SUMMARY NUMBER THEORY II, SOMMERSEMESTER 2015

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1 SUMMARY NUMBER THEORY II, SOMMERSEMESTER 2015 LARS KINDLER, FREIE UNIVERSITÄT BERLIN Contents 0. Introduction and overview 2 1. Rings of integers Definitions and first properties Interlude: Separability, trace, norm, discriminant Finiteness of the integral closure in a finite separable extension The discriminant of a number field 6 2. Basic properties of Dedekind domains Fractional ideals, factorization and the class group Factorization of prime ideals in finite extensions of Dedekind domains The norm of an ideal Geometry of numbers Minkowski theory Lattices Minkowski s theorem on lattice points The Minkowski bound and finiteness of the class number Estimates for the discriminant and Hermite s theorem Integral units Cyclotomic fields Dirichlet s unit theorem Absolute values and completions Basics Topology Classification of absolute values on Q Completion Complete non-archimedean fields Complete archimedean fields Ostrowski s theorem Hensel s Lemma Extensions of valuations the complete case Unramified extensions of complete discretely valued fields Totally ramified extensions of complete discretely valued fields Absolute values on number fields and their completions 30 Comments and questions welcome at kindler@math.fu-berlin.de. Compiled July 13, 2015, 9:41. For a current version of this summary, click here. 1

2 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Introduction and overview Using the example of the ring of Gaussian integers, we exhibit the properties of number fields which will be discussed in the course of the lecture. Definition 0.1. A number field is a finite field extension K of the field of rational numbers Q. Example 0.2. K = Q(i) := Q[X]/(X 2 + 1). Here i is an element such that i 2 = 1. We define Z[i] = {x + iy x, y Z} Q(i); this is the ring of Gaussian integers. Proposition 0.3. If K = Q(i) and O K := { α K a 0, a 1 Z s.t. α 2 + a 1 α + a 0 = 0 }, then Z[i] = O K as subrings of K. The definition of O K will be generalized later to the notion of ring of integers of a number field. A major part of this course will deal with understanding the ring theoretic properties of these rings of integers. In the example of the ring of Gaussian integers, we have the following. Proposition 0.4. Z[i] is an euclidean domain, in particular a principal ideal domain and then also a unique factorization domain. In general, the ring of integers O K of a number field K is not a factorial domain. But one of the main results will show that every nonzero ideal in O K admits a unique factorization as a product of prime ideals. We will see that there is an invariant h K N such that O K is a factorial domain if and only if h K = 1. The question whether O K is factorial or not is interesting also from a point of view of classical number theory, as the following example shows. Example 0.5. The fact that Z[i] is factorial implies that a prime number p N is a sum of squares if and only if p 1 mod 4. In the course of the proof of this example, we naturally encountered a question which will be very important in this course: If K is a number field with ring of integers O K, what happens to a prime number p N considered as an element of O K? More precisely, what is the factorization of the ideal po K generated by p into prime ideals of O K? Since we know that Z[i] is factorial, for Z[i] this question is related to finding the prime elements of Z[i]; Proposition 0.6. Up to multiplication with units, the prime elements of Z[i] are (a) 1 + i. (b) a + bi, with a, b Z such that a 2 + b 2 = p with p prime and p 1 mod 4 (c) p, with p N prime and p 3 mod 4. We will also study the group of units O K of the ring O K. This group is easy to understand if K = Z[i]. Proposition 0.7. Z[i] = {±1, ±i} with the obvious multiplication.

3 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Rings of integers 1.1. Definitions and first properties. All rings will be commutative with 1. All ring homomorphisms map 1 to 1. In this chapter, we will mostly recall or develop results from Commutative Algebra which are not specific to a number theoretic situation. Definition 1.1 (Integral ring extensions). Let A be a commutative ring. (a) An A-algebra is a commutative ring B together with a homomorphism of rings ϕ : A B. This allows us to multiply elements of A and B. We will usually omit ϕ from the notation; if a A we will also write a for its image in B. If ϕ is injective, we also identify A with a subring of B and say that A B is an extension of rings. (b) If B is an A-algebra, an element b B is called integral over A if there exists a monic polynomial f(t ) A[T ], such that f(b) = 0. Here, monic means that the leading coefficient of f(t ) is 1. (c) If every b B is integral over A, then we say B is integral over A, or B is an integral extension of A. Example 1.2. If d Z, then Z Q( d), so Q( d) is a Z-algebra and the element d is integral over Z. On the other hand, 1 d is not integral over Z. Proposition 1.3. Let A B be an extension of rings. Let b 1,..., b n B. The following statements are equivalent. (a) b 1,..., b n are integral over A. (b) The A-submodule A[b 1,..., b n ] B of B is finitely generated. (c) Every element b A[b 1,..., b n ] B is integral over A. Lemma 1.4. Let M := (a ij ) 1 i,j n be a square matrix with entries in the ring A. Define the adjoint matrix M := (a ij ) 1 i,j n by a ij := ( 1) i+j det M ij, where M ij is the matrix obtained from M by deleting the i-th column and the j-th row. Then MM = det(m)i n, where I n is the n n-identity matrix. Corollary 1.5. Let A B C be extensions of commutative rings. If B is integral over A and if C is integral over B, then C is integral over A. Definition 1.6 (Integral closure). Let A B be an extension of rings and write A := {b B b is integral over A}. Then A is called the integral closure of A in B. Clearly A A. If A = A then A is said to be integrally closed in B. Proposition 1.7. With the notations from Definition 1.6, A is a subring of B and integrally closed in B. Definition 1.8. If A is an integral domain then A is called integrally closed or normal if A is integrall closed in its field of fractions Frac(A) = { a a a 0}. Example 1.9. Z is integrally closed. More generally, a unique factorization domain is integrally closed.

