An elementary proof for a generalization of a Pohst's inequality
aa r X i v : . [ m a t h . N T ] J a n AN ELEMENTARY PROOF FOR A GENERALIZATION OF A POHST’SINEQUALITY
FRANCESCO BATTISTONI AND GIUSEPPE MOLTENI
Abstract.
Let P n ( y , . . . , y n ) := Y ≤ i 11. As a consequence, the bound for the absolute discriminant of atotally real field in terms of its regulator is now proved for every degree of the field. Introduction Let y , . . . , y n be n ≥ | y | < | y | < · · · < | y n | . Define then the positive real number P n ( y , . . . , y n ) := Y ≤ i 2. This is motivatedby number theoretic reasons: in fact, let K be a number field of degree n ≥ ε be aunit of its ring of integers such that K = Q ( ε ). The discriminant d K of the field K divides thediscriminant of the minimum polynomial of ε , inducing the inequality | d K | ≤ Y ≤ i 11, and Bertin [2] produced a new proof of Remak’s estimate. Inthe same paper Bertin also gave an argument trying to prove that Pohst’s estimation holds for Mathematics Subject Classification. Key words and phrases. Totally real fields, explicit bounds.The first author was supported by the French “Investissements d’Avenir” program, project ISITE-BFC (con-tract ANR-lS-IDEX-OOOB). every n , but her procedure is not completely convincing. In this paper we prove that Pohst’sestimation holds indeed for every n .In order to achieve this result, following aforementioned works, we choose a slightly differentfunction to estimate: given P n ( y , . . . , y n ), we define the change of variables(2) x i := y i y i +1 , i = 1 , . . . , n − P n ( y , . . . , y n ) into the quantity Q n − ( x , . . . , x n − ) := n − Y i =1 n − Y j = i − j Y k = i x k ! . Since | x i | ≤ i , the polynomials Q n are non negative over the cube D n := [ − , n ,and we look for M n := max ( x ,...,x n ) ∈ D n Q n ( x , . . . , x n ) . The change of variables (2) shows that P n = M n − for every n ≥ 2. Starting from this, in thenext sections we will prove the following theorem. Theorem 1. The maximum M n of Q n in D n is ⌊ n +12 ⌋ for every n , so that P n = M n − = 2 ⌊ n/ ⌋ for every n ≥ . It is easy to verify that Q n attains its maximum at ( − , , − , , . . . ) when n is odd, while foran even n this happens at each point ([ − , k , [0 , − n/ − k ) for any choice of k = 0 , , . . . , n/ − , k means that the string [ − , 0] has to be repeated k times, the same for [0 , − n/ − k ).Our argument proving Theorem 1 can be adapted to prove also that these are the unique pointswhere Q n attains its maximum, but we leave to the interested reader a formal proof of this fact. Corollary 1. Let K be a totally real and primitive field of degree n ≥ having discriminant d K and regulator R K . Then log | d K | ≤ r γ n − · n − n · ( √ nR K ) / ( n − + j n k log 4 . Basic inequalities An elementary computation shows that the maximums for the first two polynomials Q ( x ) =(1 − x ) and Q ( x , x ) = (1 − x )(1 − x x )(1 − x ) are(3) M = 2 , M = 2 , respectively. These numbers agree with the claim of the theorem. It is clear that the deter-mination of M n via local, i.e. analytic, methods involving partial derivatives becomes quicklyinfeasible as n increases: we take a different and global, so to say, approach, where the polynomialis split in suitable blocks and the maximum for the polynomial is deduced from the maximumsof those blocks. These