An Introduction to Quadratic Forms

Klaas-Tido R¨uhl

October 30, 2006

1 Quadratic Forms and Quadratic Spaces

We deﬁne N not to contain 0. If we need the natural numbers to include 0, we will use the notation N0. Throughout this talk K will always denote the base ﬁeld.

1.1 Deﬁnition. Let n ∈ N0.

(1) A quadratic form of dimension n is a homogeneous polynomial ϕ ∈ K[X1,...,Xn] of degree 2, where X1,...,Xn are indeterminates over K. We write dim(ϕ) = n. (2) If V is an n-dimensional vector space over K, then Q : V → K is a quadratic map on V if

(i) Q(av) = a2Q(v) for all a ∈ K and v ∈ V , and (ii) V × V → K, (v, w) 7→ Q(v + w) − Q(v) − Q(w), is a symmetric K-bilinear map.

The pair (V,Q) is then called a quadratic space.

1.2 Remark. It is very important to pay attention to the fact, that a quadratic form of dimension n does not have to be a polynomial in all the X1,...Xn. The dimension is a fixed part of the 2 2 definition of a quadratic form. For example X1 + X2 ∈ K[X1,X2,X3] is a quadratic form of dimension 3. This implies, that the 0 polynomial can be a quadratic form of any dimension. But it is the only quadratic form of dimension 0. This way of defining quadratic forms is necessary to establish an easy correspondence between quadratic forms and quadratic spaces. 4

Henceforth K will be a ﬁeld with char(K) 6= 2. For a quadratic map Q : V → K, we deﬁne a symmetric K-bilinear map 1 b : V × V −→ K, (v, w) 7−→ Q(v + w) − Q(v) − Q(w). (1.1) Q 2

We see, that Q(v) = bQ(v, v) for all v ∈ V . Conversely, if b : V × V → K is a symmetric K-bilinear form, then Qb : V −→ K, v 7−→ b(v, v), is a quadratic map on V , and we have bQb = b. Since bQ is uniquely determined by Q, and since Qb is uniquely determined by b, we thus get a bijective correspondence between symmetric K-bilinear forms V × V → K and quadratic maps V → K.

1 1.3 Remark. The bijective correspondence between symmetric bilinear forms and quadratic maps depends heavily on the deﬁnition of bQ in (1.1). If we allowed char(K) = 2, this definition would not be possible. As a consequence there does not exist an analogous bijective correspondence over ﬁelds with characteristic 2. 4

Now let (V,Q) be a quadratic space of dimension n, and let B = {v1, . . . , vn} be a K-basis of V . We obtain a symmetric matrix A = (bQ(vi, vj))i,j=1,...,n, and we call A the matrix associated to Q with respect to B. With the help of this matrix A = (aij)i,j we obtain the n-dimensional Pn quadratic form i,j=1 aijXiXj ∈ K[X1,...,Xn], which we call the quadratic form associated to Q with respect to B.

Obviously an n-dimensional quadratic form ϕ ∈ K[X1,...,Xn] deﬁnes a quadratic map

n K −→ K, (x1, . . . , xn) 7−→ ϕ(x1, . . . , xn).

We will also denote this map by ϕ. This makes (Kn, ϕ) a quadratic space. If A is the matrix associated to ϕ with respect to the standard basis {e1, . . . , en}, then we obviously have t t bϕ(ei, ej) = eiAej, where for any matrix B its transpose is denoted by B . Furthermore, if Pn A = (aij)i,j, we obtain ϕ = i,j=1 aijXiXj. In general we denote by Aϕ the matrix associated to a quadratic form ϕ with respect to the standard basis.

1.4 Deﬁnition. Let n ∈ N0.

(1) Let (V1,Q1) and (V2,Q2) be two quadratic spaces over K. A homomorphism f : V1 → V2 of K-vector spaces is called metric, if Q1(v) = Q2(f(v)) for all v ∈ V1.

(2) The spaces (V1,Q1) and (V2,Q2) are called equivalent or isometric, if there exists a metric ∼ isomorphism V1 → V2. We then write (V1,Q1) = (V2,Q2). (3) Two n-dimensional quadratic forms ϕ and ψ are equivalent or isometric, if there exists an t ∼ invertible matrix T ∈ GLn(K) such, that Aϕ = TAψT . We write ϕ = ψ.

