Galois module structure for Artin-Schreier theory over bicyclic extensions

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1 Wellesley College Wellesley College Digital Scholarship and Archive Honors Thesis Collection 2017 Galois module structure for Artin-Schreier theory over bicyclic extensions Lauren Heller Follow this and additional works at: Recommended Citation Heller, Lauren, "Galois module structure for Artin-Schreier theory over bicyclic extensions" (2017). Honors Thesis Collection This Dissertation/Thesis is brought to you for free and open access by Wellesley College Digital Scholarship and Archive. It has been accepted for inclusion in Honors Thesis Collection by an authorized administrator of Wellesley College Digital Scholarship and Archive. For more information, please contact

2 Galois module structure for Artin-Schreier theory over bicyclic extensions Lauren Cranton Heller advisor: Professor Andrew Schultz submitted in partial fulfillment of the prerequisite for honors Department of Mathematics Wellesley College April 2017

3 i Abstract If K/F is a Galois field extension with Galois group of order p n where p char(f ), then Gal(K/F ) acts on the quotient K /K p. The structure of this F p [Gal(K/F )]-module is known for Gal(K/F ) isomorphic to Z/p n Z or the Klein 4-group. We use Artin-Schreier theory to produce a similar decomposition for characteristic p extensions with Gal(K/F ) isomorphic to Z/pZ Z/pZ. Acknowledgements Andrew Schultz for sharing his work and his time with me; Alexander Diesl and Karen Lange for serving on my committee; Bridget Schreiner for her mathematical and non-mathematical camaraderie; Simona Boyadzhiyska for encouraging me to write a thesis; Donald Elmore for his excellent advice; and Lisa, Thomas, and Megan Heller for their constant support

4 Contents 1 Introduction Kummer theory Artin-Schreier theory Main theorem Preliminaries Trace maps Properties of modules Klein 4-group case Notation The map T Construction of Y Construction of X Structure of J Decomposition for p odd Notation Supertrace Construction of Y Construction of X Structure of J ii

5 Chapter 1 Introduction Galois theory assigns to each field extension an automorphism group, and this process is well understood. Thus one might ask the inverse question: given a group, what sorts of field extensions give rise to it? We will tackle a small part of this problem by specifying the structure of a module that can be constructed from an extension and its Galois group. In particular, we are interested in bicyclic extensions, those with Galois group of the form Z/pZ Z/pZ for a prime p, where the characteristic of the ground field matches the exponent p of the group. Such extensions are described by Artin-Schreier theory, which we will cover briefly in this chapter. It is a variant of the more common Kummer theory, which covers extensions with Galois group i Z/pn i Z where the characteristic is not p. The decomposition of the analagous Kummer-theoretic module is known when the Galois group is Z/p n Z or Z/2Z Z/2Z (the Klein 4-group), and the latter proof by Chemotti ([1]) inspired much of this work. Since Kummer theory is more classical than Artin-Schreier theory, we will begin there. 1.1 Kummer theory If F is a field containing p distinct p-th roots of unity for some prime p char(f ) and p a is the formal root of an irreducible polynomial x p a F [x], then the Galois group of the field extension F ( p a)/f is isomorphic to Z/pZ. Kummer theory says that the converse is also true: if K/F is a Galois field extension with Gal(K/F ) = Z/pZ for some p char(f ) and F contains the p-th roots of unity, then there exists a F such that K = F ( p a) where p a is a root of x p a. Let F p = {f p : f F }, ie. the units in F with p-th roots also in F. This is a normal subgroup of the multiplicative group of units F, and there is a bijection between finite subgroups of F /F p and extensions over F with finite Abelian Galois group of exponent dividing p. Given a subgroup U F /F p we can obtain such an extension by adjoining to F the roots of all polynomials of the form x p a for af p U. Conversely, given an extension K/F with Gal(K/F ) finite, Abelian, and of exponent dividing p, we can obtain a subgroup of F /F p by taking all elements af p such that a F K p. Kummer theory says that these processes are inverses of each other. In particular, the extension F ( p a)/f is associated with the subgroup generated by af p in 1

6 CHAPTER 1. INTRODUCTION 2 F /af p, namely the cosets with representatives 1, a, a 2,..., a p 1. The Galois group of an extension K/F acts on elements of K, so K is a module over the group ring Z[Gal(K/F )], which consists of formal Z linear combinations of elements in Gal(K/F ) and acts by the following: k α i σ i = σ α i i (k) If α i σ i is in the ideal pz[gal(k/f )], then [k] α i σ i = [0] in K /K p for all k K, so this action descends to an action on K /K p by Z[Gal(K/F )] pz[gal(k/f )] = F p [Gal(K/F )]. As stated above, the structure of K /K p as an F p [Gal(K/F )]-module is known when Gal(K/F ) = Z/p n Z for char(f ) p (Theorems 1 and 2 in [3]) and when Gal(K/F ) = Z/2Z Z/2Z for char(f ) 2 ([1]). 1.2 Artin-Schreier theory If F is a field of characteristic p, then polynomials of the form x p a F [x] are not separable. (Given a single root θ a they can be factored as (x θ a ) p.) Instead, Galois groups isomorphic to Z/pZ arise as extensions F (θ a )/F where θ a is a root of an irreducible and separable polynomial x p x a F [x]. Theorem 1.1. If θ a is a root of the polynomial x p x a F [x] and char F = p, then θ a + 1 is another root. Proof. Suppose θ a is a root of x p x a. Then (θ a + 1) p (θ a + 1) f = θ p a + 1 p θ a 1 a = θ p a θ a a = 0 because F (θ a ) has characteristic p. Thus the roots of x p x a are θ a, θ a + 1,..., θ a + p 1, so F (θ a ) is a splitting field and we no longer need to require that F contain p-th roots of unity. The relevant normal subgroup is contained in K +, the group of K under addition. Definition 1. If K is a field, let (K) = {k p k : k K}. We can now state the central result of the theory. Theorem 1.2 (Artin-Schreier). A Galois field extension K/F of characteristic p has cyclic Galois group of order p if and only if it is the splitting field of an irreducible polynomial x p x a for some a F. Furthermore, there is a bijection between finite subgroups of the quotient F + / (F ) and finite Abelian extensions over F with exponent dividing p.

7 CHAPTER 1. INTRODUCTION 3 The bijective correspondence is given by U F (θ a : a + (F ) U) where θ a is a root of the polynomial x p x a, and inversely, K/F {a + (F ) : a F (K)}. In particular, the extension F (θ a )/F is associated with the subgroup generated by a + (F ) in F + / (F ), namely { (F ), a + (F ), 2a + (F ),..., (p 1)a + (F )}. Moving from traditional Artin-Schreier theory to the subject of our research, the group ring F p [Gal(K/F )] corresponding to an extension K/F with Galois group of order p n for p = char(f ) acts on the quotient K + / (K), this time additively: ( ) [ αi σ i [k] = αi σ i (k)] The structure of K + / (K) as an F p [Gal(K/F )]-module is known when Gal(K/F ) = Z/p n Z (Prop. 6.2 in [2]). 1.3 Main theorem The primary result of this thesis is to provide the first module decomposition for K + / K when Gal(K/F ) is a noncyclic p-group. Theorem 1.3. Let K/F be a Galois field extension of characteristic p with Gal(K/F ) = Z/pZ Z/pZ. Then K + / (K) = X Y as an F p [Gal(K/F )]-module for Y a direct sum of free modules and X an indecomposable module with F p -dimension p and presentation x L, x R : (σ 1)x L = (τ 1)x R. After some background information in Chapter 2, this will be restated and proven for p = 2 in Chapter 3 and for odd p in Chapter 4.

