THE CLASS NUMBER THEOREM

Transcription

1 THE CLASS NUMBER THEOREM TIMUR AKMAN-DUFFY Abstract. In basc number theory we encounter the class group (also known as the deal class group). Ths group measures the extent that a rng fals to be a prncpal deal doman. If the group s fnte, we call the order of the group the class number. Not every rng wll have a class number, but the class number theorem states that class group of the rng of of algebrac ntegers n an algebrac number feld s fnte. In ths paper we present a proof of ths theorem by buldng the foundatons needed. Although there are less lengthy proofs to the class number theorem, ths method establshes a sold understandng the class number theorem and number theoretc aspects of the class group. Contents 1. Localzaton 2 2. Dedeknd Rngs 2 3. Algebrac Number Felds 3 4. Fractonal Ideals and Class Groups Class Groups 5 5. Determnants, Traces, and Norms 6 6. Lattces 7 7. The Class Number Theorem 11 Acknowledgments 14 References 14 The class group s a useful noton n number theory. It can be thought as provdng a measurement on a rng, whch tells us how far the rng s from beng a prncpal deal doman and, consequently, degree of the falure of unque factorzaton (snce every prncpal deal doman s a unque factorzaton doman). In the case that ths group s fnte, the order of the group s defned as the class number. A class number of 1 ndcates that a rng s a prncpal deal doman, wth larger numbers ndcatng a hgher degree of falure for unque factorzaton. Understandng the class group yelds useful nsghts nto basc number theory and the structure of varous types of rngs. The prmary goal of ths paper s to defne and prove the class number theorem, whch states, the class group of the rng of algebrac ntegers n an algebrac number feld s fnte. In dong so, we wll also defne the objects that class groups study and the tools needed for the proof. Ths paper does not offer any orgnal materal, but ams to clearly explan the class number theorem and closely related topcs to a person wth a knowledge of abstract algebra. 1

2 2 TIMUR AKMAN-DUFFY We begn by defnng localzaton, whch wll be useful n later proofs and concepts. From there, we wll construct Dedeknd rngs and explan algebrac number felds. These wll be a prme focus of the paper and are the subject of the class number theorem. We wll then descrbe determnants and lattces, whch are the tools we need to fnally prove the class number theorem. 1. Localzaton Defnton 1.1. Let R be an ntegral doman. A nonempty subset S of R s a multplcatve set f S does not contan 0 and S contans the product of any two elements n S. Proposton 1.2. Let R be an ntegral doman and S a multplcatve set n R. There s a rng, denoted R S, whch contans a subrng somorphc to R and also contans multplcatve nverses of every element of S. R S s generated by R and s 1, where s S. The above statement may be made clearer f R S s defned as such: (1.3) R S = {r/s : r R, s S} wth the equvalence relaton on R S defned as (r, s) s equvalent to (q, t) f rt = qs. We defne 1 as 1 = s/s and we defne 0 as 0 = 0/s. Notce R S s an ntegral doman. Defnton 1.4. R S, as constructed above, s called the localzaton of R at S. It should be noted that r/s s the equvalence class of (r, s). We can localze an ntegral doman wth respect to any multplcatve set, but for our purposes we wll only focus on localzaton at prme deals. The followng proposton dsplays an mportant correspondence between prme deals of ntegral domans and ther localzatons. Proposton 1.5. Let R be an ntegral doman, and S be a multplcatve set n R. There s a one-to-one correspondence between the prme deals of R whch have empty ntersecton wth S and the prme deals of R S. The prme deal p of R corresponds to the deal pr S of R S. When we consder localzaton at a prme deal p, we are allowng R to be an ntegral doman and S to be the set of elements s R such that s / p. Notce that, snce p s a prme deal, S s a multplcatve set. For the sake of convenence, we wll wrte R p to represent R R p. Proposton 1.6. Let p be a prme deal of R. Then R p has only one maxmal deal, whch s pr p The proof follows from Proposton Dedeknd Rngs Defnton 2.1. A dscrete valuaton rng, or DVR, s s a prncpal deal doman wth only one maxmal deal. Notce that every feld s a DVR, but not every DVR must be a feld. For the rest of the paper, every DVR we refer to wll not be a feld.

