Galois Groups As Permutation Groups
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APPLICATIONS of GALOIS THEORY 1. Finite Fields Let F Be a Finite Field
CHAPTER IX APPLICATIONS OF GALOIS THEORY 1. Finite Fields Let F be a finite field. It is necessarily of nonzero characteristic p and its prime field is the field with p r elements Fp.SinceFis a vector space over Fp,itmusthaveq=p elements where r =[F :Fp]. More generally, if E ⊇ F are both finite, then E has qd elements where d =[E:F]. As we mentioned earlier, the multiplicative group F ∗ of F is cyclic (because it is a finite subgroup of the multiplicative group of a field), and clearly its order is q − 1. Hence each non-zero element of F is a root of the polynomial Xq−1 − 1. Since 0 is the only root of the polynomial X, it follows that the q elements of F are roots of the polynomial Xq − X = X(Xq−1 − 1). Hence, that polynomial is separable and F consists of the set of its roots. (You can also see that it must be separable by finding its derivative which is −1.) We q may now conclude that the finite field F is the splitting field over Fp of the separable polynomial X − X where q = |F |. In particular, it is unique up to isomorphism. We have proved the first part of the following result. Proposition. Let p be a prime. For each q = pr, there is a unique (up to isomorphism) finite field F with |F | = q. Proof. We have already proved the uniqueness. Suppose q = pr, and consider the polynomial Xq − X ∈ Fp[X]. As mentioned above Df(X)=−1sof(X) cannot have any repeated roots in any extension, i.e. -
Determining the Galois Group of a Rational Polynomial
Determining the Galois group of a rational polynomial Alexander Hulpke Department of Mathematics Colorado State University Fort Collins, CO, 80523 [email protected] http://www.math.colostate.edu/˜hulpke JAH 1 The Task Let f Q[x] be an irreducible polynomial of degree n. 2 Then G = Gal(f) = Gal(L; Q) with L the splitting field of f over Q. Problem: Given f, determine G. WLOG: f monic, integer coefficients. JAH 2 Field Automorphisms If we want to represent automorphims explicitly, we have to represent the splitting field L For example as splitting field of a polynomial g Q[x]. 2 The factors of g over L correspond to the elements of G. Note: If deg(f) = n then deg(g) n!. ≤ In practice this degree is too big. JAH 3 Permutation type of the Galois Group Gal(f) permutes the roots α1; : : : ; αn of f: faithfully (L = Spl(f) = Q(α ; α ; : : : ; α ). • 1 2 n transitively ( (x α ) is a factor of f). • Y − i i I 2 This action gives an embedding G S . The field Q(α ) corresponds ≤ n 1 to the subgroup StabG(1). Arrangement of roots corresponds to conjugacy in Sn. We want to determine the Sn-class of G. JAH 4 Assumption We can calculate “everything” about Sn. n is small (n 20) • ≤ Can use table of transitive groups (classified up to degree 30) • We can approximate the roots of f (numerically and p-adically) JAH 5 Reduction at a prime Let p be a prime that does not divide the discriminant of f (i.e. -
ON R-EXTENSIONS of ALGEBRAIC NUMBER FIELDS Let P Be a Prime
ON r-EXTENSIONS OF ALGEBRAIC NUMBER FIELDS KENKICHI IWASAWA1 Let p be a prime number. We call a Galois extension L of a field K a T-extension when its Galois group is topologically isomorphic with the additive group of £-adic integers. The purpose of the present paper is to study arithmetic properties of such a T-extension L over a finite algebraic number field K. We consider, namely, the maximal unramified abelian ^-extension M over L and study the structure of the Galois group G(M/L) of the extension M/L. Using the result thus obtained for the group G(M/L)> we then define two invariants l(L/K) and m(L/K)} and show that these invariants can be also de termined from a simple formula which gives the exponents of the ^-powers in the class numbers of the intermediate fields of K and L. Thus, giving a relation between the structure of the Galois group of M/L and the class numbers of the subfields of L, our result may be regarded, in a sense, as an analogue, for L, of the well-known theorem in classical class field theory which states that the class number of a finite algebraic number field is equal to the degree of the maximal unramified abelian extension over that field. An outline of the paper is as follows: in §1—§5, we study the struc ture of what we call T-finite modules and find, in particular, invari ants of such modules which are similar to the invariants of finite abelian groups. -
