PART I ALGEBRAIC NUMBER THEORY

Transcription

1 PART I ALGEBRAIC NUMBER THEORY

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3 3. Introduction Let Q { a b ; a, b Z, b 0}. Then we can regard Z as Z {α Q; fα 0 for some fx X + b Z[X]}. We can associate with Z the Riemann zeta function ζs n s Na s n a 0 where each nonzero ideal a of Z is of the form a nz and Na n #Z/nZ. The above series converges for Res >, and ζs can be meromorphically continued to a function which is analytic everywhere except for a simple pole at s, where it has residue, so ζs s + gs, where g is an entire function of s. The Riemann zeta function also has a functional equation and an Euler product π s s Γ ζs hs h s ζs p p s. By studying ζs as a function of s, one obtains information about primes and integers, e.g. the existence of infinitely many primes. Example Let gs Then m square-free m s p + p s p p s ζs p s ζs so 3 lim s gs s π 6 of the integers are square-free. 6 π Now let rn denote number of ways in which n can be written as a sum of two squares.

4 4 Then rn 4[#divisors of nwhich are mod 4] [#divisors of nwhich are 3 mod 4] Define if n mod 4 χn if n 3 mod 4 0 if n is even. which is a character mod 4. Then rn 4 χd, so d n n rn n s 4 χd n s 4 m s χdd s 4ζsLs, χ d n m d n And now consider the field Qi. The solutions of the equation with integer coefficients in this field are x + bx + c 0 x b ± b 4c if b 4c m. If b is odd, the LHS is mod 4, but the RHS is 0, mod 4. Therefore, b has to be even, b b, and so x b + m i. Hence the ring of integers is O {a + bi; a, b Z}. The ideals of this ring are all of the form a a + bi and Na #O/a a + b. Thus and ζ Qi s a 0 Na s 4 a,b 0,0 a + b s 4 n rn n s ζsls, χ lim s s ζ Qis L, χ π 4 which can be written as π 4 4.

5 5 For a general imaginary quadratic number field K Q D, we have a class number h K and a discriminant d K, and, for D > 3, Res s ζ K s πh k dk. As we will see, for an arbitrary number field K, Res s ζ K s encodes important arithmetic information about the field K.. Facts about fields - review We assume a basic familiarity with Galois theory, so here we are only reviewing the main facts that we will need later on. Let A C be an integral domain and let k be its quotient field. For α C let B denote the set of elements of kα satisfying some monic polynomial with coefficients in A. Then B kα and B is the integral closure of A in kα. If f k[x] is an irreducible polynomial over k and fα 0, then kα is a vector space over k of dimension n degf and {, α, α,..., α n } is a k-basis of kα. If α is some other root of the polynomial f, then kα kα via the map that sends α α and leaves k fixed. In this case α and α are called conugates over k and kα, kα are called conugate fields. If β a α, a conugate of β is β a α. Theorem. Primitive Element Suppose the complex numbers α and β are algebraic over k. Then there is an element θ C, algebraic over k, such that kα, β kθ Definition. An algebraic extension K kθ of k is normal over k if all the conugates of θ are in K.

6 6 Theorem. An algebraic extension K/k is normal iff for every α K all the conugates of α are also in K. Theorem 3. Let α,..., α n be the zeros of the polynomial f k[x] and β,..., β m be the roots of the polynomial g k[x]. Then kα,..., α n, β,..., β m is normal over k. Definition. Let K/k be a Galois extension of fields of Galois group G. We say that an intermediate subfield L of this extension, k L K is fixed by an element g G if g fixes every element of L.We say that L is fixed by H G if L is fixed by every element of H. The field consisting of all the elements of K that are fixed by H is called the fixed field corresponding to H and is denoted by K H. Theorem 4. Fundamental Theorem of Galois Theory Suppose K is a normal algebraic extension of k. Then there is a correspondence between the intermediate fields k L K and the subgroups H G, given in one direction by L GalK/L and in the other direction by H K H. Corollary 5. [K : K H ] #H and [K H : K] #G #H. Definition 3. If L and L are two subfields of kθ containing k, their composite is the smallest subfield of kθ containing both of them and it is denoted by L L. Note that, if L kα,,, then L L kα, α. Also, if H, H are subgroups of a group G, then will denote by < H, H > the smallest subgroup of G containing both H and H. For the foreseeable future, we will consider K/k to be a Galois extension with Galois group G. Recall that a Galois extension is by definition normal.

7 7 Theorem 6. Suppose L and L are two intermediate fields of the extension K/k and H GalK/L,,. Then L L iff H H.Also, GalK/L L H H and GalK/L L < H, H >. Theorem 7. Suppose kα K and let α,..., α n denote the algebraic conugates of α, where n [kα : k]. Then the isomorphism kα kα i can be extended to an automorphism of K. If g G maps gα α i, and H GalK/kα, then the [K : kα] elements of G that send α α i are exactly the elements in gh. Finally, GalK/kα i ghg. Theorem 8. Suppose L is an intermediate field of the extension K/k and H GalK/L. Then L/k is normal iff H is a normal subgroup of G and in this case GalL/k G/H. Example Let K Q, 3. Then [K : Q] 4 and K is a normal extension of Q. Let A, B, C 3, and D 3. Let A, B denote the automorphism of K/Q that sends A B and leaves C and D fixed and let C, D denote the automorphism of K/Q that sends C D and leaves A and B fixed. Then G {, A, B, C, D, A, BC, D}. The subgroups of G are, H {, A, B}, H {, C, D}, H 3 {, A, BC, D}, and G itself. Then K H Q 3, K H Q, and K H3 Q 6. We can also write K as K Q + 3. Let a + 3. The conugates of a are a itself, b + 3, c 3, and d 3. The automorphism of K that sends a b has also the effect of sending c d a c has also the effect of sending b d a d has also the effect of sending b c. So G {, a, bc, d, a, cb, d, a, db, c} and K a,bc,d Q 3 K a,cb,d Q K a,db,c Q 6.

