SUMS OF SQUARES OF LINEAR FORMS

JOSÉ F. FERNANDO, JESÚS M. RUIZ, CLAUS SCHEIDERER

Abstract. Let k be a real field. We show that every non-negative homogeneous quadratic polynomial f(x_{1}, …, x_{n}) with coefficients in the polynomial ring k[t] is a sum of 2n · τ(k) squares of linear forms, where τ(k) is the supremum of the levels of the finite non-real field extensions of k. From this result we deduce bounds for the Pythagoras numbers of affine curves over fields, and of excellent two-dimensional local henselian rings.

Introduction

Let A be a commutative ring with unit. We consider quadratic forms with coefficients in A, that is, homogeneous polynomials of degree two

f = f(x_{1}, …, x_{n}) = ∑_{i,j=1}^{n} a_{ij} x_{i}x_{j}

with a_{ij} ∈ A. If such f is a sum of squares of linear forms, one may ask for the minimal number of squares needed to represent f. So we write

ℓ(f) := inf{r: there are linear forms h_{1}, …, h_{r} ∈ A[x_{1}, …, x_{n}] with f = ∑_{i=1}^{r} h_{i}^{2}}

(with the convention ℓ(f) = ∞ if f is not a sum of squares of linear forms), and put

g_{n}(A) := sup{ℓ(f): f is a sum of squares of linear forms in A[x_{1}, …, x_{n}]}.

For n = 1, g_{1}(A) is nothing but the Pythagoras number of the ring A, usually denoted p(A): Every sum of squares of elements of A is a sum of p(A) squares, and p(A) is the minimal number with this property (or p(A) = ∞).

Our primary interest is in Pythagoras numbers for certain classes of rings. However, by an elementary observation originally due to Pfister, bounds for the g_{n}- invariants of a ring A give bounds for the Pythagoras numbers of finite A-algebras (Lemma 2.3 below). This led us to the study of the invariants g_{n}.

Our main result concerns the case where A = k[t], the polynomial ring in one variable over a real field k. Let

τ(k) := sup{s(E): E/k finite, non-real},

where s(E) denotes the level of the field E, i.e. the smallest number n such that −1 is a sum of n squares in E. Since the level of a non-real field is a power of 2, according to Pfister’s famous theorem, the invariant τ(k), if finite, is a power of 2 as well. It was first proved by Pfister that

1 + τ(k) ≤ p(k(t)) ≤ 2τ(k)

(see [L] p. 397). Cassels’ theorem implies p(k[t]) = p(k(t)), and so this gives g_{1}(k[t]) = p(k[t]) ≤ 2τ(k). We will prove that

g_{n}(k[t]) ≤ 2n · τ(k)

holds for all n (Theorem 1.2). Also, we show that every positive semidefinite quadratic form over k[t] is a sum of squares of linear forms. (See 1.1 for the notion of positive semidefiniteness.)

For k = ℝ (or more generally, for k a real closed field), the bound g_{n}(k[t]) ≤ 2n had been proved around 1970 by Jakubović [J] and by Rosenblum and Rovnyak [RR] (note that τ(k) = 1 in this case). Different proofs were later given by Djoković [D] and by Choi, Lam and Reznick [CLR]. More recently, the first of us [F1] extended the result to the case where k = Quot ℝ{x}, the quotient field of the ring of convergent power series in one variable (again, τ(k) = 1 in this case). While the proof in [F1] was inspired by Djoković’s ideas from [D], our proof of the general case takes up the approach of [CLR].

Our main result, together with its proof, is presented in Section 1. Section 2 contains several applications. An immediate one is to Pythagoras numbers of affine curves (2.4). Somewhat less obvious is the application to the Pythagoras numbers of two-dimensional excellent local henselian rings. Indeed, assuming that the residue field k of such a ring A is real, we show g_{n}(A) ≤ 2n·τ(k) if A is regular (Proposition 2.7). Without any regularity assumption, this gives p(A) < ∞ whenever τ(k) < ∞ (Theorem 2.9). We do not know whether conversely p(A) < ∞ implies τ(k) < ∞, but at least we can show that p(A) < ∞ implies τ(k′) < ∞ for some finite real field extension k′ of k (2.11). The essential tool for the proofs of these results, apart from our main theorem, is the Artin approximation theorem. Finally, we discuss the lines between our results and the well-known weak and strong questions from [CDLR].

