Definition of heirs

As already mentioned before the definability question for orderings inR[X₁, ..., X_n] could be solved (Theorem 1.18). One crucial step in the proof of this uses the fact that an ordering is definable if and only if for every real closed extension R⁰ ⊇R it has a unique heir. The notion of heirs originates in model theory and can be used for orderings because of the correspondence between orderings and types as explained in Section 1.3 and in the Appendix. The aim of this section is to generalize heirs of types to heirs of subsets of R[X₁, ..., X_n] such that one can speak of an heir of a quadratic module or a preordering.

First we motivate the definition of an heir of a general subset of R[X₁, ..., X_n].

As in Chapter 1, R is a real closed field, X = (X₁, ..., X_n) and Y, Z finite tuples of variables (of variable length) as well as L =Lor ={+,−,·,0,1, <} the language of ordered rings. In this setting we know that the notion of being weakly semialgebraic and of being definable is equivalent for someQ⊆R[X]. However we use the notion of definability when developing the concept of heirs because everything is also true if L is an arbitrary first-order language and M some L-structure.

If Q ⊆ R[X] is definable and R⁰ ⊇ R is a real closed extension field then we can define in a canonical way a set associated toQ, namely

Q⁰ :={f(X, c⁰)|f(X, Y)∈Z[X, Y], c⁰ ∈R^0Y and R⁰ |=ϑf(c⁰)}

where ϑ_f(Y) is an L(R)-formula definingD_R(f, Q).

Recall that because of the definability ofQ we have for every f(X, Y) ∈Z[X, Y] a formulaϑf(Y)∈FmlL(R) such that the set DR(f, Q) ={c∈R^Y |f(X, c) ∈Q} is defined by the formula ϑ_f(Y). We note that Q⁰ does not depend on the particular formulaϑ_f(Y) defining D_R(f, Q), it only depends on D_R(f, Q).

What properties does the set Q⁰ have?

One property is the following:

(H⁺) For all f(X, Y)∈Z[X, Y] and every ϕ(Y)∈FmlL(R) we have:

D_R⁰(f, Q⁰)∩ϕ(R^0Y)6=∅ ⇒D_R(f, Q)∩ϕ(R^Y)6=∅ Another one is:

(H⁻) For all f(X, Y)∈Z[X, Y] and every ϕ(Y)∈FmlL(R) we have:

D_R⁰(f, R⁰[X]\Q⁰)∩ϕ(R^0Y)6=∅ ⇒D_R(f, R[X]\Q)∩ϕ(R^Y)6=∅

That the properties (H⁺) and (H⁻) are fulfilled for Q⁰ as defined above is clear because in this case D_R⁰(f, Q⁰) = ϑ_f(R^0Y),D_R⁰(f, R⁰[X]\Q⁰) =¬ϑ_f(R^0Y) and R⁰ is an elementary extension of R.

Now we consider an arbitrary subset Q⁰ ⊆R⁰[X].

It is easy to see that if (H⁺) is satisfied for Qand Q⁰ then (H⁺) is also satisfied for Q and every subset of Q⁰. Similarly if (H⁻) is satisfied for Q and Q⁰ then (H⁻) is also satisfied for Q and every set containing Q⁰.

Therefore it is interesting to look at the smallest (resp. largest) subset ofR⁰[X] such that (H⁻) (resp. (H⁺)) is satisfied forQ and this set.

Lemma 3.1

Suppose R⁰ ⊇R is real closed and Q⊆R[X].

The set

h(Q, R⁰) := {f(X, c⁰)| f(X, Y)∈Z[X, Y] such that there is a ϕ(Y)∈FmlL(R) with c⁰ ∈ϕ(R^0Y)and ϕ(R^Y)⊆D_R(f, Q)}

is the smallest subset of R⁰[X] such that (H⁻) is satisfied for Qand this set.

