RREDUCED IDEALS FROM THE REDUCTION ALGORITHM
HA THANH NGUYEN TRAN AND DUONG HOANG DUNG
Abstract.
The reduction algorithm is used to compute reduced ideals of a numberfield. However, there are reduced ideals that can never be obtained from this algorithm.In this paper, we will show that these ideals have inverses of larger norms among reducedones. Especially, we represent a sufficient and necessary condition for reduced ideals ofreal quadratic fields to be obtained from the reduction algorithm. Introduction
Reduced ideals of a number field F have inverses of small norms and they form a finiteand regularly distributed set in the infrastructure of F . Therefore, they can be used tocompute the regulator and the class number of a number field [2–7]. One usually appliesthe reduction algorithm (see Algorithm 10.3 in [6]) to find them. Ideals obtained fromthis algorithm are called 1-reduced [8]. There exist reduced ideals that are not 1-reduced.For example, the real quadratic field F = Q ( √
73) has nine reduced ideals but only sevenof them are 1-reduced. The ideals D and D are reduced but not 1-reduced. Figure 1.
The reducedideals of F = Q ( √ Figure 2.
The 1-reducedideals of F = Q ( √ F . The result isrepresented in Section 3. Mathematics Subject Classification.
Key words and phrases. reduced ideal; reduction algorithm; LLL reduced basis. a r X i v : . [ m a t h . N T ] A ug HA THANH NGUYEN TRAN AND DUONG HOANG DUNG
In case of real quadratic fields, we prove a sufficient and necessary condition for areduced ideal to be 1-reduced. Explicitly, each reduced ideal contains a unique element f satisfying the conditions given in Remark 4.1. We will prove the following theorem. Theorem 1.1.
Let F be a real quadratic field and let I = Z + Z f be a reduced fractionalideal of F where f is in Remark 4.1. Then I is 1-reduced if N ( f − / ≤ − / . The upper bound in this theorem is actually tight. In other words, there exist 1-reduced ideals such that N ( f − /
2) = − /
4. For instance, this equality is obtainedin the quadratic field F = Q ( √ I = 1 · Z + √ · Z ,and in the quadratic field F = Q ( √
3) for the 1-reduced ideal I = 1 · Z + √ · Z .Especially, the later one is the only case of which the inverse ideal has smallest norm,that is N ( I − ) = (cid:112) | ∆ F | / F is the discriminant of F .In addition, our results partly answer the question mentioned in Section 12 in [6]which asks about the number of reduced but not 1-reduced ideals. Indeed, the results ofCorollary 4.1 and Proposition 3.1 imply that reduced but not 1-reduced ideals are theones have inverses of large norms among reduced ideals. We further more, are interestedin finding properties of reduced but not 1-reduced ideals, estimating how many of themas well their distribution in the topological group Pic F .Note that we do not consider 1-reduced ideals of imaginary quadratic fields. That isbecause in these fields, reduced ideals are always 1-reduced.2. Preliminaries
In this section, let F a number field of degree n and the discriminant ∆ F . Assume that F has r real embeddings σ , · · · , σ r and r complex embeddings (up to conjugation) σ r +1 , · · · , σ r + r . Thus n = r + 2 r . Denote by Φ = ( σ , · · · , σ r , σ r +1 , · · · , σ r + r ).Each fractional ideal I of F can be viewed as the lattice Φ( I ) in F R = R r × C r . Now let u = ( u i ) ∈ R r + r . Then we define N ( u ) = (cid:81) i u i . We also identify each element f ∈ I with Φ( f ) = ( σ i ( f )) i ∈ F R and use the following metric in F R . (cid:107) uf (cid:107) = r (cid:88) i =1 u i | σ i ( f ) | + 2 r + r (cid:88) i = r +1 u i | σ i ( f ) | . Reduced ideals.