It is easy to check, that equivalence of quadratic spaces and quadratic forms is indeed an equivalence relation. Furthermore it is obvious, that ϕ and ψ are equivalent if and only if (Kn, ϕ) and (Kn, ψ) are equivalent.

1.5 Remark. Now we have established a correspondence between quadratic forms, quadratic spaces (with a fixed basis) and symmetric matrices. We can assign to every quadratic space with a fixed basis a unique symmetric matrix, which in addition defines a unique quadratic form. Conversely for every quadratic form we obtain a unique quadratic space and a unique symmetric matrix. Furthermore we have seen, that this correspondence is compatible with equivalence. In the following sections this will enable us to switch between these three classes of objects, which will be of great help during many of the following proofs. 4

1.6 Deﬁnition. We say, that a quadratic map Q : V → K represents an element a ∈ K if there exists v ∈ V \{0} with Q(v) = a. Analogously an n-dimensional quadratic form ϕ is said to represent a if there exists w ∈ Kn \{0} such that ϕ(w) = a.

2 Orthogonality

2.1 Deﬁnition. Let (V,Q) be a quadratic space of dimension n over K.

2 (1) Two vectors v, w ∈ V are called orthogonal, if bQ(v, w) = 0. We use the notation v⊥w.

(2) A basis {v1, . . . , vn} of V is an orthogonal basis of (V,Q) if bQ(vi, vj) = 0 for all i 6= j.

(3) Two subspaces U1,U2 ⊂ V are orthogonal, if we have v⊥w for all v ∈ U1 and w ∈ U2. (4) For a subspace U ⊂ V we deﬁne

U ⊥ := {v ∈ V | v⊥u ∀ u ∈ U } .

If U ∩ U ⊥ = {0}, we call U ⊥ the orthogonal complement of U in V . (5) The Radical of (V,Q) is deﬁned as Rad(V ) = Rad(V,Q) := V ⊥ ⊂ V . (6) The space (V,Q) is called nondegenerate or regular, if Rad(V ) = {0}. Otherwise (V,Q) is called degenerate. (7) An n-dimensional quadratic form ϕ over K is called nondegenerate or regular, if (Kn, ϕ) is nondegenerate. Accordingly the form ϕ is called degenerate, if (Kn, ϕ) is degenerate.

⊥ From the fact that bQ is a bilinear form it follows, that U is a subvector space of V for every subspace U ⊂ V . In particular this implies, that Rad(V ) is a subspace of V . Observe, that the 0-dimensional quadratic space ({0}, 0) is nondegenerate by deﬁnition. Let (V,Q) be a quadratic space over K, and let U ⊂ V be a subspace of V . Denote by U ∗ = {f : U → K | f is K-linear} the dual of U. Then we obtain a K-linear map

∗ qU : V −→ U , v 7−→ U → K, u 7→ bQ(v, u) .

⊥ Since qU (v) = 0 if and only if bQ(v, u) = 0 for all u ∈ U, it becomes apparent that ker(qU ) = U . ∗ In particular if follows, that (V,Q) is nondegenerate if and only of qV : V → V is an isomorphism.

2.2 Deﬁnition. (1) Let (U1,Q1) and (U2,Q2) be quadratic spaces over K. We deﬁne the orthogonal sum of these two spaces to be (U1 ⊕ U2,Q1 + Q2), where simply

Q1 + Q2 : U1 ⊕ U2 −→ K, v1 + v2 7−→ Q1(v1) + Q2(v2).

We also use the notation (U1,Q1)⊥(U2,Q2) := (U1 ⊕ U2,Q1 + Q2).

(2) Let ψ ∈ K[X1,...,Xn] and χ ∈ K[Y1,...,Ym] be quadratic forms over K. The orthogonal sum ψ⊥χ is the quadratic form ψ + χ ∈ K[X1,...,Xn,Y1,...,Ym]. (3) If ϕ and ψ are quadratic forms over K such that there exists a quadratic form χ over K with ϕ =∼ ψ⊥χ, then ψ is a subform of ϕ.