4 SUMMARY NUMBER THEORY II, SOMMERSEMESTER We now consider the following situation: Let A be a n integrally closed domain with field of fractions K. Let L/K be a finite field extension and B the integral closure of A in L. Definition 1.10 (Ring of integers of a number field). If in the above situation, A = Z, and L/Q a finite extension, then B is written O L and called the ring of integers of L. One of our first main results will be that if L/K is a separable extension and A noetherian, then B is finitely generated as an A-module. This is not true in general. Note however, that if char K = 0, then every finite extension of K is separable. We have to develop some more field theory before we can prove this. Proposition Let A be an integrally closed domain, K = Frac(A) its field of ractions, L/K a finite extension and B the integral closure of A in L. (a) Every element x L can be written as x = b, b B, a A. a In particular, L = Frac(B). (b) An element x L lies in B if and only if its monic minimal polynomial p x (T ) K[T ] lies in A[T ]. (c) If M/L is a finite extension and if C B :=integral closure of B in M, C A :=integral closure of A in M, then C B = C A. April Interlude: Separability, trace, norm, discriminant. Reminder. (a) If L/K is a finite algebraic extension of fields, then an element x L is called separable over K, if its minimal polynomial p x (T ) K[T ] has only simple roots over an algebraic closure of L. Equivalently, (p x (T ), p x(t )) = K[T ]. (b) The extension L/K is called separable, if every element of L is separable over K. (c) If L/K is finite and separable, then there exists θ L such that L = K(θ) = K[θ], i.e. if n = [L : K] then that 1, θ,..., θ n 1 is a basis of L over K. (d) If L/K is finite and separable, fix an embedding ι : K K into an algebraic closure of K. Then there exist precisely [L : K] embeddings γ : L K such that j K = ι. Definition Let A be a ring and B an A-algebra, free of finite rank as an A-module. For example A = K and B = L, L/K a finite extension of fields. For b B, write mult b : B B for the A-linear map x bx. (a) Tr B/A (b) := Tr(mult b ) is called the trace of b. additive map Tr B/A : B A. This induces an

5 SUMMARY NUMBER THEORY II, SOMMERSEMESTER (b) N B/A (b) := det(mult b ) is called the norm of b. This induces a map N B/A : B A restricting to a homomorphism of groups N B/A : B A. The norm and trace map are particularly nice if B/A is a finite separable extension of fields. Proposition Fix an algebraic closure K of K and an embedding K K. Assume that L/K is separable and that σ 1,..., σ n are the different embeddings L K fixing K. Moreover, let a L and f a (T ) := det(t id L mult a ) be the characteristic polynomial of mult a. (a) f a (T ) = n i=1 (T σ i(a)). (b) Tr L/K (a) = n i=1 σ i(a). (c) N L/K (a) = n i=1 σ i(a). Corollary If M/L/K are finite separable extensions, then N M/K = N L/K N L /L and Tr M/K = Tr L/K Tr M/L. Definition 1.15 (Discriminant). Let A be a ring and B an A-algebra, free of rank n as an A-module (e.g. B/A a finite extension of fields). Let α 1,..., α n B. The discriminant of α 1,..., α n is defined to be d(α 1,..., α n ) := det ( Tr B/A (α i α j ) 1 i,j n ). Proposition Let A be a ring and B an A-algebra, free of rank n as an A-module. Let α 1,..., α n B and β i := n j=1 a ijα j, a ij A. Then d(β 1,..., β n ) = det ((a ij ) 1 i,j n ) 2 d(α 1,..., α n ) Proposition Assume that L/K is a finite, separable extension of fields and that α 1,..., α n is a K-basis of L. Let σ 1,..., σ n be the distinct K-embeddings of L into a fixed algebraic closure of K. (a) d(α 1,..., α n ) = det ((σ i α j ) 1 i,j n ) 2. (b) If α 1,..., α n has the shape 1, θ,..., θ n 1 for some θ L, then d(1, θ,..., θ n 1 ) = i<j(σ i (θ) σ j (θ)) 2. Theorem 1.18 (Discriminant of separable extensions). If L/K is a finite separable extension of fields and α 1,..., α n a K-basis for L, then d(α 1,..., α n ) 0 and the bilinear form (x, y) Tr L/K (xy) on the K-vector space L is nondegenerate, i.e. for every x L\{0} there exists y L such that Tr L/K (xy) Finiteness of the integral closure in a finite separable extension. Let A be an integrally closed domain, K = Frac(A) its field of fractions, L/K a finite separable extension and B the integral closure of A in L. In this situation, the norm and trace of L/K map B to A. Proposition If b B then N L/K (b) A and Tr L/K (b) A. Moreover, b B if and only if N L/K (b) A.

6 SUMMARY NUMBER THEORY II, SOMMERSEMESTER We return to the question whether the integral closure B of A is a finitely generated A-module. Theorem Let A be a noetherian integrally closed domain, K = Frac(A) its field of fractions and B the integral closure of A in a finite separable extension L/K. (a) B is a finitely generated A-module. (b) If A is a principal domain, then B is a free A-module of rank [L : K]. (More generally: Every finitely generated B-submodule of L is free of rank [L : K].) Definition 1.21 (Integral bases). In the situation of Theorem 1.20 let n = [L : K]. A set of elements α 1,..., α n B such that B n = α i A i=1 is called integral basis of B. In general, an integral basis does not need to exist, but if A is a principal ideal domain (e.g. Z), then Theorem 1.20 guarantees the existence of an integral basis. In the course of the proof of Theorem 1.20 we use the following nice lemma. Lemma In the situation of Theorem 1.20, if α 1,..., α n B is a K-basis of L and d := d(α 1,..., α n ), then d B α 1 A α n A B The discriminant of a number field. We make the notions from the previous paragraph more concrete in the number theoretic setup. Let L be a number field, O L its ring of integers. As L/Q is finite and separable, and as Z is a principal ideal domain, Theorem 1.20 is a free Z-module of rank [L : Q], so there exists an integral basis α 1,..., α n. Proposition In this situation, the discriminant d(α 1,..., α n ) is independent of the choice of the integral basis. Definition We define d L := d(o L ) := d(α 1,..., α n ) Z to be the discriminant of the number field L. We have the following criterion for a set of elements α 1,..., α n O L to be an integral basis. Proposition If L is a number field and α 1,..., α n O L, then the elements α 1,..., α n for an integral basis if and only if d(α 1,..., α n ) = d L. For future reference we also record the following generalization of the previous facts. Proposition Let L be a number field, O L its ring of integers and M a finitely generated O L -submodule of L. Then M is a free Z-module. If α 1,..., α n is a Z-basis of M, then the integer d(α 1,..., α n ) is independent of the choice of α 1,..., α n and denoted d(m). If M M are two finitely generated O L -submodules of L, then d(m) = (M : M) 2 d(m ).