maximums will be deduced from the following basic inequalities. Lemma 1. Let x, y, z be real numbers in [0 , . Then the following inequalities hold: (4) (1 − x )(1 + xy ) ≤ , (5) (1 − x )(1 + xy ) ≤ (1 + x )(1 − xy ) , (6) (1 − y )(1 + xy )(1 + yz )(1 − xyz ) ≤ (1 + y )(1 − xy )(1 − yz )(1 + xyz ) , (7) (1 − y )(1 + xy )(1 + yz )(1 − xyz ) ≤ . Proof. (4) is obvious, since 1+ xy ≤ x . (5) is reduced via direct computations to − x + xy ≤ y (1 + xy z )( x + z ) ≤ y (1 + xy z )(1 + xz ) . N ELEMENTARY PROOF FOR A GENERALIZATION OF A POHST’S INEQUALITY 3 This inequality is trivially satisfied if y = 0, otherwise we can remove y (1 + xy z ) obtaining(1 − x )(1 − z ) ≥ 0, which is true under our hypotheses.(7) appears in [3]; for sake of completeness we recall here a quicker proof. Compute all theproducts, remove the common terms, factor out y and move the terms to left hand side or righthand side according to the sign of the coefficient. In this way the inequality is proved to beequivalent to( y z + y z ) x + ( y z + yz + 1) x + xyz + z ≤ ( y z + yz ) x + ( y z + yz + y + z ) x + yz + 1 . Here, the first four monomials are smaller than the corresponding monomial to the right, sinceeach one contains a variable with higher exponent. The remaining inequality says that x + xyz + z ≤ xy + xz + yz + 1, which is a consequence of the fact that (1 − x )(1 − y )(1 − z ) ≥ (cid:3) Graphical schemes We call graphical scheme of dimension n any triangular n × n array C with symbols “+”or “ − ” in each entry C i,j with 1 ≤ i ≤ j ≤ n . The following are some examples of graphicalschemes in dimension n = 3 and n = 5, respectively:+ + −− ++ , − − + + − + − − ++ + −− −− .We associate with C the function F C : [0 , n → R defined as F C ( z , . . . , z n ) := n Y i =1 n Y j = i − C i,j j Y k = i z k ! , and we denote its ( i, j ) factor as F C i,j := 1 − C i,j j Y k = i z k . Given two graphical schemes C and C ′ of dimension n , we say that C ≤ C ′ if F C ( z , . . . , z n ) ≤ F C ′ ( z , . . . , z n ) for every choice of ( z , . . . , z n ) ∈ [0 , n . The following lemma describes fourbasic moves that when performed on a given scheme produce a larger (in the previous sense)scheme. Lemma 2. Let C be a graphical scheme of dimension n .P) ( Point ) Assume C i,j = + . Let C ′ be the graphical scheme defined by C ′ r,s = ( − ( r, s ) = ( i, j ) C r,s otherwise.Then C ≤ C ′ . Moreover, F C i,j ≤ .H) ( Horizontal segment ) Assume C i,j = + and C i,j + k = − , with k ≤ n − j . Let C ′ be thegraphical scheme defined by C ′ r,s = − ( r, s ) = ( i, j )+ ( r, s ) = ( i, j + k ) C l,k otherwise.Then C ≤ C ′ . Moreover, F C i,j · F C i,j + k ≤ . V) ( Vertical segment ) Assume C i,j = − and C i + k,j = + with k ≤ j − i . Let C ′ be thegraphical scheme defined by C ′ r,s = + ( r, s ) = ( i, j ) − ( r, s ) = ( i + k, j ) C l,k otherwise. F. BATTISTONI AND G. MOLTENI Then C ≤ C ′ . Moreover, F C i,j · F C i + k,j ≤ . S) ( Square ) Assume C i,j = − , C i,j + k = + , C i + l,k = + and C i + l,j + k = − . Let C ′ be thegraphical scheme defined by C ′ r,s = + ( r, s ) = ( i, j ) − ( r, s ) = ( i, j + k ) − ( r, s ) = ( i + l, j )+ ( r, s ) = ( i + l, j + k ) C l,k otherwise.Then C ≤ C ′ . Moreover, F C i,j F C i + l,j F C i,j + k F C i + l,j + k ≤ . We introduce a notation for these moves:P) Point : P( i ; j ) denotes the change of ji + into ji − ,H) Horizontal : H( i ; j, j ′ ) denotes the change of j j ′ i + − into j j ′ i − + ,V) Vertical : V( i, i ′ ; j ) denotes the change of ji − i ′ + into ji + i ′ − ,S) Square : S( i, i ′ ; j, j ′ ) denotes the change of j j ′ i − + i ′ + − into j j ′ i + − i ′ − + . Proof. P) We have F C i,j = 1 − j Y k = i z k ≤ j Y k = i z k and since every other factor of F C remains unchanged, we get F C ≤ F C ′ . The statement F C i,j ≤ F C i,j · F C i,j + k ≤ F C i,j · F C i,j + k = − j Y l = i z l ! j Y l = i z l j + k Y l = j +1 z l ≤ j Y l = i z l ! − j Y l = i z l j + k Y l = j +1 z l = F C ′ i,j · F C ′ i,j + k and this proves F C ≤ F C ′ since every other factor is unchanged.V) is proved in a similar way to case H).S) F C i,j F C i + l,j F C i,j + k F C i + l,j + k ≤ F C i + l,j · F C i,j · F C i + l,j + k · F C i,j + k = − j Y v = i + l z v ! i + l − Y v = i z v j Y v = i + l z v ! j Y v = i + l z v j + k Y v = j +1 z v − i + l − Y v = i z v j Y v = i + l z v j + k Y v = j +1 z v ≤ j Y v = i + l z v ! − i + l − Y v = i z v j Y v = i + l z v ! − j Y v = i + l z v j + k Y v = j +1 z v i + l − Y v = i z v j Y v = i + l z v j + k Y v = j +1 z v = F C ′ i + l,j · F C ′ i,j · F C ′ i + l,j + k · F C ′ i,j + k and this proves F C ≤ F C ′ since every other factor is unchanged. (cid:3) N ELEMENTARY PROOF FOR A GENERALIZATION OF A POHST’S INEQUALITY 5 Properties of the schemes generated by sign vectors Identifying numbers ± ± , we can generate a graphical scheme C ( ε ) from eachsigns vector ε := ( ε , . . . , ε n ), ε k ∈ {± } , by setting C ( ε ) i,j := Q jk = i ε k for every ( i, j ). Forexample, the vector ε := (1 , , − , , − 1) generates the scheme C ( ε ) = + + − − ++ − − + − − ++ −− . The interest for this construction comes from the following remark. We can split D n = [ − , n into 2 n different chambers D n, ε , each one associated with a different signs vector ε = ( ε , . . . , ε n ),where D n, ε := { ( x , . . . , x n ) ∈ [ − , n : x i ε i ≥ , ∀ i } . Once we have chosen D n, ε , the change of variables z i := ε i x i transforms D n, ε into [0 , n , and Q n ( x , . . . , x n ) into Q n ( ε z , . . . , ε n z n ) = n Y i =1 n Y j = i − j Y k = i ε k j Y k = i z k ! , which is exactly the polynomial F C ( ε ) associated with the scheme C ( ε ) generated by the signsvector ε . This gives us a strategy to prove Theorem 1: we will prove that for each scheme C ( ε )there is a list of moves P , V , H and S which transform C ( ε ) into C − , the n -dimensional schemegenerated by the signs ε − := ( − , · · · , − 1) (see next Theorem 2): by Lemma 2 these movesincrease the value of the associated polynomial, hence the maximum of each F C ( ε ) is lower thanthe one of F C − . In other words, this means that the maximum of Q n in every chamber D n, ε isthe one of F C − , at most. Thus, the conclusion easily follows from the next lemma giving themaximum for F C − . Lemma 3. Let C − be the n -dimensional scheme generated by the signs ε − := ( − , · · · , − .Then F C − ( z , . . . , z n ) ≤ ⌊ n +12 ⌋ ∀ ( z , . . . , z n ) ∈ [0 , n . Proof. The graphical scheme C − has the form − + − + · · ·− + − · · ·− + · · ·− · · ·· · · where every row starts with a sign − and continues with alternating signs. We know that theclaim for n = 1 and n = 2 is true thanks to (3). Let n ≥ 3. If n is odd, the scheme C − has theform − + − + · · · − + −− + − · · · + − + C − ,n − while for n even has the form − + − + · · · − + − + − · · · + − C − ,n − where in both cases C − ,n − is the n − n − F C − ,n − ≤ ⌊ ( n − / ⌋ .Let us look at the first two rows of C − : here, the first two columns form a triangular array F. BATTISTONI AND G. MOLTENI in dimension 2: hence F C , F C , F C , ≤ ⌊ ( n − / ⌋ consecutive squares − ++ − , plus, in case n is odd, an extra vertical segment − + . Entries V) andS) of Lemma 2 prove that the contribution of each such square and of the vertical segment arebounded by 1. Hence, in every case the contribution of the first two rows is estimated by 2, and F C − ≤ · F C − ,n − ≤ · ⌊ n − ⌋ = 2 ⌊ n +12 ⌋ . (cid:3) To succeed in this task we need to further investigate some properties of the schemes generatedby sign vectors; they are contained in next three lemmas. Lemma 4. Let C ( ε ) be a scheme generated by the sign vector ε of dimension n ≥ . Let i < i ′ , j < j ′ with i ′ < j . The product of the four signs C ( ε ) i,j , C ( ε ) i ′ ,j , C ( ε ) i,j ′ and C ( ε ) i ′ ,j ′ is . Inother words, the number of minus signs in every square j j ′ ii ′ is even.Proof. In fact, we have C ( ε ) i,j C ( ε ) i ′ ,j C ( ε ) i,j ′ C ( ε ) i ′ ,j ′ = j Y k = i ε k j Y k = i ′ ε k j ′ Y k = i ε k j ′ Y k = i ′ ε k = j ′ Y k = j +1 ε k j ′ Y k = j +1 ε k = 1 . (cid:3) Let C be a graphical scheme. We say that the sign C i,j is correct if C i,j = ( − i − j +1 , otherwisewe say that C i,j is wrong . It is clear that the only graphical scheme having only correct signs is C − , i.e., the one generated by the signs vector ε − := ( − , . . . , − Lemma 5. Let C ( ε ) be a scheme generated by the sign vector ε of dimension n and for i ≤ j ≤ n let H ( i, j ) := P j − u = i C ( ε ) i,u (the sum of entries in C ( ε ) appearing to the left of C ( ε ) i,j ), and V ( i, j ) := P jv = i +1 C ( ε ) v,j (the sum of entries in C ( ε ) appearing below C ( ε ) i,j ). Suppose that C ( ε ) i,j = − , then H ( i, j ) = − V ( i, j ) .Proof. In fact, C ( ε ) i,u = Q uk = i ε k and by hypothesis C ( ε ) i,j = Q jk = i ε k = − 1. Thus, for i ≤ u ≤ j − C ( ε ) i,u = u Y k = i ε k = − C ( ε ) i,j u Y k = i ε k = − j Y k = i ε k u Y k = i ε k = − j Y k = u +1 ε k = − C ( ε ) u +1 ,j . Hence, each term appearing below C ( ε ) i,j is opposite to a convenient term appearing to the leftof C ( ε ) i,j , and vice versa. (cid:3) We introduce the following quantities, again under the hypothesis that i ≤ j . H w ± ( i, j ) := { k : i ≤ k ≤ j − , C i,k = ± , C i,k is wrong } ,V w ± ( i, j ) := { k : i + 1 ≤ k ≤ j, C k,j = ± , C k,j is wrong } ,H w ( i, j ) := H w + ( i, j ) − H w − ( i, j ) , V w ( i, j ) := V w + ( i, j ) − V w − ( i, j ) . Lemma 6. Let C ( ε ) be a scheme generated by the sign vector ε and assume that C ( ε ) i,j = − and that i + j is odd. Then V ( i, j ) = 2 V w ( i, j ) − and H ( i, j ) = 2 H w ( i, j ) − . We know that V ( i, j ) and H ( i, j ) are opposite in sign by Lemma 5, therefore H w ( i, j ) + V w ( i, j ) = 1 and inparticular, at least one between H w ( i, j ) and V w ( i, j ) is positive.Proof. Since i + j is odd, there exist j − i signs C ( ε ) l,j below C ( ε ) i,j , and the quantity j − l + 1is odd for ( j − i + 1) / j − i − / C ( ε ) i,j appear at positions ( l, j ) where j − l + 1 is odd, and every other sign here whichis not a wrong + is necessarily a − (actually a correct − , but this in not