3 THE CLASS NUMBER THEOREM 3 Theorem 2.2. Let R be a DVR (that s not a feld) and let π R such that the unque maxmal deal p = πr. R has the followng propertes. () R s a notheran rng. () Every nonzero x R has the form x = uπ k, where k s a nonnegatve nteger and u s a unt n R. () Every nonzero deal has the form Rπ k, where k s a nonnegatve nteger. (v) R s ntegrally closed. In other words, f we look at the feld of fractons of R, call t S, then every element of S that s ntegral over R s an element of R. (v) p s the only nonzero prme deal of R. Proof. () PIDs are notheran. () R s a PID, and thus a UFD. Thus, snce R s not a feld, t can only have one prme element up to unt multples. () Follows from () and the fact that all deals n R are prncpal. (v) Every UFD s ntegrally closed. (v) Snce R s not a feld, π and p are not zero. Snce p s a nonzero maxmal deal, t s prme. By (), all deals have the form Rπ k, so there are no other prme deals. Defnton 2.3. A rng R s a Dedeknd rng f t s a notheran ntegral doman such that the localzaton R p s a DVR for every prme deal p of R. Theorem 2.4. If R s a Dedeknd rng, then R S s a Dedeknd rng. Dedeknd rngs have many nterestng propertes, especally wth regards to deals. Theorem 2.5. Every nonzero prme deal of a Dedeknd rng R s a maxmal deal. Proof. Assume not. Then there exst dstnct, nonzero prme deals p 1, p 2, such that p 1 p 2. By Proposton 1.5, p 1 R p2 p 2 R p2 s a chan dstnct prme deals n R p2. But R p2 s a DVR, and thus only has one nonzero prme deal. Contradcton. Theorem 2.6. Let U be a nonzero deal of a Dedeknd rng R. Then U = p a1 1...pan n, where p 1,..., p n are dstnct prme deals unquely determned by U and a 1,..., a n are postve ntegers unquely determned by U. We do not necessarly have unque factorzaton nto rreducble elements n Dedeknd domans (we wll use an example to show ths n the next secton). However, the prevous Theorem shows that, by usng prme deals, we can establsh a noton of unque factorzaton n Dedeknd rngs. Ths gves the number theoretc feel to Dedeknd rngs. 3. Algebrac Number Felds Our goal n ths paper s to prove that the class group s fnte for the rng of algebrac ntegers n an algebrac number feld. In ths secton, we wll defne these terms and ther relaton to Dedeknd rngs. Defnton 3.1. Gven a rng R and a rng extenson S, an element of S s ntegral over R f t s the root of a monc polynomal wth coeffcents n R. Defnton 3.2. A rng R s ntegrally closed f every element of S that s ntegral over R s an element of R.

4 4 TIMUR AKMAN-DUFFY Defnton 3.3. The ntegral closure of a rng R s the set of all elements of S that are ntegral over R. Defnton 3.4. An algebrac number feld s a fnte dmensonal extenson feld, call t L, of the ratonal number feld Q. Defnton 3.5. An element of an algebrac number feld, l L, s an algebrac nteger f l s ntegral over the rng of ntegers Z. Proposton 3.6. Consder a Dedeknd rng R wth quotent feld K. Let L be a fnte dmensonal extenson feld of K. The ntegral closure of R n L wll result n a Dedeknd rng. Notce that, snce Z s a Dedeknd rng, the rng of algebrac ntegers n an algebrac number feld s also a Dedeknd rng. It s a fact that the rng of algebrac ntegers can be computed f gven a prmtve element. For the proof, please refer to page 47 of Janusz [1]. We wll now show that a Dedeknd rng does not necessarly have unque factorzaton nto rreducbles. In other words, not every Dedeknd rng s a UFD. Example 3.7. Consder Z[ 5]. By Proposton 3.6, t s a Dedeknd rng. Notce that, n Z[ 5], 21 = 3 7 = ( )(1 2 5). We want to show that these both are rreducble factorzatons of 21. Let us defne a norm such that N(a + b 5) = a 2 + 5b 2. We know an element x s a unt f N(x) = 1. Thus, we can clearly see that 1 and 1 are the only unts. Obvously 3 and 7 are rreducble. Let s assume s not rreducble (the exact same argument wll work for 1 2 5). Then = xy. But N( ) = 21. So N(x)N(y) = 21. Thus the possble values for N(x) are 1, 3, 7, or 21. But, by the defnton of the norm, N(x) 3, 7. Thus ether x or y must be a unt. Contradcton. Hence Z[ 5] s a Dedeknd rng that s not a UFD. 4. Fractonal Ideals and Class Groups In order to understand class groups, we must defne what a fractonal deal s. Unless otherwse specfed, let R be a Dedeknd rng and K be the quotent feld of R. Defnton 4.1. A fractonal deal of a Dedeknd Rng R s a nonzero fntely generated R-submodule of K. Defnton 4.2. If M s a fractonal deal of R, then M 1 s the set {x K : xm R}. M 1 s called the nverse of M. Defnton 4.3. A regular element s a nonzero element of a rng that s nether a rght nor a left zero dvsor. Defnton 4.4. A regular deal s an deal of a rng that contans a regular element of the rng. Proposton 4.5. A fractonal deal s, up to a constant, a regular deal.

