MATH 242: Algebraic number theory

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1 MATH 4: Algebraic number theory Matthew Morrow Contents 1 A review of some algebra Quadratic residues and quadratic recirocity 4 3 Algebraic numbers and algebraic integers 1 4 Algebraic number fields First examle: Gaussian integers Second examle: Z[ω] where ω e πi/ Third examle: Z[ 5] Introductory tools for studying number fields: Norm, Trace, Discriminant, and Integral bases Norm and Trace Discriminant Integral bases Alications of integral bases to number fields Main theoretic roerties of D F The class grou, its finiteness, and cancellation of ideals Dedekind domains and Unique factorisation of ideals Norms of ideals Exlicitly constructing ideals in D F and generators of Cl F 37 7 Calculations of class grous of quadratic extensions, and alications d d d d d d d d 17, d d d, 3, 5, 6, 7, 11, 13, 17, 1, 9, 33, 37, Cyclotomic extensions and Fermat s Last Theorem Q(ζ) and its ring of integers Fermat s Last Theorem Ramification theory Ramification in quadratic extensions and quadratic recirocity Ramification in cyclotomic extensions A new roof of quadratic recirocity 6 1

2 MATH 4 Algebraic Number Theory 1 A review of some algebra In this course all rings R are commutative with unity. Algebraic number theory historically began as a study of factorization, and so we begin by reviewing roerties of factorization in general commutative rings. You are exected to be familiar with this material, which may be found in any standard algebra textbook, since it will gradually be needed in the course. Definition 1.1. Let R be a ring (commutative with unity!). (i) R is an integral domain if and only if whenever a, b R satisfy ab 0, then a 0 or b 0. (ii) We write a b to mean that a divides b, i.e. that there exists c R such that ac b. (iii) An integral domain R is a rincial ideal domain (PID) if and only if every ideal I of R is rincial, i.e. I a for some a R, where a ar is the rincial ideal generated by a; this notation will be used throughout. (iv) An element u of R is called a unit if and only if it has a multilicative inverse. R denotes the grou of units. An element of R is called irreducible if and only if it is not a unit and whenever ab for some a, b R then a or b is a unit. Two elements a, b R are called associates if and only if there is a unit u R such that a bu. An integral domain R is a unique factorization domain (UFD) if and only if every non-unit of R is a finite roduct of irreducible elements and whenever two such roducts 1... r, q 1... q s are equal, then r s and, u to reordering, i and q i are associates. (v) An integral domain R is a Euclidean domain (ED) if and only if there is a ma with the following two roerties: ν : R \ {0} Z 0 For all a, b R with b 0, there exist q, r R such that a bq + r, where either ν(r) < ν(b) or r 0. For all non-zero a, b R, ν(a) ν(ab). The ma ν is then called a Euclidean norm on R. (vi) R is a ED R is a PID R is a UFD. Examle 1.. The following examles will frequently aear and should be familiar to you: (i) The ring of integers Z is a ED, with Euclidean norm ν(n) n. Hence it is also a PID and a UFD. (ii) If F is a field then the olynomial algebra F [X] is a ED with Euclidean norm ν(f(x)) deg f(x) d, if f(x) a 0 + a 1 X + + a d X d with a d 0. Hence it is also a PID and a UFD. Next we review the rings Z/nZ, which also occur frequently in algebraic number theory: Definition 1.3. Given n Z, the ring Z/nZ is the quotient of Z by the rincial ideal nz. Given x, y Z, we say x is congruent to y mod n, and write x y mod n if and only if n x y, which is equivalent to saying that x + nz y + nz, or that x and y have the same class in Z/nZ. We may write x mod n to mean the class of x in Z/nZ, when it is not likely to cause confusion. Note: I will never use the notation Z n for Z/nZ. If x Z, then x mod n is a unit in the ring Z/nZ if and only if x is corime to n. In articular, if N is a rime number, then (Z/Z) is a grou of order 1: any element is equal to exactly one of 1,,..., 1 mod. The following results will be used:

3 MATH 4 Algebraic Number Theory 3 Theorem 1.4. Let N be a rime number. (i) ( Fermat s little theorem ) If a Z is corime to, then a 1 1 mod. (ii) (Z/Z) is a cyclic grou. Proof. (i): [0.1, FRA]. (ii): [3.6, FRA]; more generally, any finite subgrou of the multilicative grou of a field is cyclic. Corollary 1.5. Let N be a rime number. Then there exists g Z with the following roerties: (i) If a Z is corime to, then a g r mod for some r 0. (ii) If r Z is such that g r 1 mod, then 1 r. Proof. Let g Z be such that g mod generates the cyclic grou (Z/Z) ; then (i) and (ii) are restatements of that fact that (Z/Z) is cyclic of order 1. The classical name for an integer g with the roerties of the following corollary is a rimitive root modulo, but we will use this notation very little. 3

4 4 MATH 4 Algebraic Number Theory Quadratic residues and quadratic recirocity In this section we study so-called quadratic residues modulo an odd rime number, which is essentially an analysis of when the equation X a mod has an integer solution, for a fixed value of a Z. These results and ideas will frequently reaear during the course when we study exlicit examles, while the key theorem, namely the Law of Quadratic Recirocity, will lay a key role in describing the arithmetic of quadratic number fields in section 9.1. Definition.1. Let 3 be a rime number, and suose that a Z is corime to. Then a is said to be a quadratic residue modulo if and only if there exists x Z satisfying x a mod. (Note: The condition that a is corime to is imortant; 0 is not a quadratic residue modulo, even though 0 0 mod.) Examle.. Here are basic examles: (i) 3 is a quadratic residue modulo 13, since mod 13. (ii) 0 0 mod 3, 1 1 mod 3, and 1 mod 3. So if x is an integer then x 0 or 1 mod 3. Therefore is not a quadratic residue mod 3. The following characterisation of quadratic residues is needed for later roofs: Lemma.3. Let 3 be a rime number and suose that a Z is corime to. Then a is a quadratic residue modulo if and only if a 1 1 mod. Proof. : Suose first that a is a quadratic residue modulo ; then a x mod for some x Z. Then x is also corime to, and so x 1 1 mod by Fermat s little theorem. Therefore a 1 (x ) 1 x 1 1 mod. : Conversely, assume that a 1 1 mod. Let g be a rimitive root modulo (see the end of section 1); then a g r for some r 0. Our assumtion imlies that (g r ) 1 1 mod. As g mod generates the cyclic grou (Z/Z) of order 1, this imlies that 1 r 1. That is, r/ is an integer, and so a (g r/ ) mod ; i.e. a is a quadratic residue modulo. In order to develo a way of maniulating quadratic residues, and to clarify their roerties, the following iece of notation is absolutely fundamental: Definition.4. The Legendre symbol, for 3 a rime number and a Z, is defined by ( ) a 1 if a and a is a quadratic residue modulo, 1 if a and a is not a quadratic residue modulo, 0 if a. Examle.5. To clarify the definition we offer the following examles: (i) 3 is a quadratic residue modulo 13, by the revious examle, so ( 3 13) 1. (ii) ( 5 is not a quadratic residue modulo 7 (since 1 6 1, 5, mod 7), so 5 ) (iii) ( ) since 3 divides 6. 4