7 SUMMARY NUMBER THEORY II, SOMMERSEMESTER In particular, the index (M : M) is finite. April Basic properties of Dedekind domains 2.1. Fractional ideals, factorization and the class group. We saw that if K is a number field and O K its ring of integers, i.e. the integral closure of Z in K, then O K is a free Z-module of rank [K : Q]. We want to study the basic ring theoretic properties of O K. Proposition 2.1. The ring O K is a noetherian, integrally closed domain and every nonzero prime ideal p O K is maximal. Definition 2.2 (Dedekind domains). A noetherian integrally closed integral domain A which is not a field, and such that every nonzero prime ideal of A is maximal, is called a Dedekind domain. Example 2.3. Principal ideal domains are Dedekind domains. Rings of integers of a number field are Dedekind domains. Reminder. Let A be a ring. Given a chain of prime ideals p 0 p 1... p n we say that its length is n. Then the Krull dimension of A is defined to be dim A = sup{n there is an ascending chain of prime ideals of length n}. For any nonzero ring A, dim A 0, but there are notherian rings of infinite Krull dimension. If A is a Dedekind domain, then dim A = 1. Proposition 2.4. Let A be a Dedekind domain. (a) If A is local, then A is a principal ideal domain. (b) A is a principal ideal domain if and only if A is a unique factorization domain. There are Dedekind domains which are not unique factorization domains; even among the rings of integers of number fields. Instead we have the following. Theorem 2.5 (Prime factorization in Dedekind domains). Let A be a Dedekind domain and I A a nonzero ideal. Then there is a unique way to write I = p v pvr r with p i A pairwise distinct maximal ideals and v i 1. If you have taken a course on commutative algebra, one way to prove Theorem 2.5 is to use the primary decomposition of an ideal. We will give a direct proof. Lemma 2.6. Let A be a Dedekind domain and I A a nonzero ideal. Then there are nonzero prime ideals p 1,..., p r such that I p 1... p r.

8 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Definition 2.7 (Fractional ideals). Let A be an integral domain, K = Frac A. An A-submodule M K is called fractional ideal of A if there exists some x A \ {0} such that xm := {xm K m M} A. Note that if A is noetherian, then an A-submodule M K is a fractional ideal if and only if M is a finitely generated A-module. Definition 2.8. Let A be an integral domain and I A an ideal. (a) I is a fractional ideal. These ideals will often be called integral ideals to distinguish them from fractional ideals. (b) Define I 1 := {x K xi A}. If A is noetherian, then I 1 Hom A (I, A) is a fractional ideal. (c) If M, N K are fractional ideals, then define { n } M N := x i y i x i M, y i N, n 0. i=1 This is a fractional ideal. Proposition 2.9. Let A be a Dedekind domain and p A a nonzero prime ideal. (a) For all nonzero ideals I A we have (b) pp 1 = A. I Ip 1. This allows us to prove Theorem 2.5. Definition 2.10 (Group of fractional ideals). Let A be a Dedekind domain. We write Id(A) for set of nonzero fractional ideals of A. Corollary Let A be a Dedekind domain. The product of fractional ideals makes Id(A) into an abelian group with neutral element A = (1). In fact, every nonzero fractiona ideal I Id(A) can be written uniquely as I = p v pvr r with p i nonzero prime ideals and v i Z \ {0}. In other words, Id(A) is the free abelian group generated by nonzero prime ideals. Remark We have seen that in a Dedekind domain A, Id(A) carries the structure of an abelian group. The converse is a theorem of Noether: If A is an integral domain such that Id(A) is a group then A is a Dedekind domain. The group Id(A) has an interesting subgroup. Definition 2.13 (Principal fractional ideals). Let A be a Dedekind domain with field of fractions K. (a) For x K write (x) := xa K for the fractional ideal generated by x. We say that (x) is a principal fractional ideal. (b) We write P (A) := {(x) x K } for the set of nonzero principal fractional ideals of A.

9 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Proposition Let A be a Dedekind domain with fraction field K. Then P A Id(A) is a subgroup and there is an exact sequence of abelian groups 1 A K x (x) Id(A) Id(A)/P (A) 1. Definition 2.15 (Class group). Let A be a Dedekind domain. The abelian group Cl(A) := Id(A)/P (A) is called the class group of A. We write h A := Cl(A) N { } for the order of Cl(A) and call h A the class number of A. If K is a number field and A = O K, then we also write Id(K) := Id(O K ), P (K) := P (O K ), Cl(K) := Cl(O K ) and h K := h OK. The group Cl(K) is the class group of K and its order h K is the class number of K. Proposition If A is Dedekind domain, then A is a principal domain if and only if Cl(A) = 1. Thus Cl(A) measures how far A is away from being a principal ideal domain and hence a unique factorization domain (see Proposition 2.4). We will see later on that if K is a number field then h K <, i.e. Cl(K) is a finite group. Example There are Dedekind domains with infinite class group. If d N is a square free positive integer, then it is known that for K = Q( d) we have h K > 1 unless d = 1, 2, 3, 7, 11, 19, 43, 67, 163 (Heegener, 1952). On the other hand, if K = Q( d) it is conjectured (but unknown), that there are infinitely many d with h K = Factorization of prime ideals in finite extensions of Dedekind domains. Let A be a noetherian integral domain with fraction field K. Let L/K be a finite separable field extension. Then Theorem 1.20 shows that the integral closure B of A in L is a finitely generated A-module. Proposition If A is a Dedekind domain with field of fractions K and L/K a finite separable extension. Then B is a Dedekind domain. Remark The assumption that L/K be separable is not necessary, this is the theorem of Krull-Akizuki. Proposition Let A be a Dedekind domain with field of fractions K, L/K a finite separable extension and B the integral closure of A in L. Let p A be a nonzero prime idea. (a) pb B. (b) If P B is a nonzero prime ideal, then P pb if and only if P K = p. In this case, the field extension A/p B/P is finite. (c) The map Id(A) Id(B), I IB, is injective. Definition In the situation of Proposition 2.20, there is a unique decomposition p = P e(p 1/p) 1... P e(pg/p) g. (1) with P i nonzero prime ideals of B and e(p i /p) 1. (a) p is called unramified in L if e(p 1 /p) =... = e(p g /p) = 1. Otherwise, p is said to be ramified in L. (b) The number e(p i /p) is called ramification index of P i over p.

