THE THEORY OF NUMBERS IN DEDEKIND RINGS

Transcription

1 THE THEORY OF NUMBERS IN DEDEKIND RINGS JOHN KOPPER Abstract. This aer exlores some foundational results of algebraic number theory. We focus on Dedekind rings and unique factorization of rime ideals, as well as some celebrated consequences such as a artial roof of Fermat s last theorem due to Kummer, and the law of quadratic recirocity. Contents 1. Introduction 1 2. Dedekind Rings 3 3. Fermat s Last Theorem 7 4. Ramification Galois Theory Alied to Ramification 12 Acknowledgments 16 References Introduction It is not clear that the methods of algebra necessarily lend themselves to solving the roblems of number theory. There are a few examles, though, that will serve as motivation for the rest of this aer, and indeed historically motivated a significant amount of the develoments of algebraic number theory. These examles will exhibit some of the ower of algebraic abstraction when alied to the ring Z. Our first examle is known as the two squares theorem. Theorem 1.1. (Two squares)a rime number is the sum of two squares if and only if 1 (mod 4). In other words, we have a statement about solutions to the Diohantine equation x 2 + y 2 = z. In general, Diohantine equations are difficult to solve not because they involve multilication, and not because they involve addition, but because they involve both. So in order to deal with this equation we may wish to factor the left side using comlex numbers: (x + iy)(x iy) = z. We are thus led to consider the ring Z[i]. It is a fact that this ring, called the Gaussian integers is a rincial ideal domain; in fact it is a Euclidean domain, and this is not difficult to rove: for any Gaussian integers z, w we may calculate z/w first in the field of comlex numbers. Now it is easy to see geometrically that any comlex number is within a distance of 2/2 from some Gaussian integer. Let y be a Gaussian integer within 2/2 of z/w. Then wy is a Gaussian integer such that z wy < w. Date: Aug 24,

2 2 JOHN KOPPER Now let () denote the ideal generated by a rime integer within the ring Z[i]. Then () = (q 1 )(q 2 ) for rime ideals (q 1 ) and (q 2 ) if and only if Z[i]/() = Z[i]/((q 1 )(q 2 )) = Z[i]/(q 1 ) Z[i]/(q 2 ) is the roduct of two fields. But rewriting Z[i] as Z[X]/(X 2 + 1) we see that Z[i]/() = F [X]/(X 2 +1) is a roduct of fields if and only if X 2 +1 has a root (mod ). Now (F ) = Z/( 1)Z, so 1 is a square in F if and only if 1 ( 1)/2 (mod 1) is even. But this occurs if and only if 4 ( 1) 1 (mod 4). Assume 1 (mod 4). We can take norms so that 2 = N() = N(q 1 )N(q 2 ). Then since is rime and N(q 1 ) > 1, we have = N(q 1 ). But q 1 is a Gaussian integer q 1 = x + iy, and N(q 1 ) = x 2 + y 2, so is the sum of two squares. This roves the theorem. One may wish to rove similar theorems about solutions to Diohantine equations of the form x 2 + ny 2 = for rimes. It would be natural to aly a similar method and consider the ring Z[ n]. To do so, we need to characterize those rimes which slit into two distinct rincial rime ideals. It is easy to characterize those which slit into two rime ideals using the same method as above: it deends only on whether n is a square (mod ). Using the law of quadratic recirocity (5.14) this becomes a congruence condition (mod 4n). However, it can be very difficult to characterize those rimes which slit into rincial rime ideals. In the examle of the two squares theorem it was easy because Z[i] is a Euclidean domain, hence a rincial ideal domain. Thus the same argument will work for x 2 + 2y 2 = because Z[ 2] is also Euclidean. Unfortunately, this is not generally the case with Z[ n]. In fact, there is no guarantee that Z[ n] is a unique factorization domain, much less a rincial ideal domain, as we shall see in Section (2). The question of which rimes are of the form x 2 + ny 2 then becomes quite difficult. We will not be able to solve this roblem in this aer. Nevertheless, many of the methods used in solving the x 2 + y 2 = z 2 case esecially the use of ideals will inform much of our discussion. Another examle, and a articularly salient motivation in the historical develoment of algebraic number theory is the diohantine equation x n + y n = z n and the question, for which n are there integer solutions for x, y and z? The answer is Fermat s last theorem (1.2), a theorem which requires little introduction and less motivation. The theorem itself rovides surrisingly little insight into other roblems, even Diohantine roblems. Much more valuable are the roof and the attemts to rove it. Essentially all of the mathematics in this aer, including the modern definitions, are somehow insired by 20th century attemts to rove Fermat s last theorem. In Section (2) we will introduce the notion of Dedekind rings and unique factorization of ideals. This leads naturally to the definition of the ideal class grou and Kummer s artial roof of Fermat s last theorem in Section (3). Section (4) will be another alication of the theory of Dedekind rings in the context of ramification theory. We will then use the abstraction of ramification combined with Galois theory to rove the law of quadratic recirocity in Section (5). To begin, however, we will first introduce some of the historical background of Fermat s last theorem and elementary methods of trying to rove it.

