THE UNIT GROUP OF A REAL QUADRATIC FIELD

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1 THE UNIT GROUP OF A REAL QUADRATIC FIELD While the unit group of an imaginary quadratic field is very simple the unit group of a real quadratic field has nontrivial structure Its study involves some geometry and analysis but also it relates to Pell s equation and continued fractions topics from elementary number theory Review Let F = Q( n) be a real quadratic field Thus n > is not a square and we take n squarefree Recall various facts about F The nontrivial automorphism of F is the conjugation function : F F a + b n = a b n The trace function of F is the abelian group homomorphism Specifically tr : F Q tr(x) = x + x tr(a + b n) = (a + b n) + (a b n) = a The norm function of F is the homomorphism Specifically N : F Q N(x) = x x N(a + b n) = (a + b n)(a + b n) = a b n Sometimes we also define N(0) = 0 The unit group O F of F consists precisely of the elements of O F such that N(x) = ± The discriminant of F is D F = { n if n = (mod 4) 4n if n = 3 (mod 4) and the integer ring of F is [ DF + D F O F = Z Let {x j } j Z + be a sequence in O F all of whose elements satisfy N(x j ) α for some constant α Then for some pair of distinct positive integers j and j x j = ux j u O F Indeed two sequence elements must have the same absolute norm ]

2 THE UNIT GROUP OF A REAL QUADRATIC FIELD Geometric Results Lemma Let Λ be a lattice in R and let E be a measurable subset of R such that µ(e) > µ(r /Λ) Then there exist distinct points x x E such that x x Λ Proof Let (e f) be a Z-basis of Λ so that corresponding fundamental parallelogram of R /Λ is Thus area(π) < µ(e) = λ Λ Π = {ξe + ηf : ξ η [0 ]} µ(e (λ + Π)) = λ Λ µ((λ + E) Π) Consequently the intersection (λ + E) (λ + E) Π must be nonempty for some distinct λ λ Λ In particular λ + x = λ + x for some x x E Thus x x = λ λ Λ {0} as desired Proposition Let Λ be a lattice in R and let E be a compact measurable subset of R that is symmetric about 0 and convex and such that µ(e) 4µ(R /Λ) Then E contains a nonzero point of Λ Proof The lemma says that for any ε > 0 the set E = +ε E contains distinct points x and x such that x x Λ But also using symmetry about 0 the nonzero difference x x = (x + ( x))/ is a convex linear combination of points of ( + ε)e and hence a point of ( + ε)e That is letting Λ = Λ {0} Λ ( + ε)e Note that Λ (+ε)e is compact (finite for that matter) since it is the intersection of a discrete set and a compact set Thus the nested intersection over all ε remains nonempty (Λ ( + ε)e) ε>0 (The finite intersection property of compact sets rephrases the definition of compactness If α K α = then α Kc α = K o gives an open cover of K o hence Kα c = K o for some α by the nestedness hence K α = contradiction) But the nonempty intersection is (again using the compactness of E at the last step) ε>0(λ ( + ε)e) = Λ ( + ε)e = Λ E = Λ E ε>0 This completes the argument The lemma and the proposition are special cases of results due to Minkowski about the geometry of numbers

3 THE UNIT GROUP OF A REAL QUADRATIC FIELD 3 3 The Canonical Embedding Definition 3 The canonical embedding of F is the ring homomorphism ι : F R x (x x) Recall that our real quadratic field is F = Q( n) and that the discriminant of F is { n if n = (mod 4) D F = 4n if n = 3 (mod 4) The abelian group structure of the integer ring of F is a direct sum O F = D F + D F Z Z The images of the basis elements under the fundamental embedding are ι() = ( ) ( DF + ) ( D F DF + D F ι = D F + D F D F D F D F D F Thus ι(o F ) is a lattice in R whose fundamental parallelogram has area [ ] det = D F That is µ(r /ι(o F )) = D F Consequently Proposition says that any compact box B that is symmetric about the origin has area 4 D F contains a nonzero point of ι(o F ) We will quote this fact later 4 The Logarithmic Embedding Definition 4 The logarithmic embedding of O F l : O F R u (log u log u ) Thus the logarithmic embedding takes the form l = h ι O F where h is the continuous group homomorphism h : (R ) R (x y) (log x log y ) ) is the group homomorphism Note that ι O is also a homomorphism of multiplicative groups Because uu = F N(u) for u O F the image ι(o F ) lies in the hyperbola Also note that the calculation H = {(x y) (R ) : xy = ±} xy = ± = log x + log y = log xy = log = 0 shows that h restricts to a continuous homomorphism from the hyperbola H to the line of slope L = {(z w) R : z + w = 0} That is ι(o F ) H l(o F ) L