10 SUMMARY NUMBER THEORY II, SOMMERSEMESTER (c) The degree of the extension A/p B/P i is written f(p i /p) and called the residue degree of P i over p. (d) p is called inert if pb is a prime ideal of B, i.e. if g = e(p 1 /p) = 1 (e) p is called (totally) split if e(p i /p) = f(p i /p) = 1 for i = 1,..., g. Example From Proposition 0.6 we see that in Z Z[i] the primes p Z with p 3 mod 4 are inert, p 1 mod 4 split completely and p = 2 is ramified with ramification index 2. The numbers, g, e(p i /p), f(p i /p) from Definition 2.21 satisfy a nice relation. Theorem 2.23 (efg formula). Let A be a Dedekind domain with field of fractions K, L a separable extension of K and B the integral closure of A in L. If p A is a nonzero prime ideal, then the numbers e(p i /p), g and f(p i /p) obtained from the factorization (1) satisfy g [L : K] = e(p i /p)f(p i /p). i=1 Moreover, if L/K is a Galois extension, then and and hence e(p 1 /p) =... = e(p g /p) =: e f(p 1 /p) =... = f(p g /p) =: f [L : K] = efg. The proof of the theorem relies on a general form of the famous Chinese Remainder Theorem. Lemma 2.24 (Chinese Remainder Theorem). Let R be a ring, I 1,..., I n R ideals such that I i + I j = R for all pairs i j, 1 i, j n. Then there is an isomorphism of rings n R/ I i n = R/I i. i=1 Note that if I + J = R then IJ = I J. There is a convenient criterion for when a prime ideal p A is ramified. Proposition Let L/K be a finite extension of number fields and O K, O L the rings of integers. Assume that O L is a free O K -module (e.g. if h K = 1, e.g. K = Q). Then a nonzero prime p O K ramifies in L if and only if p d L/K, where d L/K is the ideal generated by the discriminant of an O K -basis of O L (the discriminants of two O K -bases of O L differ by the square of a unit of O K (Definition 1.24). In particular, if K = Q, then p Z ramifies in L if and only if p d K, where d K Z is the discriminant of the number field K. We need two lemmas. i=1 April 27

11 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Lemma Let A be a ring and B a finite A-algebra, free as an A-module. If {e 1,..., e m } is an A-basis of B, then for all ideals I A, the residue classes {ē 1,..., ē m } form a basis of the free A/I-module B/IB. In particular, d(ē 1,..., ē m ) d(e 1,..., e m ) mod I. Lemma Let k be a perfect field (e.g. char(k) = 0 or #k < ) and B a k-algebra such that dim k B <. Then B is reduced (i.e. b n = 0 implies b = 0) if and only if for any basis e 1,..., e n of the k-vector space B we have d(b 1,..., b n ) The norm of an ideal. Definition 2.28 (Norm of ideals). Let A be a Dedekind domain with field of fractions K, L/K a finite separable extension and B the integral closure of A in L. For a nonzero fractional ideal I Id(B) we can write I = P v Pvr r with v i Z\{0} and P i distinct nonzero prime ideals in B. Write p i := P i K and f i := [B/P i : A/p i ]. The norm of the fractional ideal I is defined to be N L/K (I) := r i=1 p v if i i Id(A). We establish properties analogous to properties of the norm N L/K : L K. Proposition We continue to use the notation from Definition (a) N L/K : Id(B) Id(A) is a homomorphism of groups. (b) If M/L is a finite separable extension then N L/K N M/L = N M/K. (c) If I A is a nonzero ideal then N L/K (IB) = I [L:K]. (d) If L/K is Galois and P B a nonzero prime ideal, then with p := P A we have pb = (P 1... P g ) e and N L/K (P)B = (P 1... P g ) ef = σp. σ Gal(L/K) (e) For x L, we have N L/K (xb) = N L/K (x)a, in other words, the diagram L x (x) Id(B) N L/K K y (y) Id(A) N L/K commutes and hence N L/K induces a homomorphism N L/K : Cl(B) Cl(A). We make this norm more concrete in the number theoretic situation in the case that K = Q.

12 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Definition 2.30 (Absolute norm). Let K be a number field and I O K an ideal. The norm N K/Q (I) is a nonzero ideal in Z, hence a nonzero integer up to sign. We will write N(I) := N K/Q (I) N and call N(I) the absolute norm or numerical norm of I. Proposition Let K be a number field and O K its ring of integers. (a) If I O K is a nonzero ideal, then N(I) = O K /I N, and N(I) 2 = d(i)/d K. (b) If I J are fractional ideals then J/I = N(IJ 1 ), The following simple observation is crucial. Proposition Let K be a number field and µ R >0 arbitrary. There are only finitely many nonzero ideals I A with N(I) < µ. Together with the next proposition, we find a criterion for when the class group Cl(K) is finite. Proposition Let K be a number field and µ R >0. The following statements are equivalent. (a) For all nonzero ideals I A there exists a I \ {0} such that N K/Q (a) < N(I) µ. (b) Every class of fractional ideals in Cl(K) has a representative I A such that N(I) < µ. In the next chapter, we will find for every K a bound µ as in Proposition 2.33, a. and hence prove that h K = # Cl(K) <. May 4 3. Geometry of numbers Minkowski theory We will leave the realm of commutative algebra and prove theorems about number fields using that they can be embedded into the field of complex numbers, which is not true for fraction fields of general Dedekind domains Lattices. Definition 3.1. Let V be a finite dimensional real vector space. A lattice Λ V is a subgroup of the form Λ = e 1 Z +... e r Z for e 1,..., e r V linearly independent elements. If r = dim V, then Λ is said to be a full lattice. Note that in this case the inclusion Λ V induces an isomorphism of R-vector spaces Λ Z R = V. If λ V is a lattice, then λ is a free abelian group. The converse is not true, as the follwoing example shows. Example 3.2. V = R, Λ = Z + 2Z V is a free abelian group as 2 is irrational, but 1, 2 are of course not R-linearly independent. Given a subgroup Λ V, how do we decide whether Λ is a lattice or not?