5 MATH 4 Algebraic Number Theory 5 The Legendre symbol is a convenient tool for discussing and maniulating quadratic residues; the following are some of its key roerties: Lemma.6. Let 3 be a rime number and let a, b Z. Then ( ) (i) a 1 mod ( Euler s lemma ); (ii) ( a ab ) ( a ) ( b ) ; ( (iii) If a b mod then a ) ( b ). Proof. If divides a or b then these identities are all trivial; so assume that does not divide a or b. (i) By Fermat s little theorem, a 1 1 mod. So divides a 1 1 (a 1 1)(a 1 + 1), whence divides a 1 1 or a 1 + 1, i.e. a 1 ±1 mod. But according to lemma.3, a is a quadratic residue modulo if and only if a 1 1 mod. (ii) Since (ab) 1 a 1 b 1, art (i) imlies that ( and since is odd, it follows that ab ) ( a ) ( ( b ab ). ) ( (iii) This is quite clear: the definition of a quadratic residue only deends on a mod. a ) ( b ) mod. As all these terms are ±1, From the roosition we obtain an imortant and useful corollary: Proosition.7 (Legendre symbol of 1). Let 3 be a rime number. Then ( ) (i) ( 1) 1 ; 1 (ii) 1 is a quadratic residue modulo if and only if is congruent to 1 modulo 4. Proof. For (i), aly art (i) of the revious roosition to a 1, and then, as in the revious roosition, use the fact that a congruence mod when both sides are either 1 or 1 is actually an equality. For (ii), note that if 1 mod 4 then ( 1) 1 1 1, whereas if 1 mod 4 then 3 mod 4 and so ( 1) 1 1. Here is an interesting alication of the results so far: Corollary.8. There are infinitely many ositive rime numbers which are congruent to 1 mod 4. Proof. Suose not, and let { 1,..., n } be the finite set of all ositive rime numbers which are 1 mod 4. Put M ( 1... n ) + 1; then M > 1, so M is divisible by some ositive rime number. Since M is odd,. ( ) Now observe that 1 ( 1... n ) 1 mod, so 1; the revious roosition therefore imlies that 1 mod 4. So i for some i, whence M 1 mod ; this contradicts M. Remark.9. A dee result, beyond this course, is the following theorem of Dirichlet: if a, n are corime integers, then there are infintely many rime numbers which are congruent to a modulo n. As we develo more tools during the course, we will see other secial cases of this result. 5

6 6 MATH 4 Algebraic Number Theory The aim of this section is to rove the next two theorems (due to Gauss, Legendre, and Eisenstein), which allow us to easily calculate any Legendre symbol and therefore determine exactly when a is a quadratic residue modulo. Their more theoretic imortance will become clear in section 9.1. The roofs will be ostoned until the end of the section. Theorem.10 (Legendre symbol of ). 3 a rime number. Then ( ) { 1 ±1 mod 8 1 ±3 mod 8. Exercise.1 (Restatement of the Leg. sym. of theorem). Suose that b is an odd integer; show that b 1 is divisible by 8, and that b 1 8 is even if b ±1 mod 8 and is odd if b ±3 mod 8. Deduce that the Legendre symbol of theorem can be rewritten as the statement that ( ) ( 1) ( 1)/8. This is sometimes a useful formulation. Examle mod 8 and so is not a quadratic residue modulo 61; checking this any other way but by using the theorem would be extremely time consuming mod 8 and so is a quadratic residue mod 73; notice that we have roved this without actually finding any integer x satisfying x mod 73. The following is the second main theorem. Desite its simle statement, it is an extraordinary result, telling us that ) if, q ) 3 are rime numbers then there is a mysterious relationshi between the Legendre symbols and. A riori, there is absolutely no reason that these should be related: although both ( q ( q symbols encode the solubility of an equation, the first is really an equation in Z/Z and the second is in Z/qZ. These two equations should have nothing to do with one another! Theorem.1 (Law of Quadratic Recirocity). Let, q 3 be distinct rimes. Then ( q ) ( ) q ( 1) 1 q 1. Exercise. (Restatement of the LQR). Show that the Law of Quadratic Recirocity can be restated in the follow way: If, q 3 are rime numbers (not necessarily distinct), then ( ) ( ) { q 1 if or q is 1 mod 4 ε, where ε q 1 if both and q are 3 mod 4 It is almost always in this form that one uses the Law of Quadratic Recirocity. Examle.13. We now demonstrate the ower of the revious two theorems by effortlessly calculating some Legendre symbols: (i) (ii) ( ) ( ) ( ) ( ( ) ( ) ( ) ) ( ) ( ) 59 3 ( ) ( ) 61 7 ( ) 5 7 ( ) 7 5 ( ) 1 5 ( ) ( ) 1 4 ( 1) (3 1)/

7 MATH 4 Algebraic Number Theory 7 (iii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Before roving the two main theorems, we rovide some samle alications to demonstrate their usefullness in questions of number theory; this week s homework includes modifications of all these results: Proosition.14. an odd rime. Then is a quadratic residue modulo if and only if is congruent to 1 or 3 modulo 8. Proof. We have ( ) ( 1 ) ( ) ( 1) ( 1)/ ( 1) ( 1)/8 ( 1) ( 1)/+( 1)/8, and One now checks that mod ( 1)( + 5) 8 ( 1)( + 5) mod. (e.g. 3 mod 8 8m 3 ( 1)(+5) 8 8m mod ). Therefore mod 8 ( ) Proosition.15. There are infinitely many ositive rimes in Z which are congruent to 1 modulo 8. Proof. As usual, we rove this by contradiction, assuming that the set of all such rimes is finite: { 1,..., m }. Let M 8( 1... m ) 1, and let M q r qrt t be the rime factorization of M. If q i 1 mod 8 for all i, then M q r qrt t 1 r rt 1 mod 8, which is false. Therefore there is i such that q i 1 mod 8. Notice that q i is odd since M is odd. Moreover, we have ( m ) (M + 1) mod q i, so is a quadratic residue modulo q i. The Legendre symbol of theorem now imlies q i ±1 mod 8, and so q i 1 mod 8. Therefore q i j for some j, whence we get the usual contradiction: q 1... m and q M imlies q. Proosition.16. Let > 3 be a Fermat rime (this means n + 1 for some n 1). Then 3 is a rimitive root modulo. 7

