On the level of principal ideal domains

J. K. Arason, R. Baeza^{∗}

1. Introduction

Let A be a commutative ring with 1. If −1 can be written as a sum of squares in A then the level s(A) of A is defined as the minimal number of squares needed. Else one says that s(A) = ∞.

If A is a field then a well known result due to Pfister says that s(A) is a power of 2 or ∞. On the other hand, in [DLP] (see also [DL]) it is shown that any natural number can be realized as the level of some ring A.

If A is a Dedekind ring then, by [B1], s(K) ≤ s(A) ≤ s(K) + 1, where K is the field of fractions. If s(K) = 1 then also s(A) = 1 because A is integrally closed. But in [B1] the question remained open, whether s(A) can have the value s(K) + 1 if s(K) > 1. In this note we answer this question affirmatively. In fact, we shall show that there are examples where A is a principal ideal domain.

At the end we also make some remarks on the sublevel of principal ideal domains. (See [DL] or Section 3 for the definition of the sublevel.)

We shall assume throughout that 2 ≠ 0. We shall use standard notations in the theory of symmetric bilinear forms as, for example, in [L].

2. Levels.

In this section we shall show that any number of the form 2^{n} + 1 is the level of some principal ideal domain.

We start with a general result of independent interest.

(2.1) Lemma. Let F be a field and let φ be a Pfister form over F. Let c and d be non-zero elements in F such that d is not represented by φ over F and c is not represented by the pure subform φ′ of φ over F. Let q(x) ∈ F [x] be a polynomial of even positive degree with leading coefficient c. Then d is not represented by φ over the ring F [x][\sqrt{−q(x)}].

∗Partially supported by Fondecyt 1090011 and Anillo ACT 56

Proof. Suppose that d is represented by φ over F [x][\sqrt{−q(x)}]. This means that there are elements u⃗(x), v⃗(x) ∈ F [x]^{2^{n}} such that

φ (u⃗(x) + v⃗(x)\sqrt{−q(x)}) = d

i.e.,

φ(u⃗(x)) − q(x)φ(v⃗(x)) = d  (1)

and

φ_{b}(u⃗(x), v⃗(x)) = 0  (2)

where φ_{b} is the symmetric bilinear form with φ(w⃗) = φ_{b}(w⃗, w⃗)) for every w⃗.

Note that v⃗(x) ≠ 0 since φ does not represent d over F. Let 2k > 0 be the degree of q(x) and l the degree of v⃗(x), i.e., the maximum of the degrees of its components. From equation (1) it then follows that deg u⃗(x) = l + k.

We denote by a⃗ the leading coefficient of u⃗(x) and by b⃗ that of v⃗(x). Looking at the coefficients by x^{2l+2k} in equation (1), we get

φ(a⃗) − cφ(b⃗) = 0

and the coefficients by x^{2l+k} in (2) give

φ_{b}(a⃗, b⃗) = 0

Let e = φ(b⃗). As φ does not represent every element in F, φ must be anisotropic over F. Since b⃗ ≠ 0 it follows that e ≠ 0. Then also φ(a⃗) = ce ≠ 0. The last equation says that a⃗ and b⃗ are orthogonal with respect to φ. Therefore the form φ contains over F the subform ⟨ce, e⟩ ≅ ⟨e⟩⟨1, c⟩. But e is represented by φ and φ is multiplicative so ⟨e⟩φ ≅ φ. Hence φ also contains ⟨1, c⟩ as a subform, i.e., ⟨1⟩ ⊥ φ′ ≅ ⟨1, c⟩ ⊥ . . .. Cancelling ⟨1⟩ we conclude that c is represented by φ′, contradicting a hypothesis. Remark. The lemma also holds for d = 0 if ‘represented by φ’ is understood as ‘non-trivially represented by φ’.

With this lemma it is easy to construct Dedekind rings A with field of fractions K such that s(A) = s(K) + 1.

(2.2) Proposition. Let n ≥ 1 be an integer and let F be a field of level > 2^{n}. Assume that there is an element c ∈ F which is a sum of 2^{n} squares in F but not fewer.

Choose a square free polynomial q(x) ∈ F [x] of positive degree and with leading coefficient c such that q(x) is a sum of less than 2^{n+1} squares in F [x].

Let A = F [x][\sqrt{−q(x)}] and K = F (x)(\sqrt{−q(x)}). Then A is a Dedekind ring with field of fractions K such that s(K) = 2^{n} and s(A) = s(K) + 1. Proof. A is a Dedekind ring because q(x) is square free. It is clear that K is the field of fractions of A. Since q(x) is a sum of m < 2^{n+1} squares in F [x], we see that −1 is a sum of m squares in K and hence s(K) < 2^{n+1}. But s(K) is a power of 2, so it follows that s(K) ≤ 2^{n}.

