On Examples of Witt Functors in Quadratic Number Fields

Paweł Gładki; Mateusz Pulikowski

doi:10.2478/amsil-2026-0011

For R a commutative ring with 1, consider the Witt ring W R of Witt equivalence classes of finitely generated projective modules M over R endowed with nondegenerate symmetric bilinear forms β : M ×M → R (see Milnor and Husemoller [6] for details). The diagonal form a₁x₁y₁ + . . . + a_nx_ny_n with a_i ∈ R, i ∈ {1, . . . , n} shall be denoted by ⟨a₁, . . . , a_n⟩, and its Witt equivalence class by [a₁, . . . , a_n]. Pfister forms, i.e. forms of the shape ⟨1, a₁⟩⊗· · · ⊗⟨1, a_n⟩, shall be denoted by ⟨⟨a₁, . . . , a_n⟩⟩, and their Witt classes by [[a₁, . . . , a_n]], a_i ∈ R, i ∈ {1, . . . , n}.

A homomorphism φ : R → R′ between two commutative rings with 1 R and R′ induces in a natural way a homomorphism between their respective Witt rings: R′ is made into an R-module with multiplication defined by R × R′ ∋ (a, a′) ↦ f (a) · a′ ∈ R′, so that if M is an R-module, it can be naturally extended to an R′-module M′ = R′ ⊗_R M with multiplication given by a′ · (b′ ⊗ m) = a′b′ ⊗ m, for a′ ∈ R′ and simple tensors b′ ⊗ m ∈ R′ ⊗_R M ; now a bilinear form β : M × M → R uniquely extends to a bilinear form β′ : M′ × M′ → R′ such that $β^{'} (a^{'} \otimes m, b^{'} \otimes n) = a^{'} b^{'} f (β (m, n)),$ \beta '\left( {a' \otimes m,\,b' \otimes n} \right) = a'b'\,f\left( {\beta \left( {m,\,n} \right)} \right), for a′, b′ ∈ R′, m, n ∈ M . If M is finitely generated and projective then so is M′, and if β is symmetric and nondegenerate, then β′ is such as well – denoting by f_# (M, β) (or f_# (M ), for short) the pair (M′, β′) we thus obtain a well-defined homomorphism of Witt rings f_# : W R → W R′ by assigning to the Witt equivalence class [M ] the class [f_# (M)].

We will be concerned with one special class of examples of such homomorphisms here. Let K be a number field and let 𝒪_K be its ring of integers. The natural homomorphism W 𝒪_K → W K induced by the map f : 𝒪_K → K, $f (a) = \frac{a}{1}$ f\left( a \right) = {a \over 1} , is injective ([5], Satz 11.1.1), but if we replace 𝒪_K with an arbitrary ring 𝒪 whose field of fractions is equal to K this may no longer be true.

Consider one particular class of such rings, namely orders of the field K, that is subrings 𝒪 of 𝒪_K which, as ℤ-modules, are of rank n = [K : ℚ]. Craven, Rosenberg and Ware ([2], Remark following Proposition 3.2) showed that W 𝒪 → W K is not injective for 𝒪 = ℤ [3i] and K = ℚ(i). This was later generalized by Ciemała and Szymiczek ([1], Example 4.5), who proved that W 𝒪 → W K is not injective for all orders 𝒪 = ℤ[fi], with f > 1, K = ℚ(i). Moreover, they also demonstrated that for an arbitrary number field K and order 𝒪 with conductor 𝔣 = {a ∈ 𝒪_K | a𝒪_K ⊆ 𝒪} such that 𝔣 ⊆ 2𝒪_K the homomorphism W 𝒪 → W K is not injective – these examples led them to conjecture that the only orders 𝒪 for which W 𝒪 → W K is injective are, in fact, the maximal orders 𝒪 = 𝒪_K . This turned out to be false, as shown by Rothkegel ([7], Theorem 2.2), who proved that for $𝕆 = 𝕑 [f \sqrt{d}]$ {\mathbb {O}} = {\mathbb {Z}}\left[ {f\sqrt d } \right] and $K = 𝕈 (\sqrt{d})$ K = {\mathbb {Q}}\left( {\sqrt d } \right) with d ≢ 1(mod 4), 2 ∤ f and f | d the homomorphism W 𝒪 → W K is injective. This was further extended by the authors: firstly, in [3], Theorem 1.1 we showed that for $𝕆 = 𝕑 [3 \sqrt[3]{6}]$ {\mathbb {O}} = {\mathbb {Z}}\left[ {3\root 3 \of 6 } \right] and $K = 𝕈 (\sqrt[3]{6})$ K = {\mathbb {Q}}\left( {\root 3 \of 6 } \right) the homomorphism W 𝒪 → W K is not injective, and then in [4], Theorem 3, we generalized this result to the case when $K = 𝕈 (\sqrt[n]{m})$ K = {\mathbb {Q}}\left( {\root n \of m } \right) , n = p^k, with k ∈ ℕ, p a prime, p ≠ 2, with m square-free, m ≠ ±1, p | m and $𝕆 = 𝕑 [p \sqrt[n]{m}]$ {\mathbb {O}} = {\mathbb {Z}}\left[ {p\root n \of m } \right] : here W 𝒪 → W K is also injective.