1 Spring 2002 – Galois Theory
1 Spring 2002 – Galois Theory Problem 1.1. Let F7 be the field with 7 elements and let L be the splitting field of the 171 polynomial X − 1 = 0 over F7. Determine the degree of L over F7, explaining carefully the principles underlying your computation. Solution: Note that 73 = 49 · 7 = 343, so × 342 [x ∈ (F73 ) ] =⇒ [x − 1 = 0], × since (F73 ) is a multiplicative group of order 342. Also, F73 contains all the roots of x342 − 1 = 0 since the number of roots of a polynomial cannot exceed its degree (by the Division Algorithm). Next note that 171 · 2 = 342, so [x171 − 1 = 0] ⇒ [x171 = 1] ⇒ [x342 − 1 = 0]. 171 This implies that all the roots of X − 1 are contained in F73 and so L ⊂ F73 since L can 171 be obtained from F7 by adjoining all the roots of X − 1. We therefore have F73 ——L——F7 | {z } 3 and so L = F73 or L = F7 since 3 = [F73 : F7] = [F73 : L][L : F7] is prime. Next if × 2 171 × α ∈ (F73 ) , then α is a root of X − 1. Also, (F73 ) is cyclic and hence isomorphic to 2 Z73 = {0, 1, 2,..., 342}, so the map α 7→ α on F73 has an image of size bigger than 7: 2α = 2β ⇔ 2(α − β) = 0 in Z73 ⇔ α − β = 171. 171 We therefore conclude that X − 1 has more than 7 distinct roots and hence L = F73 . • Splitting Field: A splitting field of a polynomial f ∈ K[x](K a field) is an extension L of K such that f decomposes into linear factors in L[x] and L is generated over K by the roots of f. -
Techniques for the Computation of Galois Groups
Techniques for the Computation of Galois Groups Alexander Hulpke School of Mathematical and Computational Sciences, The University of St. Andrews, The North Haugh, St. Andrews, Fife KY16 9SS, United Kingdom [email protected] Abstract. This note surveys recent developments in the problem of computing Galois groups. Galois theory stands at the cradle of modern algebra and interacts with many areas of mathematics. The problem of determining Galois groups therefore is of interest not only from the point of view of number theory (for example see the article [39] in this volume), but leads to many questions in other areas of mathematics. An example is its application in computer algebra when simplifying radical expressions [32]. Not surprisingly, this task has been considered in works from number theory, group theory and algebraic geometry. In this note I shall give an overview of methods currently used. While the techniques used for the identification of Galois groups were known already in the last century [26], the involved calculations made it almost imprac- tical to do computations beyond trivial examples. Thus the problem was only taken up again in the last 25 years with the advent of computers. In this note we will restrict ourselves to the case of the base field Q. Most methods generalize to other fields like Q(t), Qp, IFp(t) or number fields. The results presented here are the work of many mathematicians. I tried to give credit by references wherever possible. 1 Introduction We are given an irreducible polynomial f Q[x] of degree n and asked to determine the Galois group of its splitting field2 L = Spl(f). -
11. Counting Automorphisms Definition 11.1. Let L/K Be a Field
11. Counting Automorphisms Definition 11.1. Let L=K be a field extension. An automorphism of L=K is simply an automorphism of L which fixes K. Here, when we say that φ fixes K, we mean that the restriction of φ to K is the identity, that is, φ extends the identity; in other words we require that φ fixes every point of K and not just the whole subset. Definition-Lemma 11.2. Let L=K be a field extension. The Galois group of L=K, denoted Gal(L=K), is the subgroup of the set of all functions from L to L, which are automorphisms over K. Proof. The only thing to prove is that the composition and inverse of an automorphism over K is an automorphism, which is left as an easy exercise to the reader. The key issue is to establish that the Galois group has enough ele- ments. Proposition 11.3. Let L=K be a finite normal extension and let M be an intermediary field. TFAE (1) M=K is normal. (2) For every automorphism φ of L=K, φ(M) ⊂ M. (3) For every automorphism φ of L=K, φ(M) = M. Proof. Suppose (1) holds. Let φ be any automorphism of L=K. Pick α 2 M and set φ(α) = β. Then β is a root of the minimum polynomial m of α. As M=K is normal, and α is a root of m(x), m(x) splits in M. In particular β 2 M. Thus (1) implies (2). -