Lemma (2.1) with φ = 2^{n} × ⟨1⟩ and d = −1 implies that −1 is not a sum of 2^{n} squares in A, hence s(A) > 2^{n}. As s(K) ≤ s(A) ≤ s(K) + 1 by [B1], it follows that s(K) = 2^{n} and s(A) = 2^{n} + 1 = s(K) + 1. Remark. If F is the rational function field ℝ(t_{1}, …, t_{2^{n}}) and c = t_{1}^{2}+⋯+t_{2^{n}}^{2} then, by a well known theorem of Cassels, the hypotheses of the theorem are satisfied.

For q(x) one can take cx^{2} + 1.

By the way, this result gives rise to another algebraic proof of the fact that the ring ℤ[x_{1}, …, x_{2^{n}+1}]/(1 + x_{1}^{2} + ⋯ + x_{2^{n}+1}^{2}) has level 2^{n} + 1. (For the first algebraic proof see [B2].)

In fact, we even get principal ideal domains A with field of fractions K such that s(A) = s(K) + 1.

(2.3) Theorem. For any integer n ≥ 1 there are principal ideal domains A with field of fractions K such that s(K) = 2^{n} and s(A) = s(K) + 1. Proof. By [S, Theorem 5.1], the Dedekind ring A in the preceding proof is a principal ideal domain if the polynomial q(x) has degree 2 and if the affine conic given by y^{2} = −q(x) has no F -rational points. This is the case if, for example, q(x) = cx^{2} + 1.

3. Sublevels

Related to the level s(A) of a ring A, there is another invariant, called the sublevel σ(A) of A (see [DL]). It is the smallest natural number r such that there is a unimodular vector (a_{1}, …, a_{r+1}) ∈ A^{r+1} with a_{1}^{2} + ⋯ + a_{r+1}^{2} = 0. (Recall that a vector (a_{1}, …, a_{n}) ∈ A^{n} is said to be unimodular if there are c_{1}, …, c_{n} ∈ A with a_{1}c_{1} + ⋯ + a_{n}c_{n} = 1.)

For any ring A we have σ(A) ≤ s(A) because a_{1}^{2} + ⋯ + a_{s}^{2} = −1 implies a_{1}^{2} + ⋯ + a_{s}^{2} + 1^{2} = 0.

If A is a field or a local ring then it is easily seen that σ(A) = s(A). With some work, it can be shown that this also holds for semi-local rings A. But in general this does not hold. For example the principal ideal domain A = ℚ[x, y], 1 + x^{2} + 2y^{2} = 0, satisfies σ(A) = s(K) = 2 but s(A) = 3, where K is the field of fractions of A (see [CLRR], [DL]). We shall later generalize this example.

A natural question to ask is how large the difference s(A) − σ(A) can be. The following proposition gives a partial answer.

(3.1) Proposition. In general s(A) ≤ σ(A) + 4. If 2 is invertible in A then even s(A) ≤ σ(A) + 1.

Furthermore, for any ring A we have:

If σ(A) = 1 then s(A) = 1.

If σ(A) = 2 then s(A) ≤ 3.
If σ(A) = 3 then s(A) = 3.

Proof. In any ring A the equations

x_{1}^{2} + ⋯ + x_{r+1}^{2} = 0 and x_{1}y_{1} + ⋯ + x_{r+1}y_{r+1} = 1

imply the equation

∑_{i=1}^{r+1} ((1 + q)x_{i} − 2y_{i})^{2} = −4

where q = y_{1}^{2} + ⋯ + y_{r+1}^{2}. If 2 is invertible in A we can divide this equation by 2^{2} to see that then s(A) ≤ r + 1. In general we can rewrite the equation as ∑_{i=1}^{r+1} ((1 + q)x_{i} − 2y_{i})^{2} + 1^{2} + 1^{2} + 1^{2} = −1 and get s(A) ≤ r + 4.

If r = 1 then (x_{1}y_{2} − x_{2}y_{1})^{2} = −1 and if r = 2 then (x_{2}y_{3} − x_{3}y_{2})^{2} + (x_{3}y_{1} − x_{1}y_{3})^{2} + (x_{1}y_{2} − x_{2}y_{1})^{2} = −1. If r = 3 we use Euler’s four square formula to write (x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2})(y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2}) = (x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3} + x_{4}y_{4})^{2} + w_{2}^{2} + w_{3}^{2} + w_{4}^{2} with elements w_{2}, w_{3}, w_{4} ∈ A. It follows that w_{2}^{2} + w_{3}^{2} + w_{4}^{2} = −1. Remark. By a private communication, Detlev Hoffmann also has a proof of the general inequality s(A) ≤ σ(A) + 4. Furthermore, David Leep has announced a stronger general inequality. In [DL] there is a different proof of the inequality s(A) ≤ σ(A) + 1 in the case that 2 is invertible in A.