13 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Reminder. If V is an n-dimensional R-vector space, then V carries a topology: Choosing a basis fixes an isomorphism R n = V and we can use the euclidean topology on R n to define a topology on V. On the other hand, every R-linear automorphism of R n is a homeomorhpism, so the topology defined on V is independent of the isomorphism V = R n. Definition 3.3. A topological space X is called discrete if every subset of X is open. If Y X is a subset, then Y is said to be discrete, if Y equipped with subspace topology induced by X is discrete. This just means that for every y Y there exists an open subset U X such that U Y = {y}. If Λ R n is a subset, then Λ discrete if and only if for every λ Λ there exists ε > 0 such that B ε (λ) Λ = {λ}, where B ε (λ) := {x R n x λ < ε} is the ball of radius ε around λ with respect to some norm. Proposition 3.4. Let V be a finite dimensional R-vector space and Λ V a subgroup. Then Λ is a lattice if and only if Λ is a discrete subspace of V. Definition 3.5. Let V be an n-dimensional R-vector space and Λ V a full lattice Λ = n i=1 f iz. Given λ 0 Λ we define { } n D(λ 0 ) := λ 0 + a i f i a i R, 0 a i < 1. i=1 The set D(λ 0 ) is called a fundamental parallelepiped or more pronouncably foundamental domain for Λ and depends on the choice of the basis f 1,..., f n of Λ, but in a sense to be made precise below, its volume does not depend on the choice of the f 1,..., f n. Note that V is the disjoint union λ 0 Λ D(λ 0). We need to be able to talk about volumes of certain subsets of R n and collect the basic facts from measure theory that we will use. Reminder. Let V be an n-dimensional R-vector space and e 1,..., e n a basis. The basis gives rise to an isomorphism V = R n. We denote by µ the Lebesgue measure on R n. (a) µ ({ n i=1 a ie i a i R, 0 a i < 1}) = 1. (b) If f 1,..., f n is a second basis, then ({ n }) µ a i f i a i R, 0 a i < 1 = det(a) i=1 where A is the matrix a ij, if f j = n i=1 a ije i, i.e. the base change matrix. (c) This shows that the choice of a basis of V defines a measure on V, well-defined up to multiplication by a constant. In particular, given two measurable subsets S 1, S 2 V, the ratio µ(s 1 )/µ(s 2 ) is well-defined. (d) Let Λ := f 1 Z f n Z be a full lattice in n i=1 e ir. For every λ Λ we have µ(d(λ)) = det(a) with A as in the previous point. This volume is independent of the choice of the basis f 1,..., f n of Λ. If f 1,..., f n is a second basis of Λ,

14 SUMMARY NUMBER THEORY II, SOMMERSEMESTER then the associated base change matrix B has integer coefficients, the absolute value of its determinant is 1. Thus µ(d(λ)) = det(ba) = det(a). (e) If Λ Λ are two full lattices in R n, then there exists a basis f 1,..., f n of Λ and m 1,..., m n N such that f 1 := m 1f 1,..., f n := m n f n is a basis of Λ (this follows from the structure theorem for finitely generated abelian groups). Let D(0) be the fundamental domain of Λ based at 0 with respect to the basis f 1,..., f n and D (0) the fundamental domain of Λ based at 0 with respect to the basis f 1,..., f n. It follows that µ(d (0)) = (Λ : Λ )µ(d(0)) and that µ(d (0))/µ(D(0)) is independent of the choice of the bases Minkowski s theorem on lattice points. Proposition 3.6. Let D be a fundamental domain for the full lattice Λ V and S V a measurable subset. If µ(s) > µ(d) then there are α, β S such that 0 α β Λ. Note that the condition µ(s) > µ(d) is independent of any choices of bases according to Reminder 3.1. What additional conditions can we impose on S so that we can deduce that S Λ contains a nonzero point? Example 3.7. Silly example: Λ = Z[i] C and S = C \ Λ. Then the conditions of Proposition 3.6 are satisified, but still Λ S =. Problem here: S not convex! Definition 3.8. A subset S of a real vector space V is called convex if for all s 1, s 2 S we have {s 1 + γ(s 2 s 1 ) γ [0, 1]} S. Example 3.9. Again consider Λ := Z[i] C = R ir and S := [ 1 3, 1 2 ] ir. Clearly, S is convex and its volume is infinite, but still S Λ =. Problem: S is not symmetric around the origin. Definition A subset S of a real vector space V is called symmetric around the origin if for all s S we also have s S. Lemma Let V be an n-dimensional real vector space and S V be symmetric around the origin and convex. If Λ is a full lattice and D a fundamental domain of Λ, then Λ S contains a nonzero element if µ(s) > 2 n µ(d). We can improve the lemma to take into account the boundary of S. Theorem 3.12 (Minkowski Lattice Point Theorem). Let V be real vector space of dimension n and S V a compact, convex subset, symmetric around the origin. Let Λ V be a full lattice and D a fundamental domain for Λ. If µ(s) 2 n µ(d) then S Λ contains a nonzero element. Before we apply this theorem to the study of number fields, we mention a fun application to elementary number theory:

15 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Example 3.13 (Lagrange 1770). Every positive integer is a sum of four squares The Minkowski bound and finiteness of the class number. Definition Let K be a number field, [K : Q] = n. There are n embeddings σ i : K C (which automatically fix Q). (a) If σ i (K) R C, then σ i is called a real embedding. We will always write r for the number of real embeddings K R. (b) If σ i (K) R, then σ i is said to be a complex embedding. If σ i denotes its complex conjugation, then σ i σ i. (c) We write s for the number of pairs of complex embeddings, so that n = r + 2s. (d) We order the embeddings σ i as follows: σ 1,..., σ r are the real embeddings of K and σ r+1,..., σ r+2s are ordered such that σ r+i = σ r+s+i. (e) Using all embeddings we obtain an injective map of rings σ : K R r C s, x (σ 1,..., σ r (x), σ r+1 (x),..., σ r+s (x)). Note that in the definition of σ, we only use one complex embedding from each pair of complex embeddings. (f) We will identify C with R + ir, so that we can interpret σ as an embedding K R n. Here is our main reason for studying Minkowski s Lattice Point Theorem. Proposition Let K be a number field, O K its ring of integers. If I O K is a nonzero ideal then σ(i) R r C s = R n is a full lattice. The volume of a fundamental domain D is µ(d) = 2 s N(I) d K. Theorem 3.16 (Minkowski Bound). Let K be a number field of degree n and I O K a nonzero ideal. Then there exists an element a I \ {0} with N K/Q (a) ( 4 π ) s n! n n N(I) d K. Corollary The class group of a number field is finite. Example Q(i),Q( 5), more? The crucial ingredient in the proof of Theorem 3.16 is the following simple but tedious computation. Lemma For (x, z) R r C s, define r (x, z) := x i + 2 i=1 r+s i=r+1 This is a norm on the real vector space R r C s. For t R write X(t) := {(x, z) R r C s (x, z) t}. If µ is the Lebesgue measure on R r (R ir) s, then ( µ(x(t)) = 2 r π ) s t n 2 n!. z i. May 11

