Other than the title of this chapter suggests, the following results all hold for function fields over a complete discrete valued field with hered-itarily Euclidean residue field (which need not be real closed). 1 Remark 6.1. LetK be a complete discrete valued field with heredi-tarily Pythagorean residue field. Then K is hereditarily Pythagorean itself, which follows from (2.11 & 2.22). However, K ∼= κv((t)) is not uniquely ordered and thus not Euclidean.
1. A geometric invariant for sums of squares
LetK be a complete discrete valued field with complete discrete valu-ation ringT and hereditarily Euclidean residue fieldk. LetF/K be an algebraic function field. By (3.47 & 3.50), there exists an arithmetic surface CoverT with function fieldF. LetX denote its special fiber.
A generic point ηY of an irreducible component Y of X is of codi-mension one in C, by (3.46). Hence, OC,ηY is a discrete valuation ring, since C is regular. The residue field of the induced valuation vY :F →Z∪ {∞}is the function field of Y overk, i.e. an algebraic function field over k, by (3.30). Let X denote the set of irreducible components Y ofX such thatk(Y) is nonreal and√
−1∈/k(Y).
Lemma 6.2. Let σ ∈P
F×2. Let P ∈X be a closed point such that OC,P/mC,P is real. Assume that vY(σ) ∈ 2Z for every Y ∈ X with P ∈Y. Then there exists σ0∈O×C,P such thatσσ0 ∈DF(2).
Proof. The ringOC,P is a regular local 2-dimensional ring, by (3.46), and so isOC,P[√
−1], by (2.6). Both rings are factorial, by (2.4). We can assume thatσfactors into distinct prime elements with multiplicity one. SinceDF(2) is a multiplicative group, it is sufficient to show that 1The idea of using the recently discovered local-global principles in [HHK09]
and [CTPS] to study sums of squares in such fields are due to my advisor.
60
1. A GEOMETRIC INVARIANT FOR SUMS OF SQUARES 61 (2.5), the localization ofOC,P after the height one prime (p) induces a discrete valuationv in F, and the localization ofOC,P[√
−1] after (q) induces a discrete valuationwonF(√
−1) extendingv. By (2.21), the valuation w◦ι is the only other valuation extension ofv to F(√
−1), where ι denotes the nontrivial F-automorphism on F(√
−1). Since OC,P[√
−1] is invariant underι, it is obvious thatw◦ι is induced by the localization ofOC,P[√
−1] after the prime ideal (qι), thusqand qι are, up to a unit, the only prime factors ofp. Moreover, κv is nonreal by (2.24), asv(σ)∈/ 2Z although σ∈P
F×2. By (2.23), κv is either an algebraic function field overk, or an algebraic extension of k, or a finite extension ofK. In the last two cases, we have√
−1∈κv, by (6.1
& 4.1).
In the first case, whereκv is an algebraic function field overk, we have that t∈pOC,P(p), that is,t=pfg, wheref, g ∈OC,P withp6 |g. Since OC,P is factorial, it follows that t=phfor some h∈OC,P Let U ⊂C be an affine neighborhood ofP such that h, p ∈OC(U) ,→ OC,P and such thatpgenerates the inverse image of pOC,P. HencepOC(U) is a codimension one prime ideal that corresponds to a generic point of a irreducible componentY of the special fiberX ofCwhich containsP. Henceκv=k(Y), by (3.23). We have thatY /∈X, asv(σ)∈/ 2Z. Hence isotropic overk(P). It follows with (5.6) thath1,1,−σiFP is isotropic, as well.
If k(P) is real, then there exists σ0 ∈ P
F×2 such that σ0 ∈ O×C,P andσσ0 ∈DF(2) by (6.2). AsDFP(2) is a multiplicative subgroup of
2J. Van Geel suggested to me this crucial norm argument.
62 6. ALGEBRAIC FUNCTION FIELDS OVERR((t))
PFP×2, it is clear thath1,1,−σiFPis isotropic if and only ifh1,1,−σ0iFP
is isotropic.
OC,P is a two dimensional regular local ring with fields of fractions F.
By (2.14), there exists a valuationv onF with discretely ordered value group of rank 2, such thatOC,P ⊂Ov andmv∩OC,P =mC,P and such that the canonical embeddingOC,P/mC,P ,→O/mv is an isomorphism.