important now), thus j X l = i +1 j − l +1 odd C ( ε ) l,j = V w + ( i, j ) − (cid:16) 12 ( j − i + 1) − V w + ( i, j ) (cid:17) = 2 V w + ( i, j ) − 12 ( j − i + 1) . N ELEMENTARY PROOF FOR A GENERALIZATION OF A POHST’S INEQUALITY 7 Similarly, wrong − ’s below C ( ε ) i,j appear at positions ( l, j ) where j − l + 1 is even, and everyother sign here which is not a wrong − is necessarily a +, so that j X l = i +1 j − l +1 even C ( ε ) l,j = − V w − ( i, j ) + (cid:16) 12 ( j − i − − V w − ( i, j ) (cid:17) = − V w − ( i, j ) + 12 ( j − i − V ( i, j ) = j X l = i +1 C ( ε ) l,j = 2( V w + ( i, j ) − V w − ( i, j )) − 12 ( j − i + 1) + 12 ( j − i − 1) = 2 V w ( i, j ) − . The proof for H ( i, j ) is similar. (cid:3) The procedure We are now ready to prove the following theorem. As recalled in the previous section, it yieldsTheorem 1 as immediate corollary thanks to Lemma 2 and Lemma 3. Theorem 2. Let C = C ( ε ) be the scheme generated by any signs vector ε and let C − be thescheme generated by the sign vector ε with all negative signs. There is a list L of transformationsof type P , H , V and S which changes C into C − .Proof. Let ε = ( ε , . . . , ε n ) be the signs vector producing C . We prove the theorem by makinginduction on the dimension n .If n = 1, we only have two possibilities: either C = − and we have finished, or C = + and theclaim follows by applying P(1; 1).Now, assume that n > n . Let C ′ be the scheme obtained by removing the n -th columnfrom C : this is the scheme generated by the signs vector omitting ε n in ε . By inductivehypothesis, there exists a list L ′ of moves which applied to C ′ gives C ′− , the array of dimension n − L ′ by replacing themwith other moves which correct all wrong symbol in the n -th column and that coincide withthe old move on the common part in C ′ : in this way we will obtain a new list L of moves thatapplied to C give C − .Moreover, in order to prove that the algorithm can be correctly performed, we need to keep noteof each move we introduce, and of its effect on the n -th column. For this purpose we introducethe symbols D (1) , D (2) and so on, to denote the several new versions of the n -th column we getafter each new move is performed. At the beginning we have D (1) , which coincides with the n -th column.We start running the column D ( k ) from the bottom to the top, looking for wrong signs − . In casesuch signs do not appear, we skip this step and we go directly to the last one. On the contrary,suppose that we have found a wrong − in i -th line. We will see that in each new version ofthe column only some wrong positions are changed with respect to its previous version. As aconsequence, the wrong − in line i -th we have detected now was already there at the beginning,i.e., C i,n = − and i + n is odd. We compute both V w ( i, n ) and V w new ( i, n ), which are the sumof wrong signs appearing below the ( i, n ) position respectively in C , the original configuration,and in the column D ( k ) : at the beginning evidently numbers V w ( i, n ) and V w new ( i, n ) coincide,but as the algorithm progresses the second may change its value. However, we will check thatafter each move we will introduce is executed, the value of the index (cid:2) number of wrong + below l in n -th column (cid:3) − (cid:2) number of wrong − below l in n -th column (cid:3) for each l < i does not