It is surjective, because the classes $[\langle 1\rangle]$ and $[\langle -1\rangle]$ have dimension $1$, and its kernel is the augmentation ideal $\widetilde{GW}(F)$, generated by the differences of forms of equal dimension.

Proposition. As an abelian group, $GW(F) \cong \mathbb{Z} \oplus \widetilde{GW}(F)$, and the injective map $\mathcal{M}(F) \to GW(F)$, $[q] \mapsto [q]$, identifies the monoid with the set of classes of non-negative dimension.

Proof. The dimension homomorphism splits via $\mathbb{Z} \to GW(F)$, $1 \mapsto [\langle 1\rangle]$, giving the direct sum decomposition; cancellation is what makes the map from the monoid injective.

The Discriminant

The discriminant of a non-degenerate form is $\Delta(q) = \det G \in F^\times/(F^\times)^2$, an element of the group of square classes, as in Bilinear Forms. It satisfies $\Delta(q \perp q') = \Delta(q)\Delta(q')$ and, for a form of dimension $n$ and $c \in F^\times$, $\Delta(cq) = c^n\Delta(q)$. Hence $\Delta$ is a homomorphism on the monoid, and it descends to $GW(F)$:

$$ \Delta : GW(F) \longrightarrow F^\times/(F^\times)^2, \qquad \Delta([q] - [q']) = \Delta(q)\Delta(q')^{-1}. $$

The signed discriminant, introduced in The Rotation Group and Orientation, is

$$ d(q) = (-1)^{n(n-1)/2}\,\Delta(q), \qquad n = \dim q. $$

It differs from $\Delta$ by the sign $(-1)^{n(n-1)/2}$, which is the sign of the reversal of the $n$ factors in the product $e_1 \cdots e_n$ of a basis; for the hyperbolic plane $d(H) = (-1)^{1}(-1) = 1$, so the signed discriminant, unlike the unsigned one, is trivial on $H$ and therefore descends further to the Witt ring.

The Ring Structure

Definition. The tensor product of two quadratic forms $(V, q)$ and $(W, q')$ is the quadratic form $q \otimes q'$ on $V \otimes W$ with the two properties

$$ (q \otimes q')(v \otimes w) = q(v)q'(w), \qquad B_{q \otimes q'} = B \otimes B', $$

where $B \otimes B'$ is the tensor product of the polar forms. Such a form exists and is unique: the tensor product $B \otimes B'$ is a symmetric bilinear form on $V \otimes W$, and the quadratic form $Q(x) = (B \otimes B')(x, x)$ attached to it satisfies $Q(v \otimes w) = q(v)q'(w)$ on decomposable tensors, which span $V \otimes W$; conversely any form with the two properties has polar form $B \otimes B'$ and is therefore determined, since over a field of characteristic not $2$ a quadratic form is determined by its polar form.

Proposition. In a diagonal basis the tensor product is diagonal:

$$ \langle a_1, \ldots, a_m\rangle \otimes \langle b_1, \ldots, b_n\rangle \cong \bigl\langle a_i b_j : 1 \leq i \leq m,\ 1 \leq j \leq n\bigr\rangle. $$

The tensor product is commutative and associative up to isometry, distributive over the orthogonal sum, and $q \otimes \langle 1\rangle \cong q$.

Proof. Take bases with $q(e_i) = a_i$, $B(e_i, e_j) = 0$ and $q'(f_k) = b_k$, $B'(f_k, f_l) = 0$. The products $e_i \otimes f_k$ form an orthogonal basis of $V \otimes W$ with $(q \otimes q')(e_i \otimes f_k) = a_ib_k$. Bilinearity gives distributivity and the unit property; commutativity and associativity follow from the corresponding properties of the tensor product of spaces.

Theorem. The operations $+$ and $\otimes$ make $GW(F)$ a commutative ring with identity $[\langle 1\rangle]$. The subgroup $\mathbb{Z}\cdot[H]$ generated by the class of the hyperbolic plane is an ideal, and the quotient

$$ W(F) = GW(F) / \mathbb{Z}\cdot[H] $$

is the Witt ring of $F$, a commutative ring with identity.