8 8 MATH 4 Algebraic Number Theory Proof. Since the grou Z/Z has 1 n elements, any element g of the grou which satisfies g n 1 1 is a generator. Therefore it is enough to rove ( ) that 3 n 1 1 mod. ( ) But n 1 ( 1)/ and 3 ( 1)/ 3 3 ; therefore we must show that 1; i.e. that 3 is not a quadratic residue mod. ( ) 3 Since 1 mod 4, the Law of Quadratic Recirocity imlies ( ) 3 ; since 1 mod 3, we see that ( 1) n + 1 mod 3, which is 0 or modulo 3. But is rime, ( so) not divisible by 3. ( Therefore ) 3 3 mod 3; is not a quadratic residue modulo 3, so now we know 1. Therefore 1, comleting the roof. We now begin the roofs of the two main theorems: Legendre symbol of and Law of Quadratic Recirocity. Although Ireland and Rosen follows Gauss methods, we will follow a roof due to Eisenstein (there are at least 33 different roofs in existence today 1 ). The following technical lemma is essential: Lemma.17 (Eisenstein s lemma). Let be an odd rime, and a Z not divisible by ; then ( ) a ( 1) s, where s ( 1)/ k1 ka. Proof. For any integer k, we will write r(k) to denote the unique integer in the range 0,..., 1 which is congruent to k modulo ; in other words, r(k) is the remainder when k is divided by. The roof is a little tricky so we will label the stes: (1). Firstly, if 1 l 1 then r(( 1) l l) is even and non-zero. Indeed, if l is even then r(( 1) l l) r(l) l (which is even); if l is odd, then r(( 1) l l) r( l) l (which is even); and it is non-zero since r(( 1) l l) ( 1) l l 0 mod. (). So, if k {1,,..., ( 1)/} then ka is corime to and therefore r(ka) {1,..., 1}; so aragrah (1), with l r(ka), shows r(( 1) r(ka) r(ka)) {, 4,..., 1}. Thus there is a well-defined ma R : {1,..., ( 1)/} {, 4,..., 1}, k r(( 1) r(ka) r(ka)), which we next rove is injective. So suose that 1 k, k ( 1)/ and R(k) R(k ); this imlies ( 1) r(ka) r(ka) ( 1) r(k a) r(k a) mod. Therefore ( 1) r(ka) ka ( 1) r(k a) k a mod ; since a is a unit modulo, we may divide by it to deduce that ( 1) r(ka) k ( 1) r(k a) k mod. So k εk mod, where ε ( 1) r(k a) r(ka) {0, 1}. But k, k are both in the range 1 k, k ( 1)/, so k k is imossible, and k k haens if and only if k k. This roves that R is injective. (3). Since the codomain and domain of R both have cardinality ( 1)/, we see that R is actually a bijection: in other words, as k runs over the integers 1,,..., ( 1)/, then r(( 1) r(ka) r(ka)) runs over the integers, 4,..., ( 1). Therefore ( 1)/ k1 r(( 1) r(ka) r(ka)) 1 htt:// ( 1)/ l1 l, 8

9 MATH 4 Algebraic Number Theory 9 and so ( 1)/ l1 l ( 1)/ (4). With identity ( ) established, we may now comlete the roof: ( 1)/ a ( 1)/ k1 k k1 ( 1)/ k1 ( 1)/ k1 ( 1) r(ka) r(ka) mod. ( ) ka r(ka) mod ( 1) ( 1)/ k r(ka) ( 1) r(ka) r(ka) mod k1 ( 1) ( 1)/ k r(ka) k1 k mod Since ( 1)/ k1 k is corime to, it follows that a ( 1)/ ( 1) k ( ) ( ) r(ka) mod. But Euler s lemma says a ( 1)/ a a, so ( 1) k r(ka) mod. Since two owers of ( 1) are congruent modulo if and ( ) a only if they are equal, we deduce ( 1) k r(ka). Finally, the definition of the floor function and r function mean that ka ka + r(ka), ( ) ka a which imlies r(ka) mod. Therefore ( 1) k ka, as required. Now we may rove the first of the main theorems: Theorem.18 (Proof of Legendre symbol of ). an odd rime. Then ( ) { 1 ±1 mod 8 1 ±3 mod 8 Proof. If k is an integer in the range 1 k /4 then 1 4k, so actually 1 4k < (since is 4k rime) and so 0 4k/ < 1; therefore 0. If instead k is the in the range /4 < k < ( 1)/ then ( ) a similar argument shows 1. So ( 1) s (by Eisenstein s lemma) where s 4k ( 1)/ k1 k k s.t. /4 <k ( 1)/ Therefore we must show that 4 is { even ±1 mod 8 odd ±3 mod 8 So, write 8c + ε for some c Z and some ε { 3, 1, 1, 3}. 9

10 10 MATH 4 Algebraic Number Theory If ε ±1 then 4 4m ± 1/ m ± 1/4 { (4m 1) (m 1) ε 1 4m m ε 1 m 0 mod Secondly, if ε ±3 then 4 4m ± 3/ m ± 3/4 { (4m ) (m 1) ε 3 (4m + 1) m ε 3 { m 3 ε 3 m + 1 ε 3 1 mod And now we rove the second main theorem: Theorem.19 (Proof of the Law of Quadratic Recirocity). Let, q be distinct odd rimes. Then ( ) ( ) q ( 1) 1 q 1. q Proof. We will study the following diagram in the (x, y)-lane: Let Λ be the set of lattice oints, i.e. oints with integer coordinates, which are strictly inside the rectangle ACHF, i.e. Λ {(x, y) Z : 0 < x <, 0 < y < q}. 10

11 MATH 4 Algebraic Number Theory 11 We are going to count certain geometrically-defined subsets of Λ in different ways and comare the results. Let Λ(EF G) Λ denote the set of lattice oints strictly inside the triangle EF G, and Λ(EF G) e (res. Λ(EF G) o ) those oints with even (res. odd) x-coordinate. Use the same style of notation for other triangles DEF, etc. and quadrilaterals CEGH, etc. The ma R : Λ Λ, (x, y) ( x, q y) is a bijection, geometrically given by rotation of 180 about the oint E (/, q/). Restricted to Λ(EF G) o, it gives a bijection R : Λ(EF G) o Λ(BCE) e, and so Secondly, #Λ(EF G) o #Λ(BCE) e. (1) Λ(BCE) e Λ(CEGH) e Λ(BCGH) e {(x, y) Z : / < x <, x even, 0 < y < q}, which has even cardinality ( + )/4 (q 1); therefore #Λ(BCE) e #Λ(CEGH) e mod. () Next, for any fixed integer m 1, the number of lattice oints inside the triangle CF H with x-coordinate equalling m is mq/, and so ( 1)/ kq #Λ(CF H) e. (3) Putting these identities together gives k1 #Λ(EF G) #Λ(EF G) o + #Λ(EF G) e #Λ(BCE) e + #Λ(EF G) e (by (1)) #Λ(CEGH) e + #Λ(EF G) e mod (by ()) #Λ(CF H) e ( 1)/ kq, (by (3)) k1 and so ( 1) #Λ(EF G) ( 1) ( 1)/ k1 kq ( q ). Reeating the argument with and q swaed imlies ( 1) #Λ(DEF ) ( ) q. so Finally, as required. 1 q 1 #Λ(DEF G) #Λ(DEF ) + #Λ(EF G), ( 1) 1 q 1 ( 1) #Λ(DEF G) ( 1) #Λ(DEF ) ( 1) #Λ(EF G) ( q ) ( ), q This comletes the roofs of the main theorems on quadratic residues and quadratic recirocity. 11