Obviously the subspaces U1 and U2 of U1 ⊕U2 are orthogonal, hence the term “orthogonal sum”. This fact carries over to quadratic forms. More speciﬁcally (Kn+m, ψ⊥χ) =∼ (Kn, ψ)⊥(Km, χ). We immediately see, that Aψ 0 Aψ⊥χ = . 0 Aχ 2.3 Proposition. Let (V,Q) be a nondegenerate quadratic space over K, and let U ⊂ V be a subspace of V .

(1) All metric homomorphisms f :(V,Q) → (V 0,Q0) are injective

3 (2) For all subvector spaces U ⊂ V we have (i) (U ⊥)⊥ = U, ⊥ (ii) dimK (U) + dimK (U ) = dimK (V ), and (iii) Rad(U) = Rad(U ⊥) = U ∩ U ⊥.

⊥ The quadratic space (U, Q|U ) is nondegenerate iﬀ (U ,Q|U ⊥ ) is nondegenerate, and in this ⊥ case we have (V,Q) = (U, Q|U )⊥(U ,Q|U ⊥ ). (3) If V is the orthogonal direct sum of two subspaces, those are nondegenerate and orthogonal to each other.

[Ser73, Proposition 2, page 28]

Proof. (1): If f(v) = 0 for some v ∈ V1, then we have bQ(v, w) = bQ0 (f(v), f(w)) = 0 for all w ∈ V . Thus v ∈ Rad(V ), and as we assumed (V,Q) to be nondegenerate we must have v = 0. ∗ (2): Since (V,Q) is nondegenerate, the homomorphism qV : V → V is an isomorphism. ∗ Hence qU : V → U is surjective, since it is the composition of qV with the canonical surjection V ∗ → U ∗. As a result we get the exact sequence

0 −→ U ⊥ −→ V −→ U ∗ −→ 0,

∗ ⊥ ⊥ which implies dimK (V ) = dimK (U ) + dimK (U ) = dimK (U) + dimK (U ). Applying, what we have just proven, to U ⊥ instead of U, implies dim(U) = dim((U ⊥)⊥). Since obviously U ⊂ (U ⊥)⊥, we obtain U = (U ⊥)⊥. Now the equation Rad(U) = U ∩ U ⊥ is clear. Replacing U by U ⊥ and taking into account that U = (U ⊥)⊥ implies Rad(U) = U ∩ U ⊥ = Rad(U ⊥). The last statement of (2) immediately follows.

(3): If (V,Q) = (U1,Q|U1 )⊥(U2,Q|U2 ), then U1 and U2 are obviously orthogonal to each other. ⊥ ⊥ ⊥ Since U2 ⊂ U1 and also dim(U1 ) = dimK (V ) − dim(U1) = dim(U2), it follows that U2 = U1 . As Rad(U1) = U1 ∩ U2 = {0} we see, that U1 is nondegenerate. By (2) the same is true for ⊥ U2 = U1 . 2.4 Corollary. Every subform of a nondegenerate quadratic form ϕ over K is nondegenerate.

3 Diagonal Forms

Consider an n-dimensional quadratic form ϕ over K such that the standard basis {e1, . . . , en} n is also an orthogonal basis of (K , ϕ). In this case Aϕ is a diagonal matrix with entries say a1, . . . , an ∈ K, and we use the notation

2 2 ha1, . . . , ani := ϕ = a1X1 + ··· + anXn ∈ K[X1,...,Xn]

n and call ϕ a diagonal (quadratic) form with entries a1, . . . , an. Now assume Rad(K , ϕ) 6= {0}. Then there exists some 0 6= v = λ1e1 + ··· + λnen with λi ∈ K such that

bϕ(v, ei) = λibϕ(ei, ei) = λiai = 0 ∀ i = 1, . . . , n.

Since v 6= 0, there must exist at least one j ∈ {1, . . . , n} such that aj = 0. Conversely every nondegenerate form can only have nonzero entries.

4 3.1 Lemma. Every quadratic space (V,Q) can be written as an orthogonal sum (V,Q) = (Rad(V ), 0)⊥(U, Q|U ) with some subvector space U ⊂ V .

Proof. Choose any basis {v1, . . . , vr} for Rad(V ) and complete it to a basis {v1, . . . , vn} of V . Let U be the subspace of V generated by vr+1, . . . , vn. Then V = Rad(V ) ⊕ U, and by the deﬁnition of Rad(V ) we have bQ(v, w) = 0 for all v ∈ Rad(V ) and all w ∈ U. This shows, that Rad(V ) and U are orthogonal, which completes the proof.