Proof. Distributivity is the distributivity of the tensor product over the orthogonal sum; the identity is $[\langle 1\rangle]$ because $\langle 1\rangle \otimes q \cong q$. For the ideal property, compute

$$ [H] \otimes [q] = [\langle 1, -1\rangle \otimes q] = [q \perp (-q)], $$

which is hyperbolic of dimension $2\dim q$, hence a sum of $\dim q$ hyperbolic planes; so the subgroup generated by $[H]$ is closed under multiplication and is an ideal.

The Witt Ring

Isometry Classes of Anisotropic Forms

Theorem. Every class in $W(F)$ has a unique representative that is anisotropic, and $[q] = 0$ in $W(F)$ if and only if $q$ is hyperbolic, that is $q \cong m\,H$ for some $m$.

Proof. By the decomposition theorem of Witt's Theorems, $q \cong q_0 \perp m\,H$ with $q_0$ anisotropic. In $W(F)$ the hyperbolic part is zero, so $[q] = [q_0]$; conversely if $[q]=0$ then $q$ differs from a hyperbolic form by a hyperbolic summand, and cancellation of hyperbolic planes removes it. Uniqueness of $q_0$ up to isometry is the uniqueness statement of the decomposition theorem.

So $W(F)$ is generated, as an additive group, by the isometry classes of anisotropic forms; two orthogonal sums of anisotropic forms are equal exactly when their anisotropic parts are isometric, and a relation appears whenever an anisotropic form splits as an orthogonal sum of two smaller forms, so the group is a quotient of the free abelian group on those classes and is not free in general.

The Dimension and the Sign

Proposition. The dimension taken modulo $2$ gives a surjective ring homomorphism

$$ \dim : W(F) \longrightarrow \mathbb{Z}/2\mathbb{Z}, $$

the augmentation, and $\mathbb{Z}/2\mathbb{Z}$ acquires the ring structure of the prime field $\mathbb{F}_2$. The fundamental ideal is

$$ I = \ker(\dim) = \{[q] \in W(F) : \dim q \text{ is even}\}. $$

Proof. The dimension is additive on orthogonal sums and multiplicative on tensor products, $\dim(q \otimes q') = \dim q \cdot \dim q'$, and the hyperbolic plane has even dimension, so the map descends to $W(F)$ and is a ring homomorphism. It is surjective because $[\langle 1\rangle]$ has dimension one.

The Discriminant Descends

Theorem. The signed discriminant descends to a well-defined function

$$ d : W(F) \longrightarrow F^\times/(F^\times)^2, $$

which is a group homomorphism on the fundamental ideal $I$ and vanishes there exactly on $I^2$, the square of the fundamental ideal. Consequently $d$ induces an isomorphism

$$ I / I^2 \cong F^\times/(F^\times)^2. $$

Proof. The signed discriminant is unchanged by the addition of a hyperbolic plane, because $d(q \perp H) = d(q)$, and it is multiplicative on forms of even dimension; it therefore descends to a well-defined function on $W(F)$ that is a group homomorphism on $I$. For $[q], [q'] \in I$ the forms have even dimension, say $2m$ and $2n$, so $N = \dim(q \otimes q') = 4mn$; the discriminant of the tensor product is $(\Delta q)^{2n}(\Delta q')^{2m}$, a square because both exponents are even, and the sign factor is $(-1)^{N(N-1)/2} = (-1)^{(N/2)(N-1)} = 1$ because $N/2 = 2mn$ is even. Hence $d(q \otimes q') = 1$ and $d$ kills $I^2$. It is surjective because $d(\langle a\rangle \perp \langle -1\rangle) = a$ runs over all square classes as $a$ runs over $F^\times$, and $\langle a\rangle\perp\langle -1\rangle \in I$; injectivity on $I/I^2$ is the standard companion statement.

Pfister Forms

The powers of the fundamental ideal are generated by a single family of forms, the tensor powers of the binary forms $\langle 1, -a\rangle$.

Definition. For $a_1, \ldots, a_n \in F^\times$ the $n$-fold Pfister form is the $n$-fold tensor product

$$ \langle\!\langle a_1, \ldots, a_n \rangle\!\rangle = \bigotimes_{i=1}^{n} \langle 1, -a_i \rangle, $$

a form of dimension $2^n$. Thus $\langle\!\langle a \rangle\!\rangle = \langle 1, -a\rangle$, and $\langle\!\langle a, b \rangle\!\rangle = \langle 1, -a, -b, ab\rangle$.

