MTH 562/662 Spring 2019 Abstract Algebra II Drew Armstrong Contents
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MTH 562/662 Spring 2019 Abstract Algebra II Drew Armstrong Contents Week 13 The Classical Problem of Algebra, Definition of Fields and Subfields/Extensions, Adjunction: The Subfield Generated by a Subset 2 Week 14 The Lattice of Subfields, Definition of the Galois Group, Splitting Fields, The Fundamental Theorem of Galois Theory 10 Problem Set 7 22 Week 15 Rings and Subrings/Extensions, Ring Homomorphisms, Ideals and Quotient Rings, Isomorphism Theorems for Rings 31 Week 16 Ideal and Subring Generated by a Subset, Fields are Simple Rings, Division With Remainder, Descartes' Factor Theorem, Z and F[x] are PIDs 38 Problem Set 8 49 Week 17 Integral Domains, Euclidean Domains, Principal Ideal Domains (PIDs) Prime and Irreducible Elements, Unique Factorization Domains (UFDs) 58 Week 18 Evaluation of Polynomials, The Minimal Polynomial Theorem, Kronecker's Theorem, Existence of Splitting Fields 71 Problem Set 9 81 Week 19 Irreducible Polynomials, Rational Root Test, The Splitting Field of x3 − 2, Cyclotomic Polynomials, Constructible Numbers 96 Week 20 Fields of Size Four and Eight, Primitive Root Theorem, Repeated Roots, Frobenius Automorphism, Existence and Uniqueness of Finite Fields 102 Problem Set 10 112 1 Week 21 Multi-Variable Polynomials, The Finiteness Theorem, Definition of Galois Extensions, The Lifting Lemma, The Splitting Field Theorem 123 Week 22 Perfect Fields, Artin's Fixed Field Lemma, Galois Extensions Over Perfect Fields, The Fundamental Theorem of Galois Theory 134 Problem Set 11 144 Epilogue: Galois' Solvability Theorem 149 Week 13 Last semester I used the story of Galois Theory to motivate the study of abstract groups. This semester I will use the same story to motivate the study abstract rings and fields. As before, we will find that linear algebra is always hiding just beneath the surface. To begin, let me refresh your memory. The classical (pre-1830) problem of algebra was to find explicit \formulas" for the roots of a polynomial equation. To be precise, suppose that some rational numbers (called \coefficients") are given: e1; e2; : : : ; en 2 Q: Then we want to find some numbers r1; r2; : : : ; rn (called \roots") such that n n−1 n−2 n x − e1x + e2x − · · · + (−1) en = (x − r1)(x − r2) ··· (x − rn): A priori, it is not obvious what kind of \numbers" the roots should be, or whether they exist at all. Soon we will prove a result called the Fundamental Theorem of Algebra which says that the roots always exist in the field C of complex numbers. Unfortunately, this theorem will not tell us how to find the roots. We would really like to have some formula or algorithm for computing the roots. To state the problem explicitly, we expand the right hand side and then equate coefficients to obtain a system of n non-linear equations in n unknowns: 8 e = r + r + ··· + r > 1 1 2 n > e = r r + r r + ··· + r r > 2 1 2 1 3 n−1 n > . <> . P ek = ri ri ··· ri > 1≤i1<i2<···<ik≤n 1 2 k > . > . > : en = r1r2 ··· rn: 2 Our goal is to somehow \invert" this system. The best we could hope for is to find some n explicit functions f1; f2; : : : ; fn from Q to C such that 8 r1 = f1(e1; e2; : : : ; en) > <> r2 = f2(e1; e2; : : : ; en) . > . > : rn = fn(e1; e2; : : : ; en): But this hope is too naive. Indeed, no such functions can exist. To see why, observe that each coefficient ek can be thought of as a function of the roots: X ek = ek(r1; r2; : : : ; rn) = ri1 ri2 ··· rik : 1≤i1<i2<···<ik≤n Furthermore, this function has the nice property of being \symmetric" under permutations of the roots. In other words, if σ 2 Sn is any permutation of the set f1; 2; : : : ; ng then we have ek(r1; r2; : : : ; rn) = ek(rσ(1); rσ(2); : : : ; rσ(n)): The easiest way to see this is to observe that the product (x − r1)(x − r2) ··· (x − rn) is symmetric under permutations of its factors. If we expand the product then each coefficient must also be a symmetric function. [Jargon. The coefficients e1; : : : ; en are called the elementary symmetric functions of the roots. This explains my use of the letter \e."] n If f : Q ! C is any function then we can think of the expression f(e1; : : : ; en) as a function of the roots, as follows: f(e1; : : : ; en)(r1; : : : ; rn) := f (e1(r1; : : : ; rn); : : : ; en(r1; : : : ; rn)) : Furthermore, it is clear that f(e1; : : : ; en) is a symmetric function of the roots. Now we see why our first hope was too naive: There can be no function fk such that rk = fk(e1; e2; : : : ; en) because fk(e1; e2; : : : ; en) is always a symmetric function of the roots, whereas rk is certainly not a sym- metric function of the roots. The solution is to weaken the requirement that fk is a \function." Instead, we will allow \multi-valued functions"1 such as square roots. We often talk about \the" square root as though it were a function p − : C ! C: p But this is not a function. Indeed, for any 0 6= α 2 C, the expression α represents two distinct complex numbers and there is no natural way to choose between them. We can use this ambiguity to solve the quadratic equation. 