1. The main result

1.1. Let A be a commutative ring with unit. See the introduction for basic notation. An element a in A is called positive semidefinite (or psd, for short), if φ(a) ≥ 0 holds for every homomorphism φ from A into an ordered field. In particular, this notion applies to polynomials over A. Using the Artin-Lang theorem from real algebra, one can rephrase the condition as follows: A polynomial f in A[x_{1}, …, x_{n}] is psd if and only if, for every homomorphism φ: A → R into a real closed field R, the polynomial f^{φ} over R (obtained by applying φ to the coefficients of f) takes non-negative values on R^{n}.

We will prove:

Theorem 1.2. Let k be a real field. Then the bound

g_{n}(k[t]) ≤ 2n · τ(k)

holds for all n. Moreover, every psd quadratic form over k[t] is a sum of squares of linear forms.

Remark 1.3. For k = ℝ, the theorem says g_{n}(ℝ[t]) ≤ 2n. As mentioned before, this has been proved by several authors, and some of these approaches were even constructive ([D] and [CLR], see also [L] XIII.5). In [CDLR] (p. 50) the authors remark that the proof from [CLR] is valid as long as k is a real closed field, but that it does not seem to generalize to other ground fields. In particular, they mention the case where k is hereditarily pythagorean (this is equivalent to τ(k) = 1), and they are speculating that the bound g_{n}(k[t]) ≤ 2n should remain valid in this case.

In [F1], the bound g_{n}(k[t]) ≤ 2n was proved for k = Quot ℝ{x}. The approach of this paper can be adjusted to work for any hereditarily pythagorean field k, which settles the guess from [CDLR]. The basic idea in [F1] was to enlarge the polynomial ring k[t] to k(\sqrt{−1})[t], and to factor matrices over this larger ring, in order to produce the desired sums of squares decompositions. This was inspired by [D], and it used Djoković’s theorem on dominant diagonalization.

It is possible to carry this method one step further: One can employ the polynomial ring D[t] over D = (−1, −1)_{k}, the skew field of quaternions over k, and in this way prove the bound g_{n}(k[t]) ≤ 4n for all fields k satisfying τ(k) ≤ 2. Indeed, Djoković’s construction for dominant diagonalization works for matrices over D[t].

One might speculate to stretch the method even more, and use polynomials over octonions to arrive at a bound g_{n}(k[t]) ≤ 8n for τ(k) ≤ 4. However, serious difficulties arise in this case caused by the lack of associativity, and it is not clear how to overcome them. At any rate, this case would definitely mean the end of this line, since it is well known that there are no composition algebras beyond dimension eight.

The proof given here works uniformly for any k, regardless of what τ(k) is, by reconsidering the approach from [CLR]. We point out that our proof is completely constructive as well, if one assumes that polynomials in k[t] can be factored and that sums of squares decompositions of psd polynomials in k[t] can be found.

1.4. For the proof of the theorem, we work with symmetric matrices instead of quadratic forms. Given a commutative ring A containing \frac{1}{2}, we associate to any symmetric matrix S ∈ M_{n}(A) the quadratic form (homogeneous polynomial of degree two)

F_{S} = F_{S}(x_{1}, …, x_{n}) = x^{t}Sx,

where x = (x_{1}, …, x_{n})^{t}. In this way we identify quadratic forms in n variables with symmetric n × n-matrices. A sums of squares representation

F_{S} = h_{1}^{2} + ⋯ + h_{N}^{2}

with linear forms h_{1}, …, h_{N} ∈ A[x_{1}, …, x_{n}] corresponds to a product decomposition

S = T^{t} · T

of the coefficient matrix S, with T ∈ M_{N×n}(A) (the i-th row of T consists of the coefficients of the linear form h_{i}). We will always write

ℓ(S) := inf{N: there is T ∈ M_{N×n}(A) with S = T^{t}T }

for the sums of squares length of the quadratic form F_{S}. Our proof makes use of some basic facts from quadratic form theory, such as Witt decomposition and the basic theory of Pfister forms. For all this we refer to [L].