3.2 Theorem. Every n-dimensional quadratic space (V,Q) has an orthogonal basis {v1, . . . , vn}. In particular, if there exists an a ∈ im(Q) \{0}, then we can choose v1 such that Q(v1) = a.

[Ser73, Theorem 1, page 30]

Proof. If Rad(V ) = V , then Q = 0 and any basis of V is an orthogonal basis. By the previous lemma it therefore only remains to show the theorem for the case, that dimK (V ) ≥ 1 and (Q, V ) is nondegenerate. Let a ∈ K∗ such that there exists a v ∈ V \{0} with Q(v) = a. Such an a must exists, since we excluded the case Q = 0. We will proceed by induction on n. For the case n = 1 the vector v trivially deﬁnes an orthogonal basis. Now let n > 1. The subspace Kv ⊂ V generated by v is trivially nondegenerate. By Proposi- ⊥ tion 2.3 (2) U := (Kv) is nondegenerate as well and (V,Q) = (Kv, Q|Kv)⊥(U, Q|U ). By the induction hypothesis we can ﬁnd an orthogonal basis {v2, . . . , vn} for U. If we set v1 := v, then {v1, . . . , vn} is the desired orthogonal basis. 3.3 Corollary. Every quadratic form ϕ of dimension n over K is equivalent to a diagonal form ha1, . . . , ani. If ϕ 6= 0, then a1 can be chosen to be any element of a ∈ im(ϕ) \{0}.

n Proof. By the previous theorem we can ﬁnd an orthogonal basis {v1, . . . , vn} of K such that n n ϕ(v1) = a if ϕ 6= 0. Consider the isomorphism f : K → K given by ei 7→ vi, where {e1, . . . , en} is the standard basis of Kn. Then (Kn, ϕ ◦ f) is a quadratic space with the standard basis as an orthogonal basis. Let ψ be the associated quadratic form to ϕ ◦ f with respect to the standard ∼ basis, then ψ = ϕ and ψ = ha1, . . . , ani. Furthermore a1 = a if ϕ 6= 0.

Before closing this section we introduce the discriminant of a quadratic form and prove a useful lemma, that is indispensable for the work with quadratic forms.

3.4 Deﬁnition. Let ϕ be a quadratic form over K.

(1) We deﬁne the determinant of ϕ as det(ϕ) := det(Aϕ).

∗ 2 ∗ ∗ 2 ∗ ∗ 2 (2) The discriminant of ϕ is d(ϕ) := det(Aϕ)(K ) ∈ K /(K ) ∪ {0}, where K /(K ) is the square class group of K.

∼ Let ϕ and ψ be two quadratic forms over K with ϕ = ψ, and let T ∈ GLn(K) be the isometry such t that Aψ = TAϕT . If det(T ) 6= ±1, then det(ϕ) 6= det(ψ). Thus we see, that the determinant of a quadratic form is not well deﬁned on the equivalence class of a quadratic form. But since det(ψ) = det(T )2 det(ϕ), it follows, that d(ϕ) = d(ψ). Furthermore by corollary 3.3 and the fact that det(T ) 6= 0 we have det(ϕ) 6= 0 resp. d(ϕ) 6= 0 if and only if ϕ is nondegenerate.

3.5 Lemma. If ϕ = ha, bi with a, b ∈ K∗ and c ∈ im(ϕ) \{0}, then ϕ =∼ hc, abci.

5 Proof. By corollary 3.3 we have ϕ =∼ hc, xi with some x ∈ K∗. Since d(ϕ) = d(hc, xi), there exists ∗ 2 ab ∼ 2 2 some z ∈ K with ab = z cx. Therefore x = z2c . Since obviously hc, xi = hc, z c xi, we obtain ha, bi =∼ hc, abci.

4 Isotropy and Hyperbolic Planes

4.1 Deﬁnition. Let (V,Q) be a quadratic space and ϕ be a quadratic form over K.

(1) A vector v ∈ V \{0} is called isotropic, if Q(v) = 0. A subspace U ⊂ V is called totally isotropic or plainly isotropic, if Q|U = 0. (2) The form ϕ is called isotropic, if there exists some v ∈ Kn \{0} with ϕ(v) = 0. Otherwise ϕ is called anisotropic.