8 8 3. The Ideal Class Group Let K be an algebraic number field, i.e. a finite extension of Q, of degree n and O K its ring of integers. To avoid confusion, we will call the elements of Z rational integers. Note that O K Q Z. Indeed, if r s is a rational number written in lowest terms that is a root of some equation with rational integers coefficients x m + a m x m a 0 0, then r m + a m r m s a 0 s m 0, hence s r m. But since gcdr, s, this implies that s ±, hence r s Z. Also note that if a m α m + a m α m a 0 0, a C, 0 m, and α C, then a m α satisfies the monic equation x m + a m x m a m m a 0 0, so a m α is integral over the ring Z[a 0,..., a m ]. Lemma 9. Suppose fx is a polynomial with algebraic integer coefficients and u is a root of f. Then fx is also a polynomial with algebraic integer coefficients. X u Proof: We prove this by induction over m deg f. The polynomial f is of the form fx θ m X m + θ m X m θ 0 with θ 0,..., θ m are algebraic integers. Therefore Z[θ 0,..., θ m ] is integral over Z. If m, then fx θ X u. Hence fx X u θ, which is an algebraic integer. Now assume the lemma holds for all polynomials of degree at most m. We want to show that it holds for the polynomial f. As we have seen, θ m u is integral over Z[θ 0,..., θ m ], which, in turn, is integral over Z. Thus θ m u is an algebraic integer and the polynomial gx fx θ m X ux m has algebraic integer coefficients. Also, deg g m and gu 0. So we can apply the gx induction hypothesis to g and, therefore, has algebraic integer coefficients. X u But fx X u gx X u + θ mx m, so the lemma also holds for f. Lemma 0. Suppose fx θ m X u... X u m has algebraic integer coefficients. Then θ m u... u is an algebraic integer for each, m.

9 9 Proof: If m, the product is actually equal, up to sign, to the constant term of f, and therefore it is an algebraic integer. For < m, the product is equal, up to sign, fx to the constant term of the polynomial gx. By the X u +... X u m previous lemma, g has algebraic integer coefficients, so θ m u... u is an algebraic integer. Definition 4. If α and β are two elements of some algebraic number field, we say that α divides β and write α β if there is an element a of the ring of integers of the field, such that β aα. m r Theorem. Let fx α i X i and gx β X be two polynomials i0 with algebraic integer coefficients and let h fg. Suppose that δ 0 divides all the coefficients of h. Then δ α i β for all i and. 0 Proof: Let fx α m X u... X u m and gx β r X v... X v r. Then hx δ α mβ r X u... X u m X v... X v r has algebraic integer δ coefficients. So δ α m β r. By the previous lemma, α m β r δ any product of the u i s any product of the v s is an algebraic integer. 0 r, of the v s. α m β r δ integer. α i α m Thus, Now, by the Viéte relations for all 0 i m and all is a sum of products of the u i s and β α i β δ α mβ r δ α i α m β β r β r is a sum of products is equal to a sum of terms of the form product of some u i s product of some v s and, therefore, an algebraic Corollary. Gauss Suppose f, g Q[X] and fg has integer coefficients. Then there exists an r Q s.t. rfx and r gx have integer coefficients.

10 0 Proof: Let h fg. Choose r Q s.t. rf has relatively prime integer coefficients. m Denote F X rfx a i X i Z[X] and GX gx Q[X]. Let d Z, r d 0 for which dgx i0 r b X has relatively prime integer coefficients. 0 Then d divides all the coefficients of the polynomial dhx F X dgx. By theorem it follows that d a i b for all i,. So, for any, d gcda 0 b, a b,..., a m b b, since the a i are relatively prime by construction. But the b s are also relatively prime, so d ±. Thus, G has integer coefficients. Corollary 3. A monic irreducible polynomial in Q[X] that has some algebraic integer for a root has rational integer coefficients. Proof: Let fx be such a polynomial and let α be its algebraic integer root. Then there exists some monic polynomial h Z[X] such that hα 0. Since f is irreducible, h fg for some g Q[X]. By Gauss s corollary there exists r Q s.t. rfx, r gx Z[X] and hx rfx r gx. Since the h is monic, it follows that the leading coefficients of rfxand r gx are ±. But, since f is monic it follows that r ±, so f Z[X]. Recall that K is an algebraic number field of degree n. { m } Definition 5. If α,..., α m K, denote by [α,..., α m ] a i α i ; a,..., a m Z the Z-submodule of K generated by α,..., α m. Theorem 4. Any Z-submodule a [α,..., α m ] of K has a Z-basis with at most n [K : Q] elements. Proof: Since a [α,..., α m ] is a subset of the field K, it is a finitely generated torsion-free abelian group, and thus a free Z-module of finite rank. Therefore it suffices to show that a can be generated by n elements.

11 Let β,..., β n be a Q-basis of K. Then each α i can be expressed as α i with a i Q. Then any element α of a is of the form m n m n α b i α i b i a i β c β m for some b,..., b m Z and where c b i a i, for all n. n a i β Let c,n gcda n,..., a mn. Then we can write c,n as a linear combination of the a in. m Choose b,,..., b,m Z such that c,n b,i a in is equal to gcda n,..., a mn. Let c, that that m b,i a i, n. Then for each α a there exists a d Z such m m b i a in d c,n, where α b i α i. So, α d n n c, β e β, for some e,... e n Q. Hence, for each i m, there exist d i Z and d i Q, n such Therefore α i d i n n c, β d i β. n n n a [α,..., α m ] α,..., α m, c, β d β,..., d m β, Now choose b,,..., b,m Z such that Repeating the procedure, we get n c, β m b,i d i n gcdd n,..., d m n. n n a d β,..., d m β, Keep going and finally obtain that n n c, β, c, β.

12 n n a c, β, c, β,..., c n,n β, so a can be generated by n elements. Definition 6. For any finitely generated Z-submodule a of K, the dimension of a is the number of elements in a Z-basis of a, and it will be denoted by dim a. Definition 7. If a [α,..., α r ] and b [β,..., β t ] are two Z-submodules of K, then their product is ab {αβ; α a, β b} [α i β ; i r, t] Clearly this definition is independent of the choice of the generators of the two modules. By the primitive root theorem, we can find θ K such that K Qθ. Then θ has n conugates θ,..., θ n, which uniquely determine the n embeddings of K into Q. K i for i n is defined by K i {α i : α K. The K i are the conugate fields of K. Definition 8. The discriminant of n elements α,..., α n of K is α... α n Dα,..., α n det..... α n... α n n Note that if α,..., α n are linearly dependent over Q, then Dα,..., α n 0. Theorem 5. Dα,..., α n Q, and, if α,..., α n O K, then Dα,..., α n Z. Proof: Let K Q θ,..., θ n, where we recall that K Qθ. Then K/Q is a Galois extension and the effect of the action of the elements of GalK/Q on the matrix that defines the discriminant is to permute it rows, thus leaving the det unchanged. Hence Dα,..., α n Q. If all the α s are algebraic integers, then the discriminant is at the same time an algebraic integer and a rational number. Therefore it has no choice but to land in Z.

