aa r X i v : . [ m a t h . A C ] O c t Arf good semigroups
Giuseppe Zito
Abstract
In this paper we study the property of the Arf good subsemigroups of N n , with n ≥ .We give a way to compute all the Arf semigroups with a given collection of multiplicitybranches. We also deal with the problem of determining the Arf closure of a set of vec-tors and of a good semigroup, extending the concept of characters of an Arf numericalsemigroup to Arf good semigroups. Introduction
In this paper we study a particular class of good subsemigroups of N n . The concept of goodsemigroup was introduced in [2]. Its definition depends on the properties of the value semi-groups of one dimensional analytically unramfied ring (for example the local rings of an al-gebraic curve), but in the same paper it is shown that the class of good semigroups is biggerthan the class of value semigroups. Therefore the good semigroups can be seen as a naturalgeneralization of the numerical semigroup and can be studied without referring to the ring the-ory context, with a more combinatorical approach. In this paper we deal only with local goodsemigroups, i.e good semigroups S ⊆ N n such that the only element of S with zero componentis the zero vector.In this paper we focus on the class of local Arf good semigroups. This is motivated by theimportance of the Arf numerical semigroups in the study of the equivalence of two algebroidbranches. Given an algebroid branch R , its multiplicity sequence is defined to be the sequenceof the multiplicities of the succesive blowups R i of R . Two algebroid branches are equivalentif and only if they have the same multiplicity sequence (cf. [4, Definition 1.5.11]). In [1] it isintroduced the concept of Arf ring and it is shown that for each ring R there is a smallest Arfoverring R ′ , called the Arf closure of R , that has also the same multiplicity sequence of R . In Keywords: Good semigroup, Arf closure, semigroup of values, algebroid curve,characters of an Arf semi-group.Mathematics Subject Classification (2010): 20M14, 13A18, 14H50,20-04. S such that S ( s ) − s is a semigroup, for each s ∈ S , where S ( s ) = { n ∈ S ; n ≥ s } .All these facts can be generalized to algebroid curves (with more than one branch) and thisnaturally leads to define the Arf good semigroups of N n by extending the numerical definitionconsidering the usual partial ordering given by the components.In the numerical case an Arf semigroup S = { s = 0 < s < s , . . . } is completely de-scribed by its multiplicity sequence, that is the sequence of the differences s i +1 − s i . Extendingthe concept of multiplicity sequence, in [2] it is also shown that to each local Arf good semi-group can be associated a multiplicity tree that characterizes the semigroup completely. A tree T of vectors of N n has to satisfy some properties to be a multiplicity tree of a local Arf goodsemigroup. For instance it must have multiplicity sequences along its branches (since the pro-jections are Arf numerical semigroups) and each node must be able to be expressed as a sum ofnodes in a subtree of T rooted in it. Thus, taking in account this - correspondence, the aim ofthis paper is to study Arf good semigroups by characterizing their multiplicity trees, finding anunambiguous way to describe them. Using this approach, we can also deal with the problem offinding the Arf closure of a good semigroup S , that is the smallest Arf semigroup containing S .The structure of the paper is the following.In Section 1, given a collection of n multiplicity sequences E , we define the set σ ( E ) of allthe Arf semigroups S such that the i -th projection S i is an Arf numerical semigroup associatedto the i -th multiplicity sequences of E . We define also the set τ ( E ) of the correspondingmultiplicity trees and we describe a tree in τ ( E ) by an upper triangular matrix ( p i,j ) , where p i,j is the highest level where the i -th and j -th branches are glued, and we give a way to deducefrom E the maximal value that can be assigned to the p i,j . This fact let us to understand whenthe set σ ( E ) is finite. We introduce the class of untwisted trees that are easier to study becausethey are completely described by the second diagonal of their matrix, and we notice that a treecan be always transformed in to an untwisted one by permuting its branches.In Section 2 we address the problem of understanding when a set of vectors G ⊆ N n de-termines uniquely an Arf semigroup of N n . Thus we define Arf ( G ) as the minimum of the set S ( G ) = { S : S ⊆ N n is an Arf semigroup and G ⊆ S } , and we find the properties that G hasto satisfy in order to have a good definition for Arf ( G ) (cf. Theorem 2.1). Finally, given a G satisfying these