3 THE THEORY OF NUMBERS IN DEDEKIND RINGS 3 Theorem 1.2. The equation x n + y n = z n has no nontrivial integer solutions for n 3. The first thing that should be noted when trying to rove this is that x, y, and z may be assumed to be relatively rime, since if d is a common divisor and x n + y n = z n then (x/d) n + (y/d) n = (z/d) n. Second, we may assume n is rime or a ower of 2. To see this, suose n = q is comosite and x n + y n = z n. Then x q + y q = z q, so x q, y q, and z q solve the Fermat equation for n =. It follows that we need only consider the cases where n is an odd rime or n = 4. Fermat himself is urorted to have solved at least the n = 3 and n = 4 cases. In Section (3) we will rove a significantly weaker version of Fermat s last theorem; first, however we will need a great deal of algebraic machinery. One of the more tantalizing asects of Fermat s last theorem is Fermat s claim to have found a marvelous roof. It is generally considered unlikely that Fermat knew an actual roof because a comlete roof wasn t ublished until a few years ago. Nevertheless, it s reasonable to assume that Fermat had discovered a artial roof. Siegel writes, [Fermat] might have conceived the idea...of working in the ring of integers of the field containing the nth roots of unity; he may have believed that such a ring is always a rincial ideal ring... For n rime, this ring is a rincial ideal ring only for finitely many values of n. C. L. Siegel (cf. Samuel [3]. 15) In articular, Fermat may have thought about cyclotomic extensions of Q, and not without good reason. We may factor Fermat s equation as (1.3) 1 (x + ζy) i = z i=0 where ζ is a rimitive th root of unity. We will now suress the subscrit. We shall soon see that the terms on the left side of Equation (1.3) are air-wise relatively rime. Suosing Z[ζ] were a unique factorization domain, we would have that (x + ζ i y) is a th ower. From this we can derive a contradiction. Unfortunately, Z[ζ] is not, in general, a unique factorization domain. In fact, the first rime for which Z[ζ ] is not a unique factorization domain is = Dedekind Rings Consider the ring Z[ 6], and the element 6: 3 2 = 6 = 6 6. But in Z[ 6], 6 doesn t divide 3 or 2. We see that Z[ 6] is not a unique factorization domain. This discovery was sufficiently disturbing to Kummer and Dedekind who each sought ways to rectify it. Dedekind s ideas are the ones which are more closely adhered to today; he introduced the notion of ideals, which in many cases retain the roerty of unique factorization into rimes. It is the object of this section to ursue the question of when secifically we can say that ideals in

4 4 JOHN KOPPER a ring will factor uniquely into rimes. But before we get to answer this question, we need some reliminary definitions and roositions. Definition 2.1. Let R be a ring and K a field containing R. Then α K is integral over R if there exists a monic olynomial P R[X] such that P (α) = 0. This aer will assume some familiarity with integrality and integral extensions. In articular, we shall assume without roof several facts about integral closures. The following roosition rovides an alternate definition of integrality which will be useful in roving the main theorem of this section. Proosition 2.2. Let R be a ring and K a field containing R. Then α K is integral over R if and only if R[α] is finitely generated as an R module. Proof. Clear. We ultimately wish to show that under the correct conditions ideals can be uniquely factored into rime ideals. These next two roositions rovide relatively weak statements about containing roducts of ideals, and will be essential in roving unique factorization. Proosition 2.3. Let R be a Noetherian integral domain. If a is a non-zero ideal of R, then it contains a roduct of non-zero rime ideals. Proof. Suose not. Let C be the collection of non-zero ideals of R which do not contain a roduct of non-zero rime ideals. By assumtion, C is not emty. Since R is Noetherian, there must be a maximal element m of C with resect to set inclusion. Clearly m is not rime. Thus there exist a, b R m such that ab m. Now m m + Ra and m m + Rb. By the maximality of m, there exist rime ideals 1,..., n, q 1,..., q k such that m + Ra 1 n m + Rb q 1 q k But ab m imlies (m + Ra)(m + Rb) m, thus 1 n q 1 q k m, a contradiction. Proosition 2.4. Let R be a ring and a rime ideal of R. Suose contains a roduct a 1 a n of ideals. Then there is some i such that a i. Proof. Suose not. Then for each i there exists a i a i. Since is rime, it follows that a 1 a 2 a n. But a 1 a 2 a n a 1 a 2 a n, a contradiction. To rove that ideals factor uniquely into rime ideals we need some assumtions on the ring in which the ideals reside; it is certainly not true in general that ideals can be uniquely factored. For examle, consider the ring Z[ 3]. Then the ideal (4) can be factored as (2)(2) and (1 + 3)(1 3). But these factorizations are distinct. The correct set of assumtions for unique factoriation of ideals are those of a Dedekind ring, defined below. Definition 2.5. Let R be an integral domain and K its field of fractions. Then R is a Dedekind ring (or Dedekind domain) if the following roerties hold. (i) R is Noetherian. (ii) R is integrally closed in K. (iii) If is a non-zero rime ideal of R then is maximal.

