MT5836 Galois Theory MRQ May 10, 2019 Contents Introduction 3 Structure of the lecture course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Recommended texts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1 Rings, Fields and Polynomials 5 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2 Field Extensions 17 The degree of an extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Algebraic elements and algebraic extensions . . . . . . . . . . . . . . . . . . . . . . . . 19 Simple extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Minimum polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3 Splitting Fields and Normal Extensions 31 Splitting fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 Existence of splitting fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Uniqueness of splitting fields and related isomorphisms . . . . . . . . . . . . . . . . . . 34 Normal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 4 Separability 40 Separable polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 Separable extensions and the Theorem of the Primitive Element . . . . . . . . . . . . 44 5 Finite Fields 47 Construction of finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 The multiplicative group of a finite field . . . . . . . . . . . . . . . . . . . . . . . . . . 49 6 Galois Groups and the Fundamental Theorem of Galois Theory 52 Galois groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 The sets F and G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 The Fundamental Theorem of Galois Theory . . . . . . . . . . . . . . . . . . . . . . . 54 Examples of Galois groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Galois groups of finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 7 Solution of Equations by Radicals 65 Radical extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Soluble groups and other group theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 Examples of polynomials with abelian Galois groups . . . . . . . . . . . . . . . . . . . 69 Galois groups of normal radical extensions . . . . . . . . . . . . . . . . . . . . . . . . . 70 A polynomial which is insoluble by radicals . . . . . . . . . . . . . . . . . . . . . . . . 72 1 Galois’s Great Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 2 Introduction The subject of Galois Theory traces back to E´variste Galois (1811–1832). He was a French mathematician whose work involved understanding the solution of polynomial equations. The standard formula √ −b± b2−4ac x = 2a for the roots of the quadratic equation ax2+bx+c = 0 is well-known. It turns out that analogous formulae exist for the roots of cubic and quartic polynomial equations. For example, the method to solve the general cubic equation was con- sidered by mathematicians based in Bologna in the early 16th century (e.g., Scipione dal Ferro (1465–1526) and those who followed him). What Galois did was to show that, in general, a quintic equation could not be solved by a similar formula. What he did not do was succeed in explaining it to anyone in a comprehensible way. For example, in 1830 he submitted his work to the Paris Academy of Sciences, but the final report states: We have made every effort to understand Galois’s proof. His reasoning is not suffi- ciently clear, sufficiently developed, for us to judge its correctness, and we can give no idea of it in this report. The author announces that the proposition which is the special object of this memoir is part of a general theory susceptible of many applications. Perhaps it will transpire that the different parts of a theory are mu- tually clarifying, are easier to grasp together rather than in isolation. We would then suggest that the author should publish the whole of his work in order to form a definitive opinion. But in the state which the part he has submitted to the Academy now is, we cannot propose to give it approval. [from Stewart, Galois Theory, Second edition, p.xxi] Galois’s ideas were eventually understood, via the letter that he wrote to Chevalier on the eveoftheduelwhichkilledhim. Thistheoryisbasicallywhatispresentedinthislecturecourse. As we now understand it, what Galois observed is the following: • To every polynomial equation, f(x) = 0, we can associate a group, the Galois group, consisting of certain permutations of the roots. • If the Galois group is soluble, then the polynomial equation can be solved by radicals (that is, by a formula of the type we are interested in). • We can construct a polynomial whose Galois group is the symmetric group S , which is 5 not soluble since it contains the non-abelian simple group A , and therefore we cannot 5 solve the corresponding polynomial equation by radicals. 