Appendix A Constructibility This book is dedicated to the synthetic (or axiomatic) approach to geometry, di- rectly inspired and motivated by the work of Greek geometers of antiquity. The Greek geometers achieved great work in geometry; but as we have seen, a few prob- lems resisted all their attempts (and all later attempts) at solution: circle squaring, trisecting the angle, duplicating the cube, constructing all regular polygons, to cite only the most popular ones. We conclude this book by proving why these various problems are insolvable via ruler and compass constructions. However, these proofs concerning ruler and compass constructions completely escape the scope of syn- thetic geometry which motivated them: the methods involved are highly algebraic and use theories such as that of fields and polynomials. This is why we these results are presented as appendices. Moreover, we shall freely use (with precise references) various algebraic results developed in [5], Trilogy II. We first review the theory of the minimal polynomial of an algebraic element in a field extension. Furthermore, since it will be essential for the applications, we prove the Eisenstein criterion for the irreducibility of (such) a (minimal) polynomial over the field of rational numbers. We then show how to associate a field with a given geometric configuration and we prove a formal criterion telling us—in terms of the associated fields—when one can pass from a given geometrical configuration to a more involved one, using only ruler and compass constructions. A.1 The Minimal Polynomial This section requires some familiarity with the theory of polynomials in one variable over a field K. The reader not familiar with this material will find some essential aspects of the theory in Appendix A of [5], Trilogy II. Proposition A.1.1 If K ⊆ L is a field extension, then is a K-vector space. Proof The scalar multiplication of k ∈ K and l ∈ L is their product in L. F. Borceux, An Axiomatic Approach to Geometry, DOI 10.1007/978-3-319-01730-3, 355 © Springer International Publishing Switzerland 2014 356 A Constructibility Under the conditions of Proposition A.1.1, it thus makes sense to speak of the (vectorial) dimension of L over K, which we shall write as dim[L : K]. Proposition A.1.2 Given finite dimensional field extensions K ⊆ L ⊆ M, one has dim[M : K]=dim[M : L]×dim[L : K]. Proof Let l1,...,lr be a basis of L as a K-vector space and m1,...,ms a basis of M as an L-vector space. It suffices to observe that the rs elements limj constitute a ∈ basis of M over K. Indeed an element μ M can be uniquely written as j λj mj , ∈ = with λj L, while each λj can itself be uniquely written as λj i ki,j li with ki,j ∈ K. The conclusion follows at once. Proposition A.1.3 Given a finite dimensional field extension K ⊆ L, every element l ∈ L is a root of a polynomial with coefficients in K. Proof Suppose that dim[L : K]=n. Then the n + 1 elements of L 1,l,l2, ..., ln cannot be linearly independent over K, proving the existence of scalars ki ∈ K such that 2 n k01 + k1l + k2l +···+knl = 0. This proves that l is a root of the polynomial 2 n p(X) = k0 + k1X + k2X +···+knX ∈ K[X]. The situation described in Proposition A.1.3 has been given a name: Definition A.1.4 An element l ∈ L in a field extension K ⊆ L is algebraic over K when it is a root of a polynomial with coefficients in K. When all elements of L are algebraic over K, the extension K ⊆ L itself is called algebraic. The main result of this section is then: Theorem A.1.5 Consider a field extension K ⊆ L and an element l ∈ L which is algebraic over K. There exists a unique polynomial p(X) ∈ K[X] such that 1. p(l) = 0; 2. the leading coefficient of p(X) is 1; 3. p(X) is irreducible in K[X]. Moreover, under these conditions: 1. p(X) divides in K[X] all other polynomials admitting l as a root; 2. the dimension of the subfield K(l) ⊆ L generated by K and l is equal to the degree of the polynomial p(X). A.1 The Minimal Polynomial 357 The polynomial p(X) is called the minimal polynomial of l over K. Proof Among all the polynomials in K[X] admitting l as a root, choose a non-zero one whose degree is minimal. Possibly