Trace forms of central simple algebras over a local field or a global field

Grégory Berhuy

U.M.R.6623 du C.N.R.S., Labo de Maths, Bureau 401M, 16 route de Gray, F-25030 Besançon, France (e-mail: berhuy@math.univ-fcomte.fr)

This work is supported by the TMR research network (ERB FMRX CT-97-0107) on “K-theory and algebraic groups ”

Introduction: Let A be a central simple algebra over a field k of characteristic different from 2. The quadratic form x ∈ A ↦ Trd_{A}(x^{2}) ∈ k is called the trace form of A, and is denoted by T_{A}. This trace form has been studied by many authors (cf.[L], [LM], [Ti] et [Se], Annexe §5 for example). In particular, these classical invariants are well-known (loc.cit.). In [B], we have determined a diagonalization of the trace form of cyclic algebras over fields such that the cube of the fundamental ideal is trivial. In particular, this gives all trace forms of central simple algebras over local fields and non formally real global fields, since any algebra over such a field is cyclic. In this article, we give a characterization of the trace form of a central simple algebra over a local field or a global field in terms of determinant and signatures. We also show that a quadratic form over a global field is isomorphic to a trace form if and only if it is true locally. Then we give a necessary and sufficient condition on trace forms to be isomorphic. Finally, we apply these results to describe explicitly the elements of Br_{2}(k) which can be written \frac{n}{2}[A], where A is a central simple algebra of even degree n, when k is a local field or a global field of characteristic different from 2.

Recalls and notation: If A is a central simple algebra over k, the exponent of A, denoted by expA, is the order of [A] in Br(k) and the index of A, denoted by indA, is the degree of the division algebra which corresponds to A. We know that expA divides the degree of A. If a, b ∈ k^{∗}, we denote by (a, b)_{k} the corresponding quaternion algebra, or simply (a, b) if no confusion is possible. We also use the same notation to design its class in the Brauer group. If L/k is any field extension, set [A]_{L} = [A ⊗ L]. Then we have [(a, b)_{k}]_{L} = (a, b)_{L}. Recall now the definition of a cyclic algebra. Let L/k be a cyclic extension of degree n, σ a generator of Gal(L/k) and a ∈ k^{∗}. The ring (a, L/k, σ) = ⨁^{n−1} ; i=0Le^{i} with the multiplication law e^{n} = a and

eλ = σ(λ)e, λ ∈ L is a central simple algebra, called a cyclic algebra. If k is a local field, we know that indA =expA. Moreover, we get an isomorphism inv_{k} : Br(k) →^{∼} ℚ/ℤ. Under this isomorphism, \frac{s}{n} (where s and n are not necessarily relatively prime) corresponds to the class of the cyclic algebra (π^{s}, L/k, σ), where π is a uniformizing parameter, L/k is the unique unramified extension (then cyclic) of degree n and σ the associated Frobenius automorphism. Moreover (π^{s}, L/k, σ) is a division algebra if and only if (s, n) = 1. We also set inv_{ℂ} ≡ 0. Finally we define

inv_{ℝ} : Br(ℝ) → ℚ/ℤ by inv_{ℝ}(−1, −1) = \frac{1}{2}. If k is a global field, we write inv_{𝔭} instead of inv_{k_{𝔭}} for all prime 𝔭 of k. Then we get the following exact

sequence 0 → Br(k) → ∑ Br(k_{𝔭}) \underrightarrow{∑ inv_{𝔭}} ℚ/ℤ → 0. Moreover, we know that expA =indA =lcm(indA ⊗ k_{𝔭}). For more details, we refer to [W] or [CS]. Concerning central simple algebras over general fields, we refer to [D], [J] or [Sc]. In the following, all the quadratic forms are non singular. If q is a quadratic form over k, dimq is the dimension, detq ∈ k^{∗}/k^{∗2} is the determinant and sign_{P} q ∈ ℤ is the signature of q relatively to the ordering P, that is the difference between the number of positive elements and the number of negative elements in any diagonalization of q. If k is a global field, each real prime (if there is any) corresponds to a real embedding, then to an ordering and vice versa. We will denote by sign_{𝔭}q the signature of q relatively to the ordering corresponding to this prime. If q ≃< a_{1}, ⋯, a_{n} >, the Hasse-Witt invariant of q is given by w_{2}(q) = ∑_{i<j}(a_{i}, a_{j}) ∈ Br_{2}(k). Thus, if L/k is any