1.5. We now take up the proof of Theorem 1.2. So let k be a real field. We always abbreviate A := k[t]. For the proof we will assume that τ := τ(k) is finite, and will later show how the same arguments can be used for the qualitative part of the theorem disregarding the τ-invariant. Here we know that every psd element of A is a sum of 2τ squares of elements of A. Note that this fact corresponds precisely to the statement of the theorem in case n = 1. We use it as the starting point for an induction on n.

Let n > 1, and fix a psd symmetric matrix S ∈ M_{n}(A). Write S as a block matrix

S = ( \begin{matrix}a & u^{t} \\ u & S′ \end{matrix} )

with a ∈ A, u ∈ A^{n−1} (vectors are columns) and a symmetric matrix S′ ∈ M_{n−1}(A). If a = 0 then u = 0, and we are done by induction. So assume a ≠ 0. The polynomial a ∈ A = k[t] is psd.

The discriminant of f_{S} with respect to the variable x_{1} is a quadratic form in (x_{2}, …, x_{n}), which corresponds to the symmetric matrix D := aS′ − uu^{t} ∈ M_{n−1}(A). Since S is psd, D is psd as well. By the inductive hypothesis, ℓ(D) ≤ 2(n − 1)τ. Completing the square we see that the quadratic form aF_{S} is a sum of 1 + ℓ(D) ≤ 2nτ squares:

aS = ( \begin{matrix}0 & 0^{t} \\ 0 & D \end{matrix} ) + \binom{a}{u}(a u^{t}).

Writing m := 2nτ from now on, we have seen that there is an identity aS = U^{t}U where U ∈ M_{m×n}(A).

To prove the theorem, we have to find an identity S = T^{t}T with T ∈ M_{m×n}(A). This will be done by induction on the degree of a. If a ∈ k^{∗}, then a is a sum of p(k) ≤ p(k(t)) ≤ 2τ(k) =: 2^{r} squares in k. So the Pfister form φ := <1, 1>^{⊗r} represents a, hence aφ ≅ φ and a(n × φ) ≅ n × φ. As m = 2^{r}n, we conclude that there is C ∈ GL_{m}(k) with aI = CC^{t}. Consequently, aS = U^{t}U implies S = (C^{−1}U)^{t}(C^{−1}U). Let now deg(a) ≥ 1, and let f be a monic irreducible polynomial dividing a. If the field k[t]/(f) is real, then reduction of the identity aS = U^{t}U modulo (f) shows that the matrix U is (coefficient-wise) divisible by f. Since a, being psd, is divisible by f^{2}, we may therefore cancel a factor f^{2} in this case.

This leaves us with the case where the field k[t]/(f) is nonreal. This is the main step of the proof. Note that f is a sum of squares in this case, and more precisely, of 2τ squares of elements of A. We will show that there exists a product decomposition U = BW with B ∈ M_{m}(A) and W ∈ M_{m×n}(A), such that B^{t}B = fI. Once this is known we are done, since combining it with aS = U^{t}U gives aS = fW^{t}W, and we can cancel f.

1.6. So let f be a monic irreducible polynomial in A = k[t], and let U ∈ M_{m×n}(A) such that U^{t}U is divisible by f. We identify matrices with linear maps, where matrices act on column vectors from the left. The submodule im(U) of A^{m} is totally isotropic modulo f, in the sense that f divides ⟨x, x⟩ = ∑_{j=1}^{m} x_{j}^{2} for every x ∈ im(U). Let M be a submodule of A^{m} which contains im(U) and which is maximal with respect to being totally isotropic modulo f. It suffices to find B ∈ M_{m}(A) with im(B) = M and B^{t}B = fI. Indeed, this will give a factorization U = BW as desired. So it is enough to prove:

Proposition 1.7. Let k be a real field, let f ∈ A = k[t] be a monic irreducible polynomial which is a sum of 2^{r} squares, and let m be a multiple of 2^{r}. For any submodule M of A^{m} which is totally isotropic modulo f and which is maximal with respect to this property, there exists a matrix B ∈ M_{m}(A) with im(B) = M and B^{t}B = fI.