Proposition. Let $\pi = \langle\!\langle a_1, \ldots, a_n\rangle\!\rangle$.

$\dim \pi = 2^n$, and the tensor product of an $m$-fold and an $n$-fold Pfister form is an $(m+n)$-fold Pfister form.
For $n \geq 2$ the discriminant $\Delta(\pi)$ is a square, indeed $\Delta(\pi) = (a_1 \cdots a_n)^{2^{n-1}}$.
$\pi \in I^n$.

Proof. The dimension statement is multiplicativity of the dimension. For the discriminant, the diagonal entries of the tensor product are the products $\prod_{i \in S}(-a_i)$ for $S \subseteq \{1, \ldots, n\}$, and the exponent of $a_i$ in the product of all of them is $\sum_{S \ni i} 1 = 2^{n-1}$, with the signs contributing $(-1)^{n2^{n-1}} = 1$ for $n \geq 2$. Finally $\langle 1, -a_i\rangle = [\langle 1\rangle] - [\langle a_i\rangle]$ has even dimension, so it lies in $I$, and $I^n$ is spanned by products of $n$ elements of $I$; the tensor product is such a product.

Proposition. The additive group $I^n$ is generated by the $n$-fold Pfister forms.

Proof. The classes $[\langle a\rangle]$ of the one-dimensional forms generate $W(F)$ as an abelian group, because a diagonal form satisfies $[\langle a_1, \ldots, a_m\rangle] = \sum_i[\langle a_i\rangle]$, the orthogonal sum being the sum in $W(F)$, and because $[\langle -1\rangle] = -[\langle 1\rangle]$, the relation $[\langle 1\rangle] + [\langle -1\rangle] = [H] = 0$ holding since $\langle 1\rangle \perp \langle -1\rangle = H$. The augmentation is onto with kernel $I$, so $I$ is generated by the classes $[\langle 1\rangle] - [\langle a\rangle] = [\langle 1, -a\rangle] = \langle\!\langle a\rangle\!\rangle$. An element of $I^n$ is a sum of products of $n$ elements of $I$, each an integral combination of one-fold Pfister forms; expanding a product of such combinations gives an integral combination of products of $n$ one-fold Pfister forms, and a product of Pfister forms is the Pfister form of the concatenated list by the first part of the proposition.

Theorem (Pfister). An $n$-fold Pfister form is either anisotropic, or hyperbolic; in the second case it is the orthogonal sum of $2^{n-1}$ hyperbolic planes.

Proof for $n = 1$. The form $\langle 1, -a\rangle$ represents zero exactly when $a = x^2$ for some $x \in F^\times$, that is exactly when $a$ is a square; then $\langle 1, -a\rangle \cong \langle 1, -1\rangle = H$, and otherwise the equation $x^2 = ay^2$ has only the trivial solution, so the form is anisotropic.

Proof for $n = 2$. The form $\langle\!\langle a, b\rangle\!\rangle = \langle 1, -a, -b, ab\rangle$ is the norm of the quaternion algebra $(a, b)_F$ with basis $1, i, j, ij$ satisfying $i^2 = a$, $j^2 = b$ and $ij = -ji$; by the norm theory of Quadratic Forms over Algebras and Norms, the norm is isotropic exactly when the algebra is not a division algebra. In that case the algebra is $M_2(F)$, and the norm becomes the determinant, which is isometric to $2H$ by the identity

$$ xw - yz = \Bigl(\frac{x + w}{2}\Bigr)^2 - \Bigl(\frac{x - w}{2}\Bigr)^2 - \Bigl(\frac{y + z}{2}\Bigr)^2 + \Bigl(\frac{y - z}{2}\Bigr)^2 $$

for the four matrix entries, the change of variables being invertible when $2 \neq 0$. Hence an isotropic $2$-fold Pfister form is hyperbolic.

The general case is the classical theorem of Pfister and is cited as standard; it is proved by induction on $n$, the inductive step using the two structural properties of Pfister forms recorded next.