5 THE THEORY OF NUMBERS IN DEDEKIND RINGS 5 It can be shown that any rincial ideal domain is a Dedeking ring, but the converse is not true. However, we will see in the next section that most rings we are concerned with are Dedekind (cf. Proosition (3.2)). We require one last definition. Definition 2.6. Let R be an integral domain and K its field of fractions. Then if a is an R submodule of K, it is called a fractional ideal of R if there exists a non-zero r R such that ra R. If a and b are both fractional ideals of R then we define the roduct ab to be the set of all finite sums of the form a i b i such that a i a and b i b. Alternatively, one can define fractional ideals as an roduct ak where a is an ideal of R and k K. Note that fractional ideals aren t necessarily ideals, but all ideals of R are fractional ideals. Fractional ideals will be essential in roving the main theorem of this section, Theorem (2.9), as well as in defining the ideal class grou which will be a valuable tool in later sections. We would like to show that fractional ideals form a grou under multilication with identity R. Lemma 2.7. Let R be a Dedekind ring which is not a field and K its field of fractions. Then for every maximal ideal m of R there exists a fractional ideal m 1 of R such that mm 1 = R. Proof. Since R is not a field, m (0). Define (2.8) m 1 = {α K : αm R}. Clearly m 1 is an R submodule, and if 0 m m then mα R for all α m 1. Thus m 1 is a fractional ideal. It remains to show mm 1 = R. It is clear that mm 1 R, and that R m 1, so m mm 1 R. But m is maximal, so either (1) mm 1 = R, or (2) mm 1 = m. It therefore remains to show that (2) is imossible. Suose, for a contradiction, mm 1 = m. It follows that for all α m 1 and all n N, αm m α 2 m m.. α n m m. Thus if r is a non-zero element of m, we have rα n m R. That is, R[α] is a fractional ideal. Since R is Dedekind and therefore Noetherian, it follows that R[α] is finitely generated as an R module. Thus α is integral over R by Proosition (2.2). But R is integrally closed, so α R. Since R m 1, this imlies m 1 = R. Now let a be a non-zero element of m. Then by Proosition (2.3) there exist non-zero rime ideals 1,..., k such that Ra 1 2 k. Further, we may assume k is minimal. Then m Ra 1 2 k, so there is some index i such that m i by Proosition (2.4). Without loss of generality, assume i = 1. Since R is Dedekind, all rime ideals are maximal. Thus m = 1. We now have Ra m 2 3 k.

6 6 JOHN KOPPER Since k is minimal, Ra 2 3 k. Therefore there exists some b 2 3 k such that b Ra. But clearly mb Ra, so mba 1 R. By definition of m 1 we see that ba 1 m 1. But b Ra imlies ba 1 R, which contradicts m 1 = R. Theorem 2.9. (Unique factorization of rime ideals) Let R be a Dedekind ring, a a non-zero fractional ideal of R, and P the collection of non-zero rime ideals of R. Then a can be uniquely factored into rime ideals as (2.10) a = P n where the n are integers and all but finitely many are zero. Proof. Let a be a non-zero fractional ideal of R. Then there is some non-zero d R such that da R. Clearly da is an ideal of R. Since a = (da)(rd) 1, it suffices to rove the theorem for ideals of R rather than all fractional ideals. Let C denote the set of all non-zero ideals of R which are not roducts of rime ideals. Suose C is not emty. Since R is Noetherian, C has a maximal element m. Note that m = R is imossible since R is the emty roduct of rime ideals. Thus m is contained in a maximal ideal. By the lemma, has an inverse 1 such that 1 = R. Now m, so m 1 1 = R. By the definition of 1 in Equation (2.8), 1 R. Thus m 1 mr = m. Suose m 1 = m. Then if α 1 we have αm m,..., α n m m for all n N. As in the roof of the lemma, this imlies R[α] is finitely generated as an R module, thus α is integral over R, thus an element of R since R is integrally closed. Then 1 = R. But this is imossible since 1 = R, but R = R. This roves that m 1 m. Since m m 1, and m is maximal in C, we see that m 1 C. Thus there exist rime ideals 1,..., n such that m 1 = 1 n. That is, m = 1 n. It remains to show uniqueness. Recall that P is the collection of all non-zero rime ideals of R. Suose we have two equivalent roducts of rime ideals, (2.11) n = m. P P We would like to show n = m for all. We may rewrite (2.11) as n m = R. P Suose there are some rime ideals { i } for which n i m i searate ositive and negative exonents so that 0. Then we can k1 1 kr r = q j1 1 qjs s, where all exonents are ositive and q i l for all indices i, l. Clearly the left hand side is contained in 1, so the right hand side is as well. By Proosition (2.4), there is some i such that 1 q i. Since R is Dedekind, both 1 and q i are maximal. Thus 1 = q i, a contradiction.