1.8. Let φ denote the quadratic form <1, 1>^{⊗r} over k (“sum of 2^{r} squares”). Write L := k[t]/(f) in the following. Since the level of L satisfies s(L) ≤ 2^{r−1}, the quadratic form φ_{L} over L is hyperbolic. Since m is a multiple of 2^{r}, say m = 2^{r}n, the form m × <1>_{L} = n × φ_{L} is hyperbolic over L as well. The reduction \overline{M} of M modulo (f) is a maximal totally isotropic subspace of L^{m} (with respect to the form n × φ_{L}). Hence its dimension is dim_{L}(\overline{M}) = \frac{m}{2}.

We need the following easy lemma, whose proof will be supplied later.

Lemma 1.9. Let A be a principal ideal domain, and let 𝔪 ≠ (0) be a prime ideal of A. Let M be a free A-module of finite rank, and let V be a subspace of the A/𝔪-vector space \overline{M} = M/𝔪M. Then there exists a direct summand P of M with \overline{P} = V.

1.10. By this lemma we find a direct sum decomposition A^{m} = P ⊕Q of A-modules with \overline{M} = \overline{P}. Since M is maximal with respect to being totally isotropic modulo f, it follows that M = P ⊕ fQ. ¿From dim_{L}(\overline{M}) = \frac{m}{2} we see rk(P) = rk(Q) = \frac{m}{2}.

Let G ∈ M_{m}(A) be the matrix which acts as the identity on P and as multiplication by f on Q. Then det(G) = f^{m/2} and im(G) = M. Since M is totally isotropic modulo f, the product G^{t}G is divisible by f, say G^{t}G = fV with V ∈ M_{m}(A). Comparing determinants gives det(V) = 1, and in particular, V ∈ GL_{m}(A). On the other hand, V is symmetric (and psd). Now we need to know that such V can be diagonalized over A:

Theorem 1.11 (Harder, Djoković). For any symmetric and invertible matrix V ∈ GL_{m}(k[t]), there exists an invertible matrix Q ∈ GL_{m}(k[t]) such that Q^{t}V Q has coefficients in k.

Proof. This result appears as Theorem 13.4.3 in Knebusch’s paper [K] and is attributed to Harder. An elementary proof (by “Hermite’s method”) can be found in [S] (Thm. 6.3.3). A constructive proof is due to Djoković ([D] Prop. 4). His method provides an algorithm for diagonalizing V. □

1.12. Let us come back to the situation in 1.10. From Theorem 1.11 we get Q ∈ GL_{m}(A) such that Q^{t}V Q = D lies in GL_{m}(k) (and is diagonal, if we want). The matrix B := GQ satisfies therefore im(B) = im(G) = M and B^{t}B = fD.

Let ψ denote the quadratic form over k which is represented by D (of dimension m = 2^{r}n). By specializing at one point, we see that actually ψ ≅ n × φ. Indeed, substitute the value t := 0 in the identity B^{t}B = fD. Since f(0) ≠ 0, we see B(0) ∈ GL_{m}(k), and so f(0) · ψ ≅ n × φ (over k). Since φ is a Pfister form which represents f(0), it follows that φ ≅ f(0) · φ, hence ψ ≅ n × φ, as we claimed.

This means that there exists a matrix C ∈ GL_{m}(k) with D = C^{t}C. Combining this with B^{t}B = fD we get fI = B̃^{t}B̃, where B̃ = BC^{−1}. Since im(B̃) = M, this finishes the proof of Theorem 1.2 in the case τ(k) < ∞. See 1.14 below for the case where τ(k) = ∞.

1.13. Proof of Lemma 1.9. We may assume M = A^{n} and 0 < dim(V) < n. The proof is by induction on dim(V). First assume dim(V) = 1 (and n ≥ 2). We are given x = (x_{1}, …, x_{n}) ∈ A^{n} with x ∉ 𝔪A^{n}, and may assume x_{1} ∉ 𝔪. Let 𝔭_{1}, …, 𝔭_{r} be those prime ideals of A which contain x_{1}. Choosing b ∈ 𝔪 for which x_{2} + b ∉ 𝔭_{i} for i = 1, …, r, the vector x′ = (x_{1}, x_{2} + b, …) is unimodular.