5 THE CLASS NUMBER THEOREM 5 Proof. We know that a fractonal deal s fntely generated. Consder a fractonal deal a whch s generated by {a 1,..., a n } of a rng R. We can pck elements b 1,..., b n K and a regular element d such that d = a 1 b 1 d a n b n d a. Thus a contans a regular element, and therefore t s a regular deal. From ths Proposton, we can see that all nonzero fractonal deals are nvertble. Notce that M need not be contaned n R. Consder the followng example: Example 4.6. For smplcty, let s consder the rng of ntegers, Z. Z s a PID, so by defnton t s a Dedeknd rng. Consder 1 2Z. By defnton ths s a fractonal deal, but t s not contaned n Z. We also note that every deal n a Dedeknd rng s fntely generated, thus every deal n a Dedeknd rng s a fractonal deal. Now we know enough about fractonal deals to defne what a class group s. However, we wll observe some basc propertes about fractonal deals frst, whch wll be relevant later. Defnton 4.7. A fractonal deal M s nvertble f MM 1 = R. We wll use what we set up n Example 4.6 to llustrate ths defnton. Example 4.8. Let M = 1 2 Z. By defnton, M 1 = {2n : n N} Hence MM 1 = Z. Thus M s nvertble. In the prevous example, t s easy to see that every nonzero fractonal deal of Z s nvertble. Ths leads us nto the followng lemma, whch we wll state wthout proof. Lemma 4.9. An ntegral doman R s a Dedeknd rng f and only f every nonzero deal of R s nvertble Class Groups. Usng the basc concepts of fractonal deals we just learned, we can defne a class group. Frst, let us defne some symbols. I(R) represents the collecton of all fractonal deals of R. We wll call t the deal group of R. P(R) represents the collecton of all prncpal deals of R. Notce t s a subgroup of I(R). We can now defne a class group. Defnton A class group, represented by C(R), s the quotent (4.11) C(R) = I(R)/P(R). Notce that f R s a PID, then C(R) = 1. In other words, the class group s of order 1. We can nterpret C(R) to be a measure of how far R s from beng a PID. Notce that, f we have a class group of order 1, we know that t s a UFD. Thus, we can also nterpret the class number as the extent to whch unque factorzaton nto rreducble elements fals n R. We wll prove that C(R) s fnte f R s the rng of algebrac ntegers n an algebrac number feld. However, C(R) does not have to be fnte for arbtrary Dedeknd rngs.