Definition . A fractional ideal I is called reduced if 1 is minimal in I . In other words,if g ∈ I and | σ i ( g ) | < i then g = 0. Definition . Let I be a fractional ideal. Then 1 is called primitive in I if 1 ∈ I andit is not divisible by any integer d ≥ Definition . Let I be a fractional ideal in F and let u ∈ ( R > ) r + r . The length of anelement g of I with respect to u is defined by (cid:107) g (cid:107) u := (cid:107) ug (cid:107) . Definition . A fractional ideal I is called 1- reduced if: • I , and EDUCED IDEALS FROM THE REDUCTION ALGORITHM 3 • there exists u ∈ (cid:81) σ R > such that (cid:107) (cid:107) u ≤ (cid:107) g (cid:107) u for all g ∈ I \{ } . Remark . • The second condition of Definition 2.4 is equivalent to saying that there exists ametric u such that with respect to this metric, the vector 1 is a shortest vectorin the lattice I . • Since the lattice L = uI := { ( u i σ i ( x )) i : x ∈ I } ⊂ F R is isometric to the latticeΦ( I ) with respect to the length function (cid:107) (cid:107) u , the second condition of Definition2.4 is equivalent to saying that u is a shortest vector of the lattice L . • If u = ( u i ) ∈ ( R > ) r + r satisfies the second condition of Definition 2.4, then sois u (cid:48) = (cid:16) u σ N ( u ) /n (cid:17) σ ∈ ( R > ) r + r and N ( u (cid:48) ) = 1. Therefore, we can always assumethat N ( u ) = 1. • If u ∈ R r + r and N ( u ) = 1 then (cid:107) u (cid:107) ≥ nN ( u ) /n = n by the arithmetic–geometric mean inequality.2.2. The reduction algorithm.
Given an ideal lattice I with a metric u such that thecovolume of this ideal is (cid:112) | ∆ F | . Compute a reduced ideal J such that ( J, N ( J ) − /n ) isclose to ( I, u ) in Pic F (see Algorithm 10.3 in [6]). Description . We compute an LLL-reduced basis b , · · · , b n of the lattice L = uI ⊂ F R . Then we compute a shortest vector x in L as follows. Any shortest vector x = (cid:80) i = ni =1 m i b i satisfies (cid:107) x (cid:107) / (cid:107) b (cid:107) ≤
1. Therefore the coordinates m i ∈ Z are bounded inde-pendent of the discriminant of F . To compute a shortest vector in the lattice in timepolynomial in log | ∆ F | , we may therefore just try all possible m i . To find a reduced ideal J such that ( J, N ( J ) − /n ) is close to ( I, u ) in Pic F , we compute a shortest vector f in thelattice ( I, u ). The fractional ideal J = f − I is then reduced. In addition, the distancebetween ( I, u ) and (
J, N ( J ) − /n ) in Pic F is at most log | ∆ F | . Remark . The ideal J obtained from the reduction algorithm above is 1-reduced. First,it is easy to show that 1 is primitive in J . Now let v = u | f | := ( u i · | σ i ( f ) | ) i ∈ ( R > ) r + r .We then have the following. (cid:107) (cid:107) v = (cid:107) v (cid:107) = (cid:107) u | f |(cid:107) = (cid:107) uf (cid:107) = (cid:107) f (cid:107) u . Any element h of J has the form h = f − g for some g ∈ I . Thus (cid:107) h (cid:107) v = (cid:107) f − g (cid:107) v = (cid:107) f − gv (cid:107) = (cid:107) f − gu | f |(cid:107) v = (cid:107) gu (cid:107) = (cid:107) g (cid:107) u . Since f is a shortest vector in the lattice I with respect to the metric u , we have (cid:107) f (cid:107) u ≤(cid:107) g (cid:107) u for all g ∈ I \{ } . Therefore (cid:107) (cid:107) v ≤ (cid:107) h (cid:107) v for all h ∈ J \{ } . Thus J is 1-reduced. 3. A result for an arbitrary field
Using Remark 2.1, we can prove the following result where γ n is the Hermite constantin dimension n [1]. HA THANH NGUYEN TRAN AND DUONG HOANG DUNG
Proposition 3.1.