Remark. Let $\pi$ be an $n$-fold Pfister form. Then $\pi \otimes \pi \cong 2^n \pi$, and $\pi$ is round: if $\pi$ represents $x \in F^\times$, then $\pi \otimes \langle 1, -x\rangle \cong \pi$. The first identity says that $\pi$ is a group form, and the second is what makes an isotropic Pfister form collapse to a sum of hyperbolic planes. Both are standard consequences of the multiplicativity of the Pfister construction.

Corollary. The two-fold Pfister form $\langle\!\langle a, b\rangle\!\rangle$ is the norm of the quaternion algebra $(a, b)_F$; Pfister's theorem in that case states that the norm of a quaternion algebra is anisotropic exactly when the algebra is a division algebra, and hyperbolic exactly when the algebra splits.

The Albert Form

The norm of a quaternion algebra is a twofold Pfister form, of dimension $4$, and it decides the division question for that algebra. One level up, for the tensor product of two quaternion algebras, the corresponding object is a form of dimension $6$; it is not a Pfister form, and it decides the same question for the product.

Definition. Let $B = (a, b)_F \otimes_F (c, d)_F$ be a biquaternion algebra over $F$, the tensor product of two quaternion algebras, of dimension $16$ over $F$. This is the algebraists' sense of the word, and it is not the complex algebra $\mathbb{B}$ of the corpus; the collision of the two names is recorded in Division Algebras. The Albert form of $B$ is the form

$$ \phi_B = \langle a, b, -ab, -c, -d, cd\rangle , $$

of dimension $6$ over $F$.

Proposition. The Albert form has trivial signed discriminant:

$$ \Delta(\phi_B) = a\,b\,(-ab)\,(-c)\,(-d)\,(cd) = -(abcd)^2 \equiv -1, \qquad d(\phi_B) = (-1)^{15}\Delta(\phi_B) = 1 . $$

Proof. The product of the six entries is $(-1)^3 (abcd)^2$, so the discriminant is the square class of $-1$ whatever the parameters; the signed discriminant is $d(q) = (-1)^{n(n-1)/2}\Delta(q)$ of the section above, with $n = 6$, so $d(\phi_B) = (-1)^{15}\cdot(-1) = 1$.

Theorem (Albert). The biquaternion algebra $B = (a, b)_F \otimes_F (c, d)_F$ is a division algebra if and only if its Albert form $\phi_B$ is anisotropic over $F$.

The statement is the dimension-$6$ companion of the corollary above, and its proof matches a zero divisor of $B$ with a nontrivial zero of $\phi_B$; it is cited as standard (Lam, Introduction to Quadratic Forms over Fields, III, Theorem 4.8). The criterion by which the anisotropy is decided over a local field is Springer's theorem, in Local Fields.

Remark (the index, read from the form). When $\phi_B$ is isotropic, the index of $B$ is read from the type of isotropy, and the three cases exhaust the values the index can take:

$B \cong M_4(F)$, of index $1$, exactly when $\phi_B$ is hyperbolic, that is $\phi_B \cong \langle 1, -1, 1, -1, 1, -1\rangle$;
$B \cong M_2(D)$ for a quaternion division algebra $D$, of index $2$, exactly when $\phi_B$ is isotropic but not hyperbolic, equivalently $\phi_B \cong \langle 1, -1, e, f, g, h\rangle$ with the four-dimensional form $\langle e, f, g, h\rangle$ anisotropic;
$B$ is a division algebra, of index $4$, exactly when $\phi_B$ is anisotropic.

Remark (isomorphism). Two biquaternion algebras over $F$ are isomorphic as $F$-algebras if and only if their Albert forms are similar, that is, isometric up to multiplication by a scalar from $F^\times$. This is the dimension-$6$ companion of the statement that two quaternion algebras are isomorphic exactly when their norm forms are isometric.

