Building a Model for Algebraically Closed Fields with Forcing Relations, Slides of Mathematical Methods

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Thierry Coquand

September 2010

Algebraic closure

In the previous lecture, we have seen how to “force” the existence of prime ideals, even in a weark framework where we don’t have choice axiom Instead of “forcing” the existence of a point of a space (a mathematical object), we are going to “force” the existence of a model (a mathematical structure)

Algebraic closure

The first step to build a closure of a field k is to show that we can build a splitting field of any polynomial P in k[X]. We have to build an extension of k where P decomposes in linear factor. As we have seen in the first lecture, there is no hope to do this effectively, even for the polynomial X^2 + 1

Algebraically closed fields

Language of ring. Theory of ring, equational Field axioms 1 6 = 0 and x = 0 ∨ ∃y. 1 = xy Algebraically closed ∃x. xn^ + a 1 xn−^1 + · · · + an = 0 For an extension of k we add the diagram of k a 6 = b stands for ¬(a = b) and ¬ϕ stands for ϕ →⊥

Algebraically closed fields

R represents a state of knowledge about the (ideal) model: we have a finite number of indeterminates X 1 ,... , Xn and a finite number of conditions P 1 = · · · = Pm = 0

Site model

Elementary covering fields R → R[a−^1 ] and R → R/〈a〉: we force a to be invertible or to be 0 algebraically closed fields: we add R → R[X]/〈p〉 where p is a monic non constant polynomial An arbitrary covering is obtained by iterating elementary coverings (in all these cases, we obtain only finite coverings)

Site model

R ∃x.ϕ(a 1 ,... , an, x) iff we have a covering fi : R → Ri and elements bi in Ri such that Ri ϕ(fi(a 1 ),... , fi(an), bi) R ϕ 0 (a 1 ,... , an) ∨ ϕ 1 (a 1 ,... , an) iff we have a covering fi : R → Ri and Ri ϕ 0 (fi(a 1 ),... , fi(an)) or Ri ϕ 1 (fi(a 1 ),... , fi(an)) for all i

Site model

R t(a 1 ,... , an) = u(a 1 ,... , an) iff we have a covering fi : R → Ri and t(fi(a 1 ),... , fi(an)) = u(fi(a 1 ),... , fi(an)) in each Ri R ⊥ iff we have a covering fi : R → Ri and 1 = 0 in each Ri

Finitely presented k-algebra

Any map R → S between two finitely presented k-algebra can be seen as a composition of two basic operations -adding a new indeterminate R → R[X] -adding a new relation R → R/〈p〉

Exploding nodes

If an element a has already an inverse in R then R/〈a〉 is trivial Similarly if a is nilpotent in R then R[a−^1 ] is trivial If R is trivial, i.e. 1 = 0 in R, then we have R ⊥ and R ϕ for all ϕ

Soundness Theorem

Lemma: If R → S and we have a covering fi : R → Ri then we can find a corresponding covering gi : S → Si with commuting maps hi : Ri → Si Lemma: If R ϕ(a 1 ,... , an) and f : R → S then S ϕ(f (a 1 ),... , f (an))

Site model

Lemma: R a = 0 iff a is nilpotent Indeed, if a is nilpotent in R[X]/〈p〉 it is nilpotent in R and if a is nilpotent in R[b−^1 ] and in R/〈b〉 then it is nilpotent in R

Completeness Theorem

We say that a formula ϕ is positive iff it does not contain ∀, →

ϕ ::= ⊥ | t = u | ϕ ∧ ϕ | ϕ ∨ ϕ | ∃x.ϕ

For a positive formula, a proof of R ϕ has a simple tree structure building a covering of R We can see this as a cut-free proof of ϕ

Completeness Theorem

Two approaches for completeness (1) Henkin-Lindenbaum (2) L¨owenheim-Skolem-Herbrand-G¨odel, gives completeness of cut-free proofs