4 4 THE UNIT GROUP OF A REAL QUADRATIC FIELD Also direct inspection shows that ker(l) = {±} and thus (the next display has a multiplicative group on the left and an additive group on the right) and thus O F R >0 l(o F ) O F = {±} (O F R >0) (Z/Z) l(o F ) 5 Unit Group Structure We show that l(o F ) is infinite cyclic First we show that it is discrete Lemma 5 For any nonnegative number r R 0 l ([ r r] ) O F is finite (Setting r = 0 in the lemma shows that ker(l) is finite Since ker(l) = {±} in the real quadratic field case we obtain nothing new here but for number fields other than real quadratic fields with the lemma modified accordingly the result is of interest) Proof Let r 0 and suppose that some element u O F l(u) [ r r] That is (log u log u ) [ r r] so that Consequently u u [e r e r ] u + u e r and uu e r satisfies But the characteristic relation of u over Z is { a = tr(u) = u + u Z u au + b = 0 b = N(u) = uu Z Thus there are only finitely many possibilities for the characteristic polynomial and hence there are only finitely many possibilities for u Recall that O F (Z/Z) l(o F ) The lemma shows that l(o F ) is a discrete subgroup of L Because L R the question now is whether l(o F ) = {0} or l(o F ) Z Proposition 5 l(o F ) Z The idea of the proof is to exhibit two elements of O F that are equal in norm but not in absolute value; there is no contradiction here because norm is not squared absolute value in a real quadratic field Their quotient is thus a unit u whose absolute value is not and so l(u) = (log u log u ) is nonzero Proof For any positive real number λ the compact box B = [ λ λ] [ D F /λ D F /λ] is symmetric about the origin and its area is 4 D F As explained in section 3 the geometry of numbers shows that there exists a nonzero integer x λ O F such that ι(x λ ) B The fact that ι(x λ ) = (x λ x λ ) lies in B says that N(x λ ) D F

5 THE UNIT GROUP OF A REAL QUADRATIC FIELD 5 Also the condition N(x λ ) holds because x λ O F so ι(x λ ) lies outside the four hyperbolic segments xy = ± These segments meet the top and bottom of the box at x = λ/ D F so λ/ D F x λ λ and consequently /λ / x λ Repeat the process with a second positive real number λ > D F λ to get λ / D F x λ and so the previous two displays combine to give < λ /( D F λ) x λ / x λ which is to say x λ < x λ Continue in this fashion to get a sequence {x λ(j) } such that x λ() < x λ() < x λ(3) < and N(x λ(j) ) D F j = 3 Thus for two distinct positive integers j j we have N(x λ(j )) = N(x λ(j) ) and x λ(j ) x λ(j) and so x λ(j ) = ux λ(j) u O F u Therefore log u = 0 This shows that l(o F ) {0} and consequently l(o F ) Z as desired 6 The Fundamental Unit Definition 6 The unique element u O F such that O F = {±} u = {±} {u i : i Z} u > is the fundamental unit of F If the fundamental unit is then since N(g) = ± altogether u = a + b n a b Q u u u u = ±a ± b n Since the fundamental unit is the largest of the four elements in the previous display in fact u = a + b n a b Q + and u is the smallest such element a + b n of O F The units u > are overall We now proceed by cases If n = 3 (mod 4) then u k = u k = (a + b n) k k = 3 O F = {a + b n : a b Z} and so the fundamental unit takes the form u = a + b n a b Z + a nb = ±