Example (over $\mathbb{R}$). Over $\mathbb{R}$ the quaternion algebra $(a, b)_\mathbb{R}$ is split, that is $M_2(\mathbb{R})$, exactly when $a > 0$ or $b > 0$, and it is $\mathbb{H}$ otherwise; so the Brauer group $\operatorname{Br}(\mathbb{R}) \cong \mathbb{Z}/2$ has two elements, and the two factors of $B$ represent the same element exactly when both are split or both are $\mathbb{H}$. The Albert form over $\mathbb{R}$ is hyperbolic, of signature $(3, 3)$, exactly in that case, and it has signature $(5, 1)$ or $(1, 5)$ otherwise. It is therefore isotropic for every choice of $a, b, c, d$, and no real biquaternion algebra is a division algebra, in agreement with the computation $\operatorname{Br}(\mathbb{R}) \cong \mathbb{Z}/2$ of Central Simple Algebras and the Brauer Group and with the real case of Division Algebras.

The Higher Ideal Quotients

Theorem (standard). There is an injective homomorphism, the Hasse invariant,

$$ e_2 : I^2/I^3 \hookrightarrow \operatorname{Br}_2(F), $$

into the two-torsion of the Brauer group of $F$, which is an isomorphism when $F$ is a local or global field and, by Merkurjev's theorem, an isomorphism for every field.

Theorem (Milnor's conjecture, Voevodsky). For every $n \geq 0$ there is an isomorphism

$$ I^n/I^{n+1} \cong k_n^M(F)/2, $$

where $k_n^M(F)$ is the $n$-th Milnor K-theory group of $F$ and the quotient is by multiplication by $2$. For $n = 0$ this recovers $I^0/I^1 \cong \mathbb{Z}/2\mathbb{Z}$, for $n = 1$ it recovers $I/I^2 \cong F^\times/(F^\times)^2$ since $k_1^M(F) = F^\times$, and for $n = 2$ it identifies $I^2/I^3$ with the two-torsion of the Brauer group modulo symbols. The theorem is deep and is cited as standard; it is the precise sense in which the Witt ring encodes the arithmetic of the field.

Orderings and the Signature

The Signature Homomorphism

Let $F$ be formally real and let $P$ be an ordering of $F$, with real closure $F_P$. Over the real closed field $F_P$ every form is classified by its signature, so $W(F_P) \cong \mathbb{Z}$; the real closure is unique up to $F$-isomorphism, and the composite

$$ \operatorname{sign}_P : W(F) \longrightarrow W(F_P) \xrightarrow{\ \cong\ } \mathbb{Z}, \qquad \operatorname{sign}_P([q]) = p - r, $$

where $(p, r)$ is the signature of the extended form $q \otimes_F F_P$, is therefore well defined.

Proposition. For each ordering $P$ the map $\operatorname{sign}_P$ is a ring homomorphism, and it kills the hyperbolic plane.

Proof. Extension of scalars commutes with orthogonal sums and with tensor products, so it induces a ring homomorphism $W(F) \to W(F_P)$; the isomorphism $W(F_P) \to \mathbb{Z}$ given by the signature is a ring homomorphism because signatures add on orthogonal sums and multiply on tensor products, $\operatorname{sign}(q \otimes q') = \operatorname{sign}(q)\operatorname{sign}(q')$. The hyperbolic plane has signature $0$, so it is in the kernel.

Theorem. Let $F$ be formally real. Then the total signature

$$ \operatorname{sign} : W(F) \longrightarrow \prod_{P} \mathbb{Z}, $$

the product over all orderings of $F$, is a ring homomorphism, and its image is a subring of a product of copies of $\mathbb{Z}$, hence reduced. The kernel contains the nilradical of $W(F)$, so the total signature is injective on $W(F)$ modulo nilpotents.

Proof. The image of a nilpotent element is nilpotent, and a product of copies of $\mathbb{Z}$ has no nonzero nilpotents, so the nilradical is in the kernel and the image, being a subring of a reduced ring, is reduced.

Remark. The refined statement, due to Pfister and the reduced theory of the Witt ring, is that for a formally real field the kernel of the total signature is exactly the nilradical, and $W(F)$ modulo nilpotents is the ring of functions on the space of orderings that arise as signatures; the theory is cited as standard. What is used here is only the direction proved: signatures are ring homomorphisms into $\mathbb{Z}$, and they detect at least the nilpotent part of the obstruction to being hyperbolic.

Examples over Ordered Fields

Example (real closed field). If $F$ is real closed it has a unique ordering and is its own real closure, so $W(F) \cong \mathbb{Z}$, generated by $[\langle 1\rangle]$, and the fundamental ideal is $I = 2\mathbb{Z}$, with $I^n = 2^n\mathbb{Z}$. The quotient $I^n/I^{n+1} \cong \mathbb{Z}/2\mathbb{Z}$ is generated by the class of the $n$-fold Pfister form, in agreement with the Milnor–Voevodsky statement.