13 3 Since K Qθ, a field basis for K is {, θ,..., θ n } and D, θ,..., θ n det θ... θ n..... θ n... θ nn i< n θ i θ 0. Assume that α,..., α n, β,..., β n K and β,..., β n α,..., α n M for some matrix M Mat n n Q. Then for any i, i n, applying the i th embedding of K into C, we get that β i,..., βi n α i,..., αi n M. Hence β... β n..... β n... β n n so Dβ,..., β n Dα,..., α n det M. α... α n..... α n... α n n M, Since K has a field basis whose discriminant is nonzero, it follows that the discriminant of any field basis is nonzero two field basis differ by an invertible matrix. On the other hand, if α,..., α n K, then α,..., α n, θ,..., θ n M for some n n matrix M with rational entries. So, if Dα,..., α n 0, it follows that det M 0, hence {α,..., α n } is a field basis for K. Note that by taking a field basis and clearing denominators, one can construct a field basis consisting entirely of elements of O K and the discriminant of such a basis is a nonzero integer. Theorem 6. Let a be an additive subgroup of O K containing n linearly independent elements linear independence over Z is the same as over Q. Then a is an n-dimensional Z-module. Proof: Let α,..., α n be the n linearly independent elements of a. They form a field basis and Dα,..., α n Z +. Suppose [α,..., α n ] a. Then there is an element α n+ a \ [α,..., α n ]. By theorem 4, [α,..., α n+ ] has a Z-basis with n elements β,..., β n. Then

14 4 α,..., α n β,..., β n M, for some n n matrix M with integral entries and det M 0. of [α,..., α n ]. Also, det M, since otherwise α n+ would be an element Thus 0 < Dβ,..., β n < Dα,..., α n and they are both integers. Keep repeating the process. And some point it has to stop, and when it does we have our n-dimensional basis. Corollary 7. O K itself is an n-dimensional module. Proof: It contains a field basis. Definition 9. A Z-basis for O K is called an integral basis of K. Note that if {α,..., α n } and {β,..., β n } are two integral bases of K, then α,..., α n β,..., β n M for some invertible matrix M GLn, Z. So M GLn, Z and M GLn, Z; therefore det M ±. Thus Dα,..., α n Dβ,..., β n. So we can define Definition 0. The discriminant of the field K, D K, is the discriminant of some integral basis of K. Let {α,..., α n } be an integral basis of K and a [β,..., β n ] be an n-dimensional subgroup of O K. Then β,..., β n α,..., α n M for some M Mat n n Z. Claim det M #O K /a. To see this, use the Elementary Divisors Theorem to find a Z-basis for a of the form d α,..., d n α n with d,..., d n nonnegative integers. Then d α,..., d n α n β,..., β n M for some M GLn, Z. So d α,..., d n α n α,..., α n MM. Then [O K : a] d... d n det M det M. But det M ± and d... d n 0, hence [O K : a] det M. If a is a Z-submodule of K, then it is clear that ao K a.

15 5 Definition. A Z-submodule a of K is called a fractional ideal of K if ao K a. If in addition a O K, then a is called an integral ideal. Clearly any fractional ideal can be written as da, where a is an integral ideal and d O K. Also, the index of any integral ideal in O K is finite. Definition. If α,..., α m are elements of K, not all zero, then the ideal generated by α,..., α m is α,..., α m [α,..., α m ]O K. The product of two ideals a α,..., α m and b β,..., β r is the ideal ab α i β ; i m, r. Principal ideals are of the form α, with α 0. Definition 3. If every ideal of O K is principal, then O K is called a principal ideal domain. Theorem 8. Given any fractional ideal a of K there exists another fractional ideal b such that ab O K. Proof: Let a α 0,..., α r and consider the polynomial fx f i X r 0 α i X and set r α X. Let 0 hx f X... f r X and gx hx fx t β l X l Let K be the normal closure of K over Q. Then the coefficients of h are fixed by all the elements of GalK/Q, and thus h Q[X] and the coefficients of g are fixed by all the elements of GalK/K, so g K[X]. Let N Q denote the gcd of the coefficients of hx. Claim: α 0,..., α r β 0,... β t N β0 Suppose the claim holds. If b N,..., β t, we have ab O K and the N theorem is proved. To prove the claim, apply theorem to N and hx. It follows that N α β l for all s and l s. So N α 0,..., α r β 0,... β t. On the other hand, the l0

16 6 coefficients of h are Z-linear combinations of the α β l s, so they are contained in α 0,..., α r β 0,... β t. But N, being the gcd of the coefficients of h, can be written as a Z-linear combination of these coefficients, so it is also contained in α 0,..., α r β 0,... β t. Corollary 9. The fractional ideals of K form an abelian group with O K as identity. Definition 4. We say that an ideal a divides an ideal b, and write a b, if there exists an integral ideal c such that b ac. Note that a b a b O K. Theorem 0. Let a and b be two ideals of K. Then a b iff a b. Proof: a b a b O K b ao K a. Definition 5. If a α,..., α r and b β,..., β t are two ideals of K, then their greatest common divisor is a, b α,..., α r, β,..., β t, the smallest ideal containing both a and b. Theorem. Let d a, b. Then d a and d b and, if c a and c b, then c d. Proof: By definition d a and d b. Now if c a and c b, then c a and c b, so c d, that is c α,..., α r, β,..., β t d. Definition 6. A prime ideal p is an integral ideal such that p O K and with the property that if an integral ideal a divides p, then either a p, or a O K Theorem. If p is a prime ideal of K and a, b are two integral ideals with p ab, then p a or p b.

17 7 Proof: Since p, a is an integral ideal that divides p, it follows that p, a p or p, a O K. In the first case, p a. In the second case b bo K bp, a bp, ab. Clearly p bp, and p ab by hypothesis. So by Theorem, p bp, ab b. Exercise Let K/k be a degree n field extension and assume that O k is a PID. Show that every ideal of K has a basis of n elements over O k. Let K Q 5. Then O K [, 5]. Which, if any, of the following three Z-modules are ideals? [ , ] [ , ] [ 3 + 5, ] Corollary 3. If p is a prime ideal and α, β O K with αβ p, then α p or β p. Proof: Let a α and b β. They are both integral ideals and p ab. Since p is prime, it follow that p a or p b, so α p or β p. Theorem 4. Every nonzero fractional ideal of K factors uniquely into a product r of the form p ai i with a i Z. For integral ideals all the a i s are nonnegative integers. Proof: Assume throughout that a is nonzero. It suffices to prove the assertion for integral ideals. To see this, assume that a is a fractional ideal. Then there is a d O K such that da is an integral ideal, b. Both d and b factor uniquely into primes, and therefore so does a b d. Now consider an integral ideal a. First we are going to prove that a factors as a product of primes. If a is prime or a O K, we are done. But suppose a can be written as the product of two integral ideals: a b b. Both b i s strictly contain