Hence, for σ0∈O×C,P, we denote byσits residue inκ(P)∼=κv. By (2.24), we get that σ0 ∈ Pκ(P)×2 = κ(P)×2 and thus the form h1,1,−σ0iis isotropic overκ(P). By (5.6), this yields thath1,1,−σ0iFP
is isotropic, and thus so is h1,1,−σiFP. Eacha∈F× induces a map
ρa :X→Z/2Z Y 7→vY(a) + 2Z.
Let (Z/2Z)X denote the group of mappings from X to Z/2Z. Recall (6.1), that K is hereditarily Pythagorean, since k is even hereditarily euclidean.
Theorem 6.4. The group homomorphism ρ:X
F×2−→(Z/2Z)X σ7→ρσ
is surjective with ker(ρ) = DF(2). In particular, PF×2/DF(2) ∼= (Z/2Z)X.
Proof. Let σ ∈ P
F×2 such that ρ(σ) = 0. We claim that then h1,1,−σiis isotropic. By (5.2) this is the case ifh1,1,−σiFP is isotropic for every P∈X. LetP ∈X be arbitrary.
In the case whereP ∈X is a closed pointh1,1,−σiFP is isotropic, by (6.3).
In the case wherePis not a closed point, then{P}=Y is an irreducible component of X, and hence OC,P/mC,P is the algebraic function field k(Y) overk, by (3.23).
IfY /∈X, the isotropy ofh1,1,−σiis easy to see. Eitherk(Y) is nonreal and contains √
−1, whereby h1,1iFη is isotropic by (5.1), or k(Y) is real. In this case, we can assume thatvY(σ) = 0 and that the residue σ∈k(Y) is a sum of squares, by (2.24). By (1.4), the first residue form h1,1,−σiis isotropic overk(Y), and hence so ish1,1,−σiFη, by (5.1).
IfY ∈X, we use the hypothesisρσ= 0, that is,vY(σ)∈2Z. Hence, we can assume thatvY(σ) = 0, whereby the first residue form ofh1,1,−σi
1. A GEOMETRIC INVARIANT FOR SUMS OF SQUARES 63
is h1,1,−σi over k(Y). This residue form is isotropic, by (1.4). This implies thath1,1,−σiFη is isotropic, by (5.1).
Thus ker(V) ⊆DF(2). The other inclusion DF(2) ⊆ker(V) is much easier to see. Suppos σ ∈ DF(2)\ker(V), then vi(σ) ∈/ 2Z for one 1 ≤ i ≤ n. We can assume thatσ = 1 +a2 for some a∈ Ovi, then in fact vi(a) = 0, since vi(σ) > 0. But then −1 = a2 ∈ κvi, which contradicts√
−1∈/OC,ηi/mC,ηi =κvi. It follows that ker(V) =DF(2).
Now we show thatρ is surjective. For this, it is enough to show that for every Y ∈ X there exists a σY ∈ PF×2 such that vY(σY) ∈/ 2Z andvY0(σY)∈2Zfor everyY06=Y ∈X.
Zero has a nontrivial representation as a sum of 3 squares in the alge-braic function fieldκvY =k(Y) over k, by (1.4), say 0 = 1 +x2+y2. Denote σ = 1 +x2+y2 if vY(1 +x2+y2) = 1, otherwise choose a uniformizing elements∈F forvY and note that vY(σ) = 1 when we setσ= (1 +s)2+x2+y2 = 2s+s2+ 1 +x2+y2. In any case, there existsσ∈P
F×2, such that vY(σ) = 1.
By (2.16) there exists z ∈ F such that 2vY0(z) > vY0(σ) for every Y0 ∈X with vY0(σ) ∈2Z, 2vY(z) > vY(σ), and 2vY0(z) < vY(σ) for every Y0 ∈ X\ {Y} with vY0(σ) ∈/∈ 2Z. Set σY := σ+z2. We see then thatvY0(σY)∈2Z for all Y0 ∈X\ {Y}, and thatvY(σY)∈/ 2Z. Hence the surjectivity ofρand hence the isomorphyPF×2/DF(2)∼=
(Z/2Z)X.
The result (6.4) shows in particular, that |X| is independent of the chosen regular model overT forF/K. This fact can also be seen by a different consideration, and in more general situations.
Remark6.5. LetT denote a discrete valuation ring with field of frac-tionsK and hereditarily Pythagorean residue fieldk. LetF/K be an algebraic function field. For a regular modelC overT ofF, let again X(C) denote the set of those irreducible components of the special fiber whose function fields are nonreal and do not contain√
−1. Then|X(C)|
does not depend on the chosen regular modelCofF.