6 6 TIMUR AKMAN-DUFFY 5. Determnants, Traces, and Norms In ths secton, we wll regard K as a feld and L as a fnte dmensonal extenson feld of K. Defnton 5.1. Each element a L gves rse to a functon r a : L L such that r a : y ya. If we select a bass u 1,..., u n for L over K, then we can construct a matrx for r a, call t [b j ], where r a (u ) = u a = j b j u j. Ths composte mappng s called the regular representaton of L over K. Obvously our choce of bass wll change the matrx. However, regardless of our bass, the trace and determnant wll reman the same. Defntons 5.2. (1) The trace of L over K s the functon T L/K (x) = trace(r x ). (2) The norm of L over K s the functon N L/K (x) = det(r x ). Remark 5.3. Unless otherwse specfed, we wll use T (x) for T L/K (x) and N(x) for N L/K (x). Proposton 5.4. Let x, y L and a K. Then: () T (x + y) = T (x) + T (y). () T (ax) = at (x). () N(xy) = N(x)N(y). (v) N(ax) = a n N(x) where n = [L : K], the dmenson of L over K. (v) Let E be a subfeld of L that contans K. Then T L/K (x) = T E/K (T L/E (x)) for all x L. Proof. () through (v) are straghtforward. (v) requres some work, and we wll prove t here. Pck a bass E over K, a 1,..., a k, and a bass L over E, b 1,..., b m. Let x L and y E. Then xb = j β j (x)b j, yb p = q α pq (y)a q, where β j (x) E and α pq (y) K. Then T E/K (y) = p α pp (y), T L/E (x) = β (x). Notce also that a b j gve a bass of L over K, xa s b t = a s β tj (x)b j = α s (βtj(x))a b j. j j Thus, T L/K (x) = α pp (β (x)) = T E/K (T L/E (x)). p

7 THE CLASS NUMBER THEOREM 7 When dealng wth Galos groups and separable extensons, useful propertes emerge. Let K L F be a chan of separable extensons such that F s Galos over K. Defne the Galos group F over K by G = G(F/K) and the Galos group F over L by H = G(F/L). Notce that H s the subgroup of G that fxes L elementwse. Let σ 1 H, σ 2 H,..., σ n H be the dstnct cosets of H n G. n = [L : K] and σ are the dstnct njectons over K of L nto a normal extenson of K. If we consder x L wth descrbed notaton above, the followng propertes hold. Proposton 5.5. () T L/K (x) = σ 1 (x) σ n (x). () N L/K = σ 1 (x)...σ n (x). Defnton 5.6. Let R be a Dedeknd rng wth quotent feld K. L s the fnte dmensonal separable extenson feld of K, and R s the ntegral closure of R n L. Let x 1,..., x n be a bass of L over K. The dscrmnant of the bass s (5.7) (x 1,..., x n ) = det[t L/K (x x j )]. Defnton 5.8. Usng the notaton gven n Defnton 5.6, f all x R, then T L/K (x x j ) R, thus (x 1,..., x n ) R. We can let x 1,..., x n range over all bases of L over K that le n R. Takng the dscrmnants of these bases generates an deal (R /R), called the dscrmnant deal of R over R. We state a useful Lemma that shows a dscrmnant deal can be determned locally. Lemma 5.9. Let R be a Dedeknd rng wth quotent feld K. Let R be the ntegral closure of R n L, and let S be a multplcatve set n R. Then (5.10) (R S/R S ) = (R /R) S. 6. Lattces What we learn n ths secton wll be ntegral n provng the class number theorem. Unless otherwse stated, V s an n-dmensonal R Eucldean space. Defnton 6.1. An abelan subgroup L of a real vector space V s called a lattce f (6.2) L = Zv Zv r. Defnton 6.3. A lattce L s called a full lattce f r from Equaton 6.2 s the dmenson of of V over R. If we are lookng at full lattces, then there s a very natural set of vectors to choose for our lattce, namely, the bass vectors for V. Defnton 6.4. Let v 1,..., v n be a bass for V and L a full lattce. Then we call the set (6.5) T = {r 1 v r n v n : 0 r < 1, 1 n}. the fundamental parallelepped for L.