18 8 a, and so are of finite index smaller than [O K : a]. The process has to stop at some point and thus we get a prime factorization of a. Assume that two such factorizations exist: r t a p ai i q b. Then for each i there exist a such that p i q. Since they are both nonzero prime ideals of K, they are maximal ideals of O K and so they must coincide. We get r t and, by eventually reordering the factors of the product, p i q i for all i. In the same fashion we get that a i b i for all i. Theorem 5. The integers of K, O K, form a UFD iff every ideal of K is principal. Proof: If every ideal of K is principal, then O K is a PID and therefore a UFD. Conversely, assume that O K is has unique factorization. Since every ideal of K factors uniquely into product of prime ideals and their inverses, it suffices to show that every prime ideal is principal. First let π be an arbitrary prime element of O K and p π the ideal generated by it. For the sake of contradiction, assume that p is not a prime ideal. Then there exist integral ideals a and b such that p ab, but p a and p b, i.e. p a and p b. Choose α a \ p and β b \ p. Then αβ p π, so π αβ. Since π is prime it follows that π α or π β. But this implies α p or β p, which contradicts the choice of α and β. Therefore π is a prime ideal. Now let p be any prime ideal of K. Choose α p, α 0. Then p α. Since O K is a UFD, α has a unique factorization into primes, α π... π r. Therefore p π... π r, and, p being prime, this implies that p π for some r. By the first part of the argument π is a also prime ideal, so p π. Definition 7. The ideal class group of K, Cl K, is the quotient group of the group of fractional ideals of K by the group of principal ideals of K. Its order, h K is called the class number of K.

19 9 Theorem 6. Given two integral ideals a and b, there exists an integral ideal c such that ac is principal and b, c. Proof: Let ab r 0 a i l i for each i. For each, r, define p li i, where l i > 0, for every i, i r. Write a d i p ai+ i. So for each p i ab, p i d iff i and for i we have p ai+ i d. r Now d,..., d, so there are γ d such that γ. α p a \ p a+, so p a, we have that p a α, but p a+ α. Define w α γ, but p a+ r p ai i, with Choose r α γ. For each α γ and p a+ α i γ i for all i. This is so because first we have γ d, hence d i γ i and so p a+ α i γ i for all i. But if p γ for any, then p would divide all the γ s, and therefore would divide their sum, which is. This is clearly impossible. Combining all this we get that for each, p a w, but p a+ w, so a w. r Let c be the integral ideal for which w ac. Then ac, ab w, r p ai i a, and thus b, c. p li i Theorem 7. Every ideal of K can be generated by two elements. Proof: It suffices to prove the assertion for integral ideals. So, let a be an integral ideal. Then a α b for some α O K and b an integral ideal. Then ab α. By theorem 6 there exists an integral ideal c such that ac w and b, c. Hence ab, ac a, so α, w a Let a be an integral ideal of K. Then O K /a is a finite ring and we have the canonical proection O K O K /a given by α α. If p happens to be a prime ideal of K, then O K /p is a finite integral domain, and thus a field. Let p be the

20 0 characteristic of the finite field O K /p. Then p 0 mod p, so p contains a rational prime, namely p, i.e. there exist a rational prime p such that p p po K The canonical proection O K O K /p induces a canonical map O K [X,..., X s ] O K /p[x,..., X s ] given by fx,..., X s fx,..., X s. Suppose that f, g O K [X,..., X s ] and that every coefficient of fg is in p. Then fg 0 in O K /p[x,..., X s ], so f 0 or ḡ 0, i.e. every coefficient of f is in p or every coefficient of g is in p. Definition 8. Let f K[X,..., X s ]. The ideal If of K generated by the coefficients of f is called the content of f. Theorem 8. If f, g K[X,..., X s ], then Ifg IfIg. Proof: Let X X,..., X s. Then fx a i X i and gx b X where i and are multi-indices. Without loss of generality, we may assume that f and g have integral coefficients. { } { } IfIg αi a i β b ; α i, β O K γi a i b ; γ i O K Ifg Let a If. This is an integral ideal, and there exists another integral ideal b such that ab α, a principal integral ideal. Also, by theorem 6 there exists an integral ideal c such that bc α principal ideal and c, a. For any α coefficient a of f we have a a, so a a, and α a bc a ca ab a O K. Therefore α α a O K. It follows that α α fx O K[X] and α I α f, If α α bc a, a ab a, a c, a Similarly, there exist β, β O K such that β β gx O K[X] and β I β g, Ig.

21 Fix p a prime ideal dividing Ifg. Then all the coefficients of fg are in p, and therefore, all the coefficients of f are in p or all the coefficients of g are in p, i.e. p If or p Ig. We will discuss the case p If, the other case being similar. α Since I α f α, If, it follows that p I α. f Define γ, γ O K as follows: if it also happens that p Ig, let γ β and γ β. Note that in this case, by the same reason as for f, p I β β g I γ γ. g if p Ig, let γ γ. In this case, we also have that p I γ γ. g α γ So, by the remark we made before, p I αγ. fg Therefore the power of p in Ifg the power of p in αγ α γ the power of p in α α + the power of p in γ γ the power of p in If + the power of p in Ig. Hence Ifg IfIg. 4. Extensions of Number Fields Let K/k be a degree n extension of number fields. Denote by K,..., K n the n conugates of K over k. Definition 9. For every element α K, the norm of α relative to k is n N K/k α α i Similarly, the norm of a polynomial f K[X,..., X s ] is N K/k f Clearly N K/k α k and N K/k f k[x,..., X s ] n f i. Proposition 9. Suppose L is an intermediate field of the extension K/k. Then, for any α K and any f K[X,..., X s ], we have N L/k NK/L α N K/k α and N L/k NK/L f N K/k f. Proof: Let K be the normal closure of K over k. Then K/k is a Galois extension with Galois group G. Denote H GalK/K and H GalK/L. Write G h hh,

22 where denotes a disoint union. Also write G h h H and H h we have G h H h h H. h H. Then For α K, N K/k α h For β L, N L/k β h hα and N K/L α h h β. h α. So N L/k NK/L α h h α h h α hα N K/k α. h h h,h h We would like to define N K/k for ideals of K. The natural definition would n be N K/k a a i, where, if a α,..., α r is an ideal of K, then a i α i,..., αi r K i, the image of a under the canonical isomorphism K K i. But we have to precisely define what we mean by the product of ideals of different n fields and we need to know that a i k. This is obviously an ideal of K that is fixed by the Galois group, but this does not mean anything. Take, for instance, in Q, which is fixed by the Galois group, but that does not make it rational. Let a,..., a m k. We will write a k a,..., a m k for the ideal of k generated by a,..., a m and a K a,..., a m K a k O K for the ideal of K generated by a,..., a m. Theorem 30. a K k a k Proof: It is obvious that a k a K k. To prove the other inclusion, pick any element b a K k. Then b α a α m a m for some α,..., α m O K. Taking norm of both sides of this equality, we get that b n n α i a α i m a m. Consider the polynomial fx,..., X m α X +...+α m X m O K [X,..., X m ] n m and let gx,..., X m N K/k f O k [X,..., X m ]. This g α i X