The set

H(Q, R⁰) :={f(X, c⁰)| f(X, Y)∈Z[X, Y] such that for every ϕ(Y)∈FmlL(R) with c⁰ ∈ϕ(R^0Y) we have ϕ(R^Y)∩D_R(f, Q)6=∅}

is the largest subset of R⁰[X] such that (H⁺) is satisfied for Qand this set.

Proof:

We first note that if (H⁻) is satisfied for Q and Q⁰ then h(Q, R⁰) must be a sub-set of Q⁰ by definition of h(Q, R⁰). For suppose to the contrary that there is some f(X, c⁰) ∈ h(Q, R⁰) which is not in Q⁰. Then we would have by (H⁻) for Q and Q⁰ that for every ϕ(Y) ∈FmlL(R) with c⁰ ∈ ϕ(R^0Y) there is some c∈ϕ(R^Y) with f(X, c)6∈Q. This contradicts the definition ofh(Q, R⁰).

It remains to show that (H⁻) is satisfied for Q and h(Q, R⁰).

Therefore we take some f(X, Y) ∈ Z[X, Y] and some ϕ(Y) ∈ FmlL(R) with D_R⁰(f, R⁰[X]\ h(Q, R⁰))∩ϕ(R^0Y) 6= ∅, i.e. there is some c⁰ ∈ ϕ(R^0Y) such that f(X, c⁰) is not in h(Q, R⁰). Being not inh(Q, R⁰) implies that there is some element c∈ϕ(R^Y) withf(X, c)6∈Qwhich means that D_R(f, R[X]\Q)∩ϕ(R^Y)6=∅. This gives the claim for the first part of the lemma.

The result for H(Q, R⁰) follows from this since H(Q, R⁰) =R⁰[X]\h(R[X]\Q, R⁰) and (H⁺) is satisfied for Qand Q⁰ if and only if (H⁻) is satisfied for R[X]\Q and R⁰[X]\Q⁰.

Lemma 3.1 2

The sets h(Q, R⁰) (resp. H(Q, R⁰)) do not just satisfy (H⁻) (resp. (H⁺)) with respect to Q. They satisfy even the following properties as we will see in the next lemma.

Having Lemma 3.1 in mind this actually means that h(Q, R⁰) is the smallest weak heir of Q on R⁰ and H(Q, R⁰) the largest dual weak heir ofQ onR⁰.

As the adjective weak already indicates we are close to the definition of an heir. The stronger property of an heir will be the one which allows to show that Q ⊆ R[X]

is definable if and only if it has a unique heir on every real closed field R⁰ ⊇ R.

This will give us another possibility to prove definability of a quadratic module or a preordering.

Definition 3.3

Q⁰ is called an heir of Q on R⁰ if (H) is satisfied for Q and Q⁰ where property (H) is given by the following:

(H)

For all k, l∈N⁰, f1(X, Y), ..., fk(X, Y), f₁⁻(X, Y), ..., f_l⁻(X, Y)∈Z[X, Y] and everyϕ(Y)∈FmlL(R)we have:

k

T

i=1

D_R⁰(f_i, Q⁰)∩

l

T

i=1

D_R⁰(f_i⁻, R⁰[X]\Q⁰)∩ϕ(R^0Y)6=∅

⇒

k

T

i=1

D_R(f_i, Q)∩

l

T

i=1

D_R(f_i⁻, R[X]\Q)∩ϕ(R^Y)6=∅

In general h(Q, R⁰) is not an heir but it is obtained from heirs as follows.

Proposition 3.4

Let R⁰ ⊇R be a real closed field andQ⊆R[X]. Then h(Q, R⁰) = \

Q⁰heir ofQonR⁰

Q⁰ and

H(Q, R⁰) = [

Q⁰heir ofQonR⁰

Q⁰ Proof:

Appendix Proposition A.11

Prop. 3.42

Corollary 3.5

If R⁰ ⊇ R is real closed and Q⊆ R[X] is a quadratic module (resp. a prerodering) then h(Q, R⁰)is also a quadratic module (resp. a preordering).

The same is in general not true for H(Q, R⁰).