7 THE THEORY OF NUMBERS IN DEDEKIND RINGS 7 In the course of the roof the theorem we showed that if a is an integral ideal then the exonents n in Equation (2.10) are non-negative. That is, a can be written as a unique roduct of rime ideals of R. This is not necessarily the case if a is a fractional ideal in the field of fractions of R. Corollary The set {a : a is a non-zero fractional ideal of R} is a grou under multilication. Let F (R) denote the grou of non-zero fractional ideals of R. It is easy to show that I(R) = {a : a is a rincial fractional ideal of R} is a subgrou of F (R). Definition With F (R) and I(R) as above, define C(R) = F (R)/I(R) to be the ideal class grou of R. We will use the letter h to denote the order of C(A). The number h is called the class number of R. An intuitive notion of the class number h is the extent to which the ring R fails to be a rincial ideal domain. Indeed since R has the roerty that every nonzero rime ideal is maximal, R is a unique factorization domain if and only if it is a rincial ideal domain. Proosition Let R be a Dedekind ring. Then R is a rincial ideal domain if and only if it is a unique factorization domain. Proof. It is well known that all rincial ideal domains are unique factorization domains. We therefore need only show that UFD imlies PID. Let I be an ideal of R. We may assume I is rime since any ideal is a roduct of rimes and a roduct of rincial ideals is rincial. Let a I. Then there is some irreducible r a such that r I. Thus (r) is rime. We have (0) (r) I. But (r) is maximal, so (r) = I. Corollary The ideal class grou C(R) is trivial if and only if R is a unique factorization domain. It is not immediately clear what the order of the ideal class grou is. However, in many circumstances it is enough to know that it is finite. This next theorem rovides sufficient conditions for this. Theorem Let K be a finite extension of Q and O K the corresonding ring of integers. Then C(O K ) is finite. Proof. The roof uses Minkowski s geometry of numbers. For the details, see Stewart and Tall [4] Fermat s Last Theorem The attemted roof of Fermat s last theorem discussed in the introduction will serve as our insiration for a artial roof due to Kummer. If we can show that Z[ζ] is a Dedekind domain, we will be able to ass to ideals in Equation (1.3) and derive a contradiction. In this section the symbol ζ n will always denote an nth root of unity. Proosition 3.1. Let ζ be a rimitive th root of unity. Then Z[ζ] is the ring of integers in Q(ζ). Proof. See Washington [5]. 1 2.

8 8 JOHN KOPPER Proosition 3.2. Let K be a number field and O K its ring of integers. Then O K is a Dedekind ring. Proof. We must verify roerties (i)-(iii) in Definition (2.5). Let n = [K : Q]. By the classification of finitely generated abelian grous, the ring O K is a Z submodule of a free Z module of rank n which is obviously torsion-free. Then if a is a non-zero ideal of O K, and α a, let f be the Z minimal olynomial for α. Then α f(0) so α a Z. Now let a a Z. Then O K /a is a quotient of O K /(a). Since O K is a free Z module, there is a Z module isomorhism O K /(a) (Z/aZ) n. Thus O K /a is finite. Now suose a 1 a 2 is a chain of ideals of O K. The set of ideals containing a 1 is in bijection with the ideals of O K /a 1, which is finite. Thus the chain has a maximal element, so O K is Noetherian. This roves (i). Further, O K /a is an integral domain. A finite integral domain is a field, and a is therefore maximal, roving roerty (ii). Finally, it follows from the transitivity of integrality that the ring of integers is integrally closed in its field of fractions, thus roerty (iii). For more details see Stewart and Tall [4]. 43. Proosition (3.2) can technically be made more general by considering an arbitrary ring R such that the field of fractions of R has characteristic zero, and any finite extension K of its field of fractions. For our uroses, however, number fields will suffice. The next two lemmas will also be essential and are reasonably easy to rove. The roofs can be found in Washington [5] chater 1. Lemma 3.3. Let u Z[ζ] be a unit. Then there exists some r Q(ζ + ζ 1 ) and n Z such that u = rζ n. Lemma 3.4. Let α Z[ζ]. Then there exists an integer a such that α a (mod ). Finally, we need one last definition. Kummer s artial roof of Fermat s last theorem requires a articular roerty of the exonent. Definition 3.5. A rime is called regular if it does not divide the class number of Q(ζ ). Many regular rimes are known. For examle, 2, 3, 5, 7, are all regular; in fact, 37 is the first irregular rime. It is not known, however, whether there are infinitely many regular rimes. Nevertheless, Kummer roved Fermat s last theorem for all regular rimes, which is certainly an imrovement on the case when the class number is 1, as discussed in the introduction. In fact, Kummer discovered a criterion for the regularity of rimes, related to the divisibility of Bernoulli numbers by. This gives a reasonably fast way to check regularity. Theorem 3.6. (Kummer) Let be an odd, regular rime. nontrivial integer solutions to x + y = z. Then there are no Proof. Recall that x, y, and z may be assumed to be relatively rime. First we shall assume 3, then deal with the case = 3 exlicitly. Further, it is imossible for x y z (mod ) to hold since then z 2z, and 3z 0. But 3z, so this cannot be. Now if x y we therefore know that x z. Thus we may rewrite the Fermat equation as x +( z) = ( y). We see that if there is a solution where