properties, we give a procedure for computing Arf ( G ) .In Section 3 we adapt the techniques learned in the previous section to the problem ofdetermining the Arf closure of a good semigroup. In [6], the authors solved this problem for n = 2 , leaving it open for larger dimensions. In this section we use the fact that a goodsemigroup S can be completely described by its finite subset Small ( S ) = { s ∈ S : s ≤ δ } ,where δ is the smallest element such that δ + N n ⊆ S , whose existence is guaranteed by theproperties of the good semigroups.Finally, in Section 4, we address the inverse problem: given an Arf semigroup S ⊆ N n , finda set of vectors G ⊆ N n , called set of generators of S , such that Arf ( G ) = S, in order to find apossible generalization of the concept of characters in the numerical case. In Theorem 4.1, wefind the properties that such a G has to satisfy and we focus on the problem of finding a minimal2ne. From this point of view we are able to give a lower and an upper bound for the minimalcardinality for a set of generators of a given Arf semigroup (Corollary 4.9). With an examplewe also show that, given an Arf semigroup S , it is possible to find minimal sets of generatorswith distinct cardinalities.The procedures presented here have been implemented in GAP ([8]). In this section we determine all the local Arf good semigroups having the same collection ofmultiplicity branches.First of all we need to fix some notations and recall the most important definitions. In thefollowing, given a vector v in N n , we will always denote by v [ i ] its i -th component.A good semigroup S of N n is a submonoid of ( N n , +) such that: (cf. [2]) • for all a, b ∈ S , min( a, b ) ∈ S ; • if a, b ∈ S and a [ i ] = b [ i ] for some i ∈ { , . . . , n } , then there exists c ∈ S such that c [ i ] > a [ i ] = b [ i ] , c [ j ] ≥ min( a [ j ] , b [ j ]) for j ∈ { , . . . , n }\{ i } and c [ j ] = min( a [ j ] , b [ j ]) if a [ j ] = b [ j ] ; • there exists δ ∈ S such that δ + N n ⊆ S (where we are considering the usual partial ordering in N n : a ≤ b if a [ i ] ≤ b [ i ] for each i = 1 , . . . , n ).In this paper we will always deal with local good semigroups. A good semigroup S is local ifthe zero vector is the only vector of S with some component equal to zero. However, it can beshown that every good semigroup is the direct product of local semigroups (cf. [2, Theorem2.5]).An Arf semigroup of N n , is a good semigroup such that S ( α ) − α is a semigroup for each α ∈ S where S ( α ) = { β ∈ S ; β ≥ α } . The multiplicity tree T of a local Arf semigroup S ⊆ N n is a tree where the nodes are vector n ji ∈ N n (where with n ji we mean that this node isin the i -th branch on the j -th level. The root of the tree is n = n i for all i because we are inthe local case and at level one all the branches must be glued) and we have S = { } [ T ′ X n ji ∈ T ′ n ji , where T ′ ranges over all finite subtree of T rooted in n .Furthermore a tree T is a multiplicity tree of an Arf semigroup if and only if its nodes satisfythe following properties (cf. [2, Theorem 5.11]): • there exists L ∈ N such that for m ≥ L , n mi = (0 , . . . , , , . . . , (the nonzero coordi-nate is in the i -th position) for any i = 1 , . . . , n ;3 n ji [ h ] = 0 if and only if n ji is not in the h -th branch of the tree; • each n ji can be obtained as a sum of nodes in a finite subtree T ′ of T rooted in n ji .Notice that from these properties it follows that we must have multiplicity sequences along eachbranch.Suppose now that E is an ordered collection of n multiplicity sequences (that will be the multi-plicity branches of a multiplicity tree). Since any multiplicity sequence is a sequence of integersthat stabilizes to 1, we can describe them by the vectors M ( i ) = [ m i, , . . . , m i,k i ] , with the convention that m i,j = 1 for all j ≥ k i − and m i,k i − = 1 ; it will be clear later whydo not truncate the sequence to the last non-one entry.If M = max( k , . . . , k n ) we write for all i = 1 , . . . , nM ( i ) = [ m i, , . . . , m i,M ] , in order to have vectors of the same length. Each M ( i ) represents a multiplicity sequence of anArf numerical semigroup, so it must satisfy the following property: ∀ j ≥ there exists s i,j ∈ N , such that s i,j ≥ j + 1 and m i,j = s i,j X k = j +1 m i,k . Denote by τ ( E ) the set of all multiplicity trees having the n branches in E and by σ ( E ) the setof the corresponding Arf semigroups. We want to find an unambiguous way to describe distincttrees of τ ( E ) .We define, for all i = 1 , . . . , n , the following vectors S ( i ) = [ s i, , . . . s i,M ] . Because we have m i,j = 1 for all j ≥ M − , it follows that s i,j = j + 1 for all j ≥ M − . Example 1.1.