Resultant and Discriminant of Polynomials
RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results are well-known 19th century mathematics, but I have not inves- tigated the history, and no references are given. 1. Resultant n m Definition 1.1. Let f(x) = anx + ··· + a0 and g(x) = bmx + ··· + b0 be two polynomials of degrees (at most) n and m, respectively, with coefficients in an arbitrary field F . Their resultant R(f; g) = Rn;m(f; g) is the element of F given by the determinant of the (m + n) × (m + n) Sylvester matrix Syl(f; g) = Syln;m(f; g) given by 0an an−1 an−2 ::: 0 0 0 1 B 0 an an−1 ::: 0 0 0 C B . C B . C B . C B C B 0 0 0 : : : a1 a0 0 C B C B 0 0 0 : : : a2 a1 a0C B C (1.1) Bbm bm−1 bm−2 ::: 0 0 0 C B C B 0 bm bm−1 ::: 0 0 0 C B . C B . C B C @ 0 0 0 : : : b1 b0 0 A 0 0 0 : : : b2 b1 b0 where the m first rows contain the coefficients an; an−1; : : : ; a0 of f shifted 0; 1; : : : ; m − 1 steps and padded with zeros, and the n last rows contain the coefficients bm; bm−1; : : : ; b0 of g shifted 0; 1; : : : ; n−1 steps and padded with zeros. In other words, the entry at (i; j) equals an+i−j if 1 ≤ i ≤ m and bi−j if m + 1 ≤ i ≤ m + n, with ai = 0 if i > n or i < 0 and bi = 0 if i > m or i < 0. -
Algebraic Combinatorics and Resultant Methods for Polynomial System Solving Anna Karasoulou
Algebraic combinatorics and resultant methods for polynomial system solving Anna Karasoulou To cite this version: Anna Karasoulou. Algebraic combinatorics and resultant methods for polynomial system solving. Computer Science [cs]. National and Kapodistrian University of Athens, Greece, 2017. English. tel-01671507 HAL Id: tel-01671507 https://hal.inria.fr/tel-01671507 Submitted on 5 Jan 2018 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. ΕΘΝΙΚΟ ΚΑΙ ΚΑΠΟΔΙΣΤΡΙΑΚΟ ΠΑΝΕΠΙΣΤΗΜΙΟ ΑΘΗΝΩΝ ΣΧΟΛΗ ΘΕΤΙΚΩΝ ΕΠΙΣΤΗΜΩΝ ΤΜΗΜΑ ΠΛΗΡΟΦΟΡΙΚΗΣ ΚΑΙ ΤΗΛΕΠΙΚΟΙΝΩΝΙΩΝ ΠΡΟΓΡΑΜΜΑ ΜΕΤΑΠΤΥΧΙΑΚΩΝ ΣΠΟΥΔΩΝ ΔΙΔΑΚΤΟΡΙΚΗ ΔΙΑΤΡΙΒΗ Μελέτη και επίλυση πολυωνυμικών συστημάτων με χρήση αλγεβρικών και συνδυαστικών μεθόδων Άννα Ν. Καρασούλου ΑΘΗΝΑ Μάιος 2017 NATIONAL AND KAPODISTRIAN UNIVERSITY OF ATHENS SCHOOL OF SCIENCES DEPARTMENT OF INFORMATICS AND TELECOMMUNICATIONS PROGRAM OF POSTGRADUATE STUDIES PhD THESIS Algebraic combinatorics and resultant methods for polynomial system solving Anna N. Karasoulou ATHENS May 2017 ΔΙΔΑΚΤΟΡΙΚΗ ΔΙΑΤΡΙΒΗ Μελέτη και επίλυση πολυωνυμικών συστημάτων με -
Math 200B Winter 2021 Final –With Selected Solutions
MATH 200B WINTER 2021 FINAL {WITH SELECTED SOLUTIONS 1. Let F be a field. Let A 2 Mn(F ). Let f = minpolyF (A) 2 F [x]. (a). Let F be the algebraic closure of F . Show that minpolyF (A) = f. (b). Show that A is diagonalizable over F (that is, A is similar in Mn(F ) to a diagonal matrix) if and only if f is a separable polynomial. 2 3 1 1 1 6 7 (c). Let A = 41 1 15 2 M3(F ). Is A diagonalizable over F ? The answer may depend 1 1 1 on the properties of F . Most students did well on this problem. For those that missed part (a), the key is to use that the minpoly is the largest invariant factor, and that invariant factors are unchanged by base field extension (since the rational canonical form must be the same regardless of the field). There is no obvious way to part (a) directly. 2. Let F ⊆ K be a field extension with [K : F ] < 1. In this problem, if you find any results from homework problems helpful you can quote them here rather than redoing them. Note that a commutative ring R is called reduced if R has no nonzero nilpotent elements. (a). Suppose that K=F is separable. Prove that the K-algebra K ⊗F K is reduced, but is not a domain unless K = F . (b). Suppose that K=F is inseparable. Show that the K-algebra K ⊗F K is not reduced. Proof. (a). Since K=F is separable, K = F (α) for some α 2 K, by the theorem of the primitive element. -
Appendices a Algebraic Geometry