16 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Estimates for the discriminant and Hermite s theorem. If we just wanted to prove the finiteness of the class group of a number field, we could have done this without establishing the Minkowski bound We give a few more important applications, which use the precise bound. Theorem Let K be a number field of degree n > 1. Then there exists a prime number p Z which ramifies in K. In the proof of the theorem we used that the discriminant d K controls the which prime numbers ramify in K. We will see later (hopefully) that to some extent it also controls how badly a prime number ramifies in K. If we use a little bit more analysis, we can say more about how the discriminant depends on the degree of the extension. Reminder. Stirling s formula says that there exists a number θ (0, 1) such that for every n n! = ( n ) n θ 2πn e 12n. e Proposition If K is a number field of degree n, then we saw ( π ) ( ) 2s n n 2 ( π ) ( ) n n n 2 d K 4 n! 4 n! where s is the number of pairs of complex embeddings of K, i.e. n = r + 2s. Stirlings formula implies that if n, then d K. Using Minkowski s lattice point theorem together with the previous proposition, we can prove the following famous theorem of Hermite. Theorem 3.22 (Hermite). Given D N, there are only finitely many number fields K such that d K D. May Integral units Until now, we spent most of our time studying the ideals and fractional ideals of the ring of integers O K of a number field K. The process of passing from elements of O K to ideals forgets information about the units O K. In this chapter we study the structure of the group O K Cyclotomic fields. Definition 4.1. Let K be a number field and n N. An element ζ K is called n-th root of unity if ζ n = 1. If ζ is an n-th root of unity with ζ r 1 for r = 1,..., n 1, then ζ is called primitive. We define µ n (K) := {ζ K n N : ζ n = 1}. This is a subgroup of K. We also write µ(k) := n 1 µ n(k), which is also a subgroup of K. Lemma 4.2. Let K be a field, n N, and ζ µ n (K) a primitive n-th root of unity. (a) For d N, ζ d is a primitive n-th root of unity if and only if (d, n) = 1. (b) µ n (K) is a finite cyclic group of order n.

17 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Remark 4.3. Of course, if K is a field and n N, it is not necessarily true that K contains a primitive n-th root of unity. For example, µ(q) = {±1}. Nonetheless, if K is an algebraic closure of K, then µ n (K) µ n (K) is a subgroup, so µ n (K) is always finite and cyclic. Definition 4.4. Let n N and let ζ C be a primitive n-th root of unity. Q(ζ) C is called the n-th cyclotomic field, and according to Lemma 4.2, if ζ C is a second primitive n-th root of unity, then Q(ζ) = Q(ζ ). The field Q(ζ) is the splitting field of the polynomial T n 1 Z[T ]. Thus Q(ζ)/Q is a Galois extension. If σ Gal(Q(ζ)/Q) then there exists d {0, 1,..., n 1}, (d, n) = 1, such that σ(ζ) = ζ d. This defines an injective homomorphism Gal(Q(ζ)/Q) (Z/nZ), σ d. Recall that the Euler ϕ-function is the map ϕ : N N, defined by ϕ(n) = #{r {1,..., n 1} (r, n) = 1} = #(Z/nZ). The n-th cyclotomic polynomial is defined by Φ n (T ) := (T ζ r ). 0<r<n,(r,n)=1 As it is fixed by Gal(Q(ζ)/Q), it lies in Q[T ]. On the other hand, as ζ is integral over Z, the same is true for the coefficients of Φ n (T ), so Φ n (T ) Z[T ]. It is monic of degree ϕ(n). As Φ n (ζ) = 0, it follows that [Q(ζ) : Q] ϕ(n). In fact Φ n (T ) is irreducible, so [Q(ζ) : Q] = ϕ(n) and Gal(Q(ζ)/Q) = (Z/nZ). We first check this for n = p r, p prime. Proposition 4.5. Let r N, and let ζ C be a primitive p r -th root of unity. Write K := Q(ζ). (a) [K : Q] = ϕ(p r ) = p r 1 (p 1). (b) Write π := 1 ζ. The ideal πo K is prime and po K = π ϕ(pr) O K. (c) discr(k/q) = ±p c for some c > 0 with sign ( 1) ϕ(pr )/2 unless p r = 2. In this case K = Q(i), discr(q(i)/q) = 4. (d) O K = Z[ζ]. If n N is not a prime power, we use induction on the number of prime divisors of n to prove: Proposition 4.6. Let ζ C be a primitive n-th root of unity. (a) [Q(ζ) : Q] = ϕ(n). (b) O K = Z[ζ]. (c) If p Z is a prime number which ramifies in Q(ζ), then p n. Moreover, if n = p r m with (p, m) = 1, then there are distinct prime ideals p 1,..., p t Z[ζ], t m, such that pz[ζ] = (p 1... p t ) ϕ(pr) Dirichlet s unit theorem. We study the structure of the abelian group O K, if K is a number field. The results from the previous section help us understand the torsion subgroup of O K.

