By Saugata Basu, Richard Pollack, Marie-Francoise Roy,

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**Sample text**

Let C be a proper cone of F. If −a∈C, then C[a] = {x + a y x, y ∈ C } F is a proper cone of F. Proof: Suppose −1 = x + a y with x, y ∈ C. If y = 0 we have −1 ∈ C which is impossible. If y 0 then −a = (1/y 2) y (1 + x) ∈ C , which is also impossible. 8: Since the union of a chain of proper cones is a proper cone, Zorn’s lemma implies the existence of a maximal proper cone C which contains C. It is then suﬃcient to show that C ∪ −C = F, and to deﬁne x ≤ y by y − x ∈ C. Suppose that −a∈C. 9, C[a] is a proper cone and thus, by the maximality of C, C = C[a] and thus a ∈ C.

Taking a value z ∈ N such that D(z)=0, the polynomial Q(z , Y ) is a square free polynomial since all its roots are distinct. We prove now that it is possible to express, for every 1 ≤ i < j ≤ p, xi + x j and xi x j rationally in terms of γi, j = xi + x j + z xi x j . Indeed let F (Z , Y ) = ∂Q/∂Y (Z , Y ) = (Y − (xk + x + Z xk x )) i< j G(Z , Y ) = k< (k, )=(i,j) (xi + x j ) H(Z , Y ) = (Y − (xk + x + Z xk x )) , k< (k, )=(i, j) i< j (Y − (xk + x + Z xk x )) . 16, f (Z , Y ), G(Z , Y ) and H(Z , Y ) are elements of R[Z , Y ].

Show that if S is a ﬁnite constructible subset of Ck, then Ext(S , C ) is equal to S. (Hint: write a formula describing S). 26) is stated without proof in [105]. 23) are given in [156] (Remark 16). 2 Real Closed Fields Real closed ﬁelds are ﬁelds which share the algebraic properties of the ﬁeld of real numbers. 1, we deﬁne ordered, real and real closed ﬁelds and state some of their basic properties. 2 is devoted to real root counting. 3 we deﬁne semi-algebraic sets and prove that the projection of an algebraic set is semi-algebraic.