We know no ring A with s(A) > σ(A) + 1. By the next proposition this is at least impossible for a Dedekind ring A.

(3.2) Proposition. If A is a Dedekind ring then there is an integer n ≥ 0 such that 2^{n} ≤ σ(A) ≤ s(A) ≤ 2^{n} + 1. Proof. Let A be a Dedekind ring and let K be is field of fractions. Writing s(K) = 2^{n} we then get the statement from the inequalities s(K) = σ(K) ≤ σ(A) ≤ s(A) ≤ s(K) + 1.

The next proposition says that there are examples of s(A) = σ(A) + 1, where A is even a principal ideal domain.

(3.3) Proposition. For any integer n ≥ 1 there is a principal ideal domain A such that σ(A) = 2^{n} and s(A) = 2^{n} + 1. Proof. Choose a field F of level > 2^{n} with an element c that is a sum of 2^{n} squares but not fewer. Write c as the sum c_{1}^{2} + ⋯ + c_{2^{n}}^{2} of squares in F. Let q(x) = cx^{2} + 2c_{1}x + 1. Then, q(x) = (c_{1}x + 1)^{2} + c_{2}^{2}x^{2} + ⋯ + c_{2^{n}}^{2}x^{2} is sum of 2^{n} < 2^{n+1} squares in F [x]. Let A = F [x][\sqrt{−q(x)}] and K = F (x)(\sqrt{−q(x)}). By Proposition (2.2), then A is a Dedekind ring with field of fractions K such that s(K) = 2^{n} and s(A) = s(K) + 1. But σ(A) ≤ 2^{n} because \sqrt{−q(x)}^{2} + (c_{1}x + 1)^{2} + c_{2}^{2}x^{2} + ⋯ + c_{2^{n}}^{2}x^{2} = 0 and a vector over A having c_{1}x + 1 and c_{2}x among its components is unimodular. As in the proof of Theorem (2.3) we see that A is a principal ideal domain.

One can show that the algebraic number ring Z[\sqrt{−6}] is a Dedekind ring A with field of fractions K such that σ(A) = s(A) = s(K) + 1. Another example, but more difficult to verify, is the ring A = ℝ(t)[x][\sqrt{−q(x)}], where q(x) = (1 + t^{2})x^{4} + 2tx^{3} + x^{2} + 1.

For a principal ideal domain A this cannot happen. Indeed, in [DL] it is shown that if A is a principal ideal domain with field of fractions K such that 2 is invertible in A then σ(A) = s(K). Their proof of this, however, does not use that 2 is invertible. Hence we have the following.

(3.3) Proposition. Let A be a principal ideal domain with field of fractions K. Then σ(A) = s(K).

In fact, the argument can be used to prove the following more general proposition.

(3.5) Proposition. Let A be a principal ideal domain with field of fractions K. Let (M, β) be a non singular symmetric bilinear form over A. If the extension of (M, β) to K is isotropic then M contains a unimodular element u such that β(u, u) = 0. Proof. As A is a principal ideal domain, M is free. We therefore may assume that M = A^{m} for some natural number m. The hypothesis then says that there is a non-zero vector u ∈ K^{m} with β(e, e) = 0. Clearing denominators, we may assume that u ∈ A^{m}. We write u = (u_{1}, …, u_{m}). Dividing by the greatest common factor of u_{1}, …, u_{m}, we then may assume that u_{1}, …, u_{m} are relatively prime. But then (u_{1}, …, u_{m}) is unimodular.

Literature

[B1] Baeza, R.: Über die Stufe von Dedekind Ringen. Archiv der Math. 33, 226-231, 1979.

[B2] Baeza, R.: Quadratic forms over semi-local rings. LNM vol. 655, Springer Verlag, Berlin-Heidelberg-New York, 1978.

[CLRR] Choi, M.D.; Lam, T.Y.; Reznick, B.; Rosenberg, A.: Sums of squares in some integral domains. Journal of Algebra, 65, 234-256, 1980.

[DL] Dai, Z.D.; Lam, T.Y.: Levels in algebra and topology. Com. Math. Helv. 59, 376-429, 1984.

[DLP] Dai, Z.D.; Lam, T.Y.; Peng, C.K.: Levels in algebra and topology. Bull. Amer. Math. Soc. (N.S.) 3, 845-848, 1980.

[L] Lam, T.Y.: Introduction to quadratic forms over fields. GSM vol. 67, AMS, 2005.

[S] Samuel, P.: On unique factorization domains. Illinois J. Math. 5, 1-17, 1961.

Faculty of Physical Sciences University of Iceland Reykjavik, Iceland

Instituto de Matemáticas, Universidad de Talca Talca, Chile