18 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Definition 4.7. Let A be an abelian group, written multiplicatively. Then A tors := {a A n 1 : a n = 1} is the torsion subgroup of A. Note that if A is finite, then A tors = A. If A is finitely generated, then A = A tors A/A tors. Lemma 4.8. If K is a number field then (O K ) tors = (K ) tors = µ(k) and µ(k) is a finite cyclic group. The full group theoretic structure of O K is described by Dirichlet s Unit Theorem. Theorem 4.9 (Dirichlet s Unit Theorem). Let K be a number field and n := [K : Q] its degree. Let r be the number of real embeddings of K and 2s the number of complex embeddings of K. Then n = r + 2s and O K is a finitely generated abelian group of rank r + s 1. Before we approach the proof, we look at a few examples. Example Let d be a square free integer and K = Q( d). We know that { Z[ d] if d 2, 3 mod 4 O K = Z[ 1+ d 2 ] if d 1 mod 4 Moreover, an element x O K is a unit if and only N K/Q (x) = ±1 (Proposition 1.19). It follows that if x = a + b d (d 2, 3 mod 4), or x = a + b 2 (1 + d) (d 1 mod 4), then the units in O K are given by the solutions to the following equations: { a 2 b 2 d = ±1 if d 2, 3 mod 4 (2a + b) 2 b 2 (2) d = ±4 if d 1 mod 4 Assume that d < 0. Then K only has complex embeddings, so Dirichlet s theorem implies that O K is finite. We can check this directly by finding the solutions to the equations (2). If d < 0 and d 2, 3 mod 4, then either O K = {±1} or d = 1, and Z[i] = {±1, ±i}. If d < 0 and d 1 mod 4, then we similarly see that O K = {±1} unless d = 3 and ( ) ( O {±1, Q( 3) = ±, ± )} On the other hand, if d > 0, then Dirichlet s theorem postulates the existence of infinitely many solutions to (2), and it is not as easy to describe all of them. We set up the proof of Dirichlet s Unit Theorem (Theorem 4.9). Let K be a number field n = [K : Q]. Let σ 1,..., σ r be the real embeddings of K and σ r+1, σ r+1,..., σ r+s, σ r+s the complex embeddings of K. As before, define σ : K R r C s, σ(x) := (σ 1 (x),..., σ r (x), σ r+1 (x),..., σ r+s (x)). This is a homomorphism of rings, but of course σ(k ) is not a subgroup of R r C s. To pass from products to sums, we use logarithms. Define L : K R r+s by L(x) = (log σ 1 (x),..., log σ r (x), log σ r+1 (x),..., log σ r+s (x) ).

19 SUMMARY NUMBER THEORY II, SOMMERSEMESTER This is a homomorphism of groups. If x O K we know that N K/Q(x) = ±1, so L(O K ) {(x 1,..., x r+s ) R r+s x x r +2x r x r+s = 0} =: H Note that the projection R r+s R r+s 1 given by forgetting the (r + s)-th coordinate induces an isomorphism H = R r+s 1, so we think of L as a homomorphism L : O K Rr+s 1. Proposition With the above setup, L(O K ) Rr+s 1 is a lattice and ker(l) = µ(k). In particular, O K is finitely generated. We make use of the following lemma. Lemma Fix m, M N. The set S(m, M) = { α C α is integral over Z, [Q(α) : Q] m, α M for all conjugates α } is finite. It is much more complicated to prove that L(O K ) is a full lattice. Proposition L(O K ) is a full lattice. This finishes the proof of Dirichlet s Unit theorem. June 1 5. Absolute values and completions Let K be a number field and p Z a prime. We studied the prime ideals p O K lying over p. Of course, the extension K/Q contains information about all primes p Z and all prime ideals in O K lying over p. Amongst other things, in this section we will develop a technique to focus on one prime number p Z. The completion of Q at p will be the field Q p of p-adic numbers, and K Q Q p = K p, p p O K where K p is a finite extension of Q p which is much easier to understand than K/Q. We start at the beginning Basics. We fix an arbitrary field K. Definition 5.1. An absolute value on K is a function : K R 0 satisfying: (a) x = 0 if and only if x = 0, (b) xy = x y for all x, y K. (c) x + y x + y for all x, y K. This is the triangle inequality. If in addition x + y max{ x, y }, then the absolute value is called non-archimedean, if not, then is called archimedean. We say that the pair (K, ) is a (non-archimedean or archimedean) valued field. Example 5.2. K = Q, R, C, the usual absolute value. This is often denoted. The absolute value is archimedean.

20 SUMMARY NUMBER THEORY II, SOMMERSEMESTER If K is any field, then defining x := 1 for x K and 0 := 0 makes (K, ) into a non-archimedean valued field. This absolute value is called the trivial absolute value and usually excluded from the discussion. Let K be a number field and σ : K C an embedding. Defining x := σ(x) defines an archimedean absolute value on K. K = Q, p Z a prime number. For x Q there is a unique way to write x = p n a b with p ab. Define v p(x) = n. Then x p := 1 p vp(x) defines a non-archimedean absolute value on Q, the p-adic absolute value. Similarly, if A is a Dedekind domain, K = Frac(A), and p A a nonzero prime ideal, define v p (x) = max{n x p n } for x A \ {0}, and v p ( a b ) = v p(a) v p (b) for a b K 1. Again, x p := defines N(p) v p(x) a non-archimedean absolute value on K, the p-adic absolute value. This applies, for example, to A = O K the ring of integers of a number field K. In the last few examples, we used certain functions v : K R. Their occurence was not a coincidence. Definition 5.3. An (additive) valuation on K is a function v : K R { } such that: (a) v(x) = if and only if x = 0. (b) v(xy) = v(x) + v(y) for all x, y K, (c) v(x + y) min{v(x), v(y)} for all x, y K. The datum of a non-archimedean absolute value on K is equivalent to the datum of an additive valuation. Proposition 5.4. If is a nontrivial non-archimedean absolute value on K, then defining v(x) := log( x ) for x K makes v into an additive valuation on K. Conversely, given an additive valuation v : K R { }, defining x := exp( v(x)) makes into a non-archimedean absolute value. The two constructions are inverse to each other. Definition 5.5. An additive valuation v is called discrete, if v(k ) R is discrete, i.e. if there exists e R >0 such that v(k ) = ez R. A non-archimedean absolute value is called discrete if K is a discrete subset of R >0, or, equivalently if the associated additive valuation is discrete. Definition 5.6. Let (K, ) be a non-archimedean valued field and write = exp( v( ). Define the following subsets. O K := {x K x 1} = {x K v(x) 0}. This is a subring of K, the valuation ring of. m K := {x K x < 1} = {x K v(x) > 0}. This is a maximal ideal in O K (see proposition below). U K := {x K x = 1} = {x K v(x) = 0}. Proposition 5.7. We continue to use the notation from the definition. (a) O K is a subring of K with the property that for x K we have x O K or x 1 O K.