Example ($\mathbb{R}$ and $\mathbb{Q}$). For $\mathbb{R}$ there is a single ordering and the signature is the difference of the numbers of positive and negative squares in a diagonal form, so $W(\mathbb{R}) \cong \mathbb{Z}$ and a form is hyperbolic exactly when its signature vanishes. The field $\mathbb{Q}$ also has a single ordering, but the signature $W(\mathbb{Q}) \to \mathbb{Z}$ is far from injective: the form $\langle 1, -2\rangle$ has signature $1 - 1 = 0$ and is not hyperbolic, because $2$ is not a square in $\mathbb{Q}$ and the form is anisotropic of dimension $2$. The same form exhibits $2$-torsion in $W(\mathbb{Q})$, since

$$ 2\langle 1, -2\rangle \cong \langle 1, 1, -2, -2\rangle \cong \langle\!\langle 2, 2\rangle\!\rangle $$

is isotropic — in the middle form the vector $(2, 0, 1, 1)$ satisfies $q = 4 + 0 - 2 - 2 = 0$, because $4 = 2 + 2$ — and an isotropic twofold Pfister form is hyperbolic; hence $2[\langle 1, -2\rangle] = 0$ while $[\langle 1, -2\rangle] \neq 0$. So the total signature sees only part of $W(F)$, and over $\mathbb{Q}$ the arithmetic of the anisotropic forms is not visible to the single ordering.

Examples

The Complex Numbers

Over $\mathbb{C}$ every non-degenerate form of dimension $n$ is isometric to $\langle 1, \ldots, 1\rangle$ of length $n$, because every element of $\mathbb{C}^\times$ is a square and the diagonal entries can be normalised to $1$. The hyperbolic plane is $\langle 1, -1\rangle \cong \langle 1, 1\rangle$, so $2[\langle 1\rangle] = 0$, and

$$ W(\mathbb{C}) \cong \mathbb{Z}/2\mathbb{Z}, \qquad GW(\mathbb{C}) \cong \mathbb{Z}. $$

Indeed the monoid of isometry classes is $\mathbb{N}$, generated by $[\langle 1\rangle]$, whose Grothendieck group is $\mathbb{Z}$; the hyperbolic plane is $\langle 1, -1\rangle \cong \langle 1, 1\rangle = 2[\langle 1\rangle]$, so the quotient is $\mathbb{Z}/2\mathbb{Z}$.

The fundamental ideal is $I = 0$, since $[\langle 1, 1\rangle] = 2[\langle 1\rangle] = 0$ in a group of exponent two; hence $I^2 = 0$ and the filtration collapses: over a field in which every element is a square the Witt ring has order two and all the higher invariants vanish.

The Real Numbers

Over $\mathbb{R}$ the invariants are the rank and the signature $\sigma(q) = p - r$ of Sylvester's law, as in Quadratic Forms and Polarisation. The signature is additive under orthogonal sums and multiplicative under tensor products, so it defines a surjective ring homomorphism

$$ \sigma : W(\mathbb{R}) \longrightarrow \mathbb{Z}, \qquad [q] \longmapsto \sigma(q). $$

Theorem. The signature is an isomorphism $W(\mathbb{R}) \cong \mathbb{Z}$.

Proof. Every non-degenerate form over $\mathbb{R}$ decomposes as $p\langle 1\rangle \perp r\langle -1\rangle$ by Sylvester's law, so its class in $W(\mathbb{R})$ is $p[\langle 1\rangle] + r[\langle -1\rangle]$. The hyperbolic plane has signature $0$ and lies in the kernel. Since $\langle -1\rangle = -[\langle 1\rangle]$ in $W(\mathbb{R})$ (because $\langle 1\rangle \perp \langle -1\rangle = H$), the class of $q$ is $\sigma(q)[\langle 1\rangle]$, so the group is generated by $[\langle 1\rangle]$ with relations coming from $\sigma$; the map $\sigma$ is thus an isomorphism of abelian groups. It is a ring homomorphism because the signature is multiplicative under tensor products, as the diagonalisation $\langle a\rangle\otimes\langle b\rangle \cong \langle ab\rangle$ shows on generators.