decrease. This proves that the number V w new ( i, n ) we compute in anytime is for sure ≥ V w ( i, n ).We note that the number V ( i, n ) is odd, by Lemma 6. In particular, it cannot be 0.Suppose that V ( i, n ) > 0. Then V w ( i, n ) > V w new ( i, n ) is positive as wellby the previous remark. This means that in some position below ( i, n ) there is a wrong + incolumn n . Let i ′ be the first (i.e., smallest) index i ′ > i such that in the ( i ′ , n ) position there isa wrong +. We add to L ′ the move V( i, i ′ ; n ): this move is independent of the other moves, and F. BATTISTONI AND G. MOLTENI converts the wrong − and + in those positions into two correct symbols. This move does notchange the value of (cid:2) number of wrong + below l in n -th column (cid:3) − (cid:2) number of wrong − below l in n -th column (cid:3) for each l < i , because the move simply exchanges a + with a − both in positions below the l -th position.Suppose that V ( i, n ) < 0. Then V w ( i, n ) < H w ( i, n ) is positive, both by Lemma 6. Thus,in the i -th horizontal line to the left of C i,n , and hence in C ′ , there is an excess of wrong +’swith respect to wrong − ’s. By induction there are moves in L ′ changing all these wrong entries.Moves of type H or S cannot be the unique moves in L ′ affecting these positions, since theyexchange both a wrong + and a wrong − and therefore cannot remove the excess. Also a moveof type V( i, i ′ ; j ) ji − i ′ + is not sufficient to remove the excess, since it removes only a wrong − from that line, a fact which actually increases the excess. Thus, at least a move P( i ; j ) ji + or amove V( i ′ , i ; j ) ji ′ − i + is in L ′ . Let us take j to be the greatest index < n such that this happens.In the first case we substitute P( i ; j ) with H( i ; j, n ) j ni + − which has the same effect on the C ′ part of the configuration. In the second case we note that the signs at ( i ′ , j ), ( i, j ) and ( i, n )positions are j ni ′ − i + − . By Lemma 4 the fourth corner C i ′ ,n of the square in C is a wrong +. Wewill show in a moment that this is a + also in D ( k ) , i.e. it appears also at this stage of thealgorithm. Letting this fact for granted for the moment, we proceed substituting V( i ′ , i ; j ) in L ′ with S( i ′ , i ; j, n ) j ni ′ − + i + − which again has the same effect on the C ′ part of the configuration.Both moves change (cid:2) number of wrong + below l in n -th column (cid:3) − (cid:2) number of wrong − below l in n -th column (cid:3) in positions l < i . However, the first one actually simply removes a wrong − , so that it increasesthe index for all l < i , while the second one increases it when i ′ ≤ l < i (because it removes the wrong − ), and keeps unchanged its value for l < i ′ (because then also the cancellation of the wrong + at C i ′ ,n matters).We execute the move we have selected, getting the new column which is D ( k +1) , by definition.We repeat this cycle again and again, removing all wrong − ’s from the n -column in C . Finally,we add P moves to L ′ to remove any remaining wrong +’s in last column, if any exists.The description of the algorithm ends here, but we have to resume the point we have skippedbefore, i.e., the proof of the fact that the wrong + appearing at the fourth corner ( i ′ , n ) of thesquare in C also appears in D ( k ) , i.e. it appears also at that stage of