Let M (1) be the following multiplicity sequence: M (1) = [14 , , , , . Then S (1) is: S (1) = [5 , , , , . Notice that, with this notation, from the vectors S ( i ) we can easily reconstruct the sequences M ( i ) . It suffices to set m i,M = 1 and then to compute the values of m i,j using the informationcontained in the integers s i,j . 4e will use the vectors S ( i ) to determine the level, in a tree of τ ( E ) , where two brancheshave to split up.For each pair of integers i, j such that i < j and i, j = 1 , . . . , n we consider the set D ( i, j ) = { k : s i,k = s j,k } . If D ( i, j ) = ∅ we consider the integer k E ( i, j ) = min { min( s i,k , s j,k ) , k ∈ D ( i, j ) } , while if D ( i, j ) = ∅ , and then the i -th and j -th branches have the same multiplcity sequence,we set k E ( i, j ) = + ∞ . We have the following proposition. Proposition 1.2.
Consider a collection of multiplicity sequences E and let T ∈ τ ( E ) . Then k E ( i, j ) + 1 is the lowest level where the i -th and the j -th branches are prevented from beingglued in T (if k E ( i, j ) is infinite there are no limitations on the level where the branches haveto split up). Proof
The case k E ( i, j ) = + ∞ is trivial, because we have the same sequence along two consec-utive branches and therefore no discrepancies that force the two branches to split up at a certainlevel. Thus suppose k E ( i, j ) = + ∞ and, by contradiction, that the i -th and the j -th branchesare glued at level k E ( i, j ) + 1 . From the definition of k E ( i, j ) , there exists k ∈ D ( i, j ) such that k E ( i, j ) = min( s i,k , s j,k ) . Without loss of generality suppose that min( s i,k , s j,k ) = s i,k = s j,k (since k ∈ D ( i, j ) ).So in the tree we have the following nodes, ( . . . , m i,k , . . . , m j,k , . . . ) , . . . , ( . . . , m i,k E ( i,j ) , . . . , m j,k E ( i,j ) , . . . ) ,, ( . . . , m i,k E ( i,j )+1 , . . . , m j,k E ( i,j )+1 , . . . ) . We have that k E ( i, j ) = s i,k so m i,k = k E ( i,j ) X t = k +1 m i,t , while k E ( i, j ) + 1 = s i,k + 1 ≤ s j,k so m j,k ≥ k E ( i,j )+1 X t = k +1 m j,t . These facts easily imply that the first node cannot be expressed as a sum of the nodes of asubtree rooted in it, so we have a contradiction. Two branches are forced to split up only whenwe have this kind of problem, so the minimality of k E ( i, j ) guarantees that they can be glued atlevel k E ( i, j ) (and obviously at lower levels). Example 1.3.
Suppose that we have M (1) = [14 , , , , and M (2) = [7 , , , , .
5o we have the vectors S (1) and S (2) : S (1) = [5 , , , , and S (2) = [6 , , , , . We have D (1 ,
2) = { , } , then k (1 ,
2) = min { min(5 , , min(4 , } = min { , } = 4 . Thenthe branches have to be separated at the fifth level. (14 , , , ,
1) (0 , ,
0) (14 , , , , ,
1) (0 , , Notice that the first tree in the previous picture fulfills the properties of the multiplcity treesof an Arf semigroup. The second one cannot be the multiplicity tree of an Arf semigroupbecause the third node (5 , cannot be expressed as a sum of nodes in a subtree rooted in it.Now we prove a general lemma that will be useful in the following. Lemma 1.4.
Consider v , v and v in N n . If i, j ∈ { , , } with i = j we define: • MIN ( v i , v j ) = + ∞ if v i = v j ; • MIN ( v i , v j ) = min { min( v i [ k ] , v j [ k ]) , k ∈ { , . . . , n } : v i [ k ] = v j [ k ] } . Then there exists a permutation δ ∈ S such thatMIN ( v δ (1) , v δ (2) ) = MIN ( v δ (2) , v δ (3) ) ≤ MIN ( v δ (1) , v δ (3) ) . Proof
Suppose by contradiction that the thesis is not true. Then, renaming the indices if neces-sary, we have MIN ( v , v ) < MIN ( v , v ) ≤ MIN ( v , v ) . From the definition of MIN ( v , v ) = l , it follows that there exists a k ∈ { , . . . , n } such that v [ k ] = v [ k ] and min( v [ k ] , v [ k ]) = l , . We have two cases: • If v [ k ] = l , ⇒ v [ k ] > l , . Then we must have v [ k ] = l , , in fact otherwise we wouldhave MIN ( v , v ) ≤ l , < MIN ( v , v ) . But then l , < MIN ( v , v ) ≤ min( v [ k ] , v [ k ]) = l , , and we have a contradiction. 6 If v [ k ] = l , ⇒ v [ k ] > l , . Then we must have v [ k ] = l , , in fact otherwise we wouldhave MIN ( v , v ) ≤ l , < MIN ( v , v ) . But then l , < MIN ( v , v ) ≤ min( v [ k ] , v [ k ]) = l , , and we have a contradiction. Remark 1.5.