8 8 TIMUR AKMAN-DUFFY Consder such a parallelepped. Notce that each vector of the bass v can be wrtten as v = α j u j. j where u 1,..., u n s the orthonormal bass of V. Defnton 6.6. The volume of L s defned as (6.7) vol(t ) = det[α j ]. It s a fact that, regardless of what bass s chosen for a gven V, the volume of the resultng fundamental parallelepped s does not vary. Ths can be seen n Janusz [1], page 63. Lemma 6.8. Let L be a full lattce n V. Let T be a fundamental parallelepped of L. Then the translates, λ + T, where λ L, are parwse dsjont and cover V. Proof. Let v 1,..., v n be our bass for V. Then we can wrte any v V as v = s v, where s R. We can express s as s = n + r, where n Z and 0 r < 1. Thus v = n v + r v, so we have expressed v as the sum of an element of L and and element of T. Thus the translates cover V. Now let s prove they are parwse dsjont. Suppose λ 1 + T and λ 2 + T have a common pont. Then λ 1 λ 2 L T. But L T = 0, snce Zv r v. Thus λ 1 = λ 2, and we conclude the translates are parwse dsjont. Defnton 6.9. A sphere n V s the set (6.10) U(m) = {r 1 v r n v n : r r 2 n m 2 } where v 1,..., v n s a bass for V. Defnton A subset of V s bounded f t s contaned n some sphere. Theorem An addtve subgroup L of V s a lattce f and only f every sphere contans only a fnte number of ponts n L. Proof. Let L be a lattce wth bass vectors v 1,..., v r. (In the case that r < n, we can extend ths set to the bass v 1,..., v n and replace L wth Zv.) By defnton, any sphere s contaned n U(m) usng bass vectors v 1,..., v n. So let s examne L U(m). Consder the vector n v L U(m). Then each n Z and n m by Defnton 6.9. Thus, only fnte many n are contaned n the ntersecton, so the sphere only contans a fnte number of ponts n L. We wll use nducton on the dmenson n of V. Let V = Rv 1, whch has dmenson 1. There are two cases: () L = 0. Then t s a lattce. () L 0. Let r 1 v 1 be a nonzero element of L. Thus U( r 1 ) has at least one element and a fnte number of elements from L. Now select r > 0 such that v = rv 1 L and r s mnmal wth respect to ths. Thus L Zv.

9 THE CLASS NUMBER THEOREM 9 Now pck sv L, such that s R. We can wrte s = m + q, where m Z and 0 q < 1. We know qv = qrv 1 L. But r s mnmal, so 0 qr < r mples q = 0. Thus L Zv, and therefore L = Zv. Now we wll consder n > 1. By the nducton hypothess, we can assume that L s not contaned n any proper subspace of V. So we can wrte RL = V, whch means we can select a bass of V contaned n L. We wll represent ths bass as v 1,..., v n. Let V 0 be the subspace spanned by v 1,..., v n 1. By nducton, we know L 0 = L V 0 s a full lattce n V 0. Now pck a bass of L 0 represented by w 1,..., w n 1. We can make ths a bass of V by addng one element, w 1,..., w n 1, v n. Hence, any element of L can be wrtten as such: n 1 (6.13) λ = r w + r n v n. Snce t s a lattce, f r n Z, then r Z for all. If we subtract an element of L 0 from such λ we may obtan an element such that r < 1 f 1 n 1. Notce that there are a only fnte number of λ L where r n s bounded above. Pck λ such that 0 < r n and r n s mnmal. Call ths w n. Then w 1,..., w n s a bass for V. For a general element λ L, we can wrte n λ = a w, a R. 1 We want to show that ths s a lattce. If we can show that all the a are ntegers, we wll be done. Let s explore the two cases. (1) We can subtract an nteger multple of w n such that a n = 0. Thus λ L 0, and thus L s a lattce. (2) We can subtract an nteger multple of w n such that 0 < a n < 1. Then we can subtract an element of L 0 so that a < 1 for all 1 n 1. But then we wll have an element n the form of Equaton 6.13 where the coeffcent of v n s 0 < a n r n < r n. Ths contradctons our choce of a mnmal r n. Thus ths case cannot occur. Ths completes the proof. Defnton A set X n V s convex f, for all x, y X, (x + y)/2 X. Defnton A set X n V s centrally symmetrc f, for all x X, x X. Lemma Let T be a fundamental parallelepped of L. If Y s a bounded subset of V and the translates λ + Y, λ L, are dsjont, then vol(y ) vol(t ) Proof. We are gven that Y s bounded, so t s contaned n some sphere. By Theorem 6.12, ths sphere only contans a fnte number of ponts n L. Thus, there can only be fntely many λ such that (λ + T ) Y s nonempty. By Lemma 6.16, we know that all the nonempty ntersectons are parwse dsjont and cover Y, so vol(y ) = λ vol[(λ + T ) Y ]. Notce (λ + T ) Y = [T (Y λ)] + λ.