8 Chapter 2 Preliminaries 2.1 Trace maps Since the only ideals of a field are {0} and the field itself, the only nontrivial ring homomorphisms between subfields of an extension K/F are isomorphisms, which can be extended to automorphisms in Gal(K/F ), and inclusion maps. To find a map from K to a proper subfield we must relax the requirement that it preserve both additive and multiplicative structure. One such map is constructed from the elements of Gal(K/F ). Definition 2. Let K/F be a Galois extension with Galois group G and let k K. Then T K/F (k) = σ G σ(k) is known as the trace from K to F. Note that the trace map T K/F from K to F is a homomorphism on the additive group K +. Theorem 2.1. Let K/F be a Galois extension and let k K. Then T K/F (k) F. Proof. Let G = Gal(K/F ) and suppose τ G. Then applying τ to the elements of G simply permutes them, so ( ) τ(t K/F (k)) = τ σ(k) = (τ σ)(k) = σ(k) = T K/F (k). σ G σ G σ G Thus T K/F (k) is fixed by all elements of Gal(K/F ), which implies that T K/F (k) F because K/F is Galois. Though we will not prove it in full generality, if K/M/F is a tower of Galois field extensions, then T M/F T K/M = T K/F. The map T K/F also respects equivalence classes as follows: Corollary 2.2. If k 1 + (K) = k 2 + (K) then T K/F (k 1 ) + (F ) = T K/F (k 2 ) + (F ). 4

9 CHAPTER 2. PRELIMINARIES 5 Proof. If k 1 + (K) = k 2 + (K) then there exists k K such that k 1 k 2 = k p k. Thus T K/F (k 1 ) T K/F (k 2 ) = T K/F (k 1 k 2 ) = T K/F (k p k) = (T K/F (k)) p T K/F (k) (F ) because T K/F is a homomorphism and T K/F (k) F, so T K/F (k 1 ) + (F ) = T K/F (k 2 ) + (F ). (Note that T K/F (k p ) = (T K/F (k)) p because char K = p.) We will use this fact to classify elements of our module according to their images in subfields of K. For the particular field extensions of interest to us, trace maps are in fact surjective on elements, which we will first prove for cyclic extensions. Theorem 2.3. Let K/F be a Galois field extension of characteristic p with Gal(K/F ) Z/pZ. Then T K/F : K F is surjective. Proof. Let σ be a generator of Z/pZ and let θ K such that K = F (θ). We will show that { T K/F (θ i 0 i p 1 ) = 1 i = p 1. Choose σ so that σ(θ) = θ + 1. By the binomial theorem T K/F (θ i ) = (1 + σ + σ σ p 1 )(θ i ) = θ i + (θ + 1) i + (θ + 2) i + + (θ + p 1) i i ( ) i i ( ) i = θ i + θ i j + θ i j 2 j + + j j j=0 j=0 p 1 ( p 1 i ( ) i = pθ i + θ )l i 1 + θ i l=1 l=1 i j=0 p 1 l θ ( ) i θ i j (p 1) j j l=1 ( ) p 1 i l j 1 + j 1 The image of the map F p F p given by l l j is cyclic. Let g be a generator. Then p 1 l=1 ( ) i l j = j because the order of g divides p 1, so ( i 0 = (g 1) j ( ) i p 1 j ) p 1 k=1 k=1 ( i g k = g j ) p 1 g k l=1 k=1 p 1 ( ) i g k = (g 1) l j. j l=1 ( ) i l j j We have g = 1 if and only if j = p 1, so the coefficient of θ i j in T K/F (θ i ) is 0 modulo p for j p 1. For j = p 1 p 1 ( ) ( ) i i p 1 l j = l p 1 = j p 1 l=1 l=1 ( ) i p 1 p 1 l=1 ( ) i 1 = (p 1) p 1 by Fermat s Little Theorem. This is equal to 0 for i p 1 (ie. (p 1)-th powers do not appear in the original sum). If i = p 1 then it is equal to p 1 1 mod p. Thus

10 CHAPTER 2. PRELIMINARIES 6 T K/F (θ i ) = 0 for i p 1 and the only nonzero term of T K/F (θ p 1 ) is the constant 1. Suppose f F. Then fθ p 1 K and so T K/F is surjective. T K/F ( fθ p 1 ) = ft K/F (θ p 1 ) = f( 1) = f Corollary 2.4. Let K/F be a Galois field extension of characteristic p with Gal(K/F ) Z/p n 1 Z Z/p n 2 Z Z/p n k Z for some positive integers n 1,..., n k. Then T K/F is surjective. Proof. Let m = n n k (the power of p in the degree of K/F ) and consider the tower K = L 0 L 1 L m = F of intermediate fields corresponding to the subgroups Z/p n 1 Z Z/p n 2 Z/p n k Z pz/p n 1 Z Z/p n 2 Z Z/p n k Z. p n 1 1 Z/p n 1 Z Z/p n 2 Z Z/p n k Z 0 Z/p n 2 Z Z/p n k Z 0 pz/p n 2 Z Z/p n k Z. 0 0 Z/p n k Z of Gal(K/F ), so that Gal(L i /L i+1 ) Z/pZ for all i. Then T Li /L i+1 the theorem above, and is surjective for all i by so T K/F is also surjective. T K/F = T L0 /L m = T Lm 1 /L m T L1 /L 2 T L0 /L Properties of modules For the purposes of this section, let G = Z/pZ Z/pZ = σ, τ for some prime p, let M, N be modules over the group ring F p [G], and denote by M G the submodule of M fixed under action by G. We will prove general results about these modules to be used later. Lemma 2.5. We have M G N G = {0} if and only if M N = {0}.

11 CHAPTER 2. PRELIMINARIES 7 Proof. Clearly M N = {0} implies M G N G = {0}. We will prove the converse by contrapositive. Let y M N be nonzero. Since G is finite, there exists a smallest positive integer i such that (σ 1) i y = 0. Consider (σ 1) i 1 y. By choice of i, this element is nonzero and fixed by σ. Similarly, let j be the smallest positive integer such that (τ 1) j (σ 1) i 1 y = 0. Then (τ 1) j 1 (σ 1) i 1 y is nonzero and fixed by σ and τ. Since M and N are closed under action by G, we have (τ 1) j 1 (σ 1) i 1 y M G N G. Therefore M G N G {0}. Lemma 2.6. Suppose M and N are submodules of a single F p [G]-module and M N = {0}. Then (M N) G = M G N G. Proof. Suppose x M G N G. Then there exist m M G and n N G such that x = m + n, so σ(x) = σ(m + n) = σ(m) + σ(n) = m + n = x and similarly τ(x) = x. Thus M G N G (M N) G. Now suppose x (M N) G. Then x M N, so there exist m M and n N such that x = m + n, so m + n = x = σ(x) = σ(m + n) = σ(m) + σ(n) where σ(m) M and σ(n) N. Since the representation m + n is unique, we must have σ(m) = m and σ(n) = n. Similarly τ(m) = m and τ(n) = n, so (M N) G M G N G. Lemma 2.7. Each element of F p [G] can be written in the form for some α ij F p. p 1 p 1 α ij (σ 1) i (τ 1) j i=0 j=0 Proof. We will first show that each element p 1 i=1 β iσ i of F p [ σ ] can be written in the form p 1 i=1 α i(σ 1) i for some α i F p. This amounts to a change in F p -basis. Note that (σ 1) i has degree i as a polynomial in σ. Thus the matrix representing the transition from {1, (σ 1), (σ 1) 2,..., (σ 1) p 1 } to {1, σ, σ 2,..., σ p 1 } is triangular and has no nonzero entries on the diagonal, so it is invertible. Since each term of an element p 1 p 1 i=1 j=1 β ijσ i τ j in F p [Gal(K/F )] can be written as the product of an element in F p [ σ ] and one in F p [ τ ], and products of elements of the form p 1 i=0 p 1 j=0 α ij(σ 1) i (τ 1) j also have this form, the lemma follows. This representation of elements in F p [G] is convenient because an element in an F p [G]-module is fixed if and only if it is killed by σ 1 and τ 1. We will also frequently use the fact that (σ 1) p = σ p 1 = 0 and (τ 1) p = τ p 1 = 0 in characteristic p. The remainder of the results in this section will be technical and very poorly motivated because they concern a specific module X that we have not yet constructed. Proofs about the properties of this X are somewhat neater in a general setting and will be referenced, as one might guess, in Sections 3.4 and 4.4, both labeled Construction of X.