9 THE THEORY OF NUMBERS IN DEDEKIND RINGS 9 the first term is congruent to the second term, there is one where it is not. We thus assume x y (mod ). Suose, for a contradiction, that there exist integers such that x + y = z. There are now two cases: when divides xyz, and when it does not. In this aer we will tackle only the latter case. Begin by rewriting x + y = z and assing to ideals: (3.7) 1 (x + ζ i y) = (z). i=0 We first show that the ideals (x + ζ i y) are airwise relatively rime. Let be a rime ideal which contains (x + ζ i y) and (x + ζ j y) for i > j. Then contains (x + ζ i y) (x + ζ j y) = yζ j (1 ζ i j ). Now ζ j is a unit, and 1 ζ i j = u(1 ζ) for some unit u, so contains y or 1 ζ. An identical argument shows that contains x or 1 ζ. Since x and y are assumed to be relatively rime, it follows that must contain (1 ζ). But 1 ζ is rime, so = (1 ζ). Then x + y = x + ζ i y + (1 ζ i )y x + ζ i y 0 in Z[ζ]/, since x + y Z. Now consider the olynomial f(x) = X 1 + X X + 1, which factors as (X ζ)(x ζ 2 ) (X ζ 1 ). Clearly f(1) =, thus. Since Z must also be rime, we see that Z = Z. Since x and y are integers, and x + y, we have x + y Z. Then (mod ) we have 0 (x + y) x + y z So divides z, contradicting z. Therefore the terms (x + ζ i y) are airwise relatively rime. Consider again Equation (3.7): both sides must factor into the same roduct of rime ideals. On the right hand side we see that any factor of (z) must have ramification index a multile of. Since the ideals (x + ζ i y) are airwise relatively rime, each must be a th ower of some ideal a i. That is, for each i we have (x + ζ i y) = a i. In articular, a i is rincial. Consider a i as an element of the ideal class grou; the equivalence class of the ideal a i is the identity. But the class grou is finite, and is assumed to not divide its order, so a i itself must be the identity. That is, a i is rincial. Write a i = (α i ). Since (x + ζ i y) = (α i ), it is clear that x + ζi y and α i are equal u to some unit u = rζn as in Lemma (3.3). Assume i = 1. Now there is some integer a such that α 1 a (mod ), so x+ζy rζn a (mod ). Further, x + ζ 1 y = rζ n ᾱ by taking comlex conjugates. So x + ζ 1 y rζ n ā (mod ). But of course ā = a and =. Thus, or ζ n (x + ζy) ζ n (x + ζ 1 y) (mod ), x + ζy ζ 2n x ζ 2n 1 y 0 (mod ). We wish to show that 1, ζ, ζ 2n, ζ 2n 1 cannot be distinct. We have k = x + ζy ζ 2n x ζ 2n 1 y for some integer k. Note that {1, ζ, ζ 2n, ζ 2n 1 } may be extended to a basis for Z[ζ] as a Z module, so long as > 3. Thus x + ζy ζ 2n x ζ 2n 1 y is an exression of k in terms of a basis. It follows that divides x, ζy,... But this contradicts our

10 10 JOHN KOPPER assumtion. Thus 1, ζ, ζ 2n, ζ 2n 1 cannot be distinct. Now 1 ζ, and ζ 2n ζ 2n 1. We are thus left with three cases. (i) Suose 1 = ζ 2n. Then x + ζy x ζ 1 y 0 (mod ), or y(ζ ζ 1 ) 0 (mod ). This imlies divides y, a contradiction. (ii) Suose 1 = ζ 2n 1. Then ζ = ζ 2n. We then have (x y) (y x)ζ 0 (mod ), so (x y) (y x)ζ = α for some integer α Z[ζ]. We see that (x y) and (y x) are coefficients for α with resect to a basis for Z[ζ] as a Z module. Thus divides x y, so x y (mod ), contrary to assumtion. (iii) Suose ζ = ζ 2n 1. Then x ζ 2 x 0 (mod ). Then x 0 (mod ), a contradiction. It now remains to show that the case = 3 is imossible. All cubes mod 9 are ±1 or 0. Since 3 x it follows that x 3 ±1 (mod 9). The same holds for y and z. But then x 3 + y 3 2, 0, 2 ±1 (mod 9). Thus z 3 cannot be a cube, which is a contradiction. 4. Ramification Consider a Dedekind ring R and its field of fractions K. Assume the characteristic of K is zero. Recall from Theorem (2.9) that any a of R may be written as a unique roduct of rime ideals as in Equation (2.10). We wish to now consider what haens to a in the integral closure O L of R in a finite extension L of K. In the remarks following Proosition (3.2) we noted that O L is in fact a Dedekind ring itself. Thus if is a rime ideal of R, then O L is an ideal of O L and has a unique factorization n (4.1) O L = where each P i is rime in O L, and the e i are all ositive integers. It is not difficult to characterize the P i comletely. Proosition 4.2. With notation as above, the P i are recisely the rime ideals of O L whose intersection with R is. The P i are commonly said to lie over. Proof. Suose first that P is a rime factor of O L. Then O L P i, so P i R is an ideal of R containing. But it is obvious that P i R is rime, and so must be identically. On the other hand, if P has the roerty that P R =, then P. Thus P is a factor of ; by uniqueness of factorization, it must aear in the factorization. Definition 4.3. With notation as above, the integer e i in Equation (4.1) is the ramification index of P i over R. If e i > 1 for some i then is said to ramify in O L and L. If is a rime and P i lies above, then it is clear that, with notation as above, O L /P is a finite field extension of R/. We call degree of this extension, f i, the residual degree of P i over R. i=1 Theorem 4.4. Let R, K, O L, L be as above, and a rime ideal of R. Let P 1,..., P m be the rimes of O L lying over. Then m (4.5) e i f i = [L : K]. i=1 P ei i