12 1 MATH 4 Algebraic Number Theory 3 Algebraic numbers and algebraic integers Now we introduce the field of algebraic numbers and the ring of algebraic integers; these will be the main objects of study in this course. For the moment we study them one at a time, before introducing number fields in the next section. Definition 3.1. An algebraic number is a comlex number α for which there exist a 0,..., a n 1 Q (n 1) such that α n + a n 1 α n a 1 α + a 0 0. An algebraic integer is a comlex number α for which there exist a 0,..., a n 1 Z (n 1) such that α n + a n 1 α n a 1 α + a 0 0. The set of algebraic numbers is denoted Q alg, and the set of algebraic integers is denoted Z alg (Ireland and Rosen use the notation Ω). Examle 3.. Some remarks and examles: (i) Any α Q is an algebraic number (take n 1, a 0 α), and every α Z is an algebraic integer (again take n 1, a 0 α). (ii) Every algebraic integer is an algebraic number. (iii) 5 is an algebraic integer, since (iv) Let ω 1+ 3 ; then ω + ω + 1 0, so ω is an algebraic integer (even though it looks like a fraction ). (v) If n 1 then ζ : e πi/n is an algebraic integer, since ζ n 1 0. (vi) If α is an algebraic number and r Q then rα is an algebraic number. Moreover, it is ossible to find a non-zero r Z such that rα is an algebraic integer; this is on the homework. (vii) π and e are not algebraic numbers, though these are difficult theorems which we will not discuss in this course. The following rovides a useful abstract test for whether a comlex number is an algebraic number/integer. We will use it many times in the course. Consider subsets V C such that V is a vector sace over Q, i.e. v, w V v + w V v V, r Q rv V. For examle, V Q, V {r + s 5 : r, s Q}. Note also that V even contains a finitely generated abelian grou M {a + b 5 : a, b Z}. Lemma 3.3. Let α C. Then (i) α is an algebraic number if and only if there exists a non-zero finite dimensional vector sace V C over Q such that αv V. (ii) α is an algebraic integer if and only if there exists a non-zero finitely generated grou M C such that αm M. 1

13 MATH 4 Algebraic Number Theory 13 Proof. (i). : Assume α is an algebraic number. So there exist a 0,..., a n 1 Q such that Let α n + a n 1 α n a 1 α + a 0 0. V {r 0 + r 1 α + + r n 1 α n 1 : r 0,..., r n 1 Q}, and notice that V is a non-zero vector sace over Q of dimension at most n. We claim that αv V. So let v V, and write v r 0 + r 0 α + + r n 1 α n 1 ; then αv r 0 α + r 1 α + + r n 1 α n r 0 α + r 1 α + + r n α n 1 r n 1 (a n 1 α n a 1 α + a 0 ) a 0 + (r 0 a 1 )α + (r 1 a )α + + (r n a n 1 )α n 1 V. : Assume that there is a non-zero finite dimensional Q vector sace V C such that αv V. Let v 1,..., v d V be a basis of V as Q vector sace. For each i, αv i V, so we may write αv i n r i,j v j j1 for some r i,1,..., r i,d Q. In other words, the Q-linear ma V V, v αv has matrix R (r i,j ) 1 i,j d with resect to the basis v 1,..., v d. Now think of R as a matrix over C; it satisfies d v 1 j1 r 1,jv 1 v 1 R.. α.. v d d j1 r d,jv d v d So α is a comlex eigenvalue of R and therefore det(αi d d R) 0; exanding the determinant gives a monic olynomial of degree d with coefficients in Q of which α is a root. Therefore α is an algebraic number. (ii). : Imitate art (i), relacing V by the finitely generated abelian grou M {m 0 + m 1 α + + m n 1 α n 1 : m 0,..., m n 1 Z}. : Imitate art (ii); the matrix R will have coefficients in Z and so the olynomial obtained by exanding det(αi d d A) will also have coefficients in Z. We use the lemma to rove: Proosition 3.4. Q alg is a subfield of C. Z alg is a subring of C. Proof. Let α, β be algebraic numbers. According the revious lemma, there are non-zero finite dimensional vector saces V, W C over Q such that αv V and βw W. Let V W : {v 1 w v t w t : v i V, w i W } be the set of all finite sums of roducts vw for v V, w W. Observe the following: (i) Let v 1,..., v n san V and w 1,..., w m san W. Then V W is sanned by (v i w j ) i,j ; so V W is finite dimensional over Q. Also, V W 0. (ii) If v V and w W then αvw (αv)w V W and βvw v(βw) V W. Therefore, by linearity, if z V W then αz, βz V W. 13

14 14 MATH 4 Algebraic Number Theory So if z V W we see that αz, βz V W, so (α + β)z αz + βz V W ; also, since βz V W, we have α(βz) V W, i.e. (αβ)z V W. By the condition of the revious lemma, α + β and αβ are algebraic numbers. Secondly, if α, β are actually algebraic integers, then the revious lemma rovides us with non-zero finitely generated abelian grous M, N C such that αm M, βn N; we may then exactly reeat the argument above for algebraic numbers, relacing V W by MN {v 1 w v t w t : v i M, w i N}. This shows that α, β are also algebraic integers. It remains only to check that if α is a non-zero algebraic number then so is α 1. There exist a 0,..., a n 1 Q such that α n + a n 1 α n a 1 α + a 0 0. Let i 0 be the smallest index such that a i 0. Then 0 α n + a n 1 α n a i α i α i (α n i + a n 1 α n i a i ) and so α n i + a n 1 α n i a i 0. Multily this by α i n a 1 i therefore α 1 is an algebraic number. to get (α 1 ) i n + a i+1 a 1 i (α 1 ) i n a 1 i 0; Exercise 3.1. Let f(x) be a monic olynomial whose coefficients are algebraic integers. In this roblem you will show that any comlex root of f(x) is also an algebraic integer. (i) Prove that if α 0,..., α d 1 are algebraic integers, then there is a non-zero finitely generated abelian grou M C such that α i M M for i 0,..., d 1. (ii) Let α be a comlex root of f(x). Show that there exists a non-zero finitely generated abelian grou N C such that αn N. Deduce that α is an algebraic integer. (iii) Show that Q alg is an algebraically closed field, i.e. every non-constant olynomial with coefficients in Q alg has a root in Q alg. (Hint: you can either rerove analogues of the revious two results for algebraic numbers or, more quickly, use question (6) to reduce to the case of algebraic integers. You may quote the fact that C is an algebraically closed field.) Examle 3.5. (i) Every element of Z[ 5] can be written as a + b 5 for some a, b Z; since 5 and a, b Z are all algebraic integers, we see that every element of Z[ 5] is an algebraic integer. For examle, α : 3 5 is an algebraic number, and it is not hard to find an equation of which it is a root: so α 4α ( 3 5) ( 3 5) 49, (ii) + 3 is an algebraic integer since it is the sum of two algebraic integers. Let s find an integer olynomial it satisfies: So ( + 3) 4 10( + 3) ( + 3) ( + 3) ( + 3) 1 14