Let me briefly make this plausible in the case where the genus of F is at least one. In this case, by [Liu06, p. 457], any regular model can be successively contracted along exceptional divisors to a partic-ular regpartic-ular model, the so-called minimal model. A contraction of an arithmetic surface is a morphism to another arithmetic surface that is an isomorphism everywhere except in one irreducible component of the special fiber, which is mapped to a closed point. This irreducible
64 6. ALGEBRAIC FUNCTION FIELDS OVERR((t))
component is called anexceptional divisor, and it follows e.g. by the so-calledCastelnuovo criterion for arithmetic surfaces[Liu06, 9.3.8] that such an exceptional divisor is, as a prevariety over k, necessarily iso-morphic toP1L for some finite extensionL/k. The function field of this component overkis therefore isomorphic to the rational function field overL, and as such is either real or contains√
−1, by (4.1). Thus the components that we are interested in, are the same as in the unique minimal model, whereby they are independent of the chosen regular model.
For an algebraic function field F over a discrete valued field (K, v) with hereditarily pythagorean residue field κv, we define the invariant χ(F) = |X(C)|, where C is an arbitrary regular model of F over the valuation ring. Under the additional assumption thatκv is euclidean, it was shown in (6.4) thatχ(F) = log2(P
F×2/DF(2)).
2. Discrete valuations and the Pythagoras number For any field K, let
p0(K) =
p(K) ifKis real, s(K)+1 ifKis nonreal.
Proposition 6.6. Let K be a field that is complete with respect to a discrete nondyadic valuation and F/K an algebraic function field.
Thenp(F) = max{p0(κv)|v∈Ω(F)}.
Proof. By (2.26) & (2.25), one has thatp(F)≥max{p0(κv)|v∈Ω}.
Letn= max{p0(κv)|v ∈Ω}. Letσ∈PF×2. We want to show that σ∈DF(n). Supposing otherwise, we have thatq=n× h1i ⊥ h−σiis anisotropic over F. By (5.8), there exists v ∈Ω(F) such thatqFv is anisotropic.
Assume firstly thatv is real. By (2.24) there existsσ0 ∈PF×2 with v(σ0) = 0 and σ0 ∈ Pκ×2v such that q ∼=n× h1i ⊥ h−σ0i. The first residue form ofn× h1i ⊥ h−σ0ioverκvwith respect to any uniformizer πisn× h1i ⊥ h−σ0i, and the second residue form is of dimension zero.
By (5.1), the first residue form is anisotropic. Hence, p(κv)> n gives rise to a contradiction in the case wherev is real.
Assume now that v is nonreal. Then n× h1i is a subform of the first residue form of q with respect to an arbitrary uniformizer. Again, by (5.1), the first residue form is anisotropic. Hence, −1 ∈/ Dκv(n) and thus p0(κv)> n, yielding the contradiction in the case wherev is
nonreal.
3. EXAMPLES 65
Theorem 6.7. Let k be a hereditarily Euclidean field andn∈N. Let F be a algebraic function field overk((t1)). . .((tn)). Then2≤p(F)≤3.
Proof. We show thatp(F)≤3 by induction overn. The claim is true forn= 0, since then we even havep(F) = 2 by (1.4).
Now, we consider the case wheren >0. WriteK =k((t1)). . .((tn−1)) and considerF as an algebraic function field overK((tn)).
By (6.6) we have thatp(F) = max{p0(κv)|v∈Ω(F)}. If v∈Ω(F) is such that v|K((tn)) is trivial, then κv is a finite extension of the hered-itarily Pythagorean field K((tn)) by (2.23) and thus p0(κv) ≤ 2, by definition and 4.1.
Otherwisev|K((tn)) is equivalent to the discrete valuation with respect to whichK((tn)) is complete by (2.19). In this case (2.23) yields thatκv is either an algebraic extension ofK or an algebraic function field over K. In the first case,p0(κv)≤2, by 4.1 asKis hereditarily Pythagorean.
In the second case we havep(κv)≤3 by the induction hypothesis, and thusp0(κv)≤3, sinces(κv)≤p(κv) for nonrealv, and hences(κv) = 2, as levels of nonreal fields are two-powers, by [Pfi95, 3.1.3].
Remark6.8. By (2.26),p(F) = 2 implies thats(κv) = 1 for all nonreal discrete valuations v on F. Conversely, if s(κv) = 1 for all nonreal discrete valuations v on F, then in particular χ(F) = 0, and thus p(F) = 2, by (6.4).