11 THE THEORY OF NUMBERS IN DEDEKIND RINGS 11 Proof. The ring O L /O L is certainly a finitely generated R/ vector sace of dimension n = [L : K]. By the Chinese remainder theorem, we have (R/) n = OL /O L = O L /P ei i. Now O L /P i is a field extension of R/ of degree f i isomorhic to F q for some q. Thus O L /P ei i = F q [X]/X ei, which is thus (isomorhic to) an R module of rank e i f i. We wish to focus our attention on number fields. In this articular examle, we will consider quadratic extensions. Let d be a square-free integer, an odd rime and let S denote the ring of integers in Q( d). We first calculate S exlicitly. Proosition 4.6. Let d be a square-free integer. Then the ring of integers S in Q( d) is { Z[ d] d 2, 3 (mod 4) (4.7) [ 1 Z 2 (1 + ] d) d 1 (mod 4). Proof. Suose w S is integral over Z. Then w = a+b d for some a, b Q. Then w is a root of the olynomial (X w)(x σ(w)) = (X a b d)(x a + b d) where σ is the nontrivial element of Gal(Q( d)/q). This olynomial must have integer coefficients. Thus, 2a Z a 2 b 2 d Z Suose first that a Z. Then a = n/2 for some integer n. We have that n 2 4db 2 4Z. Since n 2 0 or 1 (mod 4), it follows that 4db 2 0 or 1 (mod 4) resectively. The second case occurs if and only if b 2 is not an integer. But b 2 is not an integer because this would imly a 2 is an integer, contrary to assumtion. However, d is square-free so if 4b 2 is not an integer then 4b 2 d is not either. Thus [ 4b 2 Z. 1 Evidently 2b is an integer. Thus 2a and 2b are integers, so S = Z 2 (1 + ] d). Further, since this is the case if and only if 4db 2 1 (mod 4), we necessarily have d 1 (mod 4). On the other hand, if a Z then db 2 Z. But d is square-free, so b Z. Thus in this case S = Z[ d], and d is necessarily congruent to 2 or 3 (mod 4). We may now revisit the roof and see that all imlications are actually double imlications, thus roving the roosition. Equied with this knowledge, we can now characterize recisely those rimes which ramify in quadratic extensions. Examle 4.8. Let d be a square-free integer and S be the ring of integers in Q( d) [ and an odd rime. Then consider the quotient S/(). Note first that if 1 S = Z 2 (1 + ] d) then the image of a + b 2 (1 + d) mod () is an element of Z[ d]/(). We can thus assume the case S = Z[ d]. Now Z[ d]/() = Z[X]/(X 2 d, ) = (Z/Z[X]) / (X 2 d). If X 2 d is irreducible in F [X] then the ideal it generates is rime, hence maximal. That is, S/() i

12 12 JOHN KOPPER is a field if and only if d is not a quadratic residue mod, and also if and only if () is rime in S. It is clear that the residual degree of () over Z is 2. Another ossibility is that X 2 d has two distinct roots ±ω in F. Then S/() = F /(X ω) F /(X + ω). It is clear that this occurs if and only if () slits in S. That is, if () is the roduct of two distinct rimes P 1, P 2 of S. Then f 1 = f 2 = e 1 = e 2 = 1. Finally, if X 2 d is a square mod, which occurs recisely when d or = 2, then S/() = F /(X 2 ). This is the last remaining ossibility: f = 1, e = 2. That is, ramifies in S. 5. Galois Theory Alied to Ramification This section assumes some familiarity with Galois theory. Our aim is to take the theory of ramification we have develoed and interret it in the context of Galois extensions. The methods of Galois theory lead to several alications, including a roof of quadratic recirocity which is given at the end of the section. Proosition 5.1. Let R be a Dedekind ring, K its field of fractions, and suose K has characteristic zero. Further, let L be a Galois extension of K. Define O L to be the integral closure of R in L. Then σ(o L ) = O L for all σ Gal(L/K). Proof. Suose α O L and σ is a Galois automorhism. Then α is a root of some olynomial X n + a 1 X n a n with coefficients in R. We see that σ(α) must also be a root of this olynomial, hence σ(o L ) O L. On the other hand, σ 1 (O L ) O L for the same reason. Proosition 5.2. With notation as above, let be a rime of R and P the collection of rimes lying over. Then the Galois grou Gal(L/K) acts transitively on P. Corollary 5.3. Let R, K, L, O L be as above. Then if is a maximal ideal of R, each of the rimes P lying over have the same ramifications indices e and same residual degrees f. Definition 5.4. With notation as above, then for any rime ideal P of O L we have a subgrou D P of the Galois grou given by D P = {σ Gal(L/K) : σ(p) = P}. We call D P the decomosition grou of P. Because the Galois grou acts transitively on rimes lying over, we see that the index of D P in Gal(L/K) is equal to the number of rimes lying over. Further, since R is Dedekind, and therefore so is its integral closure O L, it follows from Proosition (3.2) that P is maximal, hence O L /P is a field. Similarly, so is R/. Suose σ D P, thus σ(p) = P. Then σ restricts to an automorhism of O L such that there is an induced automorhism σ on O L /P and the following diagram commutes. R σ R R R/ σ R/ R/ Clearly R/ is fixed by each σ. Thus there is a ma ϕ : D P G with σ σ and where G is the set of automorhisms of O L /P which fix R/. Since O L /P / R/ is