Let c n = (cid:16) nγ n (cid:17) n and let I be a fractional ideal containing . Then I is not 1-reduced if N ( I − ) > (cid:112) | ∆ F | /c n .Proof. Suppose that I is 1-reduced. Then there is some u ∈ ( R > ) n such that u is ashortest vector in the lattice L = uI . Thus its length is bounded as follows. (cid:107) u (cid:107) ≤ γ n covol( L ) /n . Since N ( I − ) > (cid:112) | ∆ F | /c n , we have N ( I ) < (cid:112) c n / | ∆ F | . We can assume that N ( u ) = 1as in the Remark 2.1. It follows that covol( L ) = N ( u ) N ( I ) (cid:112) | ∆ F | = N ( I ) (cid:112) | ∆ F | < √ c n .Therefore (cid:107) u (cid:107) < γ n c /nn = n , contradicting the fact that (cid:107) u (cid:107) ≥ n as in Remark 2.1.Hence I is not 1-reduced. (cid:3) The table below shows values of c n corresponding to known values of γ n (in dimensions1 to 8 and 24). n c n / / /
64 4 Note that our result in Proposition 3.1 agrees with Theorem 1.1 in case n = 2, i.e., thenorm of the inverse of a 1-reduced ideal must be less than or equal to (cid:112) | ∆ F | / Real quadratic fields
In this section, let F be a real quadratic fields with two real embeddings σ and σ (cid:48) thatsend √ ∆ F to √ ∆ F and −√ ∆ F respectively. We denote by Φ = ( σ, σ (cid:48) ) the map from F to R .4.1. Reduced ideals of real quadratic fields.
Let I be a fractional ideal of F andlet u = ( u , u ) ∈ R . We identify each element f ∈ I with Φ( f ) = ( σ ( f ) , σ (cid:48) ( f )) ∈ R and use the standard metric in R as follows. (cid:107) ug (cid:107) = u [ σ ( f )] + u [ σ (cid:48) ( f )] . Remark . • Any reduced ideal I of F can be written as the following form I = Z + f Z for a unique f ∈ F satisfying σ ( f ) > − < σ (cid:48) ( f ) < . In particular, f can be written as(4.1) f = b + √ ∆ F a , ( a, b, c ) ∈ Z , ∆ F = b − ac and | (cid:112) ∆ F − a | < b < (cid:112) ∆ F . Moreover, the inverse of I is an integral ideal, that is I − ⊂ O F , and its norm N ( I − ) = a . (See Example 8.2 in [6] for more details.) EDUCED IDEALS FROM THE REDUCTION ALGORITHM 5
Here we view I as the lattice Φ( I ) in R as below. I ≡ Φ( I ) = (cid:20) σ ( f )1 σ (cid:48) ( f ) (cid:21) Z = (cid:34) b + √ ∆ F a b −√ ∆ F a (cid:35) Z In other words, I is identified to the free Z -module generated by two vectorsΦ(1) = ( σ (1) , σ (cid:48) (1)) = (1 ,
1) and Φ( f ) = ( σ ( f ) , σ (cid:48) ( f )) = (cid:18) b + √ ∆ F a , b − √ ∆ F a (cid:19) . Test 1-reduced property.
Assume that I is reduced and the shortest vectors of I have length strictly less than √
2. In this part, we show a method to test whether I is1-reduced or not (see [8] for more details).Let g ∈ I . We denote by Φ( g ) = ( g , g ) ∈ R where g = σ ( g ) and g = σ (cid:48) ( g ). Denoteby G = (cid:26) g ∈ I : (cid:0) g − (cid:1) (cid:0) g − (cid:1) < (cid:107) g (cid:107) < π (cid:27) = G ∪ G where G = (cid:8) g ∈ G : g − < } and G = { g ∈ G : g − < (cid:9) . For each g ∈ G , we define B ( g ) := (cid:18) − g − g − (cid:19) / . Then denote(4.2) B min = (cid:40) if G = ∅ max { B ( g ) : g ∈ G } if G (cid:54) = ∅ . (4.3) B max = (cid:40) G = ∅ min { B ( g ) : g ∈ G } if G (cid:54) = ∅ . The ideal I is then 1-reduced if and only if B max ≥ B min (see Proposition 3.5 in [8]).The Algorithm 4.1 in [8] provides a method to compute B max and B min as follows.Let { b = ( b , b ) , b = ( b , b ) } be an LLL-basis for the lattice I . We compute theintegers t ≤ t as the following. • If 0 < b < < | b | < √ t ≤ t are between − − b b and − b b . • If 1 < b < √ < | b | < t ≤ t are between − − b b and − b b .Then B max and B min are among B ( g ) where g ∈ G = { b , t b + b , t b + b , s b + b with | s | ≤ } . Especially, if we further assume that b = ( b , b ) = (1 , t = t = 0. Thus,the set G can be · · · as G = { b , b + b , b − b , b + b , b − b } . HA THANH NGUYEN TRAN AND DUONG HOANG DUNG
Proof of Theorem 1.1.