15 MATH 4 Algebraic Number Theory 15 Among all monic olynomials in Q[X] satisfied by an algebraic number, there is an otimal one, which you robably encountered in Algebra II: Proosition 3.6. Let α Q alg be non-zero. There there is a unique monic, irreducible olynomial f(x) Q[X] such that f(α) 0. Moreover, if g(x) Q[X] is any olynomial satisfying g(α) 0 then f(x) g(x). Proof. This should be familiar from Alg II, but here is a roof. Let I {h(x) Q[X] : h(α) 0}. If h 1, h I then h 1 + h I, and if h I, g Q[X] then gh I; that is, I is an ideal of Q[X]. Since α is an algebraic number, there exists a non-zero element of I, and since α 0, I Q[X]. Therefore I is a non-zero, roer ideal of Q[X]; as Q[X] is a rincial ideal domain, there exists a nonconstant olynomial f I satisfying I f. Write f a n X n + + a 0 with a n 0; then a n is a unit in Q[X] and so a 1 n f f I. Therefore, after relacing f by a 1 n f, we may assume f is monic. Obviously f(α) 0, and if g is any olynomial satisfying g(α) 0 then g I, so f divides g. For a contradiction, suose that f is not irreducible. Then there exist monic olynomials g, h Q[X] such that f gh and such that deg g, deg h < deg f. Thus 0 f(α) g(α)h(α), so one of g(α), h(α) is zero, i.e. f divides g or h; but this is imossible since deg g, deg h < deg f. It remains to rove that f is unique. Suose that F is also a monic irreducible olynomial f(x) such F (α) 0. Then we have just seen that f divides F, i.e. F fg for some g Q[X]. But F is irreducible, so this forces f or g to be a unit; f is not a constant olynomial, so it is not a unit. So g is a unit, i.e. constant olynomial; as F and f are both monic, this forces g 1, i.e. F f. Definition 3.7. If α Q alg then the unique monic, irreducible olynomial f(x) Q[X] satisfying f(α) 0 is called the minimal olynomial of α. The degree of α is defined to be the degree of its minimal olynomial. If α Q alg, write Q(α) {h(α) : h Q[X]}; similarly, if α Z alg, write Z(α) {h(α) : h Z[X]}. These are obviously both subrings of C: indeed, Q(α) is the smallest subring of C containing Q and α, while Z[α] is the smallest subring of C containing Z and α. Examle 3.8. Every element of Q( 5) (res. Z[ 5]) can be written as a + b 5 for some a, b Q (res. a, b Z). The following roosition demonstrates this is full generality. Proosition 3.9. Let α Q alg, with minimal olynomial f(x) and degree d. Then (i) As a vector sace over Q, Q(α) has basis 1, α,..., α d 1. (ii) Q(α) is a field. Proof. (i). Let β Q(α); we will show that β can be written as a sum, with rational coefficients, of 1, α,..., α d 1. Well, β h(α) for some h Q[X] and division of olynomials imlies that there exist q, r Q[X] such that h qf + r and such that deg r < d (or erhas r 0). Thus β h(α) q(α)f(α) + r(α) r(α) since f(α) 0. But r(x) a a d 1 X d 1 for some a 0,..., a d 1 Q, and so β a 0 + a 1 α + + a d 1 α d 1, as required. Secondly we must check that 1, α,..., α d 1 are linearly indeendent over Q. If we have a relation b 0 + b 1 α + + b d 1 α d 1 0 for some b 0,..., b d 1 Q, and so g(α) 0, where g(x) : b b d 1 X d 1. Therefore f(x) g(x); but deg g < deg f, so this is ossible only if g 0, i.e. b 0 b d 1 0. This comletes the roof of art (i). (ii). Let β Q(α) be non-zero; then β h(α) for some h Q[X] which is not divisible by f(x) (for else h(α) 0). Since f is irreducible, h is corime to f and so the division algorithm imlies that there exist A, B Q[X] such that Af + Bh 1. Evaluating at α, and so h(α) 1 B(α) Q(α). 1 A(α)f(α) + B(α)h(α) B(α)h(α) 15

16 16 MATH 4 Algebraic Number Theory Examle (i) The minimal olynomial of 5 is X + 5. Indeed, this olynomial is irreducible in Q[X] (why? Since it has degree two, if it were not irreducible then it would have a linear factor so it would have a root in Q, which it doesn t), it is monic, and 5 is clearly a root of it. (ii) Let ω 1+ 3 ( e πi/3 ). We saw in lecture 1 that ω 3 1 (this is clear once you know that ω e πi/3 ). But X 3 1 is not the minimal olynomial of ω for the following reason: X 3 1 (X 1)(X + X + 1), so 0 (ω 1)(ω + ω + 1). Since ω 1 0, we deduce ω + ω + 1 0, and in fact X + X + 1 is the minimal olynomial of ω (why? As in the revious case, it is enough to observe that otherwise ω Q). Tyically it is enough to study minimal olynomials of algebraic integers, not arbitrary algebraic numbers. We must review some related results from Algebra II (Fraleigh 45): If R is a UFD, recall that a olynomial f(x) R[X] is called rimitive when the gcd of its coefficients is equal to 1. Gauss lemma states that the roduct of two rimitive olynomials in R[X] is again rimitive. Moreover, Lemma Let R be a UFD, with field of fractions F. Suose that f R[X] is a monic olynomial, and that g, h F [X] are monic olynomials satisfying f gh. Then g, h actually have coefficients in R. Proof. Write each coefficient of g as a fraction in lowest terms, and let c R be the least common multile of the denominators of all coefficients. Then cg has coefficients in R and is rimitive. Similarly there is d R such that dh is a rimitive olynomial in R[X]. By Gauss lemma, cg(x)dh(x) cdf(x) must be rimitive. But every coefficient of cdf is obviously divisible by cd; therefore cd is a unit, so each of c and d is a unit. Therefore g c 1 (cg), h d 1 (dh) have coefficients in R. We may now rove: Proosition 3.1. If α Z alg then its minimal olynomial (which by definition belongs to Q[X]) actually belongs to Z[X]. Proof. Let f(x) Q[X] be the irreducible olynomial of α. Since α is an algebraic integer, there is a monic olynomial g(x) Z[X] such that g(α) 0. By roosition 3.6, f divides g in Q[X]; i.e. there exists h(x) Q[X] such that fh g. By the revious lemma, f, h Z[X]. Corollary Suose that α Q is an algebraic integer; then α Z. Proof. The minimal olynomial of α is X α, which has integer coefficients if and only if α Z. While we are studying irreducible olynomials with coefficients in Z, now is a convenient time to recall the following test: Theorem 3.14 (Eisenstein s criterion). Let f(x) a n X n + + a 0 Z[X] be a olynomial which satisfies, for some rime number, a n, a i for i 0,..., n 1, a 0. Then f is irreducible. Proof. Should have been roved in Algebra II (Fraleigh 3.15). The most imortant alication of Eisenstein s lemma is to the following examle: Examle Let 3 be rime. Then the minimal olynomial of the algebraic integer ζ e πi/ is Φ (X) X 1 + X + + X + 1. The roof sometimes aears in algebra II, and always in algebra III, but here it is in case you haven t seen it: 16