If we have three multiplicity sequences M (1) , M (2) and M (3) then, if E = { M (1) , M (2) , M (3) } then there exist a permutation δ ∈ S such that k E ( δ (1) , δ (2)) = k E ( δ (2) , δ (3)) ≤ k E ( δ (1) , δ (3)) . In fact the integers k E ( i, j ) are of the same type of the integers MIN ( v i , v j ) of the previouslemma with v i = S ( i ) .We give now a way to describe a tree of τ ( E ) . If T ∈ τ ( E ) , it can be represented by anupper triangular matrix n × nM ( T ) E = p , p , . . . p ,n p , . . . p ,n . . . . . . . . . . . . . . . . . . p n − ,n . . . , where p i,j is the highest level such that the i -th and the j -th branches are glued in T . Remark 1.6. If M ( T ) E is the matrix of a T ∈ τ ( E ) , it is clear that everytime we consider threeindices i < j < k we must have: p i,j ≥ min( p i,k , p j,k ) , p j,k ≥ min( p i,j , p i,k ) and p i,k ≥ min( p i,j , p j,k ) , when we are using the obvious fact that the relation of being glued has the transitive property.From the previous inequalities it follows that the set { p i,j , p j,k , p i,k } = { x, x, y } , with x ≤ y (independently of the order).From Proposition 1.2 we have that p i,j ∈ { , . . . , k E ( i, j ) } for all i, j = 1 , . . . , n with i < j .In the following, with an abuse of notation, we will identify a tree with its representation.We call a tree T of τ ( E ) untwisted if two nonconsecutive branches are glued at level l ifand only if all the consecutive branches between them are glued at a level greater or equal to l .We will call twisted a tree that it is not untwisted.From the definition it follows that the matrix of an untwisted tree T ∈ τ ( E ) is such that: p i,j = min { p i,i +1 , p i +1 ,i +2 , . . . , p j − ,j } for all i < j. So an untwisted tree can be completely described by the second diagonal of its matrix. Thusin the following we will indicate an untwisted tree by a vector T E = ( d , . . . , d n − ) where d i = p i,i +1 . It is easy to see that a twisted tree can be converted to an untwisted one by ac-cordingly permuting its branches. So in the following we can focus, when it is possible, onlyon the properties of the untwisted trees, that are easier to study, obtaining the twisted one bypermutation. 7 xample 1.7. Let us consider the following tree of τ ( E ) with E = { M (1) = [5 , , , , M (2) = [2 , , , , M (3) = [6 , , , } (5 , ,
6) (0 , , , , , ,
4) (0 , , , , This tree is twisted because the first and the third branches are glued at level two while thefirst and the second are not.If we consider the permutation (2 , on the branches we obtain the tree (5 , ,
2) (0 , , , , , ,
0) (0 , , , , that is untwisted, even if it belongs to a different set τ ( E ′ ) where E ′ = { M (1) = [5 , , , , M (2) = [6 , , , , M (3) = [2 , , , } , and can be represented by the vector T ′ E = (2 , .Denote by S ( T ) the semigroup determined by the tree T . In [3, Lemma 5.1] it is shownthat if T and T are untwisted trees of τ ( E ) , then S ( T ) ⊆ S ( T ) if and only if T E is coor-dinatewise less than or equal to T E . The previous result can be easily extended to the twistedtrees. Then, in the general case we have that S ( T ) ⊆ S ( T ) , where S ( T ) and S ( T ) belongto σ ( E ) , if and only if each entry of M ( T ) E is less than or equal to the corresponding entry of M ( T ) E . If k E ( i, j ) = + ∞ for all i < j , we can consider T MIN such that M ( T MIN ) E = k E (1 , k E (1 , . . . k E (1 , n )0 0 k E (2 , . . . k E (2 , n ) . . . . . . . . . . . . . . . . . . k E ( n − , n )0 0 0 . . . , that is well defined for Remark 1.5. Then S ( T MIN ) is the smallest Arf semigroup belonging to σ ( E ) . 8 emark 1.8. If in the collection E there are two branches with the same multiplicity sequencethen | σ ( E ) | = + ∞ . Example 1.9.