Trivially Rad(V,Q) is a totally isotropic subspace. Accordingly every degenerate form ϕ is isotropic.

4.2 Deﬁnition. (1) A quadratic space (V,Q) is a hyperbolic plane, if V has a basis consisting of two isotropic vectors v, w ∈ V such that bQ(v, w) 6= 0. (2) The quadratic form h1, −1i and every form, that is isometric to it, will be called a hyperbolic plane. A quadratic form, that is isometric to an orthogonal sum of hyperbolic planes is called hyperbolic.

∗ 1 1 If bQ(v, w) = a ∈ K , then bQ(v, a w) = 1 and { a v, w} is still a basis consisting of isotropic vectors. Hence we can always assume, that bQ(v, w) = 1. Then the matrix associated to Q with 0 1 respect to {v, w} is ( 1 0 ). In particular we see, that (V,Q) is nondegenerate. If we consider the 2-dimensional quadratic form χ = 2X1X2 ∈ K[X1,X2] over K, then it follows, 2 1 1 that (K , χ) is a hyperbolic plane. Choosing the basis {v + 2 w, v − 2 w} we obtain the matrix 1 0 2 0 −1 , which shows, that (K , h1, −1i) is a hyperbolic plane, too. This explains the double use ∼ for the notion “hyperbolic plane”. Notably it follows, that h1, −1i = 2X1X2. 4.3 Proposition. For every nondegenerate quadratic space (V,Q) which has an isotropic ele- 0 ment v ∈ V , there exists a decomposition (V,Q) = (U, Q|U )⊥(V ,Q|V 0 ) such that (U, Q|U ) is a hyperbolic plane.

[Ser73, Proposition 3, page 29]

0 0 Proof. Since (V,Q) is nondegenerate, there exists w ∈ V such that bQ(v, w ) = 1. We must have 0 0 0 0 v and w linearly independant since bQ(v, v) = 0. Now w := 2w −bQ(w , w )v is isotropic, as can easily be checked. Also bQ(v, w) = 2. Let U be the subspace of V generated by v and w. Since 0 0 ⊥ (U, Q|U ) is nondegenerate the same holds for (V ,Q|V 0 ) where V = U by Proposition 2.3 (2). 0 Furthermore we have (V,Q) = (U, Q|U )⊥(V ,Q|V 0 ). 4.4 Corollary. If (V,Q) is a nondegenerate quadratic space over K possessing an isotropic element, then Q(V ) = K.

[Ser73, Corollary, page 29]

6 Proof. By the previous proposition we can without loss of generality assume, that (V,Q) is a hyperbolic plane. Furthermore we can assume, that we are given a basis v, w ∈ V such that v and w are isotropic with bQ(v, w) = 1. Let a ∈ K, then a a a Q v + w = Q(v) + 2b v, w + Q w = ab (v, w) = a. 2 Q 2 2 Q

4.5 Corollary. Every nondegenerate, isotropic quadratic form ϕ over K has a decomposition ϕ =∼ ψ⊥χ, where ψ is anisotropic and χ is hyperbolic.

Proof. The statement follows by applying proposition 4.3 inductively.

4.6 Corollary. Up to equivalence there exists only one nondegenerate, isotropic quadratic form of dimension 2 over K. It is given by

h1, −1i =∼ ha, −ai ∀ a ∈ K∗.

Proof. It suﬃces to apply the last two corollaries and lemma 3.5.

5 Witt’s Cancellation Theorem

The following theorem lays the foundation for the algebraic theory of quadratic forms.

5.1 Theorem. (Witt’s Cancellation Theorem ) ∼ ∼ Let ϕ, ϕ1 and ϕ2 be quadratic forms over K such that ϕ⊥ϕ1 = ϕ⊥ϕ2, then ϕ1 = ϕ2.