10 10 TIMUR AKMAN-DUFFY Thus vol[(λ + T ) Y ] = vol[[t (Y λ)] + λ] = vol[t (Y λ)]. Y λ s a translate of Y, and by the Lemma we know that the translates of Y are dsjont. However, we do not know f vol[t (Y λ)] covers T, as λ the assumptons of of Lemma 6.16 are not met. Thus, we can only clam that vol[t (Y λ)] vol(t ). But vol(y ) = vol[t (Y λ)]. Hence λ λ vol(y ) vol(t ). Theorem 6.17 (Mnkowsk s Theorem). Let L be a full lattce n vector space V where V s dmenson n over R. Let X be a bounded, centrally symmetrc, convex subset of V. If vol(x) > 2 n vol(l), then X contans a nonzero pont of L. Proof. Let s consder X as gven above. We wll construct the set 1 2X such that { } X = 2 x : x X. The volume of the set s [ ] 1 vol 2 X = vol(2 1 X) = 2 n vol(x) > vol(l). By Lemma 6.16, f the translates of 1 2 X were dsjont, then vol[ 1 2X] vol(l). Snce ths s not the case, we can assume there exst λ 1 λ 2 such that 1 2 x + λ 1 = 1 2 y + λ 2, where x, y X. By smple manpulaton, we conclude that (x y)/2 = λ 2 λ 1 0. Thus there exsts a nonzero pont of L that s n X. Before we move on, we wll state a quck proposton about the volume of a centrally symmetrc, convex subsets n an n-dmensonal R vector space. Ths wll help us n the next secton. Proposton Let n = r + 2s, where r, s are nonnegatve. We wll represent ponts n the n-dmensonal R vector space V by the usual n-tuples. For a postve real number, t, defne a set (6.19) X t = {(x 1,..., x r, y 1, z 1,..., y s, z s ) : x x r +2 y1 2 + z ys 2 + zs 2 < t} where all the coordnates are real. Then For proof, see page 66 of Janusz [1]. vol(x t ) = 2r s π s t n. n!

11 THE CLASS NUMBER THEOREM The Class Number Theorem Fnally, we have the tools necessary to prove what we are nterested n: Theorem 7.1. The class group of the rng of algebrac ntegers n an algebrac number feld s fnte. We wll prove ths Theorem by frst provng several related propostons. Once we have these tools, the fnal proof wll be qute straghtforward. Frst, we wll defne our terms. We denote our algebrac number feld by K and the rng of algebrac ntegers n K by R. C(R) wll represent the class group of the rng of algebrac ntegers, as defned n Secton 4. E wll be a normal extenson of Q whch contans K. The Galos group of E/Q, usually wrtten as G(E/Q), wll be wrtten as G for smplcty. Smlarly, H = G(E/K). We wll let σ 1,..., σ n represent all the dstnct cosets, σh, n G. Thus, n s the degree of K over Q and σ are the embeddngs of K nto C. Notce that, for all, σ (K) ether les n the real feld or t has a complex root. Let s organze them based on where they le. We wll choose all the σ such that σ (K) les n the real feld R. We wll label them σ 1,..., σ r. (Note: r may equal 0.) For the remanng n r cosets, notce we have σ j σ j. Thus, there must be an even number of cosets remanng. We can organze them as such: σ r+1,..., σ r+s, σ r+1,..., σ r+s. Hence, n = r + 2s. Ths should look qute famlar (see Proposton 6.18). From ths, we can compute actual vectors n R n. Let s consder the functon v : K R r C s defned as, for x K, v (x) = (σ 1 (x),..., σ r (x),..., σ r+s (x)). We can defne C as a two dmensonal R vector space. Thus, we can wrte C s as a 2s dmensonal R vector space as such: For x K, we wrte (7.2) v(x) = (σ 1 (x),..., σ r (x), Re(σ r+1 (x)), Im(σ r+1 (x)),..., Re(σ r+s (x)), Im(σ r+s (x))). Now we make a computaton that wll be useful for future theorems. Lemma 7.3. Pck bass a 1,..., a n of K over Q. Let M = [v(a m )] (where v(x) s defned as above) be an n n matrx wth row m equal to v(a m ). Now defne D to be the n n matrx such that row m equals Then where 2 = 1. σ 1 (a m ),..., σ r (a m ), σ r+1 (a m ),..., σ r+s (a m ), σ r+1 (a m ),..., σ r+s (a m ). det(m) = ( 2) s det(d) Proof. The proof s essentally row and column manpulaton. In matrx D add columns r+1 and r+2 together n order to obtan, n row k, the entry 2Re(σ r+1 (a k )). We wll now multply column r + 1 by 1 2. Subtract ths from column r + 2 to obtan the entry Im(σ r+1 (a k )). Repeat ths wth all the conjugate columns. Now we have a matrx that s almost the same M, wth the only dfference beng a factor of 2 n s columns and a factor of n s columns. Thus, the determnants det(m) and det(d) dffer by a factor of ( 2) s, whch s what we wanted.