13 THE THEORY OF NUMBERS IN DEDEKIND RINGS 13 a finite extension of finite fields it is a Galois extension. Thus ϕ is a homomorhism D P Gal(O L /P / R/). Definition 5.5. Assume the above notation and that O L /P is a Galois extension of R/. Let I P denote the kernel of ϕ. The grou I P is called the inertia subgrou of P. By definition, I P is a normal subgrou of D P. In fact, D P /I P = Gal(OL /P / R/). For a roof, see Marcus [1] Proosition 5.6. A rime of R ramifies in O L if and only if I P is not trivial for any P which lies over. Proof. By definition, [O L /P : R/] = f, where f is the residual degree of P over. Since O L /P is a Galois extension of R/, the Galois grou G has order f. Thus D P /I P has order f. On the other hand, G has order equal to ref where r is the number of rimes lying over and e is the ramification index. The Galois grou G acts transitively on the rimes lying over so we have r = [G : D P ]. Thus D P = ref/r = ef. Then [D P : I P ] = f imlies I P has order e. We will now aly the above theory to the secific case of number fields that is, to finite extensions of Q. Suose K and L are number fields with L a Galois extension of K. Let O K and O L denote the ring of integers in K and L, resectively. Then K/O K = Fq for some q = n with rime, and L/O L = Fq m for some integer m. In articular, Gal(L/K) is cyclic with generator F defined by F(x) = x q. Suose now is a rime of O K such that does not ramify in O L. By Corollary (5.6) we see that D P = Gal(OL /P /O K /) for all rimes P lying over. Thus D P is cyclic with generator σ. We see that σ(x) = x q (mod P). Definition 5.7. With notation as above, the generator σ of D P is called the Frobenius automorhism of P. We can define the Frobenius automorhism for ramified rimes as the coset of I P which is maed to the generator of Gal(O L /P /O K /) Since the Frobenius automorhism σ generates the Galois grou, it follows from Proosition (5.6) that σ has order f in D P. In the context of a quadratic extension, using Theorem (4.4) we see that σ must have order 1 or 2. This is a articularly imortant examle and motivates the remainder of this aer. We will conclude with a roof of the law of quadratic recirocity, but first we need some definitions. Definition 5.8. Let be a rime and n an integer such that n. Then we define the Legendre symbol ( n ) by ( ) { n 1 n is a quadratic residue (mod ) =. 1 n is not a quadratic residue (mod ). With this notation we can rewrite the results of Examle (4.8). Proosition 5.9. Let d be a squarefree integer, an odd rime, and S the ring of integers in Q( d). Then ramifies in S iff d ( ) d slits in S iff = 1 remains rime in S iff ( d ) = 1

14 14 JOHN KOPPER For a rime Z which does not ramify in some number field K, write σ for the associated Frobenius automorhism. Since σ must have order f, we see that σ is trivial if and only if e = 2, (cf. Theorem (4.4)) so must slit in K. On the other hand, if σ is the nontrivial automorhism, then f = 2, so necessarily remains rime in K. Identifying Gal(K/Q) with {1, 1} and by aealing to the above roosition, we may write σ = ( d ). That the Legendre symbol is multilicative in the numerator follows from Proosition (5.18). Therefore to calculate ( a ) it suffices to calculate ( q ) for rimes q dividing a. Making this calculation is essentially the content of the law of quadratic recirocity. However, to rove the law we will require some facts about cyclotomic fields. Proosition Let ζ be a rimitive nth root of unity. Then Q(ζ) is a Galois extension of Q and there exists a canonical isomorhism ϕ : Gal(Q(ζ)/Q) (Z/nZ). Further, if is a rime not dividing n and P is a rime lying over with corresonding Frobenius automorhism σ, then ϕ(σ) (mod n). Proof. By defintion of σ we have that σ(x) x (mod P) for all x O Q(ζ). Now σ is an automorhism of a cyclotomic extension. Therefore σ(ζ) = ζ i, where i = ϕ(σ). We thus have: ζ ϕ(σ) ζ (mod P). We wish to show ϕ(σ) (mod n). Suose not. Assume and ϕ(σ) have already been reduced (mod n). Then we may rewrite the above as ζ (1 ζ ϕ(σ) ) P. Since P is rime and ζ is a unit, it follows that (1 ζ ϕ(σ) ) P. But consider the olynomial F (X) = (X n )/(X 1) = X n 1 + X n , which factors as (X ζ)(x ζ 2 ) (X ζ n 1 ). We see that F (1) = n. It follows that (1 ζ) (1 ζ n 1 ) P. But n, so n P Z, a contradiction. It is clear that Gal(Q(ζ)/Q) (Z/nZ) since each Galois automorhism τ is determined by the ower to which it raises ζ. On the other hand, if q and are rimes and σ q, σ are the corresonding Frobenius automorhisms, ϕ(σ q σ ) = q. So if m Z/nZ and m has the rime factorization (in Z) 1 n, then ϕ(σ 1 σ n ) = m. Thus ϕ is surjective. Theorem Let ζ be a rimitive nth root of unity. Then for all rimes which ramify in Q(ζ), we have n. Proof. Suose n. Then using the above roosition we see that the Frobenius automorhism corresonding to is well defined. That is, I P is trivial and e = 1. In Proosition (5.9) we saw that we can relate the Legendre symbol to ramification in quadratic extensions. In order to rove the law of quadratic recirocity we will use the following classification of quadratic extensions which are subfields of cyclotomic extensions. Proosition Let be an odd rime and let ζ be a rimitive th root of unity. Then the field Q(ζ) has a unique quadratic subfield F, where { Q( ) 1 (mod 4) (5.13) F = Q(. ) 3 (mod 4)