Note that the condition N ( f − / ≤ − / b − a ) + 3 a ≤ ∆ F . The condition 4.4 implies that 3 a ≤ ∆ F . It is also equivalent to the following. a − (cid:112) ∆ F − a ≤ b ≤ a + (cid:112) ∆ F − a . Therefore, we can divide the proof into three cases as below. • Case 1: a ≤ (cid:112) ∆ F / (cid:112) ∆ F / ≤ a ≤ (cid:112) ∆ F / √ a − ∆ F ≤ b ≤ a − √ a − ∆ F .By Lemma 4.1, 1 is a shortest vector of the lattice I . Therefore it is 1-reduced. • Case 2: (cid:112) ∆ F / ≤ a ≤ (cid:112) ∆ F / a − √ ∆ F − a ≤ b ≤ √ a − ∆ F . In thiscase, the two vectors Φ( f ) and Φ(1) form an LLL-reduced basis for I (see Lemma4.2). Hence, the result of Section 4.2 can be used to show that I is 1-reduced.We first compute B max and B min among B ( g ) where g ∈ G = { Φ( f ) , Φ( f ) + Φ(1) , − Φ( f ) + Φ(1) , f ) + Φ(1) , − f ) + Φ(1) } . Since the vector − f ) + Φ(1) has both coordinates greater than 1, we caneliminate it from the set G . Furthermore, we obtain that B min = B ( − Φ( f ) + Φ(1)) = ( √ ∆ F + b )(4 a − b − √ ∆ F )( √ ∆ F − b )(4 a − b + √ ∆ F ) ,B max = min { B (Φ( f )) , B (Φ( f ) + Φ(1)) , B (2Φ( f ) + Φ(1)) } .B (Φ( f )) = (2 a + b + √ ∆ F )( b + √ ∆ F − a )(2 a + b − √ ∆ F )(2 a − b + √ ∆ F ) ,B (Φ( f ) + Φ(1)) = ( √ ∆ F + b )(4 a + b + √ ∆ F )( √ ∆ F − b )(4 a + b − √ ∆ F ) ,B (2Φ( f ) + Φ(1)) = ( √ ∆ F + b )(2 a + b + √ ∆ F )( √ ∆ F − b )(2 a + b − √ ∆ F ) . By the condition b ≤ √ ∆ F , it is obvious that B min ≤ B (Φ( f ) + Φ(1)). Inaddition, B min ≤ B (2Φ( f ) + Φ(1)) since 4 a − b − √ ∆ F < a + b + √ ∆ F and4 a − b + √ ∆ F > a + b − √ ∆ F .Using the fact that b < √ ∆ F ≤ a and the condition 4.4, all the factors of thefollowing difference B (Φ( f )) − B min = 8 a √ ∆ F [∆ F − a − ( a − b ) ]( √ ∆ F − b )[4 a − ( √ ∆ F − b ) ][4 a + √ ∆ F − b ]are non negative. In other words, B min ≤ B (Φ( f )). Therefore B min ≤ B max ,then Section 4.2 says that I is 1-reduced. EDUCED IDEALS FROM THE REDUCTION ALGORITHM 7 • Case 3: (cid:112) ∆ F / ≤ a ≤ (cid:112) ∆ F / a − √ a − ∆ F ≤ b ≤ a + √ ∆ F − a .Lemma 4.3 shows that Φ( f −
1) and Φ(1) form an LLL-reduced basis for I . Byusing an argument similar to the proof of Case 2, we obtain the following. B max = B (Φ( f )) and B min = max { B ( − Φ( f ) + Φ(1)) , B ( − Φ( f ) + 2Φ(1)) , B ( − f ) + 3Φ(1)) } where B ( − Φ( f ) + 2Φ(1)) = ( √ ∆ F + b − a )(6 a − b − √ ∆ F )( √ ∆ F + 2 a − b )(6 a + √ ∆ F − b ) ,B ( − f ) + 3Φ(1)) = ( b + √ ∆ F − a )(4 a − b − √ ∆ F )( √ ∆ F + 2 a − b )(4 a + √ ∆ F − b ) . Similar to Case 2, we have B max is greater or equal to B ( − Φ( f )+Φ(1)) , B ( − Φ( f )+2Φ(1)) and B ( − f ) + 3Φ(1)) by the bounds on ∆ F and b . Thus, B max ≥ B min .Therefore I is 1-reduced by the result of Section 4.2.To complete our proof, we prove the following results. Lemma 4.1.
