Algebraic Geometry Lecture 29 – Semi-algebraic geometry
Lee Butler
1. Semi-algebraic sets
Algebraic sets are great, but from a logician’s point of view they have a few shortcomings.
Specifically, if Vis an algebraic set in knthen kn\Visn’t an algebraic set, and if we project
Vdown to kn−1then in general the result won’t be an algebraic set either. But these are the
properties one wants when one is dealing with questions of decidability, completeness, and other
things you’ve probably never heard of. And they’re also the properties that semi-algebraic sets do
have. The natural setting for semi-algebraic geometry is not algebraically closed fields, but real
closed fields.
Definition 1.1. We call a field formally real if −1 is not a sum of squares in the field. A formally
real field Ris called real closed if no proper algebraic extension of Ris formally real.
From now on we’ll fix a real closed field R. You can always think of R=Rif you want, or
if you keep accidentally writing Rinstead of R. We’ll define the semi-algebraic sets algebraically
and then deal with the logic, since it’s a little easier than doing it the other way around.
Definition 1.2. A subset E⊆Rnof affine n-space over Ris a semi-algebraic set if Eis a finite
union of sets of the form
{x∈Rn:f(x)=0, g1(x)>0, . . . , gk(x)>0}
with f, g1, . . . , gk∈R[X1, . . . , Xn].
Why, you may cry, do we only have one equality and many inequalities, that’s inequality against
equality, surely. Well if our basic sets were of the form
{x∈Rn:f1(x)=0, . . . , f`(x)=0, g1(x)>0, . . . , gk(x)>0}
then we could replace each of them by
{x∈Rn:f2
1(x) + . . . +f2
`(x) = 0, g1(x)>0, . . . , gk(x)>0}
and get back the same sets, since Ris a formally real field. We’ll call these sets basic since proofs
can often boil down to proving things about them.
Definition 1.3. Aboolean algebra of subsets of a set Xis a nonempty collection Cof subsets of
Xsuch that if A, B ∈ C then A∪B∈ C and X\A∈ C.
Note in particular that if Cis a boolean algebra of subsets of Xthen ∅, X ∈ C and for any
A, B ∈ C we will have A∩B∈ C. (Exercise!)
Lemma 1.4. The set of semi-algebraic sets of Rnforms a boolean algebra of the subsets of Rn.
Proof. Obviously if Aand Bare finite unions of basic sets then A∪Bwill be a finite union of
basic sets, at worst the sets forming Aand the sets forming B. So we just need to deal with
the complement. We’ll show that the intersection of two basic sets is a basic set, and that the
complement of a basic set is semi-algebraic, and the rest of the proof is an exercise in set theory.
First let
A={x∈Rn:f(x) = 0, g1(x)>0, . . . , gk(x)>0}
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