12 CHAPTER 2. PRELIMINARIES 8 Lemma 2.8. Let X be an F p [G]-module with generators α L and α R satisfying (σ 1)α L = (τ 1)α R such that the elements (τ 1) p 1 α L, (σ 1) p 1 α R, and (σ 1) p 1 (τ 1) p 2 α L are nonzero. Then a nonzero element of the form (σ 1) i (τ 1) j α L is in X G if and only if either i = 0 and j = p 1 or i = p 1 and j = p 2. Similarly a nonzero element (σ 1) i (τ 1) j α R is in X G if and only if either i = p 1 and j = 0 or i = p 2 and j = p 1. Proof. We have (σ 1)(τ 1) p 1 α L = (τ 1) p α R (τ 1)(τ 1) p 1 α L = (τ 1) p α L (σ 1)(σ 1) p 1 (τ 1) p 2 α L = (σ 1) p (τ 1) p 2 α L (τ 1)(σ 1) p 1 (τ 1) p 2 α L = (σ 1) p 2 (τ 1) p α R where (σ 1) p = (τ 1) p = 0, so (τ 1) p 1 α L and (σ 1) p 1 (τ 1) p 2 α L are in X G. Suppose (σ 1) i (τ 1) j α L X \ {0} is neither of these. If j = p 1 then i 0, so (σ 1) i (τ 1) j α L = (σ 1) i (τ 1) p 1 α L = (σ 1) i 1 (τ 1) p α R = 0, a contradiction. Thus j < p 1, so (σ 1) p i 1 (τ 1) p j 2 (σ 1) i (τ 1) j α L = (σ 1) p 1 (τ 1) p 2 α L 0 since at least one of p i 1 and p j 2 is nonzero. Hence (σ 1) i (τ 1) j α L / X G. Lemma 2.9. Let X be an F p [G]-module with generators α L and α R satisfying (σ 1)α L = (τ 1)α R such that (τ 1) p 1 α L, (σ 1) p 1 (τ 1) p 2 α L, (σ 1) p 1 α R X G \ {0}. Then (σ 1) i (τ 1) j α L = 0 for 0 i, j p 1 if and only if 1 i p 2 and j = p 1. Similarly (σ 1) i (τ 1) j α R = 0 if and only if i = p 1 and 1 j p 2. Proof. If 1 i p 2, then (σ 1) i (τ 1) p 1 α L = 0 because (τ 1) p 1 α L is fixed by σ. We have (τ 1) p 1 α L 0 by assumption. If 1 i p 1 and 1 j p 2, then (σ 1) i (τ 1) j α L 0, else we would have (σ 1) p 1 (τ 1) p 2 α L = (σ 1) p i 1 (τ 1) p j 2 (σ 1) i (τ 1) j α L = 0 contradicting our hypotheses. According to Lemma 2.7, all elements in an F p [G] module generated by α L and α R can be written as a sum of elements in the forms specified by Lemmas 2.8 and 2.9. Thus these lemmas can be used to identify the portions of such a sum that are trivial or fixed. We will now show that the relation (σ 1)α L = (τ 1)α R and some specific linearly independent elements are sufficient to determine the structure of X. The idea of the proof is to assume that another relation exists, apply powers of σ 1 and/or τ 1 until only known terms remain, and then contradict linear independence.

13 CHAPTER 2. PRELIMINARIES 9 Theorem Let X be an F p [G]-module with generators α L and α R, let F be the free F p [G]-module with generators α L and α R, and let ι : F X be the module homomorphism given by α L α L and α R α R. If (σ 1)α L = (τ 1)α R and the elements (τ 1) p 1 α L, (σ 1) p 1 (τ 1) p 2 α L, and (σ 1) p 1 α R are F p -linearly independent in X, then ker ι = (σ 1) α L (τ 1) α R. Proof. By hypothesis ι((σ 1) α L (τ 1) α R ) = (σ 1)α L (τ 1)α R = 0 so (σ 1) α L (τ 1) α R ker ι. Suppose that s ker ι for some s F. By Lemma 2.7 we can write p 1 p 1 p 1 p 1 s = β ij (σ 1) i (τ 1) j α L + γ ij (σ 1) i (τ 1) j α R i=0 j=0 for some β ij, γ ij F p. We have p 1 p 2 p 2 p 1 γ ij (σ 1) i (τ 1) j α L γ ij (σ 1) i (τ 1) j α R i=1 j=0 i=0 i=0 j=1 j=0 p 2 p 2 = ((σ 1) α L (τ 1) α R ) γ ij (σ 1) i (τ 1) j and since (σ 1) p = (τ 1) p = 0 in F p [G], we also have that p 2 β i,p 1 (σ 1) i (τ 1) p 1 α L i=1 and similarly i=1 p 2 p 2 = β i,p 1 (σ 1) i (τ 1) p 1 α L β i,p 1 (σ 1) i 1 (τ 1) p α R i=1 i=0 i=1 i=0 p 2 = ((σ 1) α L (τ 1) α R ) (σ 1) i 1 (τ 1) p 1 p 2 γ p 1,j (σ 1) p 1 (τ 1) j α R (σ 1) α L (τ 1) α R. Therefore, adding these, we have s (σ 1) α L (τ 1) α R if and only if i=1 p 1 p 1 p δ 0j (τ 1) j 1 α L + δ ij (σ 1) i (τ 1) j 1 α L + j=1 i=1 j=1 p δ i0 (σ 1) i 1 α R i=1

14 CHAPTER 2. PRELIMINARIES 10 is in the ideal, where δ 0j = β 0,j 1, δ i0 = γ i 1,0, and δ ij = β i,j 1 + γ i 1,j. We will show that δ ij = 0 for all i, j by contradiction. Note that the sums we added to s are also in ker ι, so this new element is as well, ie. p 1 p 1 p δ 0j (τ 1) j 1 α L + δ ij (σ 1) i (τ 1) j 1 α L + j=1 i=1 j=1 p δ i0 (σ 1) i 1 α R = 0. i=1 Let m be the maximum index such that δ ij = 0 for all 0 i, j m 1, and let n be the minimum index such that at least one of δ nm and δ mn is nonzero. Then 0 n m p with m 0 and n p. If m = p then n = 0 so this relation is δ 0p (τ 1) p 1 α L + δ p0 (σ 1) p 1 α R = 0 where at least 1 of δ 0p and δ p0 is nonzero, and if m = n = p 1 then the relation is δ 0p (τ 1) p 1 α L + δ p 1,p 1 (σ 1) p 1 (τ 1) p 2 α L + δ p0 (σ 1) p 1 α R = 0 where δ p 1,p 1 0. Both of these contradict our hypothesis of linear independence. We can thus assume that 0 n m p 1 with m 0 and n p 1. Without loss of generality, suppose δ mn 0. Applying (σ 1) p m 1 (τ 1) p n 1 kills each term with at least one of i m + 1 and j n + 1 by Lemma 2.9 unless m = p 1, in which case a term with i = 0 and j = n + 1 may also survive. The only nonzero coefficient δ ij with i m and j n is δ mn by choice of m and n. Thus the only terms remaining after application of (σ 1) p m 1 (τ 1) p n 1 are as follows: case 1: 0 < m < p 1 and n > 0 case 2: 0 < m < p 1 and n = 0 0 = (σ 1) p m 1 (τ 1) p n 1 (δ mn (σ 1) m (τ 1) n 1 α L ) = δ mn (σ 1) p 1 (τ 1) p 2 α L case 3: m = p 1 and 0 < n < p 1 0 = (σ 1) p m 1 (τ 1) p 0 1 (δ m0 (σ 1) m 1 α R ) = δ mn (σ 1) p 2 (τ 1) p 1 α R = δ mn (σ 1) p 1 (τ 1) p 2 α L 0 = (σ 1) p (p 1) 1 (τ 1) p n 1 (δ mn (σ 1) p 1 (τ 1) n 1 α L + δ 0,n+1 (τ 1) n α L ) = δ mn (σ 1) p 1 (τ 1) p 2 α L + δ 0,n+1 (τ 1) p 1 α L case 4: m = p 1 and n = 0 0 = (σ 1) p (p 1) 1 (τ 1) p 0 1 (δ mn (σ 1) p 2 α R + δ 01 (τ 1) 0 α L ) = δ mn (σ 1) p 2 (τ 1) p 1 α R + δ 01 (τ 1) p 1 α L = δ mn (σ 1) p 1 (τ 1) p 2 α L + δ 01 (τ 1) p 1 α L