3. Examples
We apply the previous results to particular algebraic function fields F/R((t)), in order to determine χ(F), whereby we decide in particular, whetherp(F) = 2 or 3.
We first fix some notation. If C is a projective plane curve over a field, then it is identified (up to scalar multiples from the base field) with a homogeneous polynomial equation f(X, Y, Z) = 0. We write C|X=1 for the affine chart given by the dehomogenizationf(1, Y, Z) = 0. Similarly, we denoteC|Y=1andC|Z=1. We write ∂X∂ Cfor the curve given by (∂X∂ f)(X, Y, Z) = 0, and similarly we define ∂Y∂ C and ∂Z∂ C.
The following example answers a question raised in [BG09, 5.15], con-cerning the Pythagoras number of a certain function field overR((t)).
Example6.9. LetF/R((t)) be the function field of the irreducible plane curveY2 = (tX−1)(X2+ 1). It was already shown by S. Tikhonov that p(F) ≥3. By (6.7), it follows that p(F) = 3. Moreover, it was
66 6. ALGEBRAIC FUNCTION FIELDS OVERR((t))
shown in [BG09] that |P
F×2/DF(2)| = 2, which is consistent with the observation thatχ(F) = 1, which we will show in the following.
We first claim that the modelC⊂P2R[[t]]ofF, given by the homogeneous equation ZY2 = (tX−Z)(X2+Z2), is regular. By [Liu06, 4.3.36], it is sufficient for regularity that the generic fiberC(0) and the special fiber C(t) are smooth curves. The generic fiber is just the curveC(0)
overR((t)) defined byZY2= (tX−Z)(X2+Z2). The special fiberC(t)
is defined by its reduction modulo t, that is, by the equationZY2 =
−Z(X2+Z2) overR. We apply Jacobi’s criterion (3.42), to show that both curves are smooth. Let us first consider the generic fiber C(0), which is given overR((t)) by the equationZY2= (tX−Z)(X2+Z2).
Consider the affine chart given by dehomogenizingZ = 1.
C(0)|Z=1: Y2= (tX−1)(X2+ 1)
∂
∂XC(0)|Z=1: 0 = 3tX2−2X+t
∂
∂Y C(0)|Z=1: 0 = 2Y.
A closed point (x, y) satisfying all three equations simultaneously, would have to satisfyy= 0, whereby 0 = (tx−1)(x2+1) and 0 = 3tx2−2x+t.
By C(0)|Z=1, we obtain thatx= 1t orx2=−1. One checks easily that both are in contradiction to ∂Y∂ C(0)|Z=1.
Now, consider the dehomogenization X= 1.
C(0)|X=1: ZY2= (t−Z)(Z2+ 1)
∂
∂ZC(0)|X=1: −Y2= 3Z2−2tZ+ 1
∂
∂Y C(0)|X=1: 0 = 2Y Z.
A closed point (y, z) satisfying these equations, would also satisfyy= 0 orz= 0, by ∂Y∂ C(0)|X=1, and hencez6= 0, byC(0)|X=1, whereby either z=torz2=−1, which contradicts the second equation in either case.
At this stage, we have verified smoothness of the generic fiber in every closed point except (0 : 1 : 0). Hence, we consider this point in the affine chart given by dehomogenizing afterY = 1, where it corresponds to the point (0,0), which obviously does not satisfy the equations
C(0)|Y=1: Z= (tX−Z)(X2+Z2)
∂
∂ZC(0)|Y=1: −1 =X2−2tZX+ 3Z2.
3. EXAMPLES 67
We have thus shown that the generic fiber is smooth. Now we consider the special fiberC(t), firstly in the affine chart given by dehomogenizing X= 1.
C(t)|X=1: −ZY2=Z(1 +Z2)
∂
∂Y C(t)|X=1: 0 = 2ZY
∂
∂ZC(t)|X=1: −Y2= 1 + 3Z2.
One sees easily, that a point (y, z) satisfying all three equations, also satisfiesyz = 0 and either y6= 0 or z6= 0. ByC(t)|X=1, the fact that z = 0 would imply y = 0. Hence, y = 0 and z 6= 0. It follows the contradiction that both z2 = −1 and 3z2 = −1. Now consider the affine chart given by dehomogizingZ= 1.