Proof:

We show that the property of being a quadratic module or a preordering transfers fromQ to every weak heir Q⁰ of Q.

First we show that Q⁰ is closed under addition. In order to do so we consider f₁(X, Y), f₂(X, Y) ∈ Z[X, Y] and f⁻(X, Y) := f₁(X, Y) +f₂(X, Y) ∈ Z[X, Y]. If there would be somec⁰ ∈ R^0Y with f₁(X, c⁰), f₂(X, c⁰)∈Q⁰ but f⁻(X, c⁰)6∈Q⁰ then the property (H_w) would give us some c∈R^Y such that f₁(X, c), f₂(X, c)∈Q but f⁻(X, c) = f₁(X, c) +f₂(X, c) 6∈ Q. This is a contradiction to the fact that Q is closed under addition.

Similarly the closure under multiplication transfers from Q toQ⁰. The fact that 1∈Q⁰ follows by (H_w) withf⁻(X, Y) := 1.

Now we prove thatR⁰[X]²Q⁰ ⊆Q⁰. With Fd(X, Z)∈Z[X, Z] we denote the general polynomial of degree d with respect to X.

We suppose that there is f₁(X, Y), f⁻(X, Y) ∈ Z[X, Y] and c⁰ ∈ R^0Y such that f1(X, c⁰)∈ Q⁰, f⁻(X, c⁰)6∈ Q⁰ and R⁰ |=ϕ(c⁰) where the L-formula ϕ(Y) is defined asϕ(Y) :=∃Z(∀X(f⁻(X, Z) = F_d(X, Z)²f₁(X, Y))). Then (H_w) implies that there is some c ∈ R^Y with f₁(X, c) ∈ Q, f⁻(X, c) 6∈ Q and R |= ϕ(c). This contradicts R[X]²Q⊆Q.

Thus in particular h(Q, R⁰) and every heir of Q is a quadratic module (resp. a pre-ordering) if Q is a quadratic module (resp. a preordering).

This is in general not true for H(Q, R⁰) because the union of quadratic modules (resp. preorderings) is in general not a quadratic module (resp. a preordering).

Corollary 3.52 With the help of heirs as defined above we can now characterize the definability of subsets of R[X] similar as for types.

Theorem 3.6

A set Q⊆ R[X] is definable if and only if it has a unique heir on R⁰ for every real closed extension fieldR⁰ ⊇R.

Proof:

Appendix Theorem A.13

Theorem 3.62 If we look once more at the canonical set

Q⁰ :={f(X, c⁰)|f(X, Y)∈Z[X, Y], c⁰ ∈R^0Y and R⁰ |=ϑf(c⁰)}

where ϑ_f(Y) is an L(R)-formula defining D_R(f, Q) which we defined at the begin-ning of this section for a definable set Q ⊆ R[X] to motivate the notion of heirs then it turns out that in the case where Qis definable, Q⁰ =h(Q, R⁰) = H(Q, R⁰) is the unique heir of Qon R⁰.

Proposition 3.4 together with Theorem 3.6 gives another way of showing definability of a set Q⊆ R[X], namely to show that h(Q, R⁰) = H(Q, R⁰) for every real closed field R⁰ ⊇R.

We use this to give an example of a finitely generated quadratic module on some real closed field which is not definable, i.e. not weakly semialgebraic.

We are considering the preordering QM_R[X]((1−X²)³) in the ring of polynomials with one indeterminate over a real closed field R for which Stengle showed in [St2]

that it is not stable over R. We deduce this result later on as a consequence of an explicit description of heirs (Corollary 3.13). Now we are going to show that this preordering is not weakly semialgebraic if R ⊃R contains infinitesimal elements.

First note that 1−X² 6∈QM_R[X]((1−X²)³).

For suppose that 1−X² = σ₀ +σ₁(1−X²)³ for some σ_i ∈ P

R[X]² (i = 0,1).

Evaluation of this expression in 1 shows that σ₀(1) = 0. Thus (1−X)|σ₀ and since this is a sum of squares even (1−X)²|σ₀. Hence (1−X)² divides the right hand side of the above expression whereas it does not divide the left hand side.