Since $W(\mathbb{R}) = \mathbb{Z}$, the fundamental ideal is the kernel of reduction mod $2$, namely $2\mathbb{Z}$, and every power $I^n = 2^n\mathbb{Z}$; the quotients $I^n/I^{n+1} \cong \mathbb{Z}/2$ match $k_n^M(\mathbb{R})/2 \cong \mathbb{Z}/2$.

A Finite Field

Theorem (standard). Let $\mathbb{F}_q$ be a finite field of odd order. Then

$$ W(\mathbb{F}_q) \cong \begin{cases} \mathbb{Z}/4\mathbb{Z}, & q \equiv 3 \pmod 4, \\ \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}, & q \equiv 1 \pmod 4. \end{cases} $$

The generator in the first case is the class of $\langle 1\rangle$, of order four, with $2[\langle 1\rangle] = [\langle 1, 1\rangle]$ the nonzero class of the anisotropic plane; in the second case $-1$ is a square, the form $\langle 1, 1\rangle$ is hyperbolic so $2[\langle 1\rangle] = 0$, and the second generator is the class of the anisotropic plane $\langle 1, a\rangle$ for any nonsquare $a$.

Example. For $\mathbb{F}_3$ the square classes are $\{1, -1\}$ with $-1$ a nonsquare; the form $x^2 + y^2$ is anisotropic, so $[\langle 1, 1\rangle] \neq 0$ and $[\langle 1\rangle]$ has order four, giving $W(\mathbb{F}_3) \cong \mathbb{Z}/4\mathbb{Z}$.

Summary

The isometry classes of non-degenerate quadratic forms over a field $F$ of characteristic not $2$ form a cancellative commutative monoid under the orthogonal sum. Its Grothendieck group is the Grothendieck–Witt group $GW(F)$, which as an abelian group is $\mathbb{Z} \oplus \widetilde{GW}(F)$ with $\mathbb{Z}$ the dimension and $\widetilde{GW}(F)$ the kernel of $\dim$; the dimension and the discriminant descend to homomorphisms on $GW(F)$.

The tensor product of quadratic forms is characterised by $(q\otimes q')(v \otimes w) = q(v)q'(w)$ and $B_{q\otimes q'} = B\otimes B'$, and in diagonal bases it multiplies entries: $\langle a_i\rangle \otimes \langle b_j\rangle = \langle a_ib_j\rangle$. Addition and tensor product make $GW(F)$ a commutative ring; the subgroup generated by the hyperbolic plane is the ideal $\mathbb{Z}[H]$ (because $H\otimes q \cong q\perp(-q)$ is hyperbolic), and the quotient is the Witt ring $W(F)$.

Every class in $W(F)$ has a unique anisotropic representative, and a form is zero in $W(F)$ exactly when it is hyperbolic. The dimension modulo $2$ is the augmentation $W(F) \to \mathbb{Z}/2$, whose kernel is the fundamental ideal $I$. The signed discriminant $d(q) = (-1)^{n(n-1)/2}\Delta(q)$ descends to $W(F)$ and induces an isomorphism $I/I^2 \cong F^\times/(F^\times)^2$, vanishing on $I^2$. The Hasse invariant $e_2$ embeds $I^2/I^3$ into the two-torsion of the Brauer group, and Milnor's conjecture, proved by Voevodsky, identifies $I^n/I^{n+1} \cong k_n^M(F)/2$ with Milnor K-theory modulo $2$. The basic computations are $W(\mathbb{C}) \cong \mathbb{Z}/2$, $W(\mathbb{R}) \cong \mathbb{Z}$ via the signature, and $W(\mathbb{F}_q) \cong \mathbb{Z}/4$ for $q \equiv 3 \pmod 4$ and $\mathbb{Z}/2 \times \mathbb{Z}/2$ for $q \equiv 1 \pmod 4$.