3 In fact, what we do is more general. We shall actually consider a pair of fields one inside the other (F ⊆ K) and then associate to this a Galois group Gal(K/F). Our work in this module will be to understand the link between the two concepts of the field extension F ⊆ K and its Galois group. As a consequence of understanding these we can then establish Galois’s above observations by specialising to the case when K is the field obtained by adjoining the roots of our polynomial f(x) to the field F. Structure of the lecture course The following topics will be covered in the lectures: • Basic facts about fields and polynomial rings: Mostly a review of material from MT3503, but some new information about irreducible polynomials. • Field extensions: Terminology and basic properties about the situation of two fields with F ⊆ K. • Splitting fields and normal extensions: Field extensions constructed by adjoining the roots of a polynomial, constructed so that the polynomial factorizes into linear factors over the larger field. • Basic facts about finite fields: Existence and uniqueness of field of order pn, together with the fact that the multiplicative group of a finite field is cyclic. • Separable extensions and the Theorem of the Primitive Element: Separability is a technical condition to avoid repeated roots of irreducible polynomials. The Theorem of the Primitive Element applies in this circumstance and allows us to assume that our field extensions have a specific form and hence to simplify various proofs. • Galois groups and the Fundamental Theorem of Galois Theory: The definition of the Galois group as the collection of invertible structure preserving maps of a field extension (this will be made more precise later). The Fundamental Theorem of Galois Theory states that the structure of the Galois group corresponds to the structure of the field extension. • Examples and Applications: Includingthelinkbetweensolutionofapolynomialequa- tion by radicals and the solubity of the Galois groups. Recommended texts • IanStewart,GaloisTheory,Chapman&Hall;3rdEdition,2004inthelibrary;4thEdition, 2015. • John M. Howie, Fields and Galois Theory, Springer Undergraduate Mathematics Series, Springer, 2006. • P. M. Cohn, Algebra, Vol. 2, Wiley, 1977, Chapter 3. [Out of print, but available in the library.] 4 Chapter 1 Rings, Fields and Polynomials This first chapter contains a review of the background material required to study Galois Theory. The majority comes from the module MT3505 Rings and Fields and consequently many proofs in this chapter are omitted or greatly abbreviated. The last part of this chapter is concerned with polynomials and polynomial rings. One important concept that we shall use throughout the module is what it means for a polynomial to be irreducible. We shall devote some time to methods for establishing that a polynomial is irreducible. Rings We start with properties of rings before specialise to fields and to polynomial rings. Definition 1.1 A commutative ring with a 1 is a set R endowed with two binary operations denoted as addition and multiplication such that the following conditions hold: (i) R forms an abelian group with respect to addition (with additive identity 0, called zero); (ii) multiplication is associative: a(bc) = (ab)c for all a,b,c ∈ R; (iii) multiplication is commutative: ab = ba for all a,b ∈ R; (iv) the distributive laws hold: a(b+c) = ab+ac (a+b)c = ac+bc for all a,b,c ∈ R; (v) there is a multiplicative identity 1 in R satisfying a1 = 1a = a for all a ∈ R. Comment: There is also a definition of a “ring”, without the assumption of the multiplica- tion being commutative or it having a multiplicative identity 1. One simply drops conditions (iii) and (v) from the definition above. Since we are interested in studying fields in this module, we shall not need to consider non-commutative rings as there will be no examples of such rings occurringinthesenotes. Thisiswhyweonlygivethemorerestricteddefinitionofacommutative ring with a 1 above as this is sufficient for our needs. In addition, note that in a commutative ring one needs only assume one of the two distributive laws since the other may be deduced from that one via commutativity. Definition 1.2 Let R be a commutative ring with a 1. An ideal I in R is a non-empty subset of R that is both an additive subgroup of R and satisfies the property that if a ∈ I and r ∈ R, then ar ∈ I. 5 Thus a subset I of R is an ideal if it satisfies the following four conditions: (i) I is non-empty (or 0 ∈ I); (ii) a+b ∈ I for all a,b ∈ I; (iii) −a ∈ I for all a ∈ I; (iv) ar ∈ I for all a ∈ I and r ∈ R. (In a non-commutative ring, one needs to assume both ar and ra belong to I, but R being commutative ensures these products are equal.) It follows from the definition that