We can count the number of untwisted trees of τ ( E ) by using their representation.If we call τ ∗ ( E ) the set of all the untwisted trees of τ ( E ) , these trees are completely determinedby the elements in the second diagonal of their matrix, that are bounded by k E ( j, j + 1) . Hencethe number of untwisted trees is: | τ ∗ ( E ) | = n − Y j =1 k E ( j, j + 1) . Suppose that E = { M (1) , M (2) , M (3) } , where M (1) = [5 , , , , M (2) = [6 , , , , M (3) = [2 , , , . We have: S (1) = [3 , , , , S (2) = [4 , , , , S (3) = [2 , , , . Then D (1 ,
2) = { } , D (2 ,
3) = { , } and k E (1 ,
2) = min(3 ,
4) = 3 and k E (2 ,
3) = min { min(2 , , min(4 , } = 2 . There are k E (1 , · k E (2 ,
3) = 6 trees in τ ∗ ( E ) .They are: (5 , , , ,
2) (0 , , , ,
0) (0 , , , , T MIN = T E = (3 , (5 , ,
2) (0 , , , , , , , ,
0) (0 , , , , T E = (3 , (5 , , , ,
2) (0 , , , , , , T E = (2 , (5 , ,
2) (0 , , , , , ,
0) (0 , , , , T E = (2 , , ,
2) (0 , ,
2) (0 , , , , , , , , T E = (1 , (5 , ,
2) (0 , , , , , , , , , , , , T E = (1 , Remark 1.10.
Because we are able to determine completely τ ∗ ( E ) for each E collection ofmultiplicity sequences we have a way to determine τ ( E ) . If δ ∈ S n is a permutation of thesymmetric group S n we can consider δ − ( τ ∗ ( δ ( E ))) ⊆ τ ( E ) . It is trivial to see that [ δ ∈ S n δ − ( τ ∗ ( δ ( E ))) = τ ( E ) . If we apply this strategy to find τ ( E ) with the E of the previous example we find that in τ ( E ) there is only one twisted tree T with M ( T ) E = . (5 , ,
2) (0 , , , , , ,
2) (0 , , , , In this section we want to understand when a set G ⊆ N n determines uniquely an Arf semigroupof N n . First of all we need to fix some notations.Given G ⊆ N n we denote by S ( G ) the following set S ( G ) = { S : S ⊆ N n is an Arf semigroup and G ⊆ S } . If the set S ( G ) has a minimum (with the partial order given by the inclusion), we will denotesuch a minimum by Arf ( G ) . Hence we have to understand when Arf ( G ) is well defined and, inthis case, how to find it. 10f i ∈ { , . . . , n } , and S ∈ S ( G ) we denote by S i the projection on the i -th coordinate.We know that S i is an Arf numerical semigroup and it contains the set G [ i ] = { g [ i ] : g ∈ G } where with g [ i ] we indicate the i -th coordinate of the vector g . We recall also that, if we havea set of integers I such that gcd( I ) = 1 , it is possible to compute the smallest Arf semigroupcontaining I , that is the Arf closure of the numerical semigroup generated by the elements of I .This computation can be made by using the modified Jacobian algorithm of Du Val (cf. [7]).We have the following theorem: Theorem 2.1.
Suppose that we have G ⊆ N n . Then Arf ( G ) is well defined if and only if thefollowing conditions hold: • gcd { g [ i ] , g ∈ G } = 1 for i = 1 , . . . , n ; • For all i, j ∈ { , . . . , n } such that i < j there exists g ∈ G such that g [ i ] = g [ j ] . Proof ( ⇒ ) Suppose that Arf ( G ) is well defined and suppose by contradiction that the twoconditions of the theorem are not simultaneuously fulfilled.We have two cases. • Case 1 : The first condition is not fulfilled.Then there exists an i such that gcd( G [ i ]) = d = 1 . When we apply the Jacobian algo-rithm to the elements of G [ i ] we will produce a sequence of the following type: [ m i, , . . . , m i,k , . . . ] where there exists a k such that m i,j = d for all j ≥ k (it happens because the Jacobianalgorithm performs an Euclidean algorithm on G [ i ] ). Denote by k the minimum k suchthat the Arf semigroup associated to the sequence [ m i, , . . . , m i,k = d, , , contains G [ i ] (such minimum exists for the properties of the algorithm of Du Val). Thenfor all z ≥ k we can consider the multiplicity sequence M ( z ) = [ m i, , . . . , m i,k = d, . . . , m