15 THE THEORY OF NUMBERS IN DEDEKIND RINGS 15 Proof. We have seen that the Galois grou of Q(ζ)/Q is isomorhic to (Z/qZ). This grou has a unique subgrou of index 2, corresonding via isomorhism to a unique index 2 subgrou H of the Galois grou. The fixed field Q(ζ) H is therefore a quadratic extension of Q. For the remainder of the roof see Samuel [3]. 75. Theorem (Law of Quadratic Recirocity) Let and q be distinct odd rimes. Then ( ) ( ) q (5.15) = ( 1) ( 1)(q 1)/4 q Proof. Let L = Q(ζ) where ζ is a rimitive qth root of unity. Then L has a unique quadratic subfield K. By Proosition (5.12), K = Q( q) or K = Q( q). Now q, so does not ramify in L, hence does not ramify in K. Thus corresonding to we have a Frobenius automorhism σ Gal(L/Q). By definition, for all x K, σ(x) x q (mod O L ), thus σ(x) x q (mod O K ). Further, σ(o K ) = σ(o L K) = O K, so σ is in the decomosition grou. This roves that σ K is the Frobenius automorhism of for K/Q. Now Gal(L/Q) = F q, and Gal(K/Q) = F q /(F q ) 2, so σ K is trivial if and only if it is in the subgrou of Gal(L/Q) corresonding to (F q ) 2. We have seen in Proosition (5.10) that σ corresonds to (mod q) in F q. Thus σ K is trivial if and only if is a square in F q. We can identify Gal(K/Q) with the grou { 1, 1}. We see that (5.16) σ K = ( ). q Now if f denotes the residual degree of in K, we know that necessarily f is the order of σ K in Gal(K/Q) (cf. the remarks following Definition (5.7)). Thus σ K is trivial if and only if remains rime in K, and σ K is nontrivial if and only if slits in K. Using Proosition (5.9) we have, ( ) σ K = q if q 1 (mod 4) (5.17) ( ) σ K = q if q 3 (mod 4). Combining Equations (5.16) and (5.17) we have ( ) ( ) ±q = σ K = q ) (( 1) q 1 = = = 2 q ( ) ( q 1 ( ) q ) q 1 2 ( 1) 1 q Finding ( q ) when q = 2 is trivial. However, there is a rather glaring omission in the theorem: we have not accounted for the case = 2. We will solve this roblem using slightly more elementary methods. The only required roosition is Euler s Criterion.

16 16 JOHN KOPPER Proosition (Euler s Criterion) Let be a rime and a an integer such that a. Then ( ) a a 1 2 (mod ) Proof. By definition ( a ) = 1 if and only if there exists some n F such that n 2 = a. Now F = Z/( 1)Z, so equivalently there exists some m Z/( 1)Z such that 2m = a. That is, 1 2 a = 0. Writing this as an equation in F we have a 1 2 = 1. On the other hand, ( a ) = 1 then a 1 = 1, and a 1 2 is a square root of 1. But we have seen that a 1 2 = 1 ( a ) = 1, thus ( a ) = 1. Theorem (Legendre Symbol for 2) If is an odd rime then ( ) { 2 1 if 1 or 7 (mod 8) = 1 if 3 or 5 (mod 8). Proof. We have that (1 + i) 2 = 2i, so Then The result follows = (1 + i) i i i 1 + i i 1 2. Acknowledgments. It is my leasure to thank my mentor, Dan Le, for his atience and guidance. I d also like to thank Nick Ramsey for utting ideas in my head. References [1] Daniel Marcus. Number Fields. Sringer-Verlag New York, Inc [2] Serge Lang. Algebraic Number Theory. Second Edition. Sringer-Verlag New York, Inc [3] Pierre Samuel. Algebraic Theory of Numbers. Translated by Allan J. Silberger. Kershaw Publishing Comany Ltd [4] Ian Stewart and David Tall. Algebraic Number Theory and Fermat s Last Theorem. Third Edition. A K Peters, Ltd [5] Lawrence C. Washington Introduction to Cyclotomic Fields. Sringer-Verlag New York, Inc. 1982