an ideal I of R is closed under multiplication: ab ∈ I for all a,b ∈ I (since an element b ∈ I is, in particular, an element of the larger set R). This means that an ideal I is, in particular, a subring. Note that, in general, I does not contain the multiplicative identity 1, since if it did r = 1r ∈ I for all r ∈ R. Thus, the only ideal of R that contains the multiplicative identity 1 is the ring R itself. The reason for being interested in ideals is that one can form quotient rings, as we shall now describe. Let R be a commutative ring and let I be an ideal of R. Then I is, in particular, a subgroup of the additive group of R and the latter is an abelian group. We can therefore form the additive cosets of I; that is, define I +r = {a+r | a ∈ I} for each r ∈ R. We know from group theory (covered in both MT2505 and MT4003) when two such cosets are equal, I +r = I +s if and only if r−s ∈ I, and that the set of all cosets forms a group via addition of the representatives: (I +r)+(I +s) = I +(r+s) for r,s ∈ R. (In arbitrary group, one requires that the subgroup is normal, but this holds because R is an abelian group under addition.) As is observed in MT3505, the assumption that I is an ideal then ensures that there is a well-defined multiplication on the set of cosets, given by (I +r)(I +s) = I +rs for r,s ∈ R, with respect to which the set of cosets I +r forms a ring, called the quotient ring and denoted by R/I. Theorem 1.3 Let R be a commutative ring with a 1 and I be an ideal of R. Then the quotient ring R/I is a commutative ring with a 1. Proof: The fact that R/I is a ring is omitted, since verifying the above operations are well- defined is relatively technical and this was all established in MT3505. That the multiplication is commutative follows from the fact that the multiplication in R is commutative: (I +r)(I +s) = I +rs = I +sr = (I +s)(I +r) for all r,s ∈ R. The multiplication identity is I +1: (I +r)(I +1) = I +r1 = I +r for all r ∈ R. (cid:3) 6 The other standard bit of terminology that we shall require relating to rings is, of course, the definition of a homomorphism. In the following, I shall use the common habit in algebra of writing maps on the right, so the image of an element a under a map φ is written aφ (rather than φ(a), as would be common in some other branches of mathematics). Definition 1.4 Let R and S be commutative rings with 1. A homomorphism φ: R → S is a map such that (i) (a+b)φ = aφ+bφ (ii) (ab)φ = (aφ)(bφ) for all a,b ∈ R. Definition 1.5 Let R and S be commutative rings with 1 and φ: R → S be a homomorphism. (i) The kernel of φ is kerφ = {a ∈ R | aφ = 0}. (ii) The image of φ is imφ = Rφ = {aφ | a ∈ R}. Theorem 1.6 (First Isomorphism Theorem) Let R and S be commutative rings with 1 and φ: R → S be a homomorphism. Then the kernel of φ is an ideal of R, the image of φ is a subring of S and ∼ R/kerφ = imφ. Proof: This is a standard result established in MT3505 (via a proof very similar to that used for groups). The isomorphism is the map given by θ: (kerφ)+a (cid:55)→ aφ for a ∈ R. One must, amongst other things, establish that this is well-defined, in the sense that the image of a coset under θ does not depend upon the choice of representative a for the coset in the quotient ring. (cid:3) A final set of ring-theoretic definitions are the following, to which we return at the end of this chapter. Definition 1.7 Let R be a commutative ring with a 1. (i) A zero divisor in R is a non-zero element a such that ab = 0 for some non-zero b ∈ R. (ii) An integral domain is a commutative ring with a 1 containing no zero divisors. Fields Galois Theory can be viewed as the study of fields and their subfields. We shall now present the basic facts about such structures. Definition 1.8 A field F is a commutative ring with a 1 such that 0 (cid:54)= 1 and every non-zero element is a unit, that is, has a multiplicative inverse. Thus a field F is a commutative ring with a 1 such that (i) there are non-zero elements and (ii) if a ∈ F with a (cid:54)= 0, then there exists some b ∈ F with ab = 1. We shall write a−1 or 1/a for the multiplicative inverse of a. 7 Example 1.9 (i) Standard examples of fields familiar from, for example, linear algebra are thefieldsQ,RandCofrationalnumbers,realnumbersandcomplexnumbers,respectively. (ii) If p is a prime number, the set F = {0,1,2,...,p−1} forms a field under addition and p multiplication modulo p. To see that every non-zero element has a multiplicative inverse, note that if 1 (cid:54) x (cid:54) p − 1, then x and p are coprime, so there exists u,v ∈ Z with ux+vp = 1 (exploiting the fact that Z is a Euclidean domain). Hence, ux ≡ 1 (mod p) and so, modulo p, u is a multiplicative inverse for x in F . p Proposition 1.10 (i) Every field is an