17 MATH 4 Algebraic Number Theory 17 Proof. Since ζ is a root of the olynomial 0 ζ 1 (ζ 1)Φ (ζ) but ζ 1 0, we see that Φ (ζ) 0. Also, the olynomial is monic, so it remains only to rove that it is irreducible in Q[X]. Write Ψ (X) Φ (X + 1); then it is enough to rove that Ψ is irreducible, for any factorization of Φ, e.g. Φ (X) f(x)g(x), would result in a factorization of Ψ (X): Ψ (X) f(x + 1)g(X + 1). Well, and so 1 ( XΨ (X) XΦ (X + 1) (X + 1) 1 X + X k) k + X + 1 1, k k 1 ( Ψ (X) X 1 + X k) k 1 +. ( ) Next use the (easy-to-rove) fact that in Z whenever k is an integer in the range 1 k 1; this k shows that Ψ (X) satisfies Eisenstein s criterion, and so is irreducible. Finally in this section of the basics of algebraic numbers, we study conjugates: Definition Algebraic numbers α, β are said to be conjugate if and only if they have the same minimal olynomial. Thus the conjugates of fixed algebraic number α are the roots of the minimal olynomial of α. Remark (i) This is an equivalence relation on the set of algebraic numbers. (ii) Two rational numbers are conjugate if and only if they are equal (as the minimal olynomial of r Q is X r). (iii) If α, β are conjugate and α is an algebraic integer then so is β (for roosition 3.1 imlies that the minimal olynomial of α has integer coefficients, so β satisfies a monic olynomial with integer coefficients). Lemma (i) Let f Q[X] be monic and irreducible of degree d. Then f has d distinct roots in C. (ii) Let α be an algebraic number of degree d, with minimal olynomial f(x). Then there exist exactly d (distinct) comlex numbers α 1,..., α d which are conjugate to α, and f(x) (X α 1 )... (X α d ). Proof. (i) Since C is algebraically closed, f(x) factors into linear terms over C: f(x) (X α 1 )... (X α d ) For a contradiction suose that α i α j for some i j; write α α i. Then f(x) (X α) h(x) where h(x) r i,j (X α r), and so f (X) (X α)h(x) + (X α) h (X), whence f (α) 0. But f (X) is a olynomial with rational coefficients and f(x) is the minimal olynomial of α; therefore f f. As deg f < deg f, this is only ossible if f 0, which is absurd as it has to degree term dx d 1. (ii) As in (i) write f(x) (X α 1 )... (X α d ). If β is a conjugate of α then it is a root of f(x) and therefore equals α i for some i. Conversely, each α i has minimal olynomial f(x) and so is a conjugate of α. 17

18 18 MATH 4 Algebraic Number Theory Examle (i) The conjugates of 5 are 5 and 5. It is tedious, though straightforward, to check that f(x) : (X 3)(X + 3)(X + 3)(X + + 3) has rational coefficients; indeed, it equals the olynomial X 4 10X +1 which we calculated in examle 3.5. Here is a way to check that f(x) is irreducible over Q. If not then f gh with g, h Q[X] monic olynomials. Since f does not have any rational roots, g, h are quadratic, hence equal to or or (X 3)(X + 3) (X 3)(X + 3) It can be directly checked that none of these have coefficients in Q. So f is irreducible. Therefore f is the minimal olynomial of + 3 and so the conjugates of + 3 are + 3, 3, + 3, 3.. Corollary 3.0. α an algebraic number, with minimal olynomial f(x) X d + a d 1 X d a 0 and conjugates α 1,..., α d. Then α 1... α d ( 1) d a 0 and α α d a d 1. Proof. Immediate from the identity f(x) (X α 1 ) (X α d ). 4 Algebraic number fields The following definition introduces the main object of study in the rest of the course. We begin by remarking that if F is a subfield of C, then it automatically contains 0, ±1, ±,... and so contains Q. Definition 4.1. An algebraic number field is a subfield F of C which has finite dimension as a vector sace over Q. The ring of integers of F is D F F Z alg, i.e., those elements of F which are algebraic integers. Remark 4.. We observe some elementary but imortant remarks and examles: (i) Notice that V F is a finite dimensional Q-vector sace contained inside C such that αv V for all α F ; according to the standard test, lemma 3.3, this means that α Q alg. So F Q alg. (ii) D F is a subring of C containing Z. Rough hilosohy: arithmetic roerties of D F encodes arithmetic in Z which we could not otherwise see. (iii) Consider the case F Q; then D Q Q Z alg Z by corollary Examle 4.3. Let α be an algebraic number; then Q(α) is a subfield of C of dimension deg α over Q by roosition 3.9. Therefore Q(α) is an algebraic number field. (In fact, the Primitive element theorem, which is sometimes in Alg III, imlies that if F is an algebraic number field then there exists an algebraic number α such that F Q(α). But we will not need to know this.) Examle 4.4. Very imortant examle! Let d Z \ {0, 1} be square-free, and ut F Q( d). Let α Q( d); you are roving on the current homework that α is an algebraic integer if and only if 18