23 3 is a homogeneous polynomial of degree n and b n ga,..., a m. Therefore b n a sum of products of the form αc... c n with α O k and c,..., c n a k. This implies that b n a n k, so an k bn. But, because of the unique factorization of the ideals this means that a k b, i.e. b a k. So we can define the equality of ideals in different fields as follows: Definition 0. If a i is an ideal in the number field L i, i,, we say that a a if they generate the same ideal in a common extension of the two fields L and L. Also, the multiplication of ideals of different number fields is done by performing the multiplication in some larger common extension of the respective fields. Example Let L Q 5 and L Q, a, + 5 and a. Then a 4, + 5, , + 5, 6 and a. So, in Q 5, it really does make sense to say that a a. In fact, + 5, because and and ± 5 are the roots of the polynomial X X + 3. Also, , so, + 5. Going back to our attempt to define the norm of an ideal, let a be an ideal of K and let g K[X] be a polynomial with Ig a. For instance, if a α,..., α r, take gx α X. Then n a i n I g i I n g i I N K/k g k. So, now we can actually define the norm of an ideal: Definition. If a is an ideal of K, its norm relative to k is N K/k a n a i. Also if L is an intermediate field of the extension K/k, then N L/k NK/L a N L/k I NK/L g I N K/L g i I N K/k g N K/k a.

24 4 And, if a and b are both ideals of K, choose f, g K[X,..., X n ] such that If a and Ig b. Then ab IfIg Ifg and N K/k ab I N K/k fg I N K/k fn K/k g I N K/k f I N K/k g N K/k an K/k b. So, if k Q, then N K/Q a N for some N Z. Definition. By convention, N K/Q a N and this is called the absolute norm of a. From now on, if K is clear from the context we will write Na for the absolute norm of a. Example We will see that in Q D, for the prime p Z, we have p p p p p if if if D p D p D p 0 with Np Np i p and N p p. D Note: p is the Legendre symbol, defined by: D 0 if p D if there exists some a such that a p D mod p if there is no such a In general, if a l, l Q and [K : Q] n, then Na l n. Also, note that if a is integral, then a N K/k a. Lemma 3. Any ideal a of K may be written as a α, β, where N K/k α, N K/k β N K/k a. Proof: Take c integral ideal with ac α. Now take N K/k c extended up to K and choose an integral ideal b of K so that b, N K/k c and ab β. Now, c N K/k c and therefore c, b. So α, β ac, ab ac, b a and we have NK/k α, N K/k β N K/k an K/k c, N K/k an K/k b N K/k a N K/k c, N K/k b

25 5 But b, N K/k c, hence b i, N K/k c for each i. Thus N K/k b, N K/k c. Note that N K/k α N K/k α. Assume that O k is a PID and recall that [K : k] n. Lemma 3. Let b [β,..., β n ] and c [γ,..., γ n ] be two ideals of K. We can write γ,..., γ n β,..., β n M for some M Mat n n k. Then det M N K/k c/b. Proof: Let α c/b. Then αb c. So there exists an n n matrix X with entries in O k such that αβ,..., β n γ,..., γ n X β,..., β n MX. Therefore α β... α β n..... α n β n... α n β n n and taking determinants we get N K/k α det β i β... β n..... β n... β n n det β i MX det M det X. Hence N K/k α det M det X, and, since det X O k, it follows that det M N K/k α. Now, there exist α, α K such that c/b α, α and N K/k c/b N K/k α, N K/k α. But this means that det M N K/k α i, i,, so det M N K/k c/b. Theorem 33. Let k be a number field with O k a PID. Suppose K/k is a degree n extension and choose w,..., w n K such that O K [w,..., w n ] k. Let a α,..., α n be an ideal of K and write α,..., α n w,..., w n M with M n Mat n n k. Let fx,..., X n N K/k α X. Then a [α,..., α n ] k iff det M divides every coefficient of f, and in this case N K/k a det M. Proof: Recall that If N K/k a.

26 6 First assume that a [α,..., α n ] k. Then by lemma 3, we have det M N K/k a/o k. But N K/k a/o k N K/k a, so det M N K/k a Furthermore, w,..., w n α,..., α n M, so det M N K/k O k a and therefore det M N K/k a. Together with above, this shows that N K/k a det M. Now suppose that a [β,..., β n ] k. Then α,..., α n β,..., β n X for some X Mat n n O k and β,..., β n w,..., w n Y for some Y Mat n n k. Hence α,..., α n w,..., w n Y X, and so M Y X. By the first part of the proof, it follows that det Y N K/k a If. Therefore det M If iff det X which is equivalent to [α,..., α n ] k [β,..., β n ] k. Example Let K Q 5 and a, + 5 K. Then also a [, + 5 ]. Here O K [, 5 ] and, + 5, 5 0 and det M N K/Q a. The associated polynomial is fx, Y N K/Q X + + 5Y X + Y + 5Y 4X + 4XY + 6Y. Note that divides every coefficient of f, and in fact If 4, 6. Corollary 34. If k Q and a [α,..., α n ], then Na O k /a det M. Proof: Both quantities are equal to det M. 5. Relative Differents and Discriminants D Take K Q. { If D, 3 mod 4, consider the integral basis, } D. Then D D D D D D D

27 So we obtained the dual pair of elements [ ] [ ], D as D, D any element of this ideal looks like D {, D [, D a + b D D a D + b 7 }. Note that ] is a fractional ideal. Note that We define the trace of any element of K as follows: trα α + α. So tr a + b D a. For α O K, we clearly have trα Z. Also notice that for the elements of, we have tr b Z. What other elements of D a D + b K have the property that their trace is an integer? That is, what is tr Z? Consider α r + s D K. Then trα r. So if the trace is an integer, we have r Z. Furthermore, tr Dα Ds. If this is an integer, then s D Z. Thus, we have: D [ ], {α K; trλα Z for all λ O K } D Also, note that D is a root of the polynomial fx X D and f D D, so N f D 4D D K. { } If D mod 4, consider the integral basis. Then So + D D D [ D +, D ] Here tr a+b D D + D D [, D ] D b + b b Z. D., + D D + D + D D Also, + D is a root of the polynomial fx X X + D and f D. So N f D D K. + D Recall that we are going to prove that p p iff p D iff p + D D, so primes that ramify are precisely the primes dividing the different or the discriminant.