A similar argument shows that also the natural generators 1−X and 1 +X are not in QM_R[X]((1−X²)³).

Proposition 3.7

Let R⊇R be a real closed field and let n= 1, ∈R. Then

f(X, ) := 1−X²+∈P :=QM_R[X]((1−X²)³)⇔ >0 not infinitesimal.

Proof:

⇒: If < 0 then f is clearly not in P because it is not nonnegative on [−1,1].

We suppose now that >0 is an infinitesimal element of R or = 0. As an element of P the polynomial f has a representation f =σ₀+σ₁(1−X²)³ for some σ_i ∈P

R[X]² (i= 0,1). If all coefficients of σ₀ and σ₁ lie in the convex hull O^{^} ofR in R then we get by applying λ a representation of 1−X² as an element of QM_O^[X]((1−X²)³). This is not possible because the residue field O^{^} is a subfield of R and we have seen above that 1−X² does not lie in this preordering. (We actually just need that O^{^} is a real closed field).

If one of the coefficients of σ₀ or σ₁ does not lie in O^{^}, i.e. has negative value with respect to the corresponding valuation v, then we take the coefficient c

with the most negative value and divide the representation of f by c². Again by applyingλ we get in the residue field 0 =eσ₀+σe₁(1−X²)³ for some eσ_i ∈ PO^{^}[X]² (i = 0,1). Since at least one of the coefficients of the polynomials appearing inσei is 1 we get a nontrivial representation of 0 in the residue field and hence a contradiction as the interior of [−1,1] is not empty and therefore supp(QM_O^[X]((1−X²)³)) = {0}.

⇐: Because nonnegative elements from R are in the additively closed set P we can suppose without loss of generality that ∈ O^{^} \ m. Since λ is order preserving we get λ() > 0 and we find some q ∈ Q with ^λ()₂ < q < λ() becauseO^{^}is archimedean. As O^{^} ⊆R we get by Schm¨udgen (Corollary 2.24) that 1− X² +q ∈ QM_O^[X]((1 −X²)³). The order preserving residue map λ:O^{^} →O^{^}/madmits an order preserving section ρ:O^{^}/m→O^{^} which is the identity on Rbecause R⊆R. Hence we get a representation of 1−X²+q as an element of QM_R[X]((1−X²)³) by applying ρ. We have q < because ρ is order preserving and thus we get what we want.

Prop. 3.72

The proposition enables us to give the promised example of a not weakly semialge-braic finitely generated preordering on some real closed field.

Example 3.8

The preordering P :=QM_R[X]((1−X²)³)⊆R[X] =R[X₁] is not weakly semialge-braic if R⊇R contains infinitesimal elements.

For the proof of this we show that the weak heir and the dual weak heir ofP do not coincide in some suitable real closed extension field R⁰ of R.

We consider therefore the polynomialf(X, Y) := 1−X²+Y ∈Z[X, Y]. Proposition 3.7 shows that

f(X, c)∈P ⇔c >0, c not infinitesimal

Let nowm⁺ be the cut corresponding to the upper edge of the maximal idealm, i.e.

m⁺ = ((m⁺)^L,(m⁺)^R) with (m⁺)^R := {x ∈ R | x > m}, and β be a realization of m⁺ in some real closed fieldR⁰ ⊇R.

Thenf(X, β)∈H(P, R⁰)\h(P, R⁰)because for every formulaϕ(Y)∈FmlL(R)with R⁰ |=ϕ(β) (i.e. for every formula from the type of β over R) there is some c₁ ∈ R with R |= ϕ(c1) and c1 > 0 not infinitesimal, i.e. f(X, c1) ∈ P, but there is also some c₂ ∈R with R|=ϕ(c₂) and c₂ >0 infinitesimal, i.e. f(X, c₂)6∈P.

Im Dokument The Membership Problem for quadratic modules with focus on the one dimensional case (Seite 90-97)