[Pﬁ95, Theorem 1.1, page 19]

Proof. Let dim(ϕ) = m, and let n = dim(ϕ1) = dim(ϕ2). Since the statement only deals with the equivalence of quadratic forms, we can by corollary 3.3 without loss of generality assume, that ϕ, ϕ1 and ϕ2 are diagonal forms. As the dimension of the radical of a quadratic space is naturally invariant under isometry, also n n the dimensions of Rad(K , ϕ1) and Rad(K , ϕ2) must be equal. Assume ϕ1 = hb1, . . . , bni and ∗ ϕ2 = hc1, . . . , cni with b1, . . . , bs, c1, . . . , cs ∈ K and bs+1 = ··· = bn = cs+1 = . . . cn = 0 for some s ∈ {0, . . . , n}. Then we can replace ϕ by ϕ⊥hbs+1, . . . , bni = ϕ⊥hcs+1, . . . , cni, and we can replace ϕ1 by hb1, . . . , bsi and ϕ2 by hc1, . . . , csi. That means we can assume, that ϕ1 and ϕ2 are nondegenerate. Finally by using a simple induction, we can further restrict our proof to the case, that dim(ϕ) = 1, say ϕ = hai for some a ∈ K∗.

The result of all these reductions is a (n + 1) × (n + 1)-matrix T such that (hai⊥ϕ1)(T v) = n+1 x0 0 n (hai⊥ϕ2)(v) for all v ∈ K . Let v = v0 with x0 ∈ K and v ∈ K . Similarly let

t xt T = y U

7 n with t ∈ K, x, y ∈ K and U ∈ Mn(K). Then

t 0 tx0 + x v (hai⊥ϕ1)(T v) = (hai⊥ϕ1) 0 yx0 + Uv t 0 2 0 = a(tx0 + x v ) + ϕ1(yx0 + Uv ) 2 0 = ax0 + ϕ2(v )

= (hai⊥ϕ2)(v).

Since char(K) 6= 2 we have t + 1 6= 0 or t − 1 6= 0. This means that one of the equations

t 0 t 0 x0 = tx0 + x v or − x0 = tx0 + x v (5.1)

xt 0 xt 0 has a solution, namely x0 = 1−t v or x0 = −t−1 v . Without loss of generality we can assume, xt 0 1 n that x0 = t−1 v is a solution of the ﬁrst equation in (5.1). Put w := 1−t x ∈ K . Then t 0 2 2 a(tx0 + x v ) = ax0 and we must have

t 0 ϕ1((yw + U)v ) = ϕ2(v).

n+1 0 n Since v ∈ K and therefore v ∈ K are arbitrary elements, and since by our assumption ϕ1 t ∼ and ϕ2 are nondegenerate, we must have (yw + U) ∈ GLn(K). Thus ϕ1 = ϕ2.

As an application we prove Witt’s chain equivalence theorem, which will be needed to show, among other things, that the Hasse invariant is well-deﬁned.

5.2 Deﬁnition. Let ϕ = ha1, . . . , ani and ψ = hb1, . . . , bni be two quadratic forms over K.

(1) The forms ϕ and ψ are simply-equivalent, if there exist two indices i and j (we do not require ∼ i 6= j) such that hai, aji = hbi, bji and ak = bk for all k ∈ {1, . . . , n} with k 6= i, j. (2) The forms ϕ and ψ are called chain-equivalent, if there exists a sequence of diagonal forms ϕ0, . . . , ϕk such that ϕ = ϕ0, ψ = ϕk and ϕi and ϕi−1 are simply-equivalent for i = 1, . . . , k. We write ϕ ≈ ψ.

It is obvious, that chain-equivalence of two quadratic forms implies, that those forms are equivalent in the usual sense. The following theorem takes care of the opposite direction.

5.3 Lemma. If ϕ = ha1, . . . , ani and ψ are two n-dimensional diagonal forms over K, such that there exists a permutation σ ∈ Sn with ψ = haσ(1), . . . , aσ(n)i, then ϕ and ψ are chain-equivalent.

Proof. The lemma follows from the fact, that Sn is generated by transpositions. 5.4 Theorem. (Witt’s Chain Equivalence Theorem) If ϕ and ψ are equivalent diagonal forms over K, then they are also chain-equivalent.

[Lam05, Theorem 5.2, page 16]

∼ n Proof. let ϕ = ha1, . . . , ani and ψ = hb1, . . . , bni. Since ϕ = ψ the K-vector spaces Rad(K , ϕ) and Rad(Kn, ψ) must have the same dimension. This means, that ϕ and ψ must have the same number of zero entries. From the previous lemma it follows, that it suﬃces to show the theorem only for the subforms of ϕ and ψ that consist only of the non-zero entries of ϕ resp. ψ. So we can without loss of generality assume, that ϕ and ψ are nondegenerate. We proceed by induction on n. The cases n = 1, 2 are trivial.