In this miniature note we add one more piece of puzzle to the big picture. Namely, although in general one expects that for a randomly selected order 𝒪 of a number field K the homomorphism W 𝒪 → W K shall not be injective, it appears that other than the few abovementioned examples by Craven-Rosenberg-Ware/Ciemała-Szymiczek for the Gaussian field ℚ(i), and the series of orders with “even” conductors – no explicit examples are to be found in literature. We aim to fill that gap here: such examples are relatively easy to build, by combining results obtained by Ciemała and Szymiczek with some (more or less) elementary number theory. We shall discuss it here in some detail. The key result used by Ciemała and Szymiczek to build their examples is the following:

Proposition 1 ([1, Theorem 4.4]).

Let K be a number field, 𝒪_K its ring of integers, 𝒪 an order and denote by U (𝒪) the group of units of 𝒪. Let (S, β) be a nondegenerate symmetric bilinear space over 𝒪 with S a free module of rank 2, and assume that in a certain basis β has the matrix $[\begin{matrix} A & C \\ C & B \end{matrix}] .$ \left[ {\matrix{ A & C \cr C & B \cr } } \right]. If all of the following conditions are met:

A, B, C ∈ 𝒪,
AB ≠ 0,
AB − C² = −u² ∈ U (𝒪), D ∈ K \ {0},
denoting by d and d′ the roots of the isotropy equation: $B^{2} X + 2 C X + A = 0,$ B^2 X + 2CX + A = 0, that is $d = \frac{- C + u}{B} = \frac{A}{- C - u}$ d = {{ - C + u} \over B} = {A \over { - C - u}} and $d^{'} = \frac{- C - u}{B} = \frac{A}{- C + u}$ d' = {{ - C - u} \over B} = {A \over { - C + u}} , d and d′ are integral over 𝒪 each of degree at least 2,

then the Witt equivalence class of (S, β) is a nonzero element in the kernel of W 𝒪 → W K.

We will use Proposition 1 to exhibit some examples of non-injective natural homomorphisms of Witt rings in quadratic fields. We turn our attention to real quadratic fields first. Let $K = 𝕈 (\sqrt{d})$ K = {\mathbb {Q}}\left( {\sqrt d } \right) with d > 0 square free and d ≢ 1(mod 4), so that $𝕆_{K} = 𝕑 (\sqrt{d})$ {\mathbb {O}}_K = {\mathbb {Z}}\left( {\sqrt d } \right) and there are infinitely many units, each of the form εⁿ, where $ε = x_{0} + y_{0} \sqrt{d}$ \varepsilon = x_0 + y_0 \sqrt d is the fundamental unit of K. For an integer f > 1 set $𝕆 = 𝕑 [f \sqrt{d}]$ {\mathbb {O}} = {\mathbb {Z}}\left[ {f\sqrt d } \right] . Let $ε^{n} = a_{n} + b_{n} \sqrt{d}$ \varepsilon ^n = a_n + b_n \sqrt d with integers a_n, b_n.

Theorem 1.

Suppose there exists an integer n ≥ 1 such that $a_{n} \equiv 0 (\mod f), b_{n} ≢ 0 (\mod f) .$ a_n \equiv 0\,\,\,\,\left( {\bmod \,f} \right),\,\,\,\,b_n \not\equiv 0\,\,\,\,\left( {\bmod \,f} \right). Then the natural homomorphism W 𝒪 → W K is not injective.

Proof.