i,z = d, , and if S ( z ) is the Arf numerical semigroup associated to M ( z ) then G [ i ] ⊆ S ( z ) . Nowit is trivial to show that S ( z ) ⊆ S ( z ) if z ≥ z . Then we have an infinite decreasingchain of Arf semigroup containing the set G [ i ] . This means that the projection on the i -thbranch can be smaller and smaller, therefore we cannot find a minimum in the set S ( G ) .Thus we have found a contradiction in this case.An example illustrating Case 1 is the following.11f we consider G = { [2 , , [4 , } , we have no information on the multiplicity sequencealong the first branch and so we can obtain the following infinite decreasing chain of Arfsemigroups containing G : (2 , ,
1) (0 , , ⊇ (2 , ,
1) (0 , , , ⊇ (2 , ,
1) (0 , , , , ⊇ (2 , ,
1) (0 , , , , , ⊇ . . . • Case 2 : The first condition is fulfilled.So in this case the second condition is not fulfilled. The fact that gcd { g [ i ] , g ∈ G } = 1 for i = 1 , . . . , n implies that we can compute the smallest Arf numerical semigroup S ( i ) containing G [ i ] for all i = 1 , . . . , n .Therefore if we denote by M i the multiplicity sequence of S ( i ) we clearly have thatArf ( G ) ∈ σ ( E ) , where E = { M i , i = 1 , . . . , n } . Suppose that it is defined by the matrix M ( T ) E = p , p , . . . p ,n p , . . . p ,n . . . . . . . . . . . . . . . . . . p n − ,n . . . . Now if we consider an element h ∈ G [ i ] we have that h ∈ S ( i ) and therefore there existsan index pos E ( i, h ) such that h = pos E ( i,h ) X k =1 m i,k . If g ∈ G we can define pos E ( g ) = [ pos E (1 , g [1]) , . . . , pos E ( n, g [ n ])] .Notice that, if we consider i, j ∈ { , . . . , n } , with i < j and g ∈ G such that pos E ( i, g [ i ]) = pos E ( j, g [ j ]) , we can easily deduce that in a multiplicity tree of an Arf semigroup of σ ( E ) containing G the i -th and j -th branches cannot be glued at a level greater than min( pos E ( i, g [ i ]) , pos E ( j, g [ j ])) .Then p i,j is at most min( pos E ( i, g [ i ]) , pos E ( j, g [ j ])) , and we also have to recall that p i,j is at most k E ( i, j ) . 12o denote by U E ( G ) = (cid:8) ( i, j ) ∈ { , . . . , n } : i < j ; pos E ( i, g [ i ]) = pos E ( j, g [ j ]) for all g ∈ G (cid:9) . For each ( i, j ) / ∈ U E ( G ) we defineMIN E ( i, j, G ) = min ( k E ( i, j ) , min { min( pos E ( i, g [ i ]) , pos E ( j, g [ j ])) : g ∈ G, pos E ( i, g [ i ]) = pos E ( j, g [ j ]) } ) . Notice that we need ( i, j ) / ∈ U E ( G ) to have the previous integers well defined.So from the previous remark it follows that an Arf semigroup S ( T ) of σ ( E ) containing G with M ( T ) E = a , a , . . . a ,n a , . . . a ,n . . . . . . . . . . . . . . . . . . a n − ,n . . . is such that a i,j is at most k E ( i, j ) for ( i, j ) ∈ U E ( G ) and a i,j is at most MIN E ( i, j, G ) for ( i, j ) / ∈ U E ( G ) . So for the Arf closure we want to choose the biggest possible values,therefore we have: p i,j = k E ( i, j ) for ( i, j ) ∈ U E ( G ) and p i,j = MIN E ( i, j, G ) for ( i, j ) / ∈ U E ( G ) . We need to prove that this integers are compatible with the transitive property of a matrixof an Arf semigroup tree. Therefore we consider a triad of integers i < j < k and wewant to show that p i,j , p j,k and p i,k are in a { x, x, y } configuration. We have the followingcases:1. ( i, j ) , ( j, k ) , ( k, i ) ∈ U E ( G ) . Then p i,j = k E ( i, j ) ,p i,k = k E ( i, k ) and p j,k = k E ( j, k ) and for the Remark 1.5 they satisfy our condition;2. ( i, j ) , ( j, k ) , ( k, i ) / ∈ U E ( G ) . We consider the vectors v l = [ pos E ( l, g [ l ]) , . . . , pos E ( l, g m [ l ])] , where l ∈ { i, j, k } and G = { g , . . . , g m } . Then, using the notations of Lemma 1.4,we have that p i,j = min( k E ( i, j ) , MIN ( v i , v j )) , p i,k = min( k E ( i, k ) , MIN ( v i , v k )) and p j,k = min( k E ( j, k ) , MIN ( v j , v k )) . Then suppose by contradiction that they are not compatible. Without loss of gener-ality we can assume that p i,j < p i,k ≤ p j,k . We have two cases 13 p i,j = k E ( i, j ) . Then we would have k E ( i, j ) = p i,j < p j,k ≤ k E ( j, k ) and k E ( i, j ) = p i,j < p i,k ≤ k E ( i, k ) , and this is absurd for the Remark 1.5; – p i,j = MIN ( v i , v j ) . Then we would haveMIN ( v i , v j ) = p i,j < p j,k ≤ MIN ( v j , v k ) and MIN ( v i , v j ) = p i,j < p i,k ≤ MIN ( v i , v k ) , and this is absurd against Lemma 1.4 applied to the vectors v i , v j and v k .3. ( i, j ) ∈ U E ( G ) and ( j, k ) , ( k, i ) / ∈ U E ( G ) (and the similar configurations). In thiscase we have that v i = v j . Then p i,j = k E ( i, j ) , p i,k = min( k E ( i, k ) , x ) , and p j,k = min( k E ( j, k ) , x ) , where x = MIN ( v i , v k ) = MIN ( v j , v k ) . We have two cases: – k E ( i, j ) = k E ( j, k ) ≤ k E ( i, k ) (or equivalently k E ( i, j ) = k E ( i, k ) ≤ k E ( j, k ) ).If x < k E ( j, k ) ≤ k E ( i, k ) then we have p j,k = p i,k = x < k E ( i, j ) and it isfine. If x ≥ k E ( j, k ) then p j,k = k E ( j, k ) = p i,j ≤ p i,k that is compatible too. – k E ( i, k ) = k ( j, k ) < k E ( i, j ) . In this case we have p i,k = p j,k < k E ( i, j ) = p i,j and it is fine.So we actually have a well defined tree.Anyway, because the second condition is not fulfilled, then there exists a pair ( i, j ) ∈{ , . . . , n } such that for all g ∈ G we have g [ i ] = g [ j ] . So ( i, j ) ∈ U E ( G ) , and, since inthis case the two sequences are the same, we obtain p i,j = k E ( i, j ) = + ∞ .Thus we have found a contradiction because Arf ( G ) is not well defined.An example illustrating Case 2 is the following. If we consider G = { [3 , , , [2 , , } ,we will have the same multiplicity sequences in the first two branches, with no cluesabout the splitting point so we can obtain the following infinite decreasing chain in S ( G ) : (2 , ,
2) (0 , , , , , , , , ⊇ (2 , ,
2) (0 , , , , , , , , , , ⊇ (2 , ,
2) (0 , , , , , , , , , , , , ⊇ . . . ⇐ ) The previous proof gives us a way to compute Arf ( G ) . We have to compute, using themodified Jacobian algorithm of Du Val, the Arf closure of each G [ i ] , finding the collection E (the first condition guarantees that it is possible to do that). Then we can find the matrixdescribing the semigroup using the set U E ( G ) and the integers MIN E ( i, j, G ) with the procedurepresent in the first part (we cannot have p i,j = + ∞ for the second condition). Example 2.2.
Suppose that we have G = { G (1) = [5 , , , G (2) = [9 , , , G (3) = [9 , , } , that satisfies the conditions of the theorem. Then we have to apply the modified Jacobian algo-rithm to the sets G [1] = { , } , G [2] = { , , } and G [3] = { , , } . We will find the following multiplicity sequences: M = [5 , , , , M = [6 , , , and M = [2 , , , . We have k E (1 ,
2) = 3 , k E (2 ,
3) = 2 and k E (1 ,
3) = 2 .So we have pos E ( G (1)) = [1 , , , pos E ( G (2)) = [2 , , and pos E ( G (3)) = [2 , , . In this case U E ( G ) = ∅ .We have MIN E (1 , , G ) = min(2 , k E (1 , , MIN E (2 , , G ) = min(1 , k E (2 , and MIN E (1 , , G ) = min(1 , k E (1 , .So the Arf closure is described by the matrix M ( T ) E = . with E = { M = [5 , , , , M = [6 , , , and M = [2 , , , } . Notice that in this case we find that the Arf closure is an untwisted tree of τ ( E ) represented bythe vector T E = (2 , . (5 , ,
2) (0 , , , , , ,
0) (0 , , , , Example 2.3.
Suppose that we have G = { G (1) = [8 , , , G (2) = [5 , , , G (3) = [10 , , } , that satisfies the conditions of the theorem. Then we have to apply the modified Jacobian algo-rithm to the sets G [1] = { , , } , G [2] = { , , } and G [3] = { , , } .