The Pfister forms $\langle\!\langle a_1, \ldots, a_n\rangle\!\rangle = \bigotimes_i \langle 1, -a_i\rangle$ have dimension $2^n$, have discriminant $1$ for $n \geq 2$, and lie in $I^n$; the group $I^n$ is generated by the $n$-fold Pfister forms, and Pfister's theorem states that an $n$-fold Pfister form is either anisotropic or hyperbolic, in the second case being the sum of $2^{n-1}$ hyperbolic planes. The twofold case is the norm of the quaternion algebra $(a, b)_F$, so the theorem there says that the norm is anisotropic exactly when the algebra is a division algebra. The Albert form $\langle a, b, -ab, -c, -d, cd\rangle$ of the biquaternion algebra $(a,b)_F \otimes_F (c,d)_F$ is the analogous object one level up: of dimension $6$, of trivial signed discriminant, not a Pfister form, and anisotropic exactly when the biquaternion algebra is a division algebra, with the type of isotropy reading off the index $1$, $2$ or $4$. A Pfister form satisfies $\pi \otimes \pi \cong 2^n\pi$ and is round: it satisfies $\pi \otimes \langle 1, -x\rangle \cong \pi$ for every nonzero value $x$ that it represents.

For a formally real field and an ordering $P$, extension to the real closure gives the signature ring homomorphism $\operatorname{sign}_P : W(F) \to \mathbb{Z}$, and the total signature is a ring homomorphism of $W(F)$ into a product of copies of $\mathbb{Z}$, so its kernel contains the nilradical of $W(F)$. Over $\mathbb{R}$ the signature is an isomorphism onto $\mathbb{Z}$ and detects hyperbolicity; over $\mathbb{Q}$ it does not, since $\langle 1, -2\rangle$ has signature $0$ and is anisotropic, and $W(\mathbb{Q})$ even has $2$-torsion, $2[\langle 1, -2\rangle] = 0$, because the isotropic twofold Pfister form $\langle 1, 1, -2, -2\rangle$ is hyperbolic.

Summary of Notation

Symbol	Meaning
$F$	Field of characteristic not $2$
$\mathcal{M}(F)$	Monoid of isometry classes of non-degenerate forms under $\perp$
$[q]$	Isometry class of a quadratic form
$q \perp q'$	Orthogonal sum
$q \otimes q'$	Tensor product of quadratic forms
$GW(F)$	Grothendieck–Witt ring
$\widetilde{GW}(F)$	Augmentation ideal, kernel of $\dim$
$\dim$	Dimension homomorphism, $GW(F) \to \mathbb{Z}$ and $W(F) \to \mathbb{Z}/2$
$\Delta(q)$	Discriminant in $F^\times/(F^\times)^2$
$d(q)$	Signed discriminant $(-1)^{n(n-1)/2}\Delta(q)$
$H$	Hyperbolic plane, $\langle 1, -1\rangle$
$W(F)$	Witt ring, $GW(F)/\mathbb{Z}[H]$
$I$	Fundamental ideal, kernel of $\dim : W(F) \to \mathbb{Z}/2$
$I^n$	Powers of the fundamental ideal
$e_2$	Hasse invariant, $I^2/I^3 \hookrightarrow \operatorname{Br}_2(F)$
$\operatorname{Br}_2(F)$	Two-torsion of the Brauer group
$k_n^M(F)$	Milnor K-theory groups
$\langle\!\langle a_1, \ldots, a_n\rangle\!\rangle$	$n$-fold Pfister form $\bigotimes_i \langle 1, -a_i\rangle$
$\pi$	A Pfister form
$B = (a,b)_F \otimes_F (c,d)_F$	Biquaternion algebra, the tensor product of two quaternion algebras
$\phi_B$	Albert form $\langle a, b, -ab, -c, -d, cd\rangle$ of $B$, of dimension $6$ and trivial signed discriminant
$P$, $\operatorname{sign}_P$	Ordering of a formally real field; the signature at $P$
$F_P$	Real closure of $F$ at the ordering $P$
$\operatorname{sign}$	Total signature, $W(F) \to \prod_P \mathbb{Z}$
$(a, b)_F$	Quaternion algebra with $i^2 = a$, $j^2 = b$
$\sigma$	Signature, $W(\mathbb{R}) \to \mathbb{Z}$
$\mathbb{F}_q$	Finite field with $q$ elements
$\mathbb{R}, \mathbb{C}$	Real and complex numbers