Set $u = ε^{n} = a_{n} + b_{n} \sqrt{d}$ u = \varepsilon ^n = a_n + b_n \sqrt d . By design, u is a unit in 𝒪_K . Because a_n ≡ 0 (mod f) and b_n ≢ 0 (mod f ), we have that u ∉ 𝒪. On the other hand, $u^{2} = ε^{2 n} = a_{2 n} + b_{2 n} \sqrt{d}$ u^2 = \varepsilon ^{2n} = a_{2n} + b_{2n} \sqrt d , and a direct computation gives b₂_n = 2a_nb_n. Since a_n ≡ 0 (mod f ), we have b₂_n ≡ 0 (mod f ), so u² ∈ 𝒪. Moreover, the norm $N (u^{2}) = {(N (u))}^{2} = 1 = a_{2 n}^{2} - b_{2 n}^{2} d = (a_{2 n} + b_{2 n} \sqrt{d}) (a_{2 n} - b_{2 n} \sqrt{d})$ N\left( {u^2 } \right) = \left( {N\left( u \right)} \right)^2 = 1 = a_{2n}^2 - b_{2n}^2 d = \left( {a_{2n} + b_{2n} \sqrt d } \right)\left( {a_{2n} - b_{2n} \sqrt d } \right) , so that the inverse of u² is its conjugate, which is an element of 𝒪. Therefore, u² is a unit in 𝒪.

Consider the bilinear space S = 𝒪² with bilinear form whose matrix is $[\begin{matrix} - u^{2} & 0 \\ 0 & 1 \end{matrix}] .$ \left[ {\matrix{ { - u^2 } & 0 \cr 0 & 1 \cr } } \right]. It clearly satisfies the conditions of Proposition 1: −u², 1, 0 ∈ 𝒪, −u² ≠ 0 is a unit in 𝒪, and the isotropy equation is X² − u² = 0 – its roots are ±u, which lie in 𝒪_K and satisfy the monic polynomial equation X² − u² = 0 with coefficients in 𝒪, since u² ∈ 𝒪.

The existence of an integer n such that a_n ≡ 0(mod f ) and b_n ≢ 0(mod f) is intimately connected to the splitting behaviour of the prime factors of f 𝒪_K in 𝒪_K and to the order of the fundamental unit modulo those primes. Let p > 0 be a rational prime and consider the order $𝕆 = 𝕑 [p \sqrt{d}]$ {\mathbb {O}} = {\mathbb {Z}}\left[ {p\sqrt d } \right] . We shall distinguish between the unramified case (when p does not divide the discriminant of K, so p ∤ 2d) and the ramified one (when p | 2d).

If p | 2d, then either p = 2, so that W 𝒪 → W K is not injective by [1, Theorem 5.2] or 2 ∤ p and p | d, in which case W 𝒪 → W K is injective by [7, Theorem 2.2]. The unramified case is more subtle:

Theorem 2.

Let p be an odd prime with p ∤ d and let $𝕆 = 𝕑 [p \sqrt{d}]$ {\mathbb {O}} = {\mathbb {Z}}\left[ {p\sqrt d } \right] , where d ≢ 1(mod 4). Let m be the order of the image of fundamental unit ε in the multiplicative group (𝒪_K/p𝒪_K )×. If 4 | m, then set n = m/4. Then $a_{n} \equiv 0 (\mod p), b_{n} ≢ 0 (\mod p) .$ a_n \equiv 0\,\,\,\,\left( {\bmod \,p} \right),\,\,\,\,b_n \not\equiv 0\,\,\,\,\left( {\bmod \,p} \right).

Proof.

Let R = 𝒪_K/p𝒪_K . The multiplicative group R^× is cyclic of order p² − 1 if p is inert, that is if $(\frac{d}{p}) = - 1$ \left( {{d \over p}} \right) = - 1 , and isomorphic to $𝔽_{p}^{\times} \times 𝔽_{p}^{\times}$ {\mathbb {F}}_p^ \times\times {\mathbb {F}}_p^ \times if p splits, that is if $(\frac{d}{p}) = 1$ \left( {{d \over p}} \right) = 1 . In either case, the order m of ε in R^× is well-defined. By hypothesis, 4 | m.

Set n = m/4. Then ε²ⁿ = ε^m/2. Since m/2 is even (because m is divisible by 4), we have ε^m/2 ≠ 1 and its square is ε^m = 1. Hence ε^m/2 is an element of order 2 in R^×. In any field of characteristic not 2, the only element of order 2 is −1. In the split case, the group is a product, and the element of order 2 is (−1, −1). In either case, ε^m/2 = −1 (where −1 denotes the element (−1, . . . , −1) in the product, or the field element −1). Thus ε²ⁿ = −1.