field extension, we get w_{2}(q)_{L} = w_{2}(q ⊗ L). Finally, we denote by ℍ the hyperbolic plane.

A. Trace forms of central simple algebras over a local field or a global field.

In this section, we first show the following results:

Theorem 1: Let k be a local field and n ≥ 2 an even integer. Then a quadratic form q over k is isomorphic to the trace form of a central simple algebra of degree n if and only if the following conditions hold:

1. dimq = n^{2}

2. detq = (−1)^{\frac{n(n−1)}{2}} Theorem 2: Let k be a global field of characterisitc different from 2, n ≥ 2 an even integer, and q a quadratic form over k. Then the following conditions are equivalent:

(1) The quadratic form q is isomorphic to the trace form of a central simple algebra over k of degree n

(2) q is isomorphic to a trace form locally everywhere, that is for all prime 𝔭 of k, q ⊗k_{𝔭} is isomorphic to the trace form of a central simple algebra of degree n over k_{𝔭}

(3) q satisfies the following conditions:

(i) dimq = n^{2}

(ii) detq = (−1)^{\frac{n(n−1)}{2}}

(iii) sign_{𝔭}q = ±n for all real prime 𝔭 of k

These two theorems only deal with the case when n is even, because we know

that T_{A} ≃ n < 1 >⊥ \frac{n(n − 1)}{2}ℍ if n is odd (cf.[Se], Annexe §5 for example).

Before proving these theorems, we recall some results about the classical invariants of trace forms of central simple algebras:

Theorem 3: Let k be a field of characteristic different from 2, and let A be a central simple algebra over k of degree n. Then we have:

1. dimT_{A} = n^{2}

2. detT_{A} = (−1)^{\frac{n(n−1)}{2}}

3. We have sign_{P} T_{A} = ±n for each ordering P, and sign_{P} T_{A} = n if and only if [A]_{k_{P}} = 0, where k_{P} is the real closure of (k, P)

4. If n = 2m ≥ 2, then w_{2}(T_{A}) = \frac{m(m − 1)}{2}(−1, −1) + m[A]

The three first statements can be found in [L], and the last one is proved in [LM] or [Ti] for example .

Lemma: Let k be a local field, n = 2m ≥ 2 and A a central simple algebra of degree n. Let (a, b) be the unique non zero element of Br_{2}(k). Then m[A] = (a, b) if and only if A is a division algebra.

Proof of the lemma: We know that expA =ind A, then A is a division algebra if and only if [A] has order n in Br(k), which is equivalent to m[A] ≠ 0, that is m[A] = (a, b) since (a, b) is the unique non zero element of Br_{2}(k).

Proof of theorem 1: According to the previous theorem, the two conditions are necessary. We show that there are sufficient. It is well-known that quadratic forms over k are determined up to isomorphism by dimension, determinant and Hasse-Witt invariant.

We have \frac{m(m − 1)}{2}(−1, −1) + w_{2}(q) = (c, d), with (c, d) = 0 or (a, b). Assume first that (c, d) = 0. In this case, set A = M_{n}(k). Then the last statement of theorem 3 gives

w_{2}(T_{A}) = \frac{m(m − 1)}{2}(−1, −1) = w_{2}(q). Thus the two quadratic forms q and T_{A} have the same invariants, so there are isomorphic. Assume now that (c, d) = (a, b), and set A = (π, L/k, σ), where π is a uniformizing parameter of k, L/k is the unique unramified extension of degree n and σ is the Frobenius automorphism. Then A is a division algebra of degree n, then m[A] = (a, b) by the lemma. Now conclude as previously.