15 CHAPTER 2. PRELIMINARIES 11 In each case δ mn 0, contradicting linear independence. Therefore no such δ mn exists, so s (σ 1) α L (τ 1) α R. Corollary If X is as given in Theorem 2.10, then the union of the sets {(τ 1) j α L : 0 j p 1} {(σ 1) i (τ 1) j α L : 1 i p 1, 0 j p 2} {(σ 1) i α R : 0 i p 1} is an F p -basis for X. Proof. Independence follows from the previous proof and spanning from Lemma 2.7. Corollary If X is as given in Theorem 2.10, then X G = (τ 1) p 1 α L, (σ 1) p 1 (τ 1) p 2 α L, (σ 1) p 1 α R F p. Proof. Suppose x X G. Then by Corollary 2.11 there exist δ ij F p such that x = p 1 p 1 p δ 0j (τ 1) j 1 α L + δ ij (σ 1) i (τ 1) j 1 α L + j=1 i=1 j=1 p δ i0 (σ 1) i 1 α R. i=1 Then (τ 1)x = 0 and (σ 1)x = 0 so by Corollary 2.11 the coefficient corresponding to each term that is not fixed is 0. This leaves x = δ 0p (τ 1) p 1 α L + δ p 1,p 1 (σ 1) p 1 (τ 1) p 2 α L + δ p0 (σ 1) p 1 α R by Lemma 2.8 so x is in the desired span. The converse also follows from Lemma 2.8. Finally, we will prove that X is indecomposable, with the help of a strange but useful lemma. Lemma Let X be the F p [Gal(K/F )]-module X generated by α L and α R subject to the relation (σ 1)α L = (τ 1)α R. If x X with (τ 1)x, (σ 1) p 1 x X G then either (σ 1) p 1 x = 0 or they are linearly independent. Similarly, if (σ 1)x, (τ 1) p 1 x X G then either (τ 1) p 1 x = 0 or they are independent. Proof. By Corollary 2.12, there exist δ 1, δ 2, δ 3, γ 1, γ 2, γ 3 F p such that and (τ 1)x = δ 1 (τ 1) p 1 α L + δ 2 (σ 1) p 2 (τ 1) p 1 α R + δ 3 (σ 1) p 1 α R (σ 1) p 1 x = γ 1 (τ 1) p 1 α L + γ 2 (σ 1) p 1 (τ 1) p 2 α L + γ 3 (σ 1) p 1 α R. Thus δ 3 (σ 1) p 1 α R is in the image of τ 1. However (σ 1) p 1 α R is not, so we must have δ 3 = 0. Similarly γ 1 (τ 1) p 1 α L is in the image of σ 1 but (τ 1) p 1 α L is not, so γ 1 = 0. Then (τ 1)x = (τ 1)(δ 1 (τ 1) p 2 α L + δ 2 (σ 1) p 2 (τ 1) p 2 α R )

16 CHAPTER 2. PRELIMINARIES 12 so there exists y ker(τ 1) such that x = δ 1 (τ 1) p 2 α L + δ 2 (σ 1) p 2 (τ 1) p 2 α R + y. By Corollary 2.11 we can write y as an F p -combination of (σ 1) p 1 α R and the terms (σ 1) i (τ 1) j α L that are in ker(τ 1) according to Lemma 2.9, all of which are also in ker(σ 1) p 1, so (σ 1) p 1 y = 0 and thus δ 1 (σ 1) p 1 (τ 1) p 2 α L = (σ 1) p 1 x Hence by linear independence we have γ 3 = 0 and γ 2 = δ 1. case 1: γ 2 = δ 1 = 0 Then γ 1 = γ 2 = γ 3 = 0, so (σ 1) p 1 x = 0. case 2: γ 2 = δ 1 0 Then = γ 2 (σ 1) p 1 (τ 1) p 2 α L + γ 3 (σ 1) p 1 α R. (τ 1)x = δ 1 (τ 1) p 1 α L + δ 2 (σ 1) p 2 (τ 1) p 1 α R = δ 1 (τ 1) p 1 α L + δ 2 (σ 1) p 1 (τ 1) p 2 α L (σ 1) p 1 x = δ 1 (σ 1) p 1 (τ 1) p 2 α L where δ 1 0, so (τ 1)x and (σ 1) p 1 x are linearly independent because (τ 1) p 1 α L and (σ 1) p 2 (τ 1) p 1 α R are. These give the 2 cases in the lemma. Theorem An F p [G]-module X with generators α L and α R that satisfy (σ 1)α L = (τ 1)α R such that (σ 1) p 1 α L, (σ 1) p 1 (τ 1) p 2 α L, and (σ 1) p 1 α R are F p -linearly independent is indecomposable. Proof. Suppose X = A B for some submodules A and B. case 1: A G = A Since α L X, there exist α A A and α B B such that α L = α A + α B. Thus and (τ 1) p 1 α L = (τ 1) p 1 (α A + α B ) = 0 + (τ 1) p 1 α B B (σ 1) p 1 (τ 1) p 2 α L = 0 + (σ 1) p 1 (τ 1) p 2 α B B. Applying the same reasoning to α R gives so dim Fp B G = 3. (σ 1) p 1 α R B

17 CHAPTER 2. PRELIMINARIES 13 case 2: A G A Let x A \ A G, let i be the maximum index such that (σ 1) i x 0, and let j be the maximum index such that (σ 1) i (τ 1) j x 0. Then (σ 1) i (τ 1) j x A G and at least one of i and j is nonzero. Without loss of generality, suppose j 0. Let δ 1, δ 2, δ 3 F p such that (σ 1) i (τ 1) j x = δ 1 (τ 1) p 1 α L + δ 2 (σ 1) p 2 (τ 1) p 1 α R + δ 3 (σ 1) p 1 α R (possible by Corollary 2.12). Then δ 3 (σ 1) p 1 α R is in the image of τ 1 while (σ 1) p 1 α R is not, so δ 3 = 0. subcase a: δ 1 0 Then i = 0, else (τ 1) p 1 α L is in the image of σ 1. Write We have (τ 1) j 1 x = x A + x B for x A A and x B B. Then (τ 1)x A + (τ 1)x B = (τ 1) j x A with (τ 1)x A A and (τ 1)x B B, so (τ 1)x B = 0 and thus (τ 1)x A = (τ 1) j x, which has δ 1 0. Thus we are in case 2 of the proof of Lemma 2.13, so (σ 1) p 1 x A is independent from (τ 1)x A. subcase b: δ 1 = 0 Then (σ 1) i (τ 1) j x = δ 2 (σ 1) p 1 (τ 1) p 2 α L with δ 2 0. Write δ 2 (τ 1) p 2 α L = x A + x B for x A A and x B B. Then (σ 1) p 1 x A + (σ 1) p 1 x B A so (σ 1) p 1 x A = δ 2 (σ 1) p 1 (τ 1) p 2 α L 0 as above. Thus (τ 1)x A is independent from (σ 1) p 1 x A by Lemma In either subcase dim Fp A G 2. 3 dim Fp X G = dim Fp (A B) G = dim Fp A G B G = dim Fp A G + dim Fp B G by Corollary 2.12 and Lemma 2.6. Case 1 gives that dim Fp A G = 0, which implies A = 0 by Lemma 2.5, so that A B is trivial. If neither A nor B is trivial then they both fall under case 2, so dim Fp A G, dim Fp B G 2, a contradiction.

18 Chapter 3 Klein 4-group case This chapter will present the decomposition of K + / (K) as an F 2 [Gal(K/F )]-module when Gal(K/F ) = Z/2Z Z/2Z. This case is slightly different from the general one due to peculiarities of p = 2, but the methods used are nonetheless similar. 3.1 Notation Let K/F be a Galois field extension of characteristic 2 with Gal(K/F ) = Z/2Z Z/2Z = σ, τ. Call the intermediate fields of this extension K 1, K 2, and K 3, and let θ a, θ b K such that K 1 = F (θ a ) = Fix τ K 2 = F (θ b ) = Fix σ K 3 = F (θ a+b = θ a + θ b ) = Fix στ. Then K = F (θ a, θ b ) and the minimum polynomials of θ a and θ b over F have the forms x 2 x a and x 2 x b, respectively, for some a, b F. Thus we can choose the generators σ and τ so that σ(θ a ) = θ a + 1 τ(θ b ) = θ b + 1. For an intermediate field M define the normal subgroup (M) = {m 2 m : m M} of M +, the group of M under addition. We will write cosets as follows: [f] = f + (K) [f] K1 = f + (K 1 ) [f] K2 = f + (K 2 ) [f] K3 = f + (K 3 ) 14