Now write $ε^{n} = a_{n} + b_{n} \sqrt{d}$ \varepsilon ^n = a_n + b_n \sqrt d in 𝒪_K . Reducing modulo p gives an element in R. The equality ε²ⁿ = −1 becomes ${(a_{n} + b_{n} \sqrt{d})}^{2} \equiv - 1 (\mod p) .$ \left( {a_n + b_n \sqrt d } \right)^2 \equiv - 1\,\,\,\,\left( {\bmod \,p} \right). Expanding, $a_{n}^{2} + d b_{n}^{2} + 2 a_{n} b_{n} \sqrt{d} \equiv - 1$ a_n^2 + db_n^2 + 2a_n b_n \sqrt d \equiv - 1 . Since the representation of elements of R in the basis ${1, \sqrt{d}}$ \left\{ {1,\,\sqrt d } \right\} is unique (mod p), we compare coefficients: $a_{n}^{2} + d b_{n}^{2} \equiv - 1, 2 a_{n} b_{n} \equiv 0 (\mod p) .$ a_n^2 + db_n^2 \equiv - 1,\,\,\,\,2a_n b_n \equiv 0\,\,\,\,\left( {\bmod \,p} \right). Because p is odd, the second congruence gives a_nb_n ≡ 0 (mod p). If b_n ≡ 0 (mod p), then the first congruence gives $a_{n}^{2} \equiv - 1 (\mod p)$ a_n^2 \equiv - 1\,\left( {\bmod \,p} \right) , which would imply that −1 is a square modulo p. But then εⁿ ≡ a_n would be a rational integer, and its norm would be $a_{n}^{2}$ a_n^2 , which is ±1. However, if b_n ≡ 0, then εⁿ ≡ ±1 (since the only units in ℤ are ±1). Then ε²ⁿ ≡ 1, contradicting ε²ⁿ ≡ −1. Therefore b_n ≢ 0 (mod p). Hence a_n ≡ 0 (mod p) from the product condition.

Thus we have a_n ≡ 0 and b_n ≢ 0 (mod p), which finishes the proof.

Example 1.

The condition that the order m of ε in (𝒪_K/p𝒪_K )^× is divisible by 4 is a concrete arithmetic condition that can be checked for given d and p. It is satisfied for many primes:

for d = 2, p = 3: $ε = 1 + \sqrt{2}$ \varepsilon = 1 + \sqrt 2 has order 8 in $𝔽_{9}^{\times}$ {\mathbb {F}}_9^ \times (since p is inert), so m = 8 is divisible by 4;
for d = 2, p = 17: 17 splits, ε has order 16 in $𝔽_{17}^{\times}$ {\mathbb {F}}_{17}^ \times , so m = 16 is divisible by 4;
for d = 3, p = 7: 7 is inert, $ε = 2 + \sqrt{3}$ \varepsilon = 2 + \sqrt 3 has order 8 in $𝔽_{49}^{\times}$ {\mathbb {F}}_{49}^ \times , which again is divisible by 4.

Thus the theorem provides infinitely many odd conductor examples for each real quadratic field: by the Chebotarev Density Theorem there are infinitely many primes for which the order of ε is a multiple of 4.

Remark 1.

The condition provided by Theorem 2 is sufficient for the existence of suitable n, but far from necessary. For example, let $K = 𝕈 (\sqrt{2})$ K = {\mathbb {Q}}\left( {\sqrt 2 } \right) and p = 7. The order of $ε = 1 + \sqrt{2}$ \varepsilon = 1 + \sqrt 2 modulo 7 is 6, which is not divisible by 4, but for n = 3 we get $ε^{3} = 7 + 5 \sqrt{2}$ \varepsilon ^3 = 7 + 5\sqrt 2 , so that a₃ = 7 ≡ 0 (mod 7) and b₃ = 5 ≢ 0 (mod 7).

The case of imaginary quadratic number fields seems to be more complicated. Recall that except for d = 1 and d = 3, the ring of integers 𝒪_K of the field $K = 𝕈 (\sqrt{- d})$ K = {\mathbb {Q}}\left( {\sqrt { - d} } \right) contains only 2 units, namely 1 and −1, so there is no hope of applying Theorem 1 here. The case of the Gaussian field $𝕈 (\sqrt{- 1})$ {\mathbb {Q}}\left( {\sqrt { - 1} } \right) , whose ring of integers contains 4 units, has been completely described by Ciemała and Szymiczek. The Eisenstein field $𝕈 (\sqrt{- 3})$ {\mathbb {Q}}\left( {\sqrt { - 3} } \right) is considerably more involved and the methods developed for the Gaussian field do not seem to be transferable here.

On Examples of Witt Functors in Quadratic Number Fields

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