With the assumption in Theorem 1.1, if one of the following holds (1) a ≤ (cid:112) ∆ F / , or (2) (cid:112) ∆ F / ≤ a ≤ (cid:112) ∆ F / and √ a − ∆ F ≤ b ≤ a − √ a − ∆ F ,then is shortest in the lattice I .Proof. By Remark 4.1, we can write I as a lattice in R as below. I = (cid:34) b + √ ∆ F a b −√ ∆ F a (cid:35) Z . The integers a, b satisfy the condition 4.1 in Remark 4.1.Let g ∈ I . Then g = (cid:18) m + n · b + √ ∆ F a , m + n · b − √ ∆ F a (cid:19) for some ( m, n ) ∈ Z . Thus,(4.5) (cid:107) g (cid:107) = 2 (cid:34)(cid:18) m + nb a (cid:19) + n ∆ F a (cid:35) . Case 1: a ≤ (cid:112) ∆ F /
4. We will show that 1 is shortest in the lattice I . Equivalently, wewill prove that if (cid:107) g (cid:107) < g ∈ I then g = 0. Indeed, if (cid:107) g (cid:107) < .
5, we have n ∆ F a < . Hence n = 0 since ∆ F ≥ a . Moreover, (cid:107) g (cid:107) = 2 m ≥ m (cid:54) = 0. Thus m = 0therefore g = 0. Case 2: (cid:112) ∆ F / ≤ a ≤ (cid:112) ∆ F / √ a − ∆ F ≤ b ≤ a − √ a − ∆ F . Similar to HA THANH NGUYEN TRAN AND DUONG HOANG DUNG
Case 1, we also show that 1 is shortest in I . Let g ∈ I such that (cid:107) g (cid:107) <
2. Then n ≤ F ≥ a and by 4 .
5. If n = 1, then n ∆ F a ≥ . Thus (cid:12)(cid:12)(cid:12)(cid:12) m + nb a (cid:12)(cid:12)(cid:12)(cid:12) < . The bounds on b imply that 0 ≤ b/ (2 a ) ≤ /
2. Therefore m = 0 and then (cid:107) g (cid:107) = 2( b + ∆ F )4 a , that is at least 2 by the lower bound on b and ∆ F . Thus n = 0 and hence g = 0 as Case1. (cid:3) Lemma 4.2.
With the assumption in Theorem 1.1, if (cid:112) ∆ F / ≤ a ≤ (cid:112) ∆ F / and a − √ ∆ F − a ≤ b ≤ √ a − ∆ F , then f is a shortest vector of the lattice I . Moreover,the two vectors Φ( f ) and Φ(1) form an LLL-reduced basis for I .Proof. Let g ∈ I \{ } . With the notations as in the proof of Lemma 4.1, we show that if (cid:107) g (cid:107) < g = ± f . Hence f is shortest in I .Since ∆ F / (4 a ) ≥ /
4, using a similar argument as in Case 2 of Lemma 4.1 leadsto n ≤
1. If n = 0 then (cid:107) g (cid:107) ≥ m ≥ m (cid:54) = 0. Thus n = ±
1. Hence (cid:12)(cid:12) m ± b a (cid:12)(cid:12) < by 4.5, then | m | < + b a . The fact that b ≤ √ a − ∆ F ≤ a implies that b a ≤ . Thus m = 0 and then g = ± f .Now let µ = (cid:104) Φ( f ) , Φ(1) (cid:105)(cid:107) Φ( f ) (cid:107) = σ ( f ) + σ (cid:48) ( f ) (cid:107) f (cid:107) = 2 abb + ∆ F . Since ∆ F ≥ a , we have √ a − ∆ F ≤ a − √ a − ∆ F . Consequently, one has b ≤ a − √ a − ∆ F . Hence 4 ab ≤ b + ∆ F , which implies that | µ | ≤ . Thus, { Φ( f ) , Φ(1) } is an LLL-reduced basis for I . (cid:3) Lemma 4.3.