Appendices A Algebraic Geometry Affine varieties are ubiquitous in Differential Galois Theory. For many results (e.g., the definition of the differential Galois group and some of its basic properties) it is enough to assume that the varieties are defined over algebraically closed fields and study their properties over these fields. Yet, to understand the finer structure of Picard-Vessiot extensions it is necessary to understand how varieties behave over fields that are not necessarily algebraically closed. In this section we shall develop basic material concerning algebraic varieties taking these needs into account, while at the same time restricting ourselves only to the topics we will use. Classically, algebraic geometry is the study of solutions of systems of equations { fα(X1,... ,Xn) = 0}, fα ∈ C[X1,... ,Xn], where C is the field of complex numbers. To give the reader a taste of the contents of this appendix, we give a brief description of the algebraic geometry of Cn. Proofs of these results will be given in this appendix in a more general context. One says that a set S ⊂ Cn is an affine variety if it is precisely the set of zeros of such a system of polynomial equations. For n = 1, the affine varieties are fi- nite or all of C and for n = 2, they are the whole space or unions of points and curves (i.e., zeros of a polynomial f(X1, X2)). The collection of affine varieties is closed under finite intersection and arbitrary unions and so forms the closed sets of a topology, called the Zariski topology. -
Galois Groups of Cubics and Quartics (Not in Characteristic 2)
GALOIS GROUPS OF CUBICS AND QUARTICS (NOT IN CHARACTERISTIC 2) KEITH CONRAD We will describe a procedure for figuring out the Galois groups of separable irreducible polynomials in degrees 3 and 4 over fields not of characteristic 2. This does not include explicit formulas for the roots, i.e., we are not going to derive the classical cubic and quartic formulas. 1. Review Let K be a field and f(X) be a separable polynomial in K[X]. The Galois group of f(X) over K permutes the roots of f(X) in a splitting field, and labeling the roots as r1; : : : ; rn provides an embedding of the Galois group into Sn. We recall without proof two theorems about this embedding. Theorem 1.1. Let f(X) 2 K[X] be a separable polynomial of degree n. (a) If f(X) is irreducible in K[X] then its Galois group over K has order divisible by n. (b) The polynomial f(X) is irreducible in K[X] if and only if its Galois group over K is a transitive subgroup of Sn. Definition 1.2. If f(X) 2 K[X] factors in a splitting field as f(X) = c(X − r1) ··· (X − rn); the discriminant of f(X) is defined to be Y 2 disc f = (rj − ri) : i<j In degree 3 and 4, explicit formulas for discriminants of some monic polynomials are (1.1) disc(X3 + aX + b) = −4a3 − 27b2; disc(X4 + aX + b) = −27a4 + 256b3; disc(X4 + aX2 + b) = 16b(a2 − 4b)2: Theorem 1.3. -
Cyclotomic Extensions
CYCLOTOMIC EXTENSIONS KEITH CONRAD 1. Introduction For a positive integer n, an nth root of unity in a field is a solution to zn = 1, or equivalently is a root of T n − 1. There are at most n different nth roots of unity in a field since T n − 1 has at most n roots in a field. A root of unity is an nth root of unity for some n. The only roots of unity in R are ±1, while in C there are n different nth roots of unity for each n, namely e2πik=n for 0 ≤ k ≤ n − 1 and they form a group of order n. In characteristic p there is no pth root of unity besides 1: if xp = 1 in characteristic p then 0 = xp − 1 = (x − 1)p, so x = 1. That is strange, but it is a key feature of characteristic p, e.g., it makes the pth power map x 7! xp on fields of characteristic p injective. For a field K, an extension of the form K(ζ), where ζ is a root of unity, is called a cyclotomic extension of K. The term cyclotomic means \circle-dividing," which comes from the fact that the nth roots of unity in C divide a circle into n arcs of equal length, as in Figure 1 when n = 7. The important algebraic fact we will explore is that cyclotomic extensions of every field have an abelian Galois group; we will look especially at cyclotomic extensions of Q and finite fields. There are not many general methods known for constructing abelian extensions (that is, Galois extensions with abelian Galois group); cyclotomic extensions are essentially the only construction that works over all fields.