Proof of theorem 2: We first show that (2) and (3) are equivalent. If 𝔭 is a complex prime, it is clear that a quadratic form over k_{𝔭} is isomorphic to a trace form if and only if (i) is satisfied. If 𝔭 is a real prime, quadratic forms are determined by dimension and signatures. By theorem 3, it follows that a quadratic form over k_{𝔭} is isomorphic to a trace form if and only if (i) et (iii) are satisfied. Finally, if 𝔭 is a finite prime, theorem 1 implies that a quadratic form over k_{𝔭} is isomorphic to a trace form if and only if (i) et (ii) are satisfied. So we get the desired equivalence. Now we show that (1) and (3) are equivalent. The three conditions are necessary by theorem 3. In order to show that there are also sufficient, we consider the two cases n ≡ 2 [4] and n ≡ 0 [4] separately.

• Assume first that n = 2m, with m odd. Since k is a global field, every central algebra of exponent 2 is similar to a quaternion algebra (it comes from the equality expA =indA).

So we can write \frac{m(m − 1)}{2}(−1, −1)_{k} + w_{2}(q) = (c, d)_{k}. Let 𝔭 be a real prime of k.

If sign_{𝔭}q = n, we have q ⊗ k_{𝔭} ≃ n < 1 >⊥ \frac{n(n − 1)}{2}ℍ.

Then we easily get w_{2}(q)_{k_{𝔭}} = w_{2}(q ⊗ k_{𝔭}) = \frac{m(m − 1)}{2}(−1, −1)_{k_{𝔭}}.

If sign_{𝔭}q = −n, we have q ⊗ k_{𝔭} ≃ n < −1 >⊥ \frac{n(n − 1)}{2}ℍ.

Then we have w_{2}(q)_{k_{𝔭}} = w_{2}(q ⊗ k_{𝔭}) = (\frac{m(m − 1)}{2} + m)(−1, −1)_{k_{𝔭}}

= \frac{m(m − 1)}{2}(−1, −1)_{k_{𝔭}} + (−1, −1)_{k_{𝔭}}, since m is odd. So we have sign_{𝔭}q = n if and only if (c, d)_{k_{𝔭}} = 0. Set A = M_{m}((c, d)_{k}). Then sign_{𝔭}T_{A} = n if and only if (c, d)_{k_{𝔭}} = 0, so q and T_{A} have the same signatures. Moreover we get m[A] = (c, d)_{k} since m is odd, so the previous quadratic forms have the same Hasse-Witt invariants. By assumption, dimensions and determinants are equal, so q and T_{A} are isomorphic.

• Assume now that n = 2m, with m ≥ 2 even. First we construct a central simple algebra over k, defining it locally everywhere and using the exact sequence recalled at the beginning.

As previously, set \frac{m(m − 1)}{2}(−1, −1)_{k} + w_{2}(q) = (c, d)_{k}. If 𝔭 is a complex prime, we have (c, d)_{k_{𝔭}} = 0. If 𝔭 is a real prime such that sign_{𝔭}q = n, we have already seen that

w_{2}(q)_{k_{𝔭}} = \frac{m(m − 1)}{2}(−1, −1)_{k_{𝔭}}, so (c, d)_{k_{𝔭}} = 0.

If 𝔭 is a real prime such that sign_{𝔭}q = −n, we have seen that

w_{2}(q)_{k_{𝔭}} = (\frac{m(m − 1)}{2} + m)(−1, −1)_{k_{𝔭}} = \frac{m(m − 1)}{2}(−1, −1)_{k_{𝔭}} since m is