the algorithm. Supposethe contrary, i.e., that the wrong + is no more there, since it has been corrected at some earlierstep of the algorithm. Then, there had been some index i ′′ > i with C i ′′ ,n = − whose correctionneeded the substitution of some move V( i ′ , i ′′ ; j ′ ) in L ′ with S( i ′ , i ′′ ; j ′ , n ) for some j ′ = j , becausethis is the only possible way the algorithm can correct the + at ( i ′ , n ) at some previous step(the case j ′ = j is for sure impossible, otherwise the wrong − at ( i ′ , j ′ ) would be corrected inthat previous step and would not be available at k -th step). This means that we have one ofthe following signs patterns in C : j j ′ ni ′ − − + i + − i ′′ + − if j ′ > j , or j ′ j ni ′ − − + i + − i ′′ + − if j ′ < j .In both cases, at ( i, j ′ ) position we have a wrong + (by Lemma (4), when the square in positions( i ′ , j ), ( i ′ , j ′ ), ( i, j ), ( i, j ′ ) is considered), and the patterns in columns j ′ and n is j ′ ni ′ − + i + − i ′′ + − N ELEMENTARY PROOF FOR A GENERALIZATION OF A POHST’S INEQUALITY 9 in both cases. Moreover, L ′ contains V( i ′ , i ′′ ; j ′ ). However, this is impossible, since the patternshows that at ( i, j ′ ) we have a wrong + which is closer to the wrong − at ( i ′ , j ′ ) than the wrong + at ( i ′′ , j ′ ): this means that when the algorithm has been applied at an early stage to producethe moves in L ′ dealing with the j ′ -th column, we should have contradicted the prescriptionaccording to which every vertical move contains the + which appears at the closest position tothe − in that move. (cid:3) An example can be useful to understand the algorithm. Let C ( ε ) be the configuration indimension 7 which is generated by signs ε = (+ , − , + , + , − , + , +). Thus, C ( ε ) = + − − − + + + − − − + + ++ + − − − + − − −− − − + ++ . Then, applying the algorithm iteratively, we get: + = ⇒ L (1) = { P(1; 1) } + −− = ⇒ L (2) = { H(1; 1 , } + −− −− + = ⇒ L (3) = { H(1; 1 , , V(2 , 3; 3) } + − −− − + −− ++ = ⇒ L (4) = { H(1; 1 , , V(2 , 3; 3) , V(1 , 4; 4) } + − − −− − − + ++ ++ −−− = ⇒ L (5) = { H(1; 1 , , V(2 , 3; 3) , S(1 , 4; 4 , } + − − − + − − − ++ + − + −− ++ −−− + = ⇒ L (6) = { H(1; 1 , , S(2 , 3; 3 , , S(1 , 4; 4 , , V(5 , 6; 6) } + − − − + + − − − + ++ + − − + − −− − + ++ −−− ++ = ⇒ L (7) = n H(1; 1 , , S(2 , 3; 3 , , S(1 , 4; 4 , , V(5 , 6; 6) , V(4 , 7; 7) , P(1; 7) o References [1] S. Astudillo, F. Diaz y Diaz, and E. Friedman. Sharp lower bounds for regulators of small-degree numberfields. J. Number Theory , 167:232–258, 2016.[2] M. J. Bertin. Sur une conjecture de Pohst. Acta Arith. , 74(4):347–349, 1996.[3] M. Pohst. Regulatorabsch¨atzungen f¨ur total reelle algebraische Zahlk¨orper. J. Number Theory , 9(4):459–492,1977.[4] M. Pohst and H. Zassenhaus. Algorithmic algebraic number theory , volume 30 of Encyclopedia of Mathematicsand its Applications . Cambridge University Press, Cambridge, 1997. Revised reprint of the 1989 original.[5] R. Remak. ¨Uber Gr¨ossenbeziehungen zwischen Diskriminante und Regulator eines algebraischen Zahlk¨orpers. Compositio Math. , 10:245–285, 1952. Laboratoire de math´ematiques de Besanc¸on, Universit´e Bourgogne Franche-Comt´e, CNRS -UMR 6623, 16, Route de Gray, 25030 Besanc¸on, France Email address : [email protected] Dipartimento di Matematica, Universit`a di Milano, via Saldini 50, 20133 Milano, Italy Email address ::