12 12 TIMUR AKMAN-DUFFY Defnton 7.4. Let U be a fractonal deal of R. The dscrmnant (U/Z) of U s the deal of Z generated by (a 1,..., a n ), where (a 1,..., a n ) run through all the bases of K over Q contaned n U. Proposton 7.5. Let our terms be as they were n the defnton above. Let M = [v(a )]. Then (7.6) (U/Z) = ( 4) s det(m) 2 Proof. Pck a matrx D as defned n Lemma 7.3. Consder DD t. By Proposton 5.5(), the, j entry of the matrx s T K/Q (a a j ). Thus Also, by Defnton 7.3, Ths leads to the result det(dd t ) = det(d) 2 = (U/Z). det(d) 2 = ( 4) s det(m) 2. (U/Z) = ( 4) s det(m) 2. Proposton 7.7. (U/Z) = N (U) 2 (R/Z) where N (U) s the number of elements of R/U. Proof. It s suffcent to prove that the the equaton holds after localzaton at every maxmal deal of Z. Let p be prme and S = Z pz. Let U S = ar S for some a R S. Pck a bass of R over Z, x 1,..., x n. We know ths s stll a bass of R S over Z S. Then ax 1,..., ax n s a Z S bass of U S. Usng Defnton 5.1, we can construct a regular representaton mappng r a : y ya wth respect to the bass x 1,..., x n. We wll call the resultng matrx ρ a. Thus T K/Q (ax ax j ) = ρ a [T K/Q (x x j )]ρ t a. We know that N K/Q (a) = det(ρ a ), so Lastly, we need the equaton (ax 1,..., ax n ) = N K/Q (a) 2 (x 1,..., x n ). N K/Q (U S ) = N K/Q (a)z S = N Z S. Combnng the three prevous equatons, we obtan the equaton (7.8) (U S /Z S ) = N (U S ) 2 (R S /Z S ). By Lemma 5.9 the result follows. Theorem 7.9. Let U be a nonzero deal n R, the rng of algebrac ntegers. Let v : K R n be as n Equaton 7.2. Then v(u) s a full lattce, and ts volume s 2 s (R/Z) 1/2 N (U). Proof. Frst, we want to show that v(u) s a full lattce. Begn by pckng a Z bass of U, a 1,..., a n. Clearly, v(a 1 ),..., v(a n ) s a Z bass of v(u). Thus, the vectors v(a 1 ),..., v(a n ) are lnearly ndependent, and we can conclude that v(u) s a full lattce. Now, we wll compute the volume. Construct a matrx M such that row equals v(a ). Then det(m) equals the volume of the lattce. By Proposton 7.5, (7.10) det(m) = 2 s (U/Z) 1/2.