even in this case. So we have again (c, d)_{k_{𝔭}} = 0. Finally, the primes for which (c, d)_{k_{𝔭}} ≠ 0 are finite. By Hilbert’s reciprocity law, the number r of these primes is finite and even, and the set S of finite primes 𝔭 such that (c, d)_{k_{𝔭}} = 0 is infinite. Let T be the set of real primes for which the signature of q is equal to −n, and let t be its cardinality. If r = 0, set B_{𝔭_{1}} = (π_{𝔭_{1}}^{tm}, L_{𝔭_{1}}/k_{𝔭_{1}}, σ_{𝔭_{1}}) for a given 𝔭_{1} ∈ S (so we have B_{𝔭} = M_{n}(k_{𝔭_{1}}) if t = 0) and B_{𝔭} = M_{n}(k_{𝔭}) for all finite prime 𝔭 ≠ 𝔭_{1}. Suppose that r > 0 and choose 𝔭_{1}, 𝔭_{2} ∈ S. Now set B_{𝔭_{1}} = (π_{𝔭_{1}}^{tm}, L_{𝔭_{1}}/k_{𝔭_{1}}, σ_{𝔭_{1}})

and B_{𝔭_{2}} = (π𝔭_{2}^{\frac{n−r}{n}}, L_{𝔭_{2}}/k_{𝔭_{2}}, σ_{𝔭_{2}}). These algebras are not division algebras because r and n are even, and tm and n are both divisible by m. Then set B_{𝔭} = M_{n}(k_{𝔭}) for all primes in S − {𝔭_{1}, 𝔭_{2}}.

So we have inv_{𝔭_{1}}[B_{𝔭_{1}}] = \frac{t}{2} in all cases and inv_{𝔭_{2}}[B_{𝔭_{2}}] = \frac{n − r}{n}. If 𝔭 is a finite prime, 𝔭 ∉ S, set B_{𝔭} = (π, L_{𝔭}/k_{𝔭}, σ_{𝔭}). The algebra B_{𝔭} is then

a division algebra and inv_{𝔭}[B_{𝔭}] = \frac{1}{n} for each finite prime 𝔭 ∉ S.

Now set B_{𝔭} = M_{m}((−1, −1)_{k_{𝔭}}) for all 𝔭 ∈ T (in this case, we have

inv_{𝔭}[B_{𝔭}] = \frac{1}{2}) and set B_{𝔭} = M_{n}(k_{𝔭}) for the other infinite primes. By construction, ∑ inv_{𝔭}[B_{𝔭}] = 0 in ℚ/ℤ. Thus there exists some

[B] ∈ Br(k), where B is a division algebra, such that [B]_{k_{𝔭}} = [B_{𝔭}] for all primes of k. Since the algebras B_{𝔭} have degree n, each index is a divisor of n, so the index of B divides n (because it is the least common multiple of the indices). Since B is a division algebra, we have indB =degB, so degB divides n. Then we set A = M_{j}(B), with j = \frac{n}{degB} .

Now we show that q ≃ T_{A}. It suffices to prove that q ⊗ k_{𝔭} ≃ T_{A⊗k_{𝔭}} for all primes of k. Then we get q ⊗ k_{𝔭} ≃ T_{A} ⊗ k_{𝔭} for all primes of k, and we obtain q ≃ T_{A}. Notice that we have [A]_{k_{𝔭}} = [B_{𝔭}]. If 𝔭 is a complex prime, the isomorphism is clear. If 𝔭 is a real prime, then we have by construction [A]_{k_{𝔭}} = [B_{𝔭}] = 0 if and only if sign_{𝔭}q = n. By theorem 3, q ⊗ k_{𝔭} and T_{A⊗k_{𝔭}} have the same signature, so there are isomorphic since the dimensions are equal. Now we consider the case of finite primes. By construction, for every finite prime 𝔭 ∈ S, B_{𝔭} is not a division algebra, so m[A]_{k_{𝔭}} = m[B_{𝔭}] = 0 by the lemma. Since 𝔭 ∈ S, we have (c, d)_{k_{𝔭}} = 0, so q ⊗ k_{𝔭} and T_{A⊗k_{𝔭}} have the same Hasse-Witt invariant. Since their dimensions and their determinants are equal, there are isomorphic. Finally if 𝔭 is a finite prime such that 𝔭 ∉ S, we have (c, d)_{k_{𝔭}} ≠ 0 and B_{𝔭} is a division algebra. If (a_{𝔭}, b_{𝔭}) denotes the unique non zero element of Br_{2}(k_{𝔭}), we have by the lemma m[A]_{k_{𝔭}} = m[B_{𝔭}] = (a_{𝔭}, b_{𝔭}) = (c, d)_{k_{𝔭}}. Now we conclude as previously and this finishes the proof of theorem 2.