28 8 Now, return to the situation in the previous section. That is, consider a degree n extension of number fields K/k. By the primitive root theorem, K kθ for some algebraic integer θ O K. Let f O k [X] be the irreducible polynomial of θ over k. We want to write θ... θ n..... θ n... θ nn β... β n..... β n... β n n So β n signed minor corresponding to θn. det of the θ matrix The signed minor corresponding to θ n is n+ times the determinant of the minor obtained by deleting the n th row and the th column of the θ matrix, which is equal to the coefficient of X in the polynomial θ... θ n. hx.... θ n... θ n n X... X n i< n θ θ i n X θ Hence β n the coefficient of X in hx h θ n the coefficient of X in n n θ n θ i X θ i Recall that f is the minimal polynomial of θ, so fx Then Therefore n X θ i β n n fx [ X θ O n k θ n] [X] and f θ n X θ n the coefficient of X in fx f θ n O k X θ O k [X]. [θ n] n θ n θ i.

29 Let gx fx n X θ b l X l O k [θ], where of course b n. Then, for each i n, we have l0 fx X θ i gi X n l0 b i l X l O k [θ i]. So, 9 and similarly β n bn f θ n β i bi f θ i, for each i, i n. In general if α,..., α n are k-linearly independent elements of K, then α,..., α n θ,..., θ n M for some M GLn, k. Hence and thus α i α... α n..... α n... α n n θ... θ n..... θ n... θ nn M θ i M β i So we proved the following result: γ M... γ n..... γ n... γ n n. Theorem 35. If α,..., α n K are linearly independent over k, then α... α n β... β n α n... α n n β n... β n n with β i K for each i n. In particular if α,..., α n θ,..., θ n, with K kθ and θ integral with minimal polynomial f over k, then β i bi n f fx θ i, where X θ b l X l. l0 Note that β,..., β n. f θ Definition 3. The relative trace of an element α K is tr K/k α n α i

30 30 Note that for all α K we have tr K/k α k and if L is an intermediate field of the extension K/k, then tr L/k trk/l α tr K/k α. In the notation of the theorem we have δ i, so β i β α i β i α i + +β n α n i δ i, i.e. tr K/k β α i δ i. Lemma 36. Suppose α,..., α n K are linearly independent over k and α i. Then λ K has the property that tr K/k λα i O k for i n iff λ [β,..., β n ] k. Proof: Suppose λ has the property that tr K/k λα i O k for all i. Then λ,..., λ n α i b,..., b n with b i O k. So λ,..., λ n b,..., b n and thus each λ i β i [ ] β i,..., βi n. k Also note that, in this case, if we write λ c β with c k, then b i tr K/k λα i tr K/k c β α i c i, so c i O k. The converse is clear. Definition 4. Let a be an ideal of K. We will denote t K/k a { λ K : tr K/k λα O k for all α a } Lemma 37. If a is an ideal of K, then t K/k a is also an ideal of K. Proof: First note that if β, γ O K and λ, λ ta, then tr λ β + λ γα tr λ βα + tr λ γα O k for any α a. So λ β + λ γ ta. Therefore we need only show that ta is finitely generated, that is to say find d O K such that d ta O K. Let λ t K/k a. Then, for any α a we have tr K/k λα O k, so tr K/Q λα Z. Hence t K/k a t K/Q a. Using the fact that Z is a PID, we may write a [α,..., α m ] Z where m [K : Q], and let here m [K : Q]. α i i, m β i i, m Since λ t K/Q a, it follows that t K/Q λα i Z for all

31 3 i m and therefore λ [β,..., β m ] Z. Choose d O K such that dβ i O K for each i. Then d ta O K. Definition 5. The ideal D K/k to K is called the relative different of K/k. If k Q, then D K D K/Q is called the different of K. Theorem 38. If a is an ideal of K, then ta a D K/k and D K/k is an integral ideal of K. If K kθ and O K minimal polynomial of θ. O k [θ], then D K/k f θ where f is the Proof: Let a, b be two ideals of K. Then a ta a tab b a tab b. Choose α a, λ ta and x b Pick any β b. Then xβ O K and therefore αxβ a. Hence tr K/k λαxβ O k. Since this holds for any β b, it follows that αλx tb, that is a tab tb. So a ta b tb. By symmetry, we see that for any two ideals a and b of K, we have a ta b tb. In particular take b O K. We have a ta O K to K to K D K/k. Pick any α, λ O K. Then trλα O k, so λ to K D K/k. Hence O K D K/k, i.e. D K/k O K O K. Now assume that K kθ and O K O k [θ]. Let theorem 35, β,..., β n f θ β i θ i. By. Recall that trθ i β δ i+, O k, so β to K for all. On the other hand, for any λ to K we have, by lemma 36, that λ [β,..., β n ] k β,..., β n. Hence D K/k to K β,..., β n f θ, so D K/k f θ. Theorem 39. Let a [α,..., α n ], where we recall that n [K : k].. β,..., β n as usual by β i α i Then a D K/k [β,..., β n ] k. Define Proof: Recall that if λ a D K/k ta, then lemma 36 implies that λ [β,..., β n ] k. On the other hand, for each i and we have trα i β δ i O k, so β ta.

32 3 Definition 6. The relative discriminant of K/k is the ideal D K/k where O K [w,..., w n ] k. det w i, Corollary 40. If O k is a PID, then N K/k D K/k D K/k. In particular, ND K D K. Proof: Let O K [w,..., w n ] k and β i By theorem 39 it follows that [β,..., β n ] k w i. β,..., β n M for some n n matrix with entries in k. Then, by theorem 33, det M N K/k D We also have M, so det β i β i Therefore N K/k D K/k w i det det w i w i K/k. det M det DK/k. If k Q, we get that ND K N K/Q D K D K. D K/k. Write w,..., w n w i N K/k D K/k Theorem 4. If K L k are number fields, then D K/k D K/L D L/k. Proof: We will prove that D K/k D K/L D L/k. Pick some α D K/L t K/LO K and some a D L/k t L/kO L L. Then, for any w O K, we have tr K/k αaw tr L/k tr K/L aαw tr L/k a trk/l αw tr L/k ao L O k. Hence aα t K/k O k D K/k. Now we want to prove the other inclusion, which is equivalent to showing that D L/k D K/k D K/L. Choose α D K/k t K/kO K and a D L/k O L. Choose w O K. Then for any b O L, we have tr L/k b trk/l αw tr K/k αbw O k, since bw O K and α t K/k O K. Thus tr K/L αw t L/k O L D L/k, and therefore tr K/L aαw a tr K/L αw ad L/k D L/k D L/k O L. Hence aα t K/L O K D K/L. Corollary 4. Suppose [K : L] m. Then D m L D K.