With the assumption in Theorem 1.1, if (cid:112) ∆ F / ≤ a ≤ (cid:112) ∆ F / and a − √ a − ∆ F ≤ b ≤ √ ∆ F , then f − is a shortest vector of the lattice I . Moreover,the two vectors Φ( f − and Φ(1) form an LLL-reduced basis for I .Proof. Let g ∈ I \{ } . With the notations as in the proof of Lemma 4.1, we show that if (cid:107) g (cid:107) < g = ± ( f − f − I .Since ∆ F / (4 a ) ≥ /
4, using a similar argument as in Case 2 of Lemma 4.1 leads to n = ±
1. Hence (cid:12)(cid:12) m ± b a (cid:12)(cid:12) < then | m | < + b a . The bounds on b and ∆ F imply that b a < | m | ≤
1. Thus g ∈ {± f, ± ( f + 1) , ± ( f − } . Since ∆ F ≥ a , the lower bound 2 a − √ a − ∆ F on b is at least √ a − ∆ F . Therefore b ≥ a − ∆ F , which implies (cid:107) f (cid:107) = 2 (cid:18) b + ∆ F a (cid:19) ≥ . EDUCED IDEALS FROM THE REDUCTION ALGORITHM 9
It is easy to see that (cid:107) ( f + 1) (cid:107) ≥ (cid:107) f (cid:107) ≥ g = ± ( f − µ = (cid:104) Φ( f − , Φ(1) (cid:105)(cid:107) Φ( f − (cid:107) . One has | µ | = (cid:12)(cid:12)(cid:12)(cid:12) σ ( f ) + σ (cid:48) ( f ) − (cid:107) f − (cid:107) (cid:12)(cid:12)(cid:12)(cid:12) = 2 a (2 a − b )(2 a − b ) + ∆ F ≤ . The last inequality is obtained since b ≥ a − ∆ F . Thus, { Φ( f − , Φ(1) } is anLLL-reduced basis for I . (cid:3) Corollary 4.1.
Let F be a real quadratic field and let I = Z + Z f be a reduced fractionalideal of F where f is in Remark 4.1. Then I is not 1-reduced if and only if N ( f − / > − / .Proof. It was shown by Example 9.5 in [6] that if N ( f − / > − /
4, then I is not1-reduced. Hence, this result is implied from Theorem 1.1. (cid:3) Corollary 4.2.
Let F be a real quadratic field and let I be a fractional ideal of F . If N ( I − ) > (cid:112) ∆ F / then I is not 1-reduced.Proof. This can be easily seen by the inequality 4.4. (cid:3) Conclusion and Open Problems
Determining when a reduced ideal is 1-reduced can be solved for quadratic fields sincetheir ideals are explicitly and nicely described (see Remark 4.1 and [8]). However, thereis no such a description for ideals of higher degree number fields. Hence, this will be achallenge for us to work in the future.In addition, finding properties, the cardinality and the distribution (in the topologicalgroup Pic F ) of the set of reduced but not 1-reduced ideals of an arbitrary number fieldis an open problem for further research. Acknowledgement
The author is financially supported by the Pacific Institute for the MathematicalSciences (PIMS).
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Department of Mathematics and Statistics, University of Calgary, Canada.
E-mail address : [email protected] School of Computing and Information Technology, University of Wollongong NSW,Australia, 2522.
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