dividing by the leading coefficient (which does not change the roots), we can assume that this leading coefficient is 1. This is our polynomial p(X). If q(X) ∈ K[X] is such that q(l) = 0, the Euclidean division of q(X) by p(X) (see Theorem A.2.1 in [5], Trilogy II) yields q(X) = p(X)d(X) + r(X), degree r(X)< degree p(X). Since q(l) = 0 = p(l), this implies r(l) = 0. By minimality of the degree of p(X), we conclude that r(X)= 0. Thus p(X) divides q(X). This also proves that p(X) is irreducible in K[X]. Indeed otherwise l would be a root of a proper factor q(X) of p(X), which is a contradiction since by what we have just proved, p(X) would then have to divide q(X), which is of strictly lower degree. The uniqueness of p(X) also follows immediately. Indeed given another irre- ducible polynomial q(X) ∈ K[X] such that q(l) = 0, p(X) divides q(X) which is irreducible, thus q(X) = kp(X) with k ∈ K. Since both have the same leading coefficient 1, necessarily k = 1. It remains to prove the result concerning dimensions. If p(X) has degree n,the sequence of elements − 1,l,l2, ..., ln 1 is linearly independent over K. Indeed otherwise (as in the proof of Proposi- tion A.1.3) there would be a polynomial of degree n − 1 with coefficients in K[X] admitting l as a root; this would contradict the choice of p(X) with minimal degree. It remains to observe that − 1,l,l2, ..., ln 1 generate K(l) as a vector space over K. Writing n n−1 p(X) = X + kn−1X +···+k1X + k0,ki ∈ K we get at once n n−1 l =−kn−1l −···−k1l − k0. This already proves that the subset n−1 M = α0 + α1l +···+αn−1l |αi ∈ K ⊆ L is not only stable under addition and subtraction, but also under multiplication, since each power ln after multiplication can be replaced by a combination of strictly lower powers; repeating this operation as many times as needed, one ends up with an expression where all powers of l are at most n − 1, thus with an element of M. 358 A Constructibility Moreover, the inverse in L of every element m ∈ M is still in M, proving that M is in fact a subfield of L; and then trivially, M = K(l). When m ∈ K, this is trivial. Otherwise, m = q(l) for a non-constant polynomial n−1 q(X) = β0 + β1(X) +···+βn−1X ,βi ∈ K. But q(X) and p(X) do not have a common non-constant factor (they are “relatively prime”), because p(X) is irreducible and has a larger degree than q(X).BytheBe- zout theorem (see Theorem A.3.2 in [5], Trilogy II), there exist polynomials r(X), s(X) in K[X] such that r(X)p(X) + s(X)q(X) = 1 with moreover degree r(X)< degree q(X), degree s(X) < degree p(X) = n. Taking the values at l we find s(l)q(l) = 1, proving that s(l) is the inverse of m = q(l). Indeed, s(l) ∈ M since degree s(X) < n. Let us conclude with the classical terminology for “non-algebraic” elements: Definition A.1.6 An element l ∈ L in a field extension K ⊆ L which is not alge- braic over K is said to be transcendental over K. As an immediate consequence of Proposition A.1.3 we have Proposition A.1.7 If a field extension K ⊆ L contains a transcendental element over K, its dimension is infinite. A.2 The Eisenstein Criterion Most impossibility proofs that we shall consider in Appendices B and C will be in terms of algebraic elements over of the field of rational numbers. Therefore, in order to easily identify the corresponding minimal polynomials, it is important to have an efficient criterion telling us when a polynomial in Q[X] is irreducible (see Theorem A.1.5). Of course, multiplying all the coefficients of a polynomial in Q[X] by a common multiple of their denominators, we obtain a polynomial with coefficients in Z, which still has the same roots. But more amazingly: Proposition A.2.1 Let p(X) ∈ Z[X] be a polynomial which is irreducible in Z[X]. Then p(X) is also irreducible in Q[X]. A.2 The Eisenstein Criterion 359 Proof It is equivalent to prove that when p[X]∈Z[X] is reducible in Q[X],itis reducible in Z[X]. Suppose therefore that p(X) = r(X)s(X) with r(X) and s(X) two non-constant polynomials in Q[X]. Multiplying by a common multiple n of the denominators of the coefficients of r(X) and s(X), we obtain np(X) = r(X)s(X), where r(X) and s(X) now have their coefficients in Z.