33 33 Proof: Take k Q in the previous theorem. We get that D K N K/Q D K N L/Q NK/L D K/L D L N K/Q D K/L N L/Q DL m DL m N K/QD K/L and the theorem follows since N K/Q D K/L is an integer. Theorem 43. Chinese Remainder Theorem Suppose a and b are integral ideals of K with a, b. If α, β O K, then the system of equations { x α mod a x β mod b has a solution in O K and that solution is unique mod ab. Proof: Theorem 44. Let P be a prime ideal of K. Then p P k is a prime ideal of k and N K/k P p f for some f n [K : k]. Proof: If a, b O k O K and ab p P, it follows that a P or b P, and since they are already elements of k, we get a p or b p. Therefore p is a prime ideal of k. Note that P p, since po K PO K P. Now let K i be a conugate field of K. i p p Then P O i P Ki, so P i p. Thus N K/kP p n, hence N K/kP p f for some f n. Lemma 45. Given a prime ideal P of K and an integral ideal a of K with P, a, there exists an element θ a such that K kθ, θ is a primitive root modp and for any m Z + and any w O K we can find an element α O k [θ] with α w mod P m. Proof: We know that O K /P is a finite cyclic group. Choose θ O K such that θ mod P is a generator of this group. Then every element in O K that is not in P will be congruent to θ i mod P for some i NP.

34 34 We want to find θ θ mod P such that π θ NP θ P \ P. To do this, set θ θ + β for some β P \ P. Then θ NP θ + β NP θ NP + NPθ NP β + β... θ NP mod P since P NP and β P. So θ NP θ θ NP θ β mod P and we can choose an appropriate β. Now, the Chinese Remainder Theorem allows us to pick θ 3 θ mod P with θ 3 0 mod a. Set θl θ 3 + lp aγ where p is a rational prime such that P p, a a Z nonzero, γ O K such that K kγ and l Z. The determinant of the matrix θl i looks like θ 3 θ i 3 + lp aγ γ i, and since γ i< γ i 0 we can choose l large enough to this determinant nonzero. In this case the column vectors are linearly independent, hence {, θl, θl,..., θl n } are linearly independent over k, hence we have K kθl. Put θ θl. Note that θ is an integer. Change π to π θ NP θ P\P and set γ i θ i for i NP. Then the m numbers β π O k [θ] run through all the residue classes mod P m as the β s 0 run through {γ i ; i NP } 0. This is because we have NP m of the form m β π and if two of them are congruent mod P m, then β β mod P for 0 each. Theorem 46. If K kθ and θ O K, let f k[x] be the minimal polynomial of θ. Then D K/k is the gcd of all such f θ. Also, if P is a prime ideal of K, p P k and p P e a with a, P, then, for the θ in lemma 45, the prime ideal P has the same power in both D K/k and f θ.

35 Proof: If θ O K and K kθ and β i 35 θ i, then for any λ D K/k to K we have, by lemma 36, that λ [β,..., β n ] k β,..., β n. So D K/k β,..., β n f θ, by theorem 35. Thus, f θ D K/k, i.e. D K/k f θ. Hence f θ D K/k B for some integral ideal B of K. The theorem will follow if we can show that with the choice of θ from the lemma 45, we have B, P. Now B N K/k B p r b for some integral ideal b of k with p, b. Given w O K, we can find α O k [θ] with α w mod P er see lemma 45. Then D K/k w αbθ r f θ w α P er Per a r b B r θ a and each of the three terms is integral, so the LHS is an integral ideal. w αbθ r Now choose b b O k. Then b b, so D K/k f is an integral θ w αbθr ideal. Hence f D θ K/k and therefore tr w αbθ r K/k f θ θ O k for all n. Again lemma 36 implies that w αbθr f θ [β,..., β n ] k β,..., β n f θ, so w αbθ r O k [θ]. But αbθ r O k [θ] and thus wbθ r O k [θ]. wbθ Hence tr r K/k f θ O k for all w O K and so bθr f θ D K/k. Therefore D K/k bθ r f θ, i.e. bθ r is an integral ideal. But f θ D D K/k f θ K/k B, so bθ r B is integral. This is true for all b b, hence B bθ r. Since p, b and P p, it follows that b, P But we also have θ, P, so θ r b, P. As B bθ r, we get B, P, exactly what we wanted. Consider the field extensions 6. Ramification Theory K k Q P p p

36 36 We have that P p for each P lying above p, i.e. such that p P k. Also, if P p, then P p, i.e. P lies above p. Thus, we may write any prime ideal of k as p P ei i, a product of the primes lying above it. Then, taking the norm of both sides and keeping in mind that N K/k P i p fi, we get p n N K/k P i ei p P eif i, hence n e i f i. In particular, the number of the P i s must be n. Possibilities are: p splits completely in K if there are n distinct P i s lying above it each with f i e i ; p remains inert in K if there is ust one prime ideal P lying above p here e, f n, so p P; 3 p ramifies in K if any of the e i >. In this case we also say that P i is a ramified prime of K relative to k. Definition 7. The degree of P is the residue class degree, i.e. the degree of the algebraic extension O K /P over Z/p. Therefore N K/Q P p f where f the degree of P. Now, O K /P contains the field O k /p in the following sense: suppose a, b O k with a b mod P; then a b P O k p, so a b mod p. Claim O K /P is a normal extension of O k /p. Proof: Say N K/k P p f and N k/q p p f0 q. Then #O K /P q f. Let θ be a generator of the multiplicative group of the field O K /P, so θ qf and no lower power of θ is. For α O K /P define σα α q. This defines an automorphism of O K /P that fixes O k /p as α + β q α q + β q and for a O k /p we have a q a mod p. Moreover, σ, σ,..., σ f Id are all distinct automorphisms of O K /P over O k /p. Note that the above field extension is automatically separable, since any finite field extension is separable.