19 CHAPTER 3. KLEIN 4-GROUP CASE 15 In particular let and J = K + / (K) J G = {[k] J : [σ(k)] = [k] = [τ(k)]}. We will indicate the F 2 [Gal(K/F )]-span of [k] by [k] and the F 2 -span by [k] F2. For a subfield M of K let [M] = {[m] J : m M}. For reference, here is a subgroup lattice for Z/2Z Z/2Z with the corresponding fixed fields and trace maps: 1 τ στ σ F (θ a, θ b ) F (θ a ) F (θ a+b ) F (θ b ) K 1 τ 1 K στ 1 K 3 σ 1 K 2 σ, τ F σ 1 F τ 1 Note that σ 1 = σ + 1 and τ 1 = τ + 1 in characteristic 2. We choose to write the former from analogy with the general case. As expected, we also have T K/F = στ + σ + τ = (σ 1)(τ 1) = T K1 /F T K/K1 and similarly T K/F = T K2 /F T K/K2. Where it does not create confusion, we may write T K/K2 for T K1 /F and T K/K1 for T K2 /F. 3.2 The map T Since K/F has 3 intermediate fields, we can classify elements of J by their images under trace maps by combining them into a single function. Restricting the domain to J G allows us to enumerate the possible outputs. Definition 3. Let be given by T : J G K+ 1 (K 1 ) K+ 2 (K 2 ) K+ 3 (K 3 ) T ([k]) = ( [T K/K1 (k)] K1, [T K/K2 (k)] K2, [T K/K3 (k)] K3 ). This T will be fixed for the remainder of Chapter 3, and we will not include class notation when writing images. Note that it is a homomorphism because the T K/Ki are. Lemma 3.1. If k 2 K 2 with [k 2 ] = [0], then there exists l F 2 such that [k 2 ] K2 = [la] K2. Proof. Let U be the subgroup of K + 2 / (K 2 ) generated by [a] K2. We will first show that K = K 2 (θ k2 : [k 2 ] K2 U) where θ k2 is a root of the polynomial x 2 x k 2. Since [a] K2 U, we have K = K 2 (θ a ) K 2 (θ k2 : [k 2 ] K2 U).

20 CHAPTER 3. KLEIN 4-GROUP CASE 16 Suppose [k 2 ] U. By definition of U there exists l F 2 such that [k 2 ] K2 = [la] K2. We have (lθ a ) 2 lθ a = l(θ 2 a θ a ) = la so there exists z F 2 such that θ k2 = θ a + z. Thus θ k2 K(θ a ), so Hence by Theorem 1.2 K 2 (θ k2 : [k 2 ] K2 U) K(θ a ) = K. U = {[k 2 ] K2 : k 2 K 2 (K)} so k 2 K 2 and [k 2 ] = 0 implies that there exists l F 2 such that [k 2 ] = [la]. Similarly k 1 K 1 (K) implies [k 1 ] K1 = [lb] K1 for some l F 2, and k 3 K 3 implies [k 3 ] K3 = [la] K3 = [lb] K3 for some l F 2. (One can see that [a] K3 = [b] K3 because [a + b] K3 = [0] K3.) For [k] J G we have so all possibilities for T ([k]) are listed below: [T K/K1 (k)] = [T K/K2 (k)] = [T K/K3 (k)] = [0] (0, 0, 0) (0, 0, a) (b, 0, 0) (b, 0, a) (0, a, 0) (0, a, a) (b, a, 0) (b, a, a) The kernel of T is precisely the classes in J with at least one representative in the ground field F, which will be extremely useful when constructing our decomposition. The type of extension problem involved in the proof is described more thoroughly in Section 4.2. Theorem 3.2. ker T = [F ]. Proof. Since F is fixed by σ and τ we have [F ] ker T. Let [γ] ker T, let θ γ be a root of the polynomial x 2 x γ, let L = K(θ γ ), and let µ be a generator for Gal(L/K) as a subgroup of Gal(L/F ). Then Gal(L/F ) has order 8 and Gal(K/F ) = Z/2Z Z/2Z as a quotient, so Gal(L/F ) is isomorphic to exactly one of the following: Z/4Z Z/2Z Z/2Z Z/2Z Z/2Z D 8 Q 8 (where D 8 and Q 8 are the dihedral group on 8 elements and the quaternions, respectively). Denote by σ and τ lifts of σ and τ in Gal(L/F ), respectively. Since [T K/K1 (γ)] K1 = [0] K1 we have k1 2 k 1 = T K/K1 (γ) = ( τ 1)γ for some k 1 K 1. We also have (( τ 1)θ γ ) 2 ( τ 1)θ γ = ( τ 1)(θ 2 γ θ γ ) = ( τ 1)γ so k 1 and ( τ 1)θ γ are roots of the same Artin-Schreier polynomial. Thus ( τ 1)θ γ = k 1 + z for some z F 2, so ( τ 2 1)θ γ = ( τ + 1)( τ 1)θ γ = ( τ + 1)(k 1 + z) = k 1 + z + k 1 + z = 0.

21 CHAPTER 3. KLEIN 4-GROUP CASE 17 Since τ 2 fixes θ a and θ b, this gives τ 2 = 1. Similarly σ 2 = 1 because [T K/K2 (γ)] K2 = [0] K2. If Gal(L/F ) = Z/4Z Z/2Z then µ, σ, and τ are distinct and all have order 2, so they are equal to (0, 1), (2, 0), and (2, 1) in some order. Only µ = (2, 0) gives the quotient Z/2Z Z/2Z. However (0, 1) = (2, 1) modulo (2, 2). Therefore Gal(L/F ) = Z/4Z Z/2Z. There is only 1 element of order 2 in Q 8, namely 1, so Gal(L/F ) = Q8. Since [T K/K3 (γ)] K3 = [0] K3 there exists k 3 K 3 such that k 2 3 k 3 = ( σ τ 1)γ. Then there exists z F 2 such that ( σ τ 1)θ γ = k 3 + z as before. Since σ and τ have order 2, they are their own inverses. Thus ([ σ, τ] 1)θ γ = ( σ τ σ 1 τ 1 1)θ γ = (( σ τ) 2 1)θ γ = ( σ τ + 1)( σ τ 1)θ γ = ( σ τ + 1)(k 3 + z ) = k 3 + z + k 3 + z = 0 so [ σ, τ] = 1 because it fixes θ a and θ b. If Gal(L/F ) = D 8 then µ generates the center, the only normal subgroup of order 2. However there are no elements of order 2 in D 8 that commute with each other and are distinct modulo the center. Therefore Gal(L/F ) = D8, so Gal(L/F ) = Z/2Z Z/2Z Z/2Z. By Theorem 1.2 there exist γ 1, γ 2, γ 3 F such that L = F (θ γ1, θ γ2, θ γ3 ). Since F (θ a ) and F (θ b ) are distinct subfields of L, we can take γ 1 = a and γ 2 = b. This gives K(θ γ ) = L = K(θ γ3 ), so [γ] is an element of the subgroup generated by [γ 3 ] in K + / (K). Since [γ 3 ] [F ], we must have [γ] [F ]. 3.3 Construction of Y Since [F ] is fixed by Gal(K/F ), it is a vector space over F 2, so linear algebra guarantees the existence of a set of elements spanning it. Since T K/F is surjective we can take preimages of these elements in order to construct a larger submodule whose components are free F 2 [Gal(K/F )]-modules with fixed parts in [F ]. Definition 4. Let I be a basis for [F ] as an F 2 vector space, and for each [y] I choose k l K such that T K/F (k y ) = y. (Such a k y exists by Corollary 2.4.) Define Y to be the F 2 [Gal(K/F )]-span of the classes of these elements in J: Y = [y] I [k y ] Theorem 3.3. If [y 1 ], [y 2 ] I and [y 1 ] [y 2 ], then [k y1 ] [k y2 ] = {[0]}. Proof. We will show that the fixed part of [k y1 ] is the F 2 -span of [y 1 ]. Since [y 1 ] [F ] [k y1 ] we have [y 1 ] F2 [k y1 ] G. Conversely, let [y] [k y1 ] G. By Lemma 2.7 y = (α 1 + α 2 (σ 1) + α 3 (τ 1) + α 4 (σ 1)(τ 1))k y1