integral domain. (ii) The set of non-zero elements in a field forms an abelian group under multiplication. We write F∗ for the multiplicative group of non-zero elements in a field. Proof: [Omitted in lectures. These facts were observed in MT3505.] Let F be a field. (i) If a,b ∈ F with a (cid:54)= 0 and ab = 0, then b = a−1(ab) = 0. Hence if ab = 0, either a = 0 or b = 0, so F contains no zero divisors. (ii) Write F∗ = F \ {0}. Part (i) tells us that F∗ is closed under multiplication. The remaining conditions to be an abelian group under this binary operation follow immediately from the definition of a field (multiplication is associative in any ring, it is commutative in any commutative ring, there is a multiplicative identity in any ring with a 1, and in a field every non-zero element has a multiplicative inverse). (cid:3) If F is any field, with multiplicative identity denoted by 1, and n is a positive integer, let us define n = 1+1+···+1. (cid:124) (cid:123)(cid:122) (cid:125) ntimes By the distributive law, mn = mn for all positive integers m and n. Since F is, in particular, an integral domain, it follows that if there exists a positive integer n such that n = 0 then necessarily the smallest such positive integer n is a prime number. Definition 1.11 Let F be a field with multiplicative identity 1. (i) If it exists, the smallest positive integer p such that p = 0 is called the characteristic of F. (ii) If no such positive integer exists, we say that F has characteristic zero. Our observation is therefore that every field F either has characteristic zero or has charac- teristic p for some prime number p. We shall say that K is a subfield of F when K ⊆ F and that K forms a field itself under the addition and multiplication induced from F; that is, when the following conditions hold: (i) K is non-empty and contains non-zero elements (or, equivalently when taken with the other two conditions, 0,1 ∈ K); (ii) a+b,−a,ab ∈ K for all a,b ∈ K; (iii) 1/a ∈ K for all non-zero a ∈ K. Theorem 1.12 Let F be a field. 8 (i) If F has characteristic zero, then F has a unique subfield isomorphic to the rationals Q and this is contained in every subfield of F. (ii) If F has characteristic p (prime), then F has a unique subfield isomorphic to the field F p of integers modulo p and this is contained in every subfield of F. Definition 1.13 This unique minimal subfield in F is called the prime subfield of F. Proof: This was proved in MT3505. One proves it as follows: (i) Suppose F has characteristic zero. Extend the notation n to all n ∈ Z by defining 0 = 0 and −n = −n forallpositiveintegersn. IfK isanysubfieldofF thenK contains0, 1andallsumsinvolving1, so n ∈ K for all n ∈ Z. Hence Q = {m/n | m,n ∈ Z, n (cid:54)= 0} is a subset of the subfield K. One now verifies, from the field axioms and the assumption that n (cid:54)= 0 when n (cid:54)= 0, that the map n (cid:55)→ n is a ring homomorphism Z → F and then extend this to a ring homomorphism Q → F given by m/n (cid:55)→ m/n. We conclude that Q is a subfield of F that is isomorphic to the field Q of rational numbers and is contained in every subfield K of F. Finally, uniqueness of Q follows from the minimality condition: if Q and Q were subfields 1 2 contained in every subfield of F then, in particular, Q ⊆ Q and Q ⊆ Q , from which we 1 2 2 1 deduce Q = Q . 1 2 (ii)Useasimilarargumenttopart(i). IfF hascharacteristicp(prime)andK isanysubfield of F, then K contains all the elements n; that is, P = {0,1,2,3,...,p−1} ⊆ K. Now observe that P is closed under addition and multiplication and the map n (cid:55)→ n is an isomorphism from the field F of p elements to P. (cid:3) p Polynomials Polynomials arise in a number of places within Galois Theory. The motivation of the subject arises in the problem of solving polynomial equations. More significantly, algebraic elements, thosearisingasrootsofpolynomialequations, willbeofgreatimportanceinourfieldextensions as discussed in Chapter 2. Definition 1.14 LetF beafield. Apolynomial overF intheindeterminateX isanexpression of the form f(X) = a +a X +a X2+···+a Xn 0 1 2 n where n is a non-negative integer and the coefficients a , a , ..., a are elements of F. 0 1 n We shall often substitute elements of a field for the indeterminate in a polynomial. Thus if α is an element of the field F, or indeed of any field that contains F as a subfield, and f(X) = a +a X +···+a Xn where the coefficients are also elements in F, we write f(α) for 0 1 n the expression f(α) = a +a α+a α2+···+a αn. 0 1 2 n We shall write F[X] for the set of all polynomials in the indeterminate X with coefficients taken from the field F. We add two such polynomials by simply adding the coefficients, (cid:88) (cid:88) (cid:88) a Xi+ b Xi = (a +b )Xi, i i i i 9