37 37 Thus, Gal O k /P/O k /p is a cyclic group of degree f with generator σ, given by σθ θ q, and the f conugates of θ over O k /p are θ qi, 0 i f. Now assume K is normal over k with Galois group GalK/k G. Suppose P e p exactly divides, and let P,..., P r be the r distinct images of P r e under the action of G. Then P e i p for each i. So, p P i a, where a, P i for all i r. But gp N K/k P p f, so any prime ideal factor of p in K is one of the gp s, g G r i.e. one of the P i s. Hence a and g G gp N K/k P p f Now let G D G D P {g G; gp P}. P i ef. Then so #G D ef. r #GD gp P i g G Note that if g G D, then g is an automorphism of O K /P over O k /p. Let a p P e and choose θ O K so that θ is a primitive root mod P, θ a and K kθ, satisfying the conclusions of the lemma 45. Then π θ θ NP θ θ qf P \ P and every element of O K is congruent to an element of O k [θ] mod P m. The irreducible polynomial for θ over k is fx g GX gθ O k [X]. Reducing mod p we see that θ is a root of f, so mod p g GX gθ is divisible by f i0 X θ qi. Thus there is a g G such that gθ θ q mod P. This g is called the Frobenius automorphism of P and is denoted by σp σ. K/k P Let σ be the Frobenius automorphism of P. Then σ G D. Indeed, suppose this was not so. Then σ P P, hence σ P, P. So, we can assume that

38 38 θ σ P. Since σθ θ q mod P, we have that θ σ θ q mod σ P, so σ θ σ P, i.e. θ P, and we arrived at a contradiction. Thus σ G D. Definition 8. G I {g G; gw w mod P for all w O K } is called the inertia group of P. Now, if g G D, then g provides an automorphism of O K /P over O k /p, so gθ θ qi mod P for some i. But also σ i θ θ qi mod P, so g σ i θ θ mod P, so g σ i w w mod P for all w O K. Hence g σ i G I, i.e. g σ i G I. It follows that G I is a normal subgroup of G D, because g D g Ig D θ g D g Iσ i θ g D θqi σ i θ qi θ mod P. Claim G D /G I is a cyclic group of order f and is generated by σg I. Proof: σg I is a generator as each g σ i G I for some i, as seen above. The only question is to determine the smallest i such that θ qi θ mod P. This is i f since θ is a primitive root mod P. Note that #G D ef, so #G I e. Also, G D /G I Gal OK P / O k p. We now have the field extensions corresponding to G G D G I : K P G I e k I p I P k I f k D p D P k D r k p Start with an ideal P of K lying above the prime ideal p of k. Denote p I P k I and p D P k D p I k D.

39 39 Then N K/kD P g G D gp P ef. At the same time, we know that N K/kD P p m D for some m {,..., ef}. Recall that ef [K : k D ]. Now if O kd /p D were a nontrivial extension of O k /p, then it would exist a nontrivial automorphism of O kd /p D that fixes O k /p. Any such automorphism would extend to a nontrivial element of Gal OK P / O k p G D /G I, so it would be induced by some g G D. But all the elements of G D fix O kd /p D contradiction. Hence O kd /p D O k /p. Therefore N kd /Qp D N k/q p q and N kd /kp D p. Also, q f N K/Q P N kd /QN K/kD P N kd /Qp m D qm, so f m and P ef N K/kD P p f D. Hence p D P e. Since the elements of G I fix O K /P, we get ust as before that O K /P O ki /p I why? and therefore N ki /k D p I p f D. It will also follow that N K/k I P p I and that p I P e. In conclusion, p splits completely in k D into r different primes. Each of this primes of k D remains inert in k I, i.e. p D p I. And each p I lying above p in k I is totally ramified in K, i.e. p I P e and P is the only prime lying above p I in K. So we have L GalK/L order of GalK/L ep L fp L rp L k G n efr e f r k D G D ef e f k I G I e e or

40 40 K P k I p I P e N K/kI P p I k D p D P e N K/kD P p f D k p P e N K/k p f Now P is ramified in K/k iff #G I P >. Recall that K K. Then #G f θ θ θ i θ gθ. i g G g Now if P θ gθ, where g, then gθ θ mod P, so θ g θ mod g P. Assume that g P P. Then we may assume that θ g P, therefore also g θ g P, so θ P contradiction. Hence g P P, so P gp, i.e. g G D, and thus g G I. why? Thus P f θ iff #G I > iff e >. So we have the following result: Theorem 47. If K/k is normal, then P D K/k iff P is ramified in K/k. Definition 9. G R { g G; gw w mod P for all w O K } G m { g G; gw w mod P m+ for all w O K }, where m 0,,... Note that G 0 G I and G G R. Theorem 48. Let g G. Then g G m iff gθ θ mod P m+. Proof: As before. Theorem 49. For all m 0, G m+ is a normal subgroup of G m.

41 4 Proof: g m+ g m θ g m θ mod P m+, so g m g m+ g m θ θ mod g m P m+. But g m P P as g m G D, and the assertion follows. Now, the exact power of P in D K/k the power of P in θ gθ g G g G g g where ig max{i; gθ θ mod P i } i such that gθ θ P i \ P i+. But this equals g G\{} igi So we have the following result: i #G #G #G 0 ig, Theorem 50. If K/k is normal, the exact power of P occurring in D K/k is #G i P : K/k. i0 Corollary 5. P D K/k iff ep >. We will now show the following theorem: Theorem 5. Let L be an extension of k not necessarily normal, and p L a prime ideal of L. Then p L D L/k iff p L is a ramified prime ideal of L relative to k. This has the following consequence: Corollary 53. If p p L k, then p D L/k iff p ramifies in L. Proof of theorem: Let K/k be normal such that K L, P a prime ideal of K dividing p L. Then G m P : K/L G m H, where H GalK/L. So, the power of P in D K/k is #G i and the power of P in D K/L is #G I H. i0 Clearly, G i H G i for all i iff G 0 H G 0, so the same power of P occurs in D K/k and D K/L iff ep ep L. Now p L occurs in D L/k iff the power of P in D K/k is > the power of P in D K/L. Therefore p L is ramified in L/k iff ep > ep L iff P D K/k iff p L D L/k.why? i0

42 4 FINAL COMMENTS this are ust facts, right? Take K/k to be a normal extension of number fields. For m, G m /G m+ is a p-group, and every non-identity element has order p. Also, G m /G m+ is abelian. #G i i0 { e e + p e + p e, if p e, if p e p is called tamely ramified if p e and wildly ramified otherwise.

43 PART II ANALYTIC NUMBER THEORY

44

45 45 Define ζs ζ Q s a Z Na s n s p n p s Our aim is to prove that the function satisfies the functional equation gs π s Γ s ζs gs g s, is analytic for all s 0,, with simple poles at this points and residue at s and residue at s 0. To do this, we first define and prove that θt n θt t θ e πnt for t > 0. t This is a consequence of the Poisson summation formula: Let fx be a function such that fx 0 sufficiently fast as x. Then F x fn + x n converges for all x and F x + F x. Thus F x has a Fourier expansion F x m Z a m e πimx, where a m 0 F xe πimx dx fn + xe πimx dx 0 n n 0 fn + xe πimx dx Making the change of variables x x + n, we see that