21 SUMMARY NUMBER THEORY II, SOMMERSEMESTER (b) U K = O K (c) m K is the only maximal ideal of O K. (d) The following are equivalent: (i) is discrete, (ii) O K is a principal ideal domain, (iii) m K is a principal ideal, (iv) there is e R >0 such that v(a) := log a = e max{n a m n K } for all a O K. Definition 5.8. A local principal ideal domain is called discrete valuation ring, or DVR for short. Remark 5.9. (a) A discrete valuation ring is a local Dedekind domain. The converse is also true. (b) Given a discrete valuation ring A with maximal ideal m, we get an absolute value on K = Frac(A) such that A is the valuation ring of this absolute value. Indeed, pick e R >0 and for x A define v(a) = e max{n x m n }. Extend this to K by v( a b ) := v(a) v(b), and define x = exp( v(x)). This is not quite canonical: We chose e, but we will see later, that while the absolute value may depend on e, the topology induced by it does not. Example Let K = Q, p Z a prime. The valuation ring of the p-adic absolute value p on Q is { a } Z (p) := b Q (b, p) = 1. Its maximal ideal is pz (p). This is a special case of the localization of a ring at a prime ideal. Definition If A is a domain with field of fractions K and p A a prime ideal, define { a } A p := b K b p. This is a subring of K containing A with maximal ideal pa p. If A is not a domain, it is still possible to define the localization, but we will not make use of it. Example Let K be a field. On the polynomial ring K[T ] define ord(f) = max{n f (T ) n }. This is the order of zero at T = 0. On K(T ) := Frac(K[T ]) define ord(f/g) = ord(f) ord(g). The associated discrete absolute value has the valuation ring { } f K[T ] (T ) = g g(0) Topology. Definition If K is a field and an absolute value on K, then defines a metric on K and hence a topology. A basis of open neighborhoods of an element a K is given by the sets for ε > 0. B ε (a) := {x K x a < ε},

22 SUMMARY NUMBER THEORY II, SOMMERSEMESTER If 1, 2 are two absolute values on K, then we say that they are equivalent if and only if they define the same topology on K. Lemma If 1, 2 are two equivalent absolute values on K and x n K a sequence in K, then x n converges in K with respect to 1 if and only if it converges with respect to 2, and the limits are the same. Lemma If 1, 2 are nontrivial absolute values on K, then the following statements are equivalent: (a) 1 and 2 are equivalent. (b) For x K, x 1 < 1 x 2 < 1. (c) There exists t R >0 such that x 2 = x t 1 for all x K. Corollary If v : K R { } is an additive absolute value, then for q > 1, e > 0 defining x := 1 gives a nonarchimedean absolute value on q ev(x) K. The equivalence class of does not depend on the choice of q, e Classification of absolute values on Q. The lemma also allows us to classify all absolute values on Q up to equivalence. Theorem 5.17 (Ostrowski). Every absolute value on Q is equivalent either to the usual, archimedean absolute value or to p for some p Z prime. Later on, we will get a similar classification for all absolute values on any number field. The proof of Ostrowski s theorem splits into two cases: The archimedean and the non-archimedean case. The following lemma allows us to distinguish more precisely. Lemma Let be an absolute value on a field. The following statements are equivalent: (a) is non-archimedean. (b) { n n Z} R 0 is bounded. Note that by definition n = (c) n 1 for all n Z. (d) n 1 for some n Z >1. } {{ } n In the lecture I messed up the proof of (d) (c), so let me give it here: Proof. Pick n 0 Z >1 such that n 0 1. For any n Z >0 we find a 0,..., a r {0,..., n 0 1} such that n = a 0 + a 1 n a r n r 0. Note that n r log(n) 0 n, so r log(n 0 ). It follows that r ( n a i n 0 i (r + 1)n }{{} 0 n log(n) ), log(n i=0 0 ) 1 as a i a i < n 0. Replacing n by n t for some t 0 and taking the t-th root yields ( n n 1 t t log(n) ) 1 t. log(n 0 ) For t, the right-hand side converges to 1, which completes the proof..

23 SUMMARY NUMBER THEORY II, SOMMERSEMESTER Thus, the two cases in Ostrowski s theorem are 1. is non-archimedean, i.e., n 1 for all n Z 2. is archimedean, i.e., n > 1 for all n > Completion. Reminder. Let K be a field with absolute value. The absolute value induces a metric on K. Recall that a Cauchy sequence in K with respect to is a sequence (x n ) n N K N such that for every ε > 0 there is N N such that x n x m < ε for all n, m N. If the sequence (x n ) n N converges, then it is automatically a Cauchy sequence. The metric space K is said to be complete if every Cauchy sequence in K converges. To every metric space one can add the missing limit points to obtain a complete metric space; in this section we will explain its construction for a field with an absolute value. Example Consider the sequence 5, 14, 41, 122,..., described precisely as a n := 2+ m i=1 3i for n 1. This is a Cauchy sequence with respect to the 3-adic absolute value: For m > n we have a m a n = 3 n n m. Consequently, a m a n 3 = 1/3 n+1, which shows that (a n ) n N is a Cauchy sequence with respect to 3. But it is not a Cauchy sequence with respect to any absolute value not equivalent to 3. Note moreover that 2a n = 4+ n i=1 2 3i = 1+3 n+1, so 2a n 1 3 = 1/3 n+1 and hence lim n a n = 1/2 with respect to 3. While this sequence does have a limit in Q with respect to 3, Q is not complete with respect to any nontrivial absolute value. It is not complete with respect to as, e.g., 2 R \ Q. For p Z a prime, p > 3, it will follow from Hensel s Lemma, which will will soon prove, that there is a Cauchy sequence with respect to p in Q, which cannot have a limit in Q, as it would have to converge to a primitive (p 1)-th root of unity. But µ(q) = {1, 1}. For p = 2 (resp. p = 3), a similar argument will show that if Q were complete, it would have to contain a root of x (resp. x 2 + x + 1), which is false. Proposition Let K be a field equipped with an absolute value. There exists a field K with absolute value 2 such that K is complete with respect to 2 together with an injective homomorphism ι : K K such that ι(x) 2 = x for all x K. The triple ( K, 2, ι) has the following universal property. If L is another field, complete with respect to the absolute value 3 and j : K L an injective homomorphism such that j(x) 3 = x for every x K, then there exists a unique homomorphism such that the diagram June 8 K ι j K L!