19 MATH 4 Algebraic Number Theory 19 (i) (Case: d or 3 mod 4) α a + b d for some a, b Z. (ii) (Case: d 1 mod 4) α a + b 1+ d for some a, b Z. Therefore, D F { Z + Z d Z[ d] d, 3 mod 4 Z + Z 1+ d Z[ d] d 1 mod 4 The case d 1 mod 4 demonstrates a very imortant rincile: if α is an algebraic integer then the ring of integers of the number field Q(α) may be strictly bigger than Z[α]. We will see that D F is a much more interesting ring to study. Lemma 4.5. If α is an algebraic number then there is a non-zero integer m such that mα is an algebraic integer. If F is a number field, then the field of fractions of D F is F. Proof. First claim on the homework. Second claim easily follows. So every number field F comes equied with this secial subring D F of which it is the field of fractions; we will see that D F is a very rich ring which can give us interesting results about ordinary integers themselves. We now send some time covering three examles in detail, focusing on factorisation roerties in rings of integers. First recall the following results from Alg II: Theorem 4.6 (Fraleigh 45, 46). Let R be an integral domain. (i) If R is a ED then it is a PID. (ii) If R is a PID then it is a UFD. A non-zero element a R is called rime if and only if the rincial ideal a generated by it is a rime ideal. If a is rime then it is irreducible; the converse is true in UFDs. Exercise 4.1. Let N be a ositive integer. (i) Let d be a ositive integer. Prove that there are only finitely many algebraic integers α of degree d such that all conjugates of α have comlex absolute value N. (ii) Let F be a number field. Deduce that there are only finitely many α D F such that all conjugates of α have comlex absolute value N. 4.1 First examle: Gaussian integers We are going to study three examles of number fields F and their rings of integers D F ; in each examle we write N(α) αα, for α F. This is a secial case of the so-called norm ma N F/Q, which we will introduce in general after the examles. Let F Q(i) where i 1. Since 1 3 mod 4, the revious examle tell us that the ring of integers of F is D F Z[i] Z + Zi, the subring of C which you have reviously encountered under the name Gaussian integers [Fraleigh 47]. You also learnt that it is a Euclidean domain, with Euclidean norm N(α) αα, where denotes comlex conjugation. In other words, N(a + bi) a + b. By the revious theorem, the ring of Gaussian integers is also a PID and a UFD. 19

20 0 MATH 4 Algebraic Number Theory 4. Second examle: Z[ω] where ω e πi/3 Several times we have encountered ω e πi/3 C We are going to study the ring of integers of the number field F Q(ω) Q( 3). According to examle 4.4, since 3 1 mod 4, the ring of integers of F is D F Z + Z Z + Zω Z[ω] Z[ 3]. You will rove in the homework that if we define N(α) αα for α F, just as in the case of Gaussian integers, then N is a Euclidean norm on D F. Therefore D F is a Euclidean domain, and hence a PID and UFD by the earlier theorem. 4.3 Third examle: Z[ 5] In the revious two examles, we saw that D F was a Euclidean domain with the Euclidean norm given by N(α) αα; this is not what tyically haens, even for similar looking extensions. Next we consider the examle F Q( 5). Since 5 3 mod 4, the ring of integers of F is D F Z[ 5] Z + Z 5 Then N(a + b 5) a + 5b ; it is imortant to notice that if α Z[ 5] then N(α) Z 0. The next lemma analyses some elements of Z[ 5]: Lemma 4.7. The only units in Z[ 5] are 1, 1. The elements 3, 7, 1+ 5, 1 5 are all irreducible. Proof. Firstly, 1, 1 are certainly units. Conversely, suose u a + b 5 Z[ 5] is a unit; then there is v 5 such that uv 1 and so 1 N(1) N(uv) N(u)N(v). Since N(u), N(v) Z 0, we deduce N(u) N(v) 1. Therefore a + 5b 1, which is only ossible if b 0 and a 1, 1; so α 1, 1, as required. This argument also shows that α Z[ 5] is any element satisfying N(α) 1, then α is a unit. Now we rove the claim that the given elements are irreducible (the argument is similar for each number): (i) 3. Suose 3 is not irreducible; then 3 αβ for some α, β Z[ 5] such that neither α or β are units. Therefore N(α), N(β) > 1, and 9 N(3) N(α)N(β). So N(α) N(β) 3. Let α a + b 5 for some a, b Z; then 3 N(α) a + 5b, which is clearly imossible. This contradication shows that 3 must be irreducible. (ii) 7. Suose 7 is not irreducible; then 7 αβ for some α, β Z[ 5] such that neither α or β are units. Therefore N(α), N(β) > 1, and 49 N(7) N(α)N(β). So N(α) N(β) 7. Let α a + b 5 for some a, b Z; then 7 N(α) a + 5b, which is clearly imossible. This contradication shows that 7 must be irreducible. 0

21 MATH 4 Algebraic Number Theory 1 (iii) Suose is not irreducible; then αβ for some α, β Z[ 5] such that neither α or β are units. Therefore N(α), N(β) > 1, and 1 N(1 + 5) N(α)N(β), so N(α) 3 or 7. But calculations (i) and (ii) showed that this is imossible. This contradication shows that must be irreducible. (iv) 1 5. Reeat the revious argument verbatim. This comletes the roof. Theorem 4.8. Z[ 5] is not a unique factorization domain. Proof. We have and also (1 + 5)(1 5) We have therefore roduced two decomositions of 1 into irreducibles, so to rove that Z[ 5] is not a UFD it is enough to show that 3 is not an associate of or of 1 5, i.e. that there is no unit u such that 3 u(1 + 5) or 3 u(1 5). As the only units are 1, 1, this is clear. Corollary 4.9. There is no Euclidean norm on Z[ 5]. Proof. If there were a Euclidean norm on Z[ 5] then it would be a Euclidean domain, hence a PID and a UFD; but this contradicts the revious theorem. 4.4 Introductory tools for studying number fields: Norm, Trace, Discriminant, and Integral bases Here we first exlain the notion of norm, trace, and discriminant for any finite field extension; we will then secialise to the case of number fields and introduce in addition the notion of an integral basis. These ideas, which use little more than some linear algebra, will be used throughout the rest of the course, both theoretically and in examles. Recall from Algebra that an extension of fields L/F is simly a air of fields F L, where F is a subfield of L. The usual algebraic oerations make L into a vector sace over F, and one defines the degree of L/F, denoted L : F, to be dim F L. One says that L/F is a finite field extension of finite extension of fields if and only if this dimension is finite. For examle, if α is an algebraic number, then Q(α)/Q is an extension of fields of degree Q(α) : Q deg α, by roosition 3.9, hence is a finite field extension. Remark 4.10 (Tower Law). If L/M and M/F are finite extension fields, and ω 1,..., ω r is a basis for L/M, then L Mω 1 Mω r. Taking dimensions as F -vector saces gives dim F L r dim F M, i.e. L : F L : M M : F. This is called the tower law. For examle, if F is a number field and α F, then F : Q F : Q(α) Q(α) : Q. As Q(α) : Q deg α we see in articular that deg α divides F : Q. Here is an examle of how this is useful: Just above we saw that α + 3 has minimal olynomial X 4 10X + 1, but here is quicker roof: 1