For the inductive step let V_{1} ⊂ V be a one-dimensional subspace, and let P_{1} be a direct summand of M such that \overline{P}_{1} = V_{1}. Choose a decomposition M = P_{1} ⊕Q_{1}, and choose a subspace V_{2} of \overline{Q}_{1} such that V = V_{1} ⊕V_{2}. By the inductive hypothesis there is a direct summand P_{2} of Q_{1} for which V_{2} = \overline{P}_{2}. So P := P_{1} ⊕ P_{2} is a direct summand of M, and \overline{P} = V.

1.14. Proof of the qualitative part of Theorem 1.2. This asserts that every psd quadratic form over A = k[t] is a sum of squares of linear forms. The arguments from the proof for τ(k) < ∞ essentially carry over, as we briefly indicate:

The proof is again by induction on the number n of variables. Given a psd quadratic form F = F (x_{1}, …, x_{n}) over A, we can assume as in 1.5 that there exists a psd polynomial 0 ≠ a ∈ A for which aF is a sum of squares. Choose m to be a power of 2 which is so large that, on the one hand, aF is a sum of m squares, and on the other hand, the leading coefficient of a as well as every monic irreducible psd polynomial dividing a is a sum of m squares. (It is well known that this last condition is equivalent to a itself being a sum of m squares, but we do not need this.) With this choice of the number m, the proof goes through as before, showing that F itself is a sum of m squares.

2. Applications

In this section, we assemble a series of examples, remarks and applications, in particular to Pythagoras numbers.

Example 2.1. If A = k is a field, quadratic forms can be diagonalized, and this gives the bound g_{n}(k) ≤ n · p(k) (which was originally observed by Pfister). For many fields, though, much better bounds are available [BLOP].

We mention here another easy case, the ring of formal power series A = k[[x]] over a real field k. As said above, one has g_{n}(k((x))) ≤ n·p(k((x))) = n·p(k). On the other hand, one gets immediately g_{n}(k[[x]]) = g_{n}(k((x))) by clearing denominators. (This is actually part of a more general statement for arbitrary real valuations, see Lemma 2.8.) Together this gives g_{n}(k[[x]]) ≤ n · p(k).

Remark 2.2. For fields of finite transcendence degree over their prime field or over a real closed field, upper bounds for τ(k) are available, although they may not always be sharp. For example, if k is a real number field then τ(k) = 4 if there is a dyadic place v for which [k_{v} : ℚ_{2}] is odd; otherwise, τ(k) = 2. (See [L], loc. cit.) If k has transcendence degree d over ℚ then τ(k) ≤ 2^{d+2}, using the Milnor conjecture as proved by Voevodsky. The better bound τ(k) ≤ 2^{d+1} holds for d ≥ 2 if Kato’s cohomological Hasse principle is true. (See [Sch] 5.26 for a more detailed discussion.)

If k has transcendence degree d over a real closed field, then τ(k) ≤ 2^{d} by a theorem of Pfister (see [L] XI.4.10).

Generally, the Milnor conjecture implies for any field k that τ(k) ≤ 2^{d}, where d = vcd_{2}(k) = cd_{2} k(\sqrt{−1}) is the virtual cohomological 2-dimension of k.

It is an open problem whether p(k) < ∞ implies p(k(t)) < ∞ (or equivalently, τ(k) < ∞) for every (real) field k.

Our applications of Theorem 1.2 are based on the following elementary lemma:

Lemma 2.3. Let A be a ring, and let B be an A-algebra which is generated by n elements as an A-module. Then p(B) ≤ g_{n}(A).

The proof is straightforward, see [CDLR] §2. From this lemma and Theorem 1.2 we immediately get:

Theorem 2.4. Let k be a real field with τ(k) < ∞. Then p(A) < ∞ holds for every one-dimensional k-algebra A of finite type.

Proof. By Noether normalization, A is a finite k[t]-algebra. □

Remarks 2.5. 1. For the record we note that if there is a homomorphism k[t] → A by which A is generated by n elements as a k[t]-module, then the bound p(A) ≤ 2n·τ(k) holds.

2. A related kind of application of Theorem 1.2 is this: Let f ∈ A[x_{1}, …, x_{n}] be a form which is a sum of squares of forms of degree d. Then f can be written as a sum of g_{N}(A) such squares, where

N = \binom{n + d − 1}{d}

(see [CDLR] p. 51).