1Recall that a \multi-valued function" is not a function. This is a terrible but common notation. 3 Example: The Quadratic Formula. For any rational coefficients e1; e2 2 Q we want to find some complex roots r1; r2 2 C such that 2 2 x − e1x + e2 = (x − r1)(x − r2) = x − (r1 + r2)x + (r1r2): In other words, we want to find some \multi-valued functions" f1; f2 such that e = r + r r = f (e ; e ) 1 1 2 () 1 1 1 2 e2 = r1r2 r2 = f2(e1; e2): As you know, the solution is ( p 2 f1(e1; e2) = (e1 + e1 − 4e2)=2 p 2 f2(e1; e2) = (e1 − e1 − 4e2)=2: p 2 The only subtlety here is that we must interpret the ambiguous expression \ e1 − 4e2" in the p 2 same way for both equations. In other words, we let \ e1 − 4e2" denote one particular 2 2 2 number α 2 C such that α = e1 − 4e2. If e1 − 4e2 6= 0 then there will be two choices and we just pick one at random.2 /// The process of choosing a random square root is called \breaking the symmetry." For higher degree equations we expect that we will need to break the symmetry by choosing random 3rd roots, 4th roots, etc. This leads us to the classical problem of algebra. The Classical Problem of Algebra. Let e1; : : : ; en 2 Q be any rational numbers. By the Fundamental Theorem of Algebra there exist some unique complex numbers r1; : : : ; rn 2 C such that n n−1 n−2 n x − e1x + e2x − · · · + (−1) en = (x − r1)(x − r2) ··· (x − rn): Our goal is to find some way to compute these roots. Specifically, we want to find an \algebraic formula" expressing the roots in terms of the coefficients, using only the \algebraic operations" p p p p +; −; ×; ÷; ; 3 ; 4 ; 5 ;:::: /// As you know, this problem turns out to be impossible when n ≥ 5. Last semester we developed the group theory necessary for the proof of impossibility.3 This semester we will fill in the other half of the proof. 2If the roots are imaginary then there is a deep sense in which they are indistinguishable. Indeed, we usually define the imaginary unit i as \the" square root of −1. But if −1 has any square root then it must have two. Which one do you want to call i ? 3 Specifically, we proved that the group Sn not \solvable" when n ≥ 5. 4 The Classical Problem was definitively solved by Galois in the 1820s. However, he died too soon to really explain it to anyone. Galois' work was eventually published in 1846 by Joseph Liouville. The first textbook on Galois theory was Camille Jordan's Trait´edes substitutions et des ´equationsalg´ebraiques (1870). At this point \Galois theory" and \group theory" were the same subject, so Jordan's work can also be viewed as the first book about groups. However, the subject was still difficult to understand. The next major advance was made by Richard Dedekind in the 1880s when he defined the concept of a field.4 Definition of Fields and Subfields/Extensions. A field is a set F together with two binary operations +; × : F × F ! F (called addition and multiplication) and two special elements 0; 1 2 F (called zero and one), which satisfy the following three axioms: (F1) (F; +; 0) is an abelian group. (F2) (F − f0g; ×; 1) is an abelian group. We will use juxtaposition to denote multiplication: ab := a × b: Furthermore, since multiplication is commutative we are free to use fractional notation to denote division: a ab−1 = b−1a = : b (F3) Distribution. For all a; b; c 2 F we have a(b + c) = ab + ac: Now let S ⊆ F be any subset. We say that S is a subfield of F (equivalently, F is a field extension of S) if the following properties are satisfied: • The special elements 0; 1 are in S. • For all a; b 2 S we have a ± b 2 S. 4Dedekind's name for this structure was K¨orper, short for Zahlk¨orper [body of numbers]. Dedekind's rival Leopold Kronecker used the term Rationalit¨atsbereich [domain of rationality]. The English term field was coined by E. H. Moore in 1893, possibly motivated by the word \domain." This creates a problem for English speakers: should we denote fields by the letter K or the letter F ? To avoid confusion I will use the blackboard bold font (i.e., K or F) to denote fields.