22 MATH 4 Algebraic Number Theory Put F Q(α) and notice that F contains and 3. For examle, α3 9α 4. Therefore Q( ) F and the tower law imlies F : Q F : Q( ) Q( ) : Q F : Q( ). It is easy to check that no element of Q( ) is a square root of 3, i.e. 3 / Q( ), and so F : Q( ) > 1. It follows that F : Q is an even number which is >. But since F : Q deg α 4 (as α does satisfy a degree 4 monic olynomial), we deduce that deg α 4; so α has minimal olynomial of degree 4, and this minimal olynomial divides, and therefore equals X 4 10X Norm and Trace Definition Let L/F be a finite extension of fields, and let α L. The norm of α, denoted N L/F (α) is defined to be the determinant of the following F -linear ma of vector saces: L L, β αβ. I.e., N L/F (α) det φ α F. Similarly, the trace of α, denote Tr L/F (α), is defined to the trace of φ α ; i.e., Tr L/F (α) Tr φ α F. In other words, let β 1,..., β n be a basis of L as an F -vector sace, write αβ i n j1 c i,jβ j with c i,j F, and ut C (c i,j ); then N L/F (α) det C, Tr L/F (α) Tr C n c i,i. Examle 4.1. Here we comute norms and traces of tyical elements in some extensions. (i) C/R has basis 1, i. If x ( + iy ) C, with x, y R, then the matrix for multilication by x + iy with x y resect to this basis is whence y x i1 N C/R (x + iy) x + y, Tr C/R (x + iy) x. (ii) Q(i)/Q has basis 1, i, and exactly the same argument as in the revious examle shows that for x, y Q. N Q(i)/Q (x + iy) x + y, Tr Q(i)/Q (x + iy) x, (iii) Q(ω)/Q, where ω + ω This has basis 1, ω, and multilication by a + bω in this basis is given by the matrix ( ) a b. b a b So N Q(ω)/Q (a + bω) a + b ab, Tr Q(ω)/Q (a + bω) a b. The next two roerties summarise the main theoretical roerties of the Norm and Trace for a finite extension of fields L/F : Proosition 4.13 (Main theoretic roerties of the Norm). N L/F : L F has the following roerties: (i) N L/F (α 1 α ) N L/F (α 1 )N L/F (α ) for α 1, α L. (ii) N L/F (a) a L:F if a F.

23 MATH 4 Algebraic Number Theory 3 (iii) N L/F (α) 0 if and only if α 0. (iv) If M is an intermediate field between L and F, then N L/F N M/F N L/M. Proof. For any α F, let φ α : L L, β αβ be the F -linear ma multilication by α, so that N L/F α det φ α. (i): For any β F, φ α1 (φ α (β) α 1 α β φ α1α (β), and so φ α1 φ α φ α1α. Therefore N L/F (α 1 )N L/F (α ) det φ α1 det φ α det(φ α1 φ α ) det(φ α1α ) N L/F (α 1 α ). (ii): If a F then the matrix for φ a, with resect to any basis α 1,..., α n, of L/F is the n n diagonal matrix with a s all along the diagonal; this has determinant a n. (iii): If α 0 then φ α 0 so N L/F (α) det φ α 0; conversely, if α 0 then 1 N L/F (1) N L/F (αα 1 ) N L/F (α)n L/F (α 1 ), so N L/F (α) 0. (iv): We only sketch the roof in the secial case that α M. Let ω 1,..., ω r be a basis for L/M; then (by the very definition of what a basis is), L Mω 1 Mω r. Since φ α (Mω i ) αmω i Mω i, for each i, we see that φ α restricts to each F -vector sace Mω i and that By a block diagonal argument, we see that φ α φ α Mω1 φ α Mωn. det(φ α ) det(φ α M ) r, i.e. N L/M (α) N M/F (α) L:M. Therefore N L/M (α) N M/F (α F :M ) N M/F (N L/M (α)). Proosition 4.14 (Main theoretic roerties of the Trace). Tr L/F : L F has the following roerties: (i) Tr L/F (α 1 + α ) Tr L/F (α 1 ) + Tr L/F (α ) for α 1, α L. (ii) Tr L/F (aα) a Tr L/F (α) for a F, α L. (iii) Tr L/F (a) L : F a if a F. (iv) If M is an intermediate field between L and F, then Tr L/F Tr M/F Tr L/M. Proof. Coy the revious roof, relacing addition by multilication. Remark 4.15 (Relation to Galois theory). Readers who know Galois theory may find the following formulae (exercise!) useful, though we will not need them: If L/F is a finite Galois extension of fields and α L, then N L/F (α) σ(α), Tr L/F (α) σ(α) σ Gal(L/F ) σ Gal(L/F ) More generally, if L/F is not necessarily Galois, then these formulae remain true if we let σ run over all F -linear field embeddings of L into a fixed normal extension of F containing L, e.g. the normal closure of L/F. 3

24 4 MATH 4 Algebraic Number Theory In the secial case of a number field, the norm and trace mas are comatible with the ring of integers: Proosition F a number field, and α D F. Then Tr F/Q (α) and N F/Q (α) belong to Z. Proof. Let M Q(α); then the revious two roositions imly that N F/Q (α) N M/Q (α) F :M, Tr F/Q (α) F : M Tr M/Q (α) Therefore it remains only to show that N M/Q (α) and Tr M/Q (α) are in Z; this is left to the homework Discriminant Using the trace mas we can construct an interesting invariant known as the discriminant: Definition L/F a finite extension of fields, and α 1,..., α n L some elements. Then the associated discriminant is (α 1,..., α n ) det(tr L/F (α i α j )) Examle Here is a comutation of a discriminants: Let d Z \ {0, 1} not be a cube, and set F Q(d 1/3 ). Then (1, d 1/3, d /3 ) is the determinant of the matrix Tr F/Q 1 Tr F/Q d 1/3 Tr F/Q d /3 So (1, d 1/3, d /3 ) 7d. Tr F/Q d 1/3 Tr F/Q d /3 Tr F/Q d Tr F/Q d /3 Tr F/Q d Tr F/Q d 4/ d 0 3d 0 The next two results are the main theoretic roerties of the disciminant: Proosition 4.19 (Discriminant detects bases). L/F a finite extension of fields, and α 1,..., α n L. If (α 1,..., α n ) 0 then α 1,..., α n are linearly indeendent over F. Conversely, if α 1,..., α n form a basis for L/F, and F has characteristic zero, then (α 1,..., α n ) 0. Proof. Suose that α 1,..., α n are linearly deendent; so there are a 1,..., a n F, not all zero, such that n i1 a iα i 0. Therefore, for any j, which can be rewritten as n 0 Tr L/F (α j a i α i ) i1 (Tr(α i α j )) a 1. a n n a i Tr(α i α j ), i1 0, where the column vector is non-zero. So det(tr(α i α j )) 0, as required. Conversely, for a contradiction suose that α 1,..., α n are a basis for L/F and that (α 1,..., α n ) 0. So the F -linear ma L L attached to the matrix (Tr(α i α j )) has a kernel, i.e. there are a 1,..., a n F, not all zero, such that n i1 a i Tr(α i α j ) 0. Set α : n i1 a iα i L; this is non-zero since α 1,..., α n form a basis and a 1,..., a n are not all zero. However, if β L, then we may write β n j1 b jα j for some b 1,..., b n F, and we deduce that Tr L/F (βα) Tr L/F ( n b j α j α) j1 n b j Tr L/F (αα j ) j1 n n b j j1 i1 a i Tr L/F (α i α j ) 0. In articular, taking β α 1, we have just shown that n Tr L/F (1) 0 in the field F, which contradicts the assumtion that char F 0. 4