Now we turn to henselian local rings. To start with, we have:

Proposition 2.6. Let A be an excellent henselian local ring. Then g_{n}(A) = g_{n}(Â) holds for all n. In particular, p(A) = p(Â).

Proof. The proof depends on the Artin approximation theorem, for which we refer to [KPP]. Suppose first that g_{n}(Â) = γ is finite, and let f ∈ A[x_{1}, …, x_{n}] be a sum of squares of linear forms. Then we have

f = ∑_{i=1}^{γ} g_{i}^{2}

for some linear forms g_{i} ∈ Â[x_{1}, …, x_{n}]. Equating coefficients on both sides and looking at the coefficients of the g_{i} as unknowns, we have a polynomial system of equations over A that has a solution over Â. Hence, by Artin approximation, it has a solution over A, and we find h_{i} ∈ A[x_{1}, …, x_{n}] such that f = ∑_{i=1}^{γ} h_{i}^{2}. We conclude that g_{n}(A) ≤ γ.

Conversely, suppose g_{n}(A) = γ < ∞, and let f ∈ Â be a sum of squares of linear forms in Â, say

f = ∑_{j=1}^{r} g_{j}^{2}

with linear forms g_{j} over Â. We try to find an identity

f = ∑_{i=1}^{γ} h_{i}^{2}  (∗)

with linear forms h_{i} over Â. Again, this amounts to solving a polynomial system over Â whose unknowns are the coefficients of the h_{i}. By Artin approximation, it is enough to solve the system modulo 𝔪̂^{k} for a suitable large k ≥ 1 (here 𝔪̂ denotes the maximal ideal of Â). Choose linear forms g̃_{j} over A such that all g̃_{j} − g_{j} have coefficients in 𝔪̂^{k}. Then f̃:= ∑_{i=1}^{r} g̃_{j}^{2} is a sum of squares of linear forms in A[x_{1}, …, x_{n}], and consequently, a sum of γ squares, say

f̃ = ∑_{i=1}^{γ} h̃_{i}^{2}

with suitable linear forms h̃_{i} over A. Since f̃− f has coefficients in 𝔪̂^{k}, this solves (∗) modulo 𝔪̂^{k}, as desired. We are done. □

Proposition 2.7. Let A be an excellent henselian local ring. Suppose that A is regular of dimension two and has a real residue field k. Then g_{n}(A) ≤ 2n · τ(k) for all n.

The proof uses the following simple observation:

Lemma 2.8. Let B be a valuation ring with real residue field and with quotient field F. Then g_{n}(B[y]) = g_{n}(F [y]).

Proof. It suffices to see that if B[y][x_{1}, …, x_{n}] contains a sum of squares of linear forms from F [y][x_{1}, …, x_{n}], then B[y][x_{1}, …, x_{n}] contains the linear forms themselves. Looking at the coefficients of the forms involved, one sees that it suffices to show that if f = ∑_{i} f_{i}^{2} ∈ B[y], with f_{i} ∈ F [y], then each f_{i} lies in fact in B[y]. To show this, let f_{i} = ∑_{j} a_{ij}y^{j} ∈ F [y], and let a_{λμ} be a coefficient with smallest valuation. Suppose by way of contradiction that a = 1/a_{λμ} ∈ 𝔪, where 𝔪 is the maximal ideal of B. Then af_{i} ∈ B[y] for all i, and af_{λ} ∉ 𝔪[y]. Since 𝔪[y] is a real prime ideal of B[y], we conclude that a^{2}f = ∑_{i}(af_{i})^{2} ∉ 𝔪[y]. Consequently, a ∉ 𝔪, which is the desired contradiction. □

Proof of Proposition 2.7. We can assume τ(k) < ∞. By Proposition 2.6 we can suppose A is complete, hence A = k[[x, y]]. Since clearly τ(k((x))) = τ(k) ([Sch] 5.13), our main theorem gives g_{n}(k((x))[y]) ≤ 2n · τ(k) for all n. From the lemma we get g_{n}(k[[x]][y]) ≤ 2n · τ(k). We end with another application of Proposition 2.6, since k[[x, y]] is the completion of k[[x]][y]_{(x,y)}. □

Finally we prove:

Theorem 2.9. Let A be any excellent henselian local ring of dimension two whose residue field k satisfies τ(k) < ∞. Then the Pythagoras number p(A) is finite.