13 THE CLASS NUMBER THEOREM 13 By Proposton 7.7, we know (U/Z) = N (U) 2 (R/Z). If we substtute ths nto Equaton 7.10, we obtan the result det(m) = 2 s (R/Z) 1/2 N (U). Theorem Let U be a nonzero deal of R. a U, a 0, such that N K/Q (a) (n!/n n )(4/π) s (R/Z) 1/2 N (U). Then there exsts an element Proof. Let the set X t be as n Equaton Then X t s a centrally symmetrc convex subset of R n, whch allows us to use what we learned n the prevous secton. By Proposton 6.18, we see that (7.12) vol(x t ) = 2r s π s t n. n! By Theorem 7.9, we know (7.13) v(u) = 2 s (R/Z) 1/2 N (U). By Theorem 6.17, f vol(x t ) > 2 n vol(u), then X t contans a nonzero pont of v(u). We see, by Equatons 7.12 and 7.13, that ths nequalty holds f ( ) 2 (7.14) t n n r = ɛ + n! N (U) R 1/2. π s As ɛ approaches zero, we wll get a pont n X t v(u). By Theorem 6.12, we know that the lattce contans a fnte number of ponts n any sphere. Thus even f ɛ = 0, the ntersecton contans a nonzero pont. Assume t s selected as n Equaton 7.14, where ɛ = 0. Pck a nonzero a U such that v(a) = (σ 1 (a),..., σ r (a), Re(σ r+1 (a)), Im(σ r+1 (a)),...) = (x 1,..., x r, y 1, z 1,..., y s, z s ) s n X t. We know, from Secton 5, that r s N K/Q (a) = σ (a) σ r+j (a) 2 = x 1... x r (y1 2 + z1)...(y 2 s 2 + zs). 2 =1 j=1 We now apply the Arthmetc-Geometrc Mean Inequalty, whch states ( n ) 1/n n 1 b b n, for nonnegatve real b, and conclude N K/Q (a) 1/n 1 n x + 2 j yj 2 + z2 j < t n. Replace t n by the value gven n Equaton 7.14, and we are gven the desred equaton, N K/Q (a) (n!/n n )(4/π) s (R/Z) 1/2 N (U). For the next Theorem, we need the followng defnton.

14 14 TIMUR AKMAN-DUFFY Defnton Two deals, U and B, are equvalent f U = Bc for some nonzero c. The equvalence class of U s represented by [U]. Theorem 7.16 (The Mnkowsk Bound). For any nonzero fractonal deal B of R there s an U [B] such that U R and N (U) (n!/n n )(4/π) s 1/2. Proof. Pck b B such that B 1 = bb 1 s an deal n R whch les n [B 1 ]. By Theorem 7.11 there exsts a nonzero a U 0 such that N K/Q (a) N (B 1 ) 1 (n!/n n )(4/π) s 1/2. Let U = ab 1 1. Then U R (snce a B 1 ). Thus N (U) = N (ab 1 1 ) = N K/Q(a) N (B 1 ) 1. Ths proves the Theorem. Notce that the prevous Theorem proves that the bound on N (U) depends only on K. Wth ths, we are ready to prove our man Theorem. Theorem The class group of the rng of algebrac ntegers n an algebrac number feld s fnte. Proof. By Theorem 7.16, every class contans an ntegral deal, call t U, such that N (U) (n!/n n )(4/π) s 1/2 = M. Now all we need to prove s that there are only fntely many ntegral deals such that the norm s below the fxed bound M. We can factor U nto powers of prme deals: U = B a1 1...Bat t, where B are dstnct prme deals and a are postve ntegers. Now let p Z = Z B, where p s a postve prme. Suppose N (U) = p fa M where f s the relatve degree of B over Z. Notce that there must be fntely many p, and each p s only dvsble by fntely many B j. Thus there exst only fntely many B. It s also clear that there are only fntely many possble a (snce each p fa M). Hence there are only fntely many U such that N (U) M. Thus every class group s represented by only a fnte number of deals. Thus the class group of the rng of algebrac ntegers n an algebrac number feld s fnte. Acknowledgments. I would lke to thank my mentors Aaron Marcus and Emly Norton for sharng ther knowledge and ther extensve patence. I greatly apprecate ther help. References [1] Gerald J. Janusz Algebrac Number Felds, Second Edton Amercan Mathematcal Socety, 1996 [2] Anthony W. Knapp Basc Algebra Brkhauser, 2006 [3] Ian Stewart Galos Theory, Thrd Edton Chapman and Hall/CRC, 2004