Remark: By theorem 2, if q is k_{𝔭}-isomorphic to the trace form of a central simple algebra A_{𝔭} over k_{𝔭} of degree n for all primes, then q is k-isomorphic to the trace form of a central simple algebra A over k of degree n. Nevertheless, it does not mean that A ⊗ k_{𝔭} ≃ A_{𝔭} for all 𝔭. Indeed , it can happens that ∑ inv_{𝔭}A_{𝔭} ≠ 0, so there is no central simple algebra over k which is similar to A_{𝔭} in Br(k_{𝔭}) for all primes, according to the exact sequence recalled at the beginning. Since the degrees are equal, this is equivalent to say that the isomorphism A ⊗ k_{𝔭} ≃ A_{𝔭} does not hold for all 𝔭.

Example: Assume that n ≡ 0 [8]. For all primes 𝔭 of k, set q_{𝔭} = T_{A_{𝔭}}, where A_{𝔭} = M_{n}(k_{𝔭}) except for two primes 𝔭_{1} and 𝔭_{2} for which we set A_{𝔭_{i}} = (π_{𝔭_{i}}, L_{𝔭_{i}}/k_{𝔭_{i}}, σ_{𝔭_{i}}). Then we have w_{2}(q_{𝔭}) = 0 if 𝔭 ≠ 𝔭_{1}, 𝔭_{2} and w_{2}(q_{𝔭_{i}}) = (a_{𝔭_{i}}, b_{𝔭_{i}}) (it suffices to apply theorem 3 and the lemma).

For all primes 𝔭, we have detq_{𝔭} = (−1)^{\frac{n(n−1)}{2}} and the number of primes 𝔭 for which w_{2}(q_{𝔭}) ≠ 0 is finite and even. So there exists a quadratic form q over k such that q ⊗ k_{𝔭} ≃ q_{𝔭}. (cf.[Sc], theorem 6.6.10). Thus this form q satisfies q ⊗ k_{𝔭} ≃ T_{A_{𝔭}}. So we have q ≃ T_{A}, but we cannot have A ⊗ k_{𝔭} ≃ A_{𝔭} because

∑ inv_{𝔭}[A_{𝔭}] = \frac{2}{n}.

We can explain the fact that it is always possible to find a suitable A observing that over a local field, there are many central simple algebras with the same trace form. So we can choose some algebras B_{𝔭} such that T_{B_{𝔭}} ≃ T_{A_{𝔭}} for all primes and satisfying ∑ inv_{𝔭}[B_{𝔭}] = 0. More precisely, we have the following proposition:

Proposition 1: Let k be a field of characteristic different from 2, and let A and B be two central simple algebras of even degree n.

(1) If k is a local field, we have T_{A} ≃ T_{B} if and only if (A is division algebra ⟺ B is a division algebra)

(2) If k = ℝ, we have T_{A} ≃ T_{B} if and only if A ≃ B

(3) If k is a global field, we have T_{A} ≃ T_{B} if and only if A ⊗ k_{𝔭} and B ⊗ k_{𝔭} satisfy the previous conditions over k_{𝔭} for all finite or real primes

Proof: If k is a local field, the Hasse-Witt invariants of T_{A} et T_{B} are equal if and only if A and B satisfy the condition (1) by the lemma. Since their dimensions and their determinants are equal, this is equivalent to T_{A} ≃ T_{B}. If k = ℝ, it suffices to notice that M_{n}(ℝ) and M_{m}((−1, −1)_{ℝ}) are the only central simple algebras of degree n and that their trace forms are not isomorphic (since the signatures differ by theorem 3). If k is a global field , the condition (3) simply expresses that T_{A} and T_{B} are isomorphic if and only if it is true locally everywhere (using the fact that these forms are automatically k_{𝔭}-isomorphic for all complex primes).