22 CHAPTER 3. KLEIN 4-GROUP CASE 18 for some α 1,..., α 4 F 2 and by hypothesis and [0] = (σ 1)[y] = (α 1 (σ 1) + α 3 (σ 1)(τ 1))[k y1 ] = α 1 T K/K2 [k y1 ] + α 3 [y 1 ] [0] = (τ 1)[y] = (α 1 (τ 1) + α 2 (σ 1)(τ 1))[k y1 ] = α 1 T K/K1 [k y1 ] + α 2 [y 1 ] where α 2 [y 1 ], α 3 [y 1 ] [F ]. However T K/K1 [k y1 ] or T K/K2 [k y1 ] in [F ] would imply [y 1 ] = T K/F [k y1 ] = T K/K1 T K/K2 [k y1 ] = [0], which is not true by linear independence of I. Therefore α 1 = α 2 = α 3 = 0, so [y] = α 4 (σ 1)(τ 1)[k y1 ] = α 4 [y 1 ] [y 1 ] F2 and thus [k y1 ] G = [y 1 ] F2 and [k y2 ] G = [y 2 ] F2. We also have [y 1 ] F2 [y 2 ] F2 = {[0]} by linear independence, so [k y1 ] [k y2 ] = {[0]} by Lemma 2.5. Therefore Y = [y] I [k y ]. This module will be fixed for the remainder of Chapter 3. Corollary 3.4. Y G = [F ]. Proof. This follows from the fact that [k y ] G = [y] F2 for each [y] I where I spans [F ]. 3.4 Construction of X According to Corollary 3.4 and Theorem 3.2, our decomposition so far only covers the kernel of T. To have any hope of spanning J G we must at least include one class [x] J G with T ([x]) (0, 0, 0), and the largest submodule would arise if [x] were in the image of both T K/K1 and T K/K2. (For our purposes this is equivalent to [x] [K 1 ] [K 2 ], but we state the following theorem this way to correspond to the first case of Proposition 7 in [1].) Theorem 3.5. T ([T K/K1 (K)] [T K/K2 (K)] J G ) {(0, 0, 0)} Proof. Consider bθ a + aθ b K. Since T ([bθ a + aθ b ]) = (a, b, a + b) = (0, 0, 0) there exists c F such that by Theorem 3.2 and thus [bθ a + aθ b ] = [c] [c + bθ a ] = [aθ b ].

23 CHAPTER 3. KLEIN 4-GROUP CASE 19 Call this equivalence class [x]. We have so [x] J G. Furthermore by Corollary 2.4. Therefore with as desired. σ([x]) = [σ(aθ b )] = [aθ b ] = [x] τ([x]) = [τ(c + bθ a )] = [c + bθ a ] = [x] [x] = [c + bθ a ] [K 1 ] = [T K/K1 (K)] [x] = [aθ b ] [K 2 ] = [T K/K2 (K)] [x] [T K/K1 (K)] [T K/K2 (K)] J G T ([x]) = (a, b, a) = (0, 0, a) (0, 0, 0) Now we use the fact that [x] is in the image of T K/K1 and T K/K2 to construct a submodule containing it, to be fixed for the remainder of Chapter 3. By taking explicit preimages, we obtain the 2 generators at the top of the following diagram: [aθ a ] τ 1 [aθ a θ b ] σ 1 [aθ b ] = [c + bθ a ] τ 1 [cθ b + bθ a θ b ] σ 1 [bθ b ] T T T (0, a, a) (0, 0, a = b) (b, 0, b) Note that the location of c on the right hand side of this diagram and not the left is arbitrary. Definition 5. Let c F be as given in the proof of Theorem 3.5, and define X to be the F 2 [Gal(K/F )]-span of the classes of aθ a θ b and cθ b + bθ a θ b in J: X = [aθ a θ b ], [cθ b + bθ a θ b ] This X will be fixed for the remainder of Chapter 3. We now show that X is the module described in Section 2.2, from which some properties immediately follow. Theorem 3.6. The classes [aθ a ], [aθ b ] = [c + bθ a ], and [bθ b ] are F 2 -linearly independent. Proof. Suppose δ 1 [aθ a ] + δ 2 [aθ b ] + δ 3 [bθ b ] = [0] for some δ 1, δ 2, δ 3 F 2. Then so δ 1 = δ 2 = δ 3 = 0. (0, 0, 0) = T ([0]) = T (δ 1 [aθ a ] + δ 2 [aθ b ] + δ 3 [bθ b ]) = δ 1 T ([aθ a ]) + δ 2 T ([aθ b ]) + δ 3 T ([bθ b ]) = δ 1 (0, a, a) + δ 2 (0, 0, a) + δ 3 (b, 0, a)

24 CHAPTER 3. KLEIN 4-GROUP CASE 20 Corollary 3.7. The submodule X is indecomposable. Proof. Take α L = [aθ a θ b ] and α R = [cθ b + bθ a θ b ] in Theorem Corollary 3.8. X G = [aθ a ], [aθ b ], [bθ b ] F2. Proof. Take α L = [aθ a θ b ] and α R = [cθ b + bθ a θ b ] in Corollary Structure of J Theorem 3.9. If X = [aθ a θ b ], [cθ b + bθ a θ b ] and Y = [y] I [k y] as defined above, then Lemma J G = X G + Y G J = X Y. Proof. Clearly X G + Y G J G. Suppose [γ] J G and let l 1, l 2, l 3 F 2 such that T ([γ]) = (l 1 b, l 2 a, l 3 a), which is possible by Lemma 3.1. Then T ([γ + l 2 aθ a + (l 3 + l 2 + l 1 )aθ b + l 1 bθ b ]) = T ([γ]) + l 2 T ([aθ a ]) + (l 3 + l 2 + l 1 )T ([aθ b ]) + l 1 T ([bθ b ]) = (l 1 b, l 2 a, l 3 a) + l 2 (0, a, a) + (l 3 + l 2 + l 1 )(0, 0, a) + l 1 (b, 0, a) = (0, 0, 0) so [γ + l 2 aθ a + (l 3 + l 2 + l 1 )aθ b + l 1 bθ b ] [F ] = Y G by Theorem 3.2 and Corollary 3.4. Since l 2 aθ a + (l 3 + l 2 + l 1 )aθ b + l 1 bθ b X G by Corollary 3.8 this gives [γ] X G + Y G. If [γ] J, then [γ] generates up to 3 additional classes after application of σ 1 and τ 1: [γ] (τ 1)[γ] (σ 1)[γ] (σ 1)(τ 1)[γ] We will show that [γ] X Y by repeatedly adding terms to [γ] in order to reduce the number of nonzero entries in this diagram. Lemma If (τ 1)[γ] = [0], then T ((σ 1)[γ]) = (0, 0, 0). Proof. Since [(σ 1)γ] [K 2 ] we have [T K/K2 ((σ 1)γ)] K2 = [0] K2. Since (τ 1)[γ] = [0] there are 2 cases for [T K/K1 ((σ 1)γ)] K1 according to Lemma 3.1: case 1: [(τ 1)γ] K1 = [0] K1 [TK/K1((σ 1)γ)] K1 = [(τ 1)(σ 1)γ] K1 = (σ 1)[(τ 1)γ] K1 = (σ 1)[0] K1 = [0] K1

25 CHAPTER 3. KLEIN 4-GROUP CASE 21 case 2: [(τ 1)γ] K1 = [b] K1 [TK/K1((σ 1)γ)]K1 = (σ 1)[b]K1 = [b] K1 [b] K1 = [0] K1 In either case [T K/K1 ((σ 1)γ)] K1 = [0] K1. We will show that [T K/K3 ((σ 1)γ)] K3 = [0] K3 by contrapositive. Suppose [T K/K3 ((σ 1)γ)] K3 [0] K3. Then [0] K3 [T K/K3 ((σ 1)γ)] K3 = [(στ 1)(σ 1)γ] K3 = [(σ 2 τ στ σ + 1)γ] K3 = [(τ στ στ 2 + 1)γ] K3 = [(στ 2 στ τ + 1)γ] K3 = [(στ 1)(τ 1)γ] K3 = [T K/K3 ((τ 1)γ)] K3 so T ((τ 1)[γ]) (0, 0, 0) and hence (τ 1)[γ] 0. T ((σ 1)[γ]) = (0, 0, 0). Therefore (τ 1)[γ] = 0 implies Lemma If [γ] J, then there exists χ X such that T ((τ 1)[γ + χ]) = (0, 0, 0). Proof. We have [T K/K1 ((τ 1)γ)] K1 = [0] K1 case 1: T ((τ 1)[γ]) = (0, 0, 0) Let χ = 0. case 2: T ((τ 1)[γ]) = (0, a, a) because (τ 1)[γ] [K 1 ], so there are 4 cases: Let χ = aθ a θ b. Then T ((τ 1)[γ + χ]) = (0, a, a) + (0, a, a) = (0, 0, 0). case 3: T ((τ 1)[γ]) = (0, 0, a) Let χ = cθ b + bθ a θ b. Then T ((τ 1)[γ + χ]]) = (0, 0, a) + (0, 0, a) = (0, 0, 0). case 4: T ((τ 1)[γ]) = (0, a, 0) Let χ = aθ a θ b + cθ b + bθ a θ b. Then T ((τ 1)[γ + χ]) = (0, a, a) + (0, 0, a) + (0, a, 0) = (0, 0, 0). Proof of Theorem 3.9. By Corollary 3.8 we have X G = [aθ a ], [aθ b ], [bθ b ] F2, so each class in X G has a nontrivial image under T. By Corollary 3.4 we have Y G = [F ], so each class in Y G has a trivial image. Thus X G Y G = {[0]}, so X Y = {[0]} by Lemma 2.5. Now let [γ] J be arbitrary. Then there exists [χ] X such that T ((τ 1)[γ + χ]) = (0, 0, 0) by Lemma Thus by Theorem 3.2 there exists f F such that (τ 1)[γ+χ] = [f] and by construction of Y there exists k f Y such that T K/F (k f ) = f. This gives (τ 1)[γ + χ + (σ 1)k f ] = (τ 1)[γ + χ] + T K/F [k f ] = [f] + [f] = [0]