8 Assume n ≥ 3. Since chain-equivalence implies equivalence, every form, that is chain-equivalent 0 to ϕ, will represent b1. We choose a diagonal form ϕ = hc1, . . . , cni that is chain-equivalent to 2 2 ϕ with b1 = c1λ1 + ··· + cpλp, λi ∈ K, such that p is minimal for all forms chain-equivalent to ϕ. We want to show, that p = 1. 2 2 Assume, that p ≥ 2. Since p is minimal no subsum of c1λ1 + ··· + cpλp can be equal to zero. In 2 2 ∼ particular d = c1λ1 + c2λ2 6= 0. By lemma 3.5 we have hc1, c2i = hd, c1c2di. Thus

0 ϕ ≈ ϕ = hc1, c2, c3, . . . , cni

≈ hd, c1c2d, c3, . . . , cni

≈ hd, c3, . . . , cn, c1c2di,

2 2 and b1 = d + c3 + . . . cn. This contradicts the minimality of p. Hence we must have p = 1 0 ∼ 0 and ϕ ≈ ϕ = hb1, c2, . . . , cni. Since ϕ = ϕ by Witt’s cancellation theorem 5.1 it follows ∼ ∼ from hb1, c2, . . . , cni = hb1, . . . , bni, that hc2, . . . , cni = hb2, . . . , bni. By the induction hypothesis hc2, . . . , cni ≈ hb2, . . . , bni. Altogether we obtain

ϕ ≈ hb1, c2, . . . , cni ≈ hb1, . . . , bni = ψ.

6 Application: Quadratic Forms over Finite Fields

Let p 6= 2 be a prime number, and let q = pr be a power of p. In this section we will give a complete classification of quadratic forms over the finite field Fq with q elements. In the following results we will only consider nondegenerate forms. By lemma 3.1 this does not siginify a restriction, since the radical of a quadratic space is invariant under equivalence.

6.1 Proposition. If ϕ is a nondegenerate quadratic form over Fq with dim(ϕ) ≥ 2, then im(ϕ) ⊃ ∗ Fq . In the case dim(ϕ) ≥ 3 we even have im(ϕ) = Fq. [Ser73, Proposition 4, page 34]

Proof. By corollary 3.3 it suﬃces to consider an arbitrary 2-dimension diagonal form ha, bi. If ha, bi is already isotropic, then we have im(ha, bi) = Fq and there is nothing to show. So assume, ∗ that ha, bi is anisotropic. The statement will be proven, if we can show for any x ∈ Fq , that ha, b, xi is isotropic. But this is a consequence of the Chevalley theorem.

∗ ∗ 2 ∗ We know, that Fq/(Fq ) has exactly two elements, since p 6= 2. Let a denote an element of Fq which is not a square.

6.2 Proposition. Every nondegenerate quadratic form ϕ of dimension n over Fq is equivalent to either h1,..., 1, 1i or h1,..., 1, ai

∗ ∗ 2 depending on whether its discriminant is the unit element of Fq/(Fq ) or not. [Ser73, Proposition 5, page 34]

Proof. The statement is trivial if n = 1. Now if n ≥ 2, then the previous proposition shows, that ϕ represents 1. By lemma 3.5 we have ϕ =∼ h1i⊥ψ with dim(ψ) = n − 1. The statement now follows by simple induction.

9 6.3 Corollary. Two nondegenerate forms over Fq are equivalent if and only if they have the same dimension and the same discriminant.

[Ser73, Corollary, page 35]

References

[Lam05] Lam, Tsit-Yuen: Introduction to Quadratic Forms over Fields. American Mathematical Society, 2005

[Pﬁ95] Pfister, Albrecht: London Math. Soc. Lecture Note Series. Bd. 217: Quadratic Forms with Applications to Algebraic Geometry and Topology. Cambridge University Press, 1995

[Ser73] Serre, Jean-Pierre: Graduate Texts in Mathematics. Bd. 7: A Course in Arithmetic. Springer-Verlag New York, Heidelberg, Berlin, 1973