B. Application to the study of the elements of Br_{2}(k).

We are now interested in the following problem: if A is a central simple algebra of even degree n, we know that \frac{n}{2}[A] ∈ Br_{2}(k). Conversely, which

classes [B] ∈ Br_{2}(k) can be written [B] = \frac{n}{2}[A], where A is a central simple algebra of degree n ? If k is a local field or a global field of characteristic different from 2, the answer is given by the following proposition:

Proposition 2: Let k be a field of characteristic different from 2 and n ≥ 2 an even integer.

(1) If k is a local field, then for all [B] ∈ Br_{2}(k) there exists a central simple algebra A over k of degree n such that \frac{n}{2}[A] = [B].

(2) If k is a global field, then we have:

(i) If n ≡ 2 [4], then for all [B] ∈ Br_{2}(k) there exists a central simple algebra A over k of degree n such that \frac{n}{2}[A] = [B]

(ii) If n ≡ 0 [4], then for all [B] ∈ Br_{2}(k), there exists a central simple algebra A over k of degree n such that \frac{n}{2}[A] = [B] if and only if [B]_{k_{𝔭}} = 0 for all real primes

Proof: If k is a local field, we can take A = M_{n}(k) if [B] = 0 and A is a division algebra if [B] = (a, b) by the lemma. Now assume that k is a global field. We can assume that B is a quaternion algebra (c, d)_{k}. If n = 2m with m odd, set A = M_{m}((c, d)_{k}). So we can assume that n = 2m, where m is even. If [B] = m[A], we have [B]_{k_{𝔭}} = m[A]_{k_{𝔭}} = 0 for all real primes since m is even and [A]_{k_{𝔭}} = 0 or (−1, −1)_{k_{𝔭}}. Conversely, assume that [B]_{k_{𝔭}} = 0 for all real primes. For each infinite prime 𝔭, set q_{𝔭} = T_{M_{n}(k_{𝔭})}. If 𝔭 is a finite prime, set q_{𝔭} = T_{A_{𝔭}}, where A_{𝔭} = M_{n}(k_{𝔭}) if (c, d)_{k_{𝔭}} = 0 and A_{𝔭} is a division

algebra otherwise. Then we get w_{2}(q_{𝔭}) = \frac{m(m − 1)}{2}(−1, −1)_{k_{𝔭}} + (c, d)_{k_{𝔭}} for all primes. Indeed, this is true by assumption if 𝔭 is real, it is trivial if 𝔭 is complex, and this is verified by choice of A_{𝔭} and by the lemma if 𝔭 is finite.

So we have w_{2}(q_{𝔭}) = (α, β)_{k_{𝔭}}, where (α, β)_{k} = \frac{m(m − 1)}{2}(−1, −1)_{k} + (c, d)_{k}. Then by Hilbert’s reciprocity law, the number of primes 𝔭 for which w_{2}(q_{𝔭}) = 0 is finite and even. Since detq_{𝔭} = (−1)^{\frac{n(n−1)}{2}} for all primes, there exists a quadratic form q over k such that q ⊗ k_{𝔭} ≃ q_{𝔭}. By choice of q_{𝔭} and by theorem 2, we have q ≃ T_{A} where A is a central simple algebra over k of degree n. Comparing Hasse-Witt invariants, we get m[A]_{k_{𝔭}} = (c, d)_{k_{𝔭}} for all primes. Then we get m[A] = (c, d)_{k} = [B] (It is a consequence of the Hasse principle for the elements of Br(k). We refer to [CS] or [W] for more details for example).

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