26 CHAPTER 3. KLEIN 4-GROUP CASE 22 so by Lemma 3.11 we have T ((σ 1)[γ + χ + (σ 1)k f ]) = (0, 0, 0). Thus by Theorem 3.2 there exists f F and k f Y such that which gives and T K/F ([k f ]) = [f ] = (σ 1)[γ + χ + (σ 1)k f ] (τ 1)[γ + χ + (σ 1)k f + (τ 1)k f ] = [f] + [f] + [0] = [0] (σ 1)[γ + χ + (σ 1)k f + (τ 1)k f ] = [f ] + [0] + [f ] = [0]. Hence [γ + χ + (σ 1)k f + (τ 1)k f ] is in J G, so it is in X G + Y G X + Y by Lemma Since [χ+(σ 1)k f +(τ 1)k f ] X +Y and (σ 1)k f, (τ 1)k f Y, we have [γ] X +Y. Therefore J = X Y. When Y is separated into its components [k y ], this is a decomposition of X Y into indecomposable submodules.

27 Chapter 4 Decomposition for p odd This chapter will present the general case, with emphasis on where it differs from Chapter 3. In particular, the construction of X requires significant additional work. 4.1 Notation Let K/F be a Galois field extension of characteristic p > 2 with Again, we have Gal(K/F ) = Z/pZ Z/pZ = σ, τ. K 1 = F (θ a ) = Fix τ K 2 = F (θ b ) = Fix σ J = K + / (K) where θ a and θ b are roots of polynomials x p x a and x p x b, respectively, and (K) = {k p k : k K} [f] = f + (K). Choose σ and τ so that σ(θ a ) = θ a + 1 and τ(θ b ) = θ b + 1. Some subgroups, fixed fields, and trace maps are illustrated below: 1 F (θ a, θ b ) (τ 1) p 1 K (σ 1) p 1 τ σ F (θ a ) F (θ b ) K 1 K 2 σ, τ F (σ 1) p 1 F (τ 1) p 1 (The intermediate field K 3 is omitted because there are now p+1 distinct proper intermediate fields.) Note that (τ 1) p 1 = 1 + τ + + τ p 1 and (σ 1) p 1 = 1 + σ + + σ p 1 in characteristic p. Again, the compositions T K/F = T K1 /F T K/K1 = T K2 /F T K/K2 hold. 23

28 CHAPTER 4. DECOMPOSITION FOR P ODD Supertrace For p = 2, our decomposition relied on classifying elements by their images under the map T, which we affectionately call the supertrace. In order to generalize this approach, we need a process by which to identify classes in J representable by elements in [F ], preferably one whose complexity does not increase with the number of intermediate fields. If L/K is a Galois field extension, then by the Fundamental Theorem of Galois Theory Gal(K/F ) = Gal(L/F )/ Gal(L/K) which can be represented by the short exact sequence 1 Gal(L/K) Gal(L/F ) Gal(K/F ) 1 where the map Gal(L/K) Gal(L/F ) is inclusion and map Gal(L/F ) Gal(K/F ) takes an automorphism σ : L L to its restriction σ : K K (an automorphism by normality). For [γ] J G, take L = K(θ γ ) = F (θ a, θ b, θ γ ) and denote lifts of generators σ and τ in Gal(K/F ) by σ and τ in Gal(L/F ), respectively. This notation will be used throughout the section. We will propose a sequence of conditions on γ in terms of trace maps that will limit the possibilities for Gal(L/F ) in the short exact sequence and allow us to conclude that [γ] [F ]. Lemma 4.1. If k 2 K 2 with [k 2 ] = [0], then there exists l F p such that [k 2 ] K2 = [la] K2. Proof. Same as for Lemma 3.1. Theorem 4.2. If [T K/K1 (γ)] K1 = [0] K1, then there exists k 1 K 1 such that [γ] = [k 1 ]. Similarly, if [T K/K2 (γ)] K2 = [0] K2, then there exists k 2 K 2 such that [γ] = [k 2 ]. Proof. In Proposition 6.2 of [2]. Lemma 4.3. We have τ p = 1 if and only if [T K/K1 (γ)] K1 = [0] K1 and σ p = 1 if and only if [T K/K2 (γ)] K2 = [0] K2. Proof. Since [γ] J G, we have [T K/K1 (γ)] = [0]. Thus [T K/K1 (γ)] K1 = [lb] K1 by Lemma 4.1, so for some l F p for some k 1 K 1. Note that (1 + τ + + τ p 1 )γ = T K/K1 (γ) = lb + k p 1 k 1 (lθ b + k 1 ) p (lθ b + k 1 ) = l(θ p b θ b) + k p 1 k 1 = lb + k p 1 k 1 = (1 + τ + + τ p 1 )γ

29 CHAPTER 4. DECOMPOSITION FOR P ODD 25 and ((1 + τ + + τ p 1 )θ γ ) p (1 + τ + + τ p 1 )θ γ = (1 + τ + + τ p 1 )(θγ p θ γ ) = (1 + τ + + τ p 1 )γ ie. lθ b + k 1 and (1 + τ + + τ p 1 )θ γ are roots of the same Artin-Schreier polynomial, so (1 + τ + + τ p 1 )θ γ = lθ b + k 1 + z for some z F p. Therefore ( τ p 1)θ γ = ( τ 1)(1 + τ + + τ p 1 )θ γ = ( τ 1)(lθ b + k 1 + z) = τ(lθ b ) + τ(k 1 ) + τ(z) lθ b k 1 z = lτ(θ b ) + k 1 + z lθ b k 1 z = l(θ b + 1) lθ b = l. We also have ( τ p 1)θ a = τ p (θ a ) θ a = θ a θ a = 0 and ( τ p 1)θ b = τ p (θ b ) θ b = θ b + p θ b = 0 so τ p 1 = 0 if and only l = 0. Therefore τ p = 1 if and only if [T K/K1 (γ)] K1 proof for σ is analagous. = [0] K1. The Lemma 4.4. Suppose there exists γ 2 K 2 such that [γ 2 ] = [γ]. Then (τ 1)[γ 2 ] K2 if and only if [ σ, τ] = 1. = [0] K2 Proof. It is sufficient to consider L = K(θ γ2 ) because K(θ γ ) = K(θ γ2 ). Since [γ 2 ] = [γ] J G we have (τ 1)[γ 2 ] = [0], so (τ 1)[γ 2 ] K2 = [la] K2 for some l F p by Lemma 4.1. Choose k 2 K 2 such that τ(γ 2 ) = γ 2 + la + k p 2 k 2. Note that because γ 2 K 2, so ( σ(θ γ2 )) p σ(θ γ2 ) = σ(θ p γ 2 θ γ2 ) = σ(γ 2 ) = σ(γ 2 ) = γ 2 σ(θ γ2 ) = θ γ2 + z 1 for some z 1 F p. Applying σ 1 to this equation gives σ 1 (θ γ2 ) = θ γ2 z 1. We also have ( τ(θ γ2 )) p τ(θ γ2 ) = τ(θ p γ 2 θ γ2 ) = τ(γ 2 ) = τ(γ 2 )