C(t)|Z=1: −Y2=X2+ 1
∂
∂XC(t)|Z=1: 0 = 2X
∂
∂Y C(t)|Z=1: 0 = 2Y.
It is obvious that no point (x, y) satisfies all three equations simulta-neously. All that is left to show, is that the special fiber is also smooth in the point (0 : 1 : 0). In the affine chart given by dehomogenizing Y = 1, this point corresponds to (0,0), which does not satisfy
∂
∂ZC(t)|Y=1: −1 =X2+ 3Z2.
Hence the special fiber is smooth, too, whereby C is a regular model forF, and therefore suitable to computeχ(F). To do so, we compute the irreducible components ofC(t), that is, the irreducible factors of the polynomialZY2+Z(X2+Z2) overR. These areZandY2+X2+Z2. The first factor defines a projective line over R, whereby its function field is real. The second factor defines a component in X, since its function fieldR(X)(√
−X2−1) is nonreal and does not contain√
−1.
Henceχ(F) = 1.
Recall that, by (1.13), a function field of a conic overR((t)) has Pytha-goras number 2 if the function field is real, and PythaPytha-goras number 3 otherwise. We conjecture that the same is true for arbitrary function fields of Fermat type over R((t)). We will show this in some special cases, that cover in particularly the function fields of all conics over R(((t)).
68 6. ALGEBRAIC FUNCTION FIELDS OVERR((t))
Proposition 6.10. Let F/R((t))be the function field of the curve de-fined by 0 =aXn+bYn+cZn witha, b∈ {±1,±t}. IfF is real, then χ(F) = 0, wherebyp(F) = 2. If F is nonreal, thenχ(F) = 1, whereby p(F) = 3.
Proof. We can assume without loss of generality thatc= 1 and b=±1. We first claim that the modelC⊂P2R[[t]] ofF, given by
Zn =aXn+bYn,
is regular. By (3.43) C(0) ⊂ P2R((t)) is regular. Consider the special fiber C(t), given by Zn = ¯aXn+ ¯bYn. If ¯a 6= 0 or n = 1, then this curveRis smooth by (3.43), and henceCis regular, by [Liu06, 4.3.36].
Otherwise, if ¯a= 0 andn >1, the special fiberZn−¯bYn= 0 is smooth everywhere, except in the point (1 : 0 : 0). Hence the open subscheme C\ {1 : 0 : 0} is regular, by [Liu06, 4.3.36]. We show that the closed point P = (1 : 0 : 0)∈Cis regular, nevertheless. The point P ∈P2R[[t]]
corresponds to the maximal ideal (Y, Z, t)∈Spec(R[[t]][Y, Z])∼=A2R[[t]]. The affine chart CX=1 ofCin this affine space is given by
Zn=bYn+a.
By [Liu06, 4.2.12], we need to verify that
Zn−bYn−a /∈(Y, Z, t)2= (Y2, Z2, t2, Y Z, Y t, Zt),
in order to show thatCis regular inP. This however, is obvious, since a =±t /∈ (X2, Z2, t2, XZ, Xt, Zt) and n > 1. Hence, C is regular in any case.
The function fieldF is nonreal if and only ifnis even anda=b=−1.
In the case where F is nonreal, the special fiber
¯
aXn+ ¯bYn =Zn
is geometrically irreducible, by (3.39), i.e. its function fieldE/Rdoes not contain √
−1. Moreover, E is nonreal, as ¯a = ¯b = −1. Thus χ(F) = 1.
Now suppose F is real. In the case where b ∈ {±t}, we consider the irreducible components of the special fiber Zn = ¯aXn over R. The polynomialZn−¯aXn decomposes into linear factors over the complex numbersC. This means, that the irreducible factors overReither define projective lines overR, or decompose into projective lines overCafter base change. In particular, their function fields overRare either real, or they contain√
−1. Hence,χ(F) = 0 ifF is real.
3. EXAMPLES 69
The problem in general is, that the fibered surface defined by Zn=aXn+bYn
overR[[t]] needs not be regular. The process of desingularization consists of an alternating sequence of blowing up a closed point followed by a normalization of the resulting scheme. In theory, blowing ups can be computed, however normalizations - to my knowledge - cannot be computed in general.
This posed a major obstruction to my attempts to see whetherχ(F) is effectively computable for a given function fieldF/R((t)). If one already starts with a regular model forF, then the answer seems to be yes, as all the conditions on the special fiber can be formulated in a first order language overR, which are effectively decidable by the work of Tarski [Tar98].