Proof. If k is nonreal then −1 is a sum of squares in A, and the assertion is clear. So we assume that k is real. Let Â be the completion of A. This is a complete noetherian local ring with dim(Â) = 2. Such a ring is known to be a finite k[[x, y]]-algebra, essentially by the Cohen structure theorem (see [Bo] ch. IX § 3 Théorème 2). ¿From Proposition 2.7 (and Lemma 2.3) we therefore get p(Â) < ∞, and thus p(A) < ∞ by Proposition 2.6. □

Remark 2.10. With this last result, the understanding of the Pythagoras numbers of excellent local henselian rings has become quite precise. To describe the situation, denote by dim_{r}(A) the real dimension of A (see [FRS]). For A an excellent local henselian ring, dim_{r}(A) = max{dim(A/𝔭): 𝔭 is a real prime ideal of A}. Clearly dim_{r}(A) ≤ dim(A), but equality need not hold in general. By the main result of [FRS], p(A) = ∞ if dim_{r}(A) ≥ 3. On the other hand, for studying p(A) one can replace A by a quotient A/I with dim(A/I) = dim_{r}(A), in such a way that bounds for p(A/I) provide bounds for p(A) (this goes back to [CDLR], see also [F2] p. 1910). Thus we are left with the case where dim(A) = dim_{r}(A) is 1 or 2.

The one-dimensional case can be considered folklore: After completion, A is a finite module over some power series ring, and the second part of Remark 2.1 applies to bound p(A) in terms of p(k). Consequently, p(A) is finite iff p(k) is so. In the same vein, for dim(A) = 2, Theorem 2.9 shows that p(A) is bounded in terms of τ(k). We deduce that p(A) is finite if τ(k) is so. Without going into details we point out that these bounds can be made explicit, involving p(k) or τ(k) and suitable multiplicities of A (see [F1]).

Conversely, if p(A) is finite then p(k) is also finite by trivial reasons. Whether or not τ(k) is finite is not clear. Here we touch the well-known open problem of deciding whether p(k) < ∞ implies p(k(t)) < ∞. The closest we can come to is this:

Proposition 2.11. Let A be local henselian and excellent, with real residue field k. If dim_{r}(A) = 2 and p(A) < ∞, there exists a finite real extension k′ of k such that τ(k′) < ∞.

Proof. We may assume that A is a real domain of Krull dimension 2. Then there is a sequence of quadratic transforms A → A′ in the field of fractions K of A, such that A′ is regular of dimension 2 and its residue field k′ is a real finite extension of k [AR]. Clearly, p(A) < ∞ implies p(K) < ∞. By [Sch] 5.16(b), applied to A′, we get p(k′(t)) ≤ p(K), and so τ(k′) < ∞. □

Remark 2.12. Although it is quite obvious, we would like to briefly sketch the line between our results and questions (Q_{1}) and (Q_{2}) from [CDLR] (p. 49), commonly referred to as “weak question” and “strong question”. Given a ring A with p(A) < ∞ and a finite A-algebra B, the weak question asks whether p(B) is necessarily finite. The strong question asks whether even p(B) ≤ n·p(A) holds, if B is generated by n elements as an A-module. To our knowledge, no single counterexample is known to date, not even to the strong question.

It is clear that the weak question has a positive answer whenever g_{n}(A) is finite for all n. Thus, our results provide such a positive answer for important new classes of rings A, in particular finitely generated k-algebras of transcendence degree one and local excellent henselian rings of dimension ≤ 2 with residue field k, both in the case where the field k satisfies τ(k) < ∞. For the same rings, although in general our results do not suffice to imply a positive answer to the strong question in its original form, they imply that this question states the correct order of magnitude for p(B), namely up to a constant factor independent of n. For example, if A = k[t] is the polynomial ring over a real field, with p(A) < ∞, then τ(k) < 2p(A), and hence p(B) < 4n · p(A) whenever B is generated by n elements as an A-module.

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