6 6 THE UNIT GROUP OF A REAL QUADRATIC FIELD Its positive powers are u k = a k + b k n = (a + b k n) k Since b k+ = a k b + b k a the sequence {b k } is strictly increasing Thus an algorithm to find the fundamental unit for n = 3 (mod 4) is: Test b = 3 until either of nb ± is a perfect square Let a be the positive integer such that a nb = ± Then u = a + b n If n = (mod 4) then O F = { (a + b n) : a b Z a = b (mod ) } and so the fundamental unit takes the form u = (a + b n) a b Z + a = b (mod ) a nb = ±4 Its positive powers are u k = k (a k + b k n) = k (a + b k n) k Since b k+ = a k b + b k a the sequence {b k } is strictly increasing Thus an algorithm to find the fundamental unit for n = (mod 4) is: Test b = 3 until either of nb ± 4 is a perfect square Let a be the positive integer such that a nb = ±4 Then u = (a + b n) Some fundamental units are shown in table n u ( + 5) (3 + 3) (5 + ) Figure Table of fundamental units A more efficient method for finding the fundamental unit uses continued fractions

7 THE UNIT GROUP OF A REAL QUADRATIC FIELD 7 7 The Continued Fraction of a Rational Number Let c and d be integers with d > 0 The Euclidean algorithm gives c d = q 0 + r 0 d q 0 Z 0 < r 0 < d d = q + r r 0 r 0 q > 0 0 < r < r 0 r 0 = q + r r r q > 0 0 < r < r r m r m = q m q m > (note) Thus c d = q 0 + q + q + + qm Working productively with the notation of the previous display is hopeless The standard remedy is to write instead c d = q 0 + q + Note that also (using the fact that q m > ) c d = q 0 + q + q + q m q + (q m )+ The expression in the previous three displays is the continued fraction expression of c/d Symbolically the first few continued fractions are q 0 + q + q 0 = q 0 q 0 + = q 0q + q q q 0 + q = q 0 + q + q q q + = q 0q q + q 0 + q q q + = q 0 + q + q 3 q q 3 + = q 0q q q 3 + q 0 q + q 0 q 3 + q q 3 + q q q 3 + q + q 3 q q q 3 + q + q 3 The numerator of the continued fraction has a standard symbol Thus for example q 0 + q + [q 0 ] = q 0 q + [q 0 q ] = q 0 q + [q 0 q q ] = q 0 q q + q 0 + q = [q 0 q q m ] q m d [q 0 q q q 3 ] = q 0 q q q 3 + q 0 q + q 0 q 3 + q q 3 +

8 8 THE UNIT GROUP OF A REAL QUADRATIC FIELD The denominator requires no symbol of its own because as the small examples have shown it is simply [q q m ] To show this in general note that the equality rewrites as q 0 + q + [q 0 q q m ] d showing that indeed q 0 + q + q + = q 0 + q m q + q + q 3+ q m d = q 0 + [q q m ] = q 0[q q m ] + d [q q m ] q + This calculation also shows the recurrence = [q 0 q q m ] q m [q q m ] [q 0 q q q m ] = q 0 [q q m ] + [q q m ] which makes computation with these symbols quick Euler gave the explicit formula [q 0 q m ] = q 0 q m + q 0 q m + q i i q i+ ij q 0 q m q i q i+ q j q j+ + That is multiply all the q s together then multiply them together omitting consecutive pairs then multiply them together omitting pairs of consecutive pairs and so on Euler s formula demands extending the numerator-symbol to include the case [ ] = A consequence of Euler s formula is symmetry so that the recurrence is also For any k 0 let [q 0 q q m ] = [q m q q 0 ] [q 0 q q q m ] = [q 0 q m ]q m + [q 0 q m ] A k = [q 0 q q k ] and B k = [q q k ] Then the kth convergent is A k /B k Since [ ] [ ] Ak Ak = q B k + k B k the convergents are easy to compute Also A k A k = A k B k B k B k [ Ak B k ] A k B k and A A 0 B B 0 = [q 0 q ] [q 0 ] [q ] [ ] = (q 0q + ) q 0 q = so that A k+ A k B k+ B k = ( )k k 0