30 CHAPTER 4. DECOMPOSITION FOR P ODD 26 and (θ γ2 + lθ a + k 2 ) p (θ γ2 + lθ a + k 2 ) = θ p γ 2 θ γ2 + l(θ p a θ a ) + k p 2 k 2 = γ 2 + k p 2 k 2 = τ(γ 2 ) ie. τ(θ γ2 ) and θ γ2 + lθ a + k 2 are roots of the same Artin-Schreier polynomial, so τ(θ γ2 ) = θ γ2 + lθ a + k 2 + z 2 for some z 2 F p. Applying τ 1 to this equation gives Therefore [ σ, τ]θ γ2 = σ τ σ 1 τ 1 (θ γ2 ) τ 1 (θ γ2 ) = θ γ2 lθ a τ 1 (k 2 ) z 2. = σ τ σ 1 (θ γ2 lθ a τ 1 (k 2 ) z 2 ) = σ τ(θ γ2 z 1 l(θ a 1) τ 1 (k 2 ) z 2 ) = σ(θ γ2 + lθ a + k 2 + z 2 z 1 l(θ a 1) τ( τ 1 (k 2 )) z 2 ) = σ(θ γ2 + l z 1 ) = θ γ2 + l + z 1 z 1 = θ γ2 + l. Since θ a and θ b are fixed by τ and σ, respectively, and [ σ, τ]θ a = θ a [ σ, τ]θ b = θ b. Thus [ σ, τ] = 1 if and only if [ σ, τ]θ γ2 = θ γ2, which occurs precisely when l = 0, so [ σ, τ] = 1 if and only if [(τ 1)γ 2 ] K2 = [0] K2. Note that if the second of the trace conditions in Lemma 4.3 is satisfied then Theorem 4.2 says that we can indeed find γ 2 K 2 such that [γ 2 ] = [γ] and Lemma 4.1 says that (τ 1)[γ 2 ] K2 = [la] K2 for some l F p. Analagously, if [T K/K1 (γ)] K1 = [0] K1, then there exists γ 1 K 1 such that [γ 1 ] = [γ] and we have [ σ, τ] = 1 if and only if [(σ 1)γ 1 ] K1 = [0] K1. This gives the following corollary: Corollary 4.5. Let γ J G. If there exist γ 1 K 1 and γ 2 K 2 such that [γ] = [γ 1 ] = [γ 2 ], then (σ 1)[γ 1 ] K1 = [0] K1 if and only if (τ 1)[γ 2 ] K2 = [0] K2. Theorem 4.6. Suppose γ J G with [T K/K2 (γ)] K2 = [0] K2 [γ] = [γ 2 ] = [γ 2]. Then (τ 1)[γ 2 ] K2 = (τ 1)[γ 2] K2. and suppose γ 2, γ 2 K 2 such that

31 CHAPTER 4. DECOMPOSITION FOR P ODD 27 Proof. Let γ 2, γ 2 be as given. Then [γ 2 γ 2] = [0] so [γ 2 γ 2] K2 = [la] K2 Lemma 4.1. Thus for some l F p by (τ 1)[γ 2 ] K2 = (τ 1)[γ 2 + la] K2 = (τ 1)[γ 2] K2 because la is fixed by τ. Theorem 4.7. Let [γ] J G. Then [γ] [F ] if and only if the following conditions hold: [T K/K1 (γ)] K1 = [0] K1 [T K/K2 (γ)] K2 = [0] K2 There exists either γ 1 K 1 such that [γ] = [γ 1 ] and (σ 1)[γ 1 ] K1 = [0] K1 or γ 2 K 2 such that [γ] = [γ 2 ] and (τ 1)[γ 2 ] K2 = [0] K2. Proof. Suppose [γ] J G satisfies these conditions and let L = K(θ γ ). We will show that Gal(L/F ) = Z/pZ Z/pZ Z/pZ. Since Gal(L/F ) = p 3, it is isomorphic to exactly 1 of the following: Z/p 3 Z Z/p 2 Z Z/pZ Z/pZ Z/pZ Z/pZ M p 3 H p 3 (The last 2 are defined below.) By Lemmas 4.3 and 4.4 (and using the same notation), we have σ 2 = τ 2 = [ σ, τ] = 1. Let µ generate Gal(L/K), which has order p, and let µ be its image, which also has order p. If Gal(L/F ) = Z/p 3 Z, then which does not give Gal(K/F ). Gal(L/F ) Gal(L/K) = Z/p3 Z Z/pZ = Z/p 2 Z If Gal(L/F ) = Z/p 2 Z Z/pZ, then Gal(L/K) is either pz/p 2 Z 0 or (lp, 1) for some l F p, which give the following quotients: Z/p 2 Z Z/pZ pz/p 2 Z 0 Z/p 2 Z Z/pZ (lp, 1) = Z/pZ Z/pZ = Z/p 2 Z Only the former agrees with Gal(K/F ), so we can take µ = (p, 0). At least 1 of σ and τ must have nonzero first coordinate in Z/pZ Z/pZ. Without loss of generality assume that σ = (m, n) where m 0 in Z/pZ, ie. m 0 (mod p). Then σ = (m + lp, n) for some l Z. However this gives σ = p 2, a contradiction. Suppose Gal(L/F ) = M p 3, the semidirect product Z/pZ Z/p 2 Z with homomorphism φ: Z/pZ Aut(Z/p 2 Z) given by φ(x)(y) = (xp + 1)y, so that (x, y)(z, w) = (x + z, (zp + 1)y + w).

32 CHAPTER 4. DECOMPOSITION FOR P ODD 28 We will compute the commutator subgroup of this semidirect product. First note that the inverse of (x, y) is ( x, (xp 1)y). Thus [(x, y), (z, w)] = (x, y)(z, w)( x, (xp 1)y)( z, (zp 1)w) = (0, (yz xw)p). Elements of this form generate the subgroup (0, p) in Z/pZ Z/p 2 Z. Since Gal(K/F ) is Abelian we must have (0, p) µ, and since they are the same size they must be equal. This gives Z/pZ Z/p 2 Z = Z/pZ Z/pZ µ as desired. Like in the last case, a generator of Z/pZ Z/pZ must have nonzero second coordinate, say m. Then preimages have second coordinates m + lp where m 0 mod p, contradicting the orders of σ and τ. Finally, suppose Gal(L/F ) = H p 3. Again, we will compute the commutator subgroup. Recall that 1 a b H p 3 = 0 1 c : a, b, c Z/pZ is a group of 3-by-3 matrices under multiplication. We have 1 a b 1 d e 0 1 c 0 1 f a b 1 d e 1 a ac b 1 d df e = 0 1 c 0 1 f 0 1 c 0 1 f af cd = which generates a subgroup of size p and a quotient isomorphic to 1 a c : a, c Z/pZ = Z/pZ Z/pZ with generators and By the computation above we have = which contradicts the fact that [ σ, τ] = 1 for all lifts of generators σ and τ.

33 CHAPTER 4. DECOMPOSITION FOR P ODD 29 Therefore Gal(L/F ) = Z/pZ Z/pZ Z/pZ, so by the same argument as in the proof of Theorem 3.2 we have [γ] [F ]. Conversely, if [γ] [F ], then there exists γ F such that [γ] = [γ ], so [T K/K1 (γ)] K1 = [T K/K1 (γ )] K1 = [0] K1 [T K/K2 (γ)] K2 = [T K/K2 (γ )] K2 = [0] K2 (σ 1)[γ ] K1 = [0] K1 for γ K 1 because γ is fixed by σ and τ. Corollary 4.5 and Theorem 4.6 show that the two parts of the third condition in Theorem 4.7 are equivalent, and that the result of the computation does not depend on the choice of representative. 4.3 Construction of Y The submodule Y is entirely analagous to that in the Klein 4-group case. Definition 6. Let I be a basis for [F ] as an F p vector space, and for each [y] I choose k y K such that T K/F (k y ) = y. (Such a k y exists by Corollary 2.4.) Define Y to be the F p [Gal(K/F )]-span of the classes of these elements in J: Y = [y] I [k y ] Theorem 4.8. If [y 1 ], [y 2 ] I and [y 1 ] [y 2 ], then [k y1 ] [k y2 ] = {[0]}. Proof. We will show that [k y1 ] G = [y 1 ] Fp. Since [y 1 ] [F ] [k y1 ] we have [y 1 ] F2 [k y1 ] G. Let [y] [k y1 ] G. By Lemma 2.7 p 1 p 1 [y] = α ij (σ 1) i (τ 1) j [k y1 ] i=0 j=0 for some α ij F p. Suppose for contradiction that there exist i, j not both p 1 such that α ij 0. Let m be the maximum index such that α mj 0 for some j and let n be the maximum index such that α mn 0. Then at least one of p m 1 and p n 1 is greater than 0, so [0] = (σ 1) p m 1 (τ 1) p n 1 [y] = m n α ij (σ 1) p+i m 1 (τ 1) p+j n 1 [k y1 ] i=0 j=0 by hypothesis, and by choice of m and n, the only nonzero coefficient is α mn. This gives [0] = α mn (σ 1) p 1 (τ 1) p 1 [k y1 ] = α mn T K/F [k y1 ] = α mn [y 1 ] [0]

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