9 THE UNIT GROUP OF A REAL QUADRATIC FIELD 9 In particular gcd(a k B k ) = for all k 0 showing that the kth convergent is in lowest terms Since A k+ B k+ A k B k = it follows that the sequence {A k /B k } is Leibniz ( )k B k B k+ 8 The Continued Fraction of an Irrational Number Let α be an irrational number Similarly to before we have α = q 0 + /α q 0 Z α > α = q + /α q > 0 α > α = q + /α 3 q > 0 α 3 > α m = q m + /α m+ q m > 0 α m+ > (specifically each q i = α i where α 0 = α) but now the process doesn t terminate Still as before α = q 0 + q + q + q m + α m+ Also as before and we have the recurrence α = [q 0 q m α m+ ] [q q m α m+ ] [q 0 q m α m+ ] = [q 0 q m ]α m+ + [q 0 q m ] = α m+ A m + A m Similarly and so A little algebra gives [q q m α m+ ] = α m+ B m + B m α = α m+a m + A m α m+ B m + B m α A m = ± B m B m (α m+ B m + B m+ ) and since α m+ > q m+ it follows that α A m < B m B m+ B m Thus A m α = lim = q 0 + m B m q + q + q m + When the continued fraction context is clear we write more concisely α = q 0 q q q m

10 0 THE UNIT GROUP OF A REAL QUADRATIC FIELD Definition 8 A quadratic irrational number is a real number of the form α = a + b n The conjugate of such a number is a b Q n Z + squarefree α = a b n A quadratic irrational number is normalized if α > and < α < 0 Proposition 8 Let α be an irrational number Then α is quadratic and normalized if and only if its continued fraction is periodic Proof of the easy direction Suppose that α = q 0 q m q 0 q m q 0 q m = q 0 q m Since q 0 repeats as the term of the continued fraction it is a positive integer giving α > Next the characteristic relation of α rewrites as the characteristic equation showing that α is quadratic fraction α = A mα + A m B m α + B m B m α + (B m A m )α A m = 0 Now consider the reverse-order periodic continued β = q m q 0 Note that β > since q m is a positive integer and note that the characteristic relation of β is almost the same as that of α β = A mβ + B m A m β + B m Consequently the characteristic relation of /β is /β = B m ( /β) A m B m ( /β) + A m so that the characteristic equation of /β is B m ( /β) + (B m A m )( /β) A m = 0 That is α and /β satisfy the same quadratic equation so /β is one of α α Since α > and < /β < 0 we see that /β = α and consequently < α < 0 That is α is normalized The proof of the other direction isn t much more difficult but it is a bit more painstaking Now consider a positive integer n that is not a perfect square Let q 0 = n α = q 0 + n Then α = q 0 n ( 0) showing that α is normalized Note also that α = α + q 0 By the (unproved direction of the) previous proposition α = q 0 q q m

11 THE UNIT GROUP OF A REAL QUADRATIC FIELD so that = q q m q 0 α q 0 Also by the proof of the proposition /α = q m q q 0 But since α = q 0 n the left sides of the two previous displays are equal hence so are the right sides and thus q q m is palindromic Thus n = α q 0 has continued fraction n = q0 q q q q q 0 For example = 3 = 3 = = Rewrite the continued fraction representation of n in a fashion that forgets its palindromic aspect n = q0 q q q m q m q 0 Then α m+ A m + A m n = α m+ B m + B m where α m+ = q 0 q q m q 0 = n + q 0 The previous two displays give ( n + q 0 )A m + A m n = ( n + q 0 )B m + B m or n(( n + q0 )B m + B m ) = ( n + q 0 )A m + A m or after equating rational and irrational components A m = q 0 A m + nb m B m = A m q 0 B m Recall that A m A m = ( )m B m B m The previous two displays give A m q 0 A m + nb m B m A m q 0 B m = ( )m or A m nbm = ( ) m That is A m + B m n is a unit of Q( n)