15e will find the following multiplicity sequences: M = [5 , , , , , M = [6 , , , , and M = [5 , , , , . We have k E (1 ,
2) = 4 , k E (2 ,
3) = 4 and k E (1 ,
3) = + ∞ .So we have pos E ( G (1)) = [2 , , , pos E ( G (2)) = [1 , , and pos E ( G (3)) = [3 , , . In this case U E ( G ) = ∅ .We have MIN E (1 , , G ) = min(1 , k E (1 , , MIN E (2 , , G ) = min(1 , k E (2 , and MIN E (1 , , G ) = min(2 , k E (1 , .So the Arf closure is described by the matrix M ( T ) E = . with E = { M = [5 , , , , , M = [6 , , , , and M = [5 , , , , } . Notice that in this case we find that the Arf closure is a twisted tree. (5 , ,
5) (0 , , , , , , , ,
3) (0 , , , , , , , , N n Denote by S a good semigroup of N n . In this section we describe a way to find the smallest Arfsemigroup of N n containing S , that is the Arf closure of S (the existence of the Arf closure isproved in [6]). We denote this semigroup by Arf ( S ) . If S is a good semigroup of N n , we denoteby S i the projection on the i -th coordinate. The properties of the good semigroups guaranteethat S i is a numerical semigroup. Thus it is clear that an Arf semigroup T containing S issuch that Arf ( S i ) ⊆ T i for all i = 1 , . . . , n , where Arf ( S i ) is the Arf closure of the numericalsemigroup S i (we can compute it using the algorithm of Du Val on a minimal set of generatorsof S i ).Therefore, in order to have the smallest Arf semigroup containing S , we must have Arf ( S ) ∈ σ ( E ) where E = { M , . . . , M n } and M i is the multiplicity sequence associated to the Arf16umerical semigroup Arf ( S i ) .Now we need to find the matrix M ( T ) E = p , p , . . . p ,n p , . . . p ,n . . . . . . . . . . . . . . . . . . p n − ,n . . . . that describes the tree of Arf ( S ) .We recall that from the properties of good semigroups, it follows that there exists a minimalvector δ ∈ N n such that δ + N n ⊆ S (we will call this vector the conductor of S ).Suppose that δ = ( c [1] , . . . , c [ n ]) . We denote bySmall ( S ) = { s : < s ≤ δ } ∩ S, the finite set of the small elements of S (the elements of S that are coordinatewise smaller thanthe conductor). In [6] it is shown that Small ( S ) describes completely the semigroup S (in thispaper we are not including the zero vector in Small ( S ) to enlight the notations of the followingprocedures). Remark 3.1.
We can recover the collection E from Small ( S ) . In fact, the multiplicity se-quence M i can be determined applying the Du Val algorithm to the set { s [ i ] , s ∈ Small ( S ) } ∪{ c [ i ] + 1 } ⊆ S i . In order to apply the Du Val algorithm we may have to add c [ i ] + 1 becausewe can have gcd( { s [ i ] , s ∈ Small ( S ) } ) = 1 . Because c [ i ] and c [ i ] + 1 belong to S i , we knowthat Arf ( S i ) has conductor smaller than c [ i ] and this implies that we only have to consider theelements that are smaller than c [ i ] + 1 .Now, we notice that p i,j ≤ min( pos E ( i, c [ i ]) , pos E ( j, c [ j ])) for all i, j ∈ { , . . . , n } , with i < j , where we are using the notations of the previous section. In fact, if pos E ( i, c [ i ]) = pos E ( j, c [ j ]) , we have already noticed that in an Arf semigroup containing δ the i -th and the j -th branches cannot be glued at a level greater than min( pos E ( i, c [ i ]) , pos E ( j, c [ j ])) , then p i,j ≤ min( pos E ( i, c [ i ]) , pos E ( j, c [ j ])) . If pos E ( i, c [ i ]) = pos E ( j, c [ j ]) then we have δ =( c [1] , . . . , c [ i ]+1 , c [ i +1] , . . . , c [ n ]) ∈ S , and pos E ( i, c [ i ]+1) = pos E ( i, c [ i ])+1 > pos E ( j, c [ j ]) .Therefore in an Arf semigroup containing δ the i -th and the j -th branches cannot be gluedat a level greater than min( pos E ( i, c [ i ]) + 1 , pos E ( j, c [ j ])) = pos E ( j, c [ j ]) == min( pos E ( i, c [ i ]) , pos E ( j, c [ j ]) , hence we have again p i,j ≤ min( pos E ( i, c [ i ]) , pos E ( j, c [ j ])) .Furthermore, we always have to take in account that p i,j ≤ k E ( i, j ) for all i, j ∈ { , . . . , n } .Now let us consider the following subset of { , . . . , n } , U E ( Small ( S )) = { ( i, j ) : pos E ( i, s [ i ]) = pos E ( j, s [ j ]) for all s ∈ Small ( S ) } . ( i, j ) ∈ { , . . . , n } \ U E ( Small ( S )) , and i < j we can consider the following integersMIN E ( i, j, Small ( S )) = min ( k E ( i, j ) , min { min( pos E ( i, s [ i ]) , pos E ( j, s [ j ])) : s ∈ Small ( S ) , pos E ( i, s [ i ]) = pos E ( j, s [ j ]) } ) . Notice that we need only to consider the vectors of Small ( S ) because if s ∈ S then s =min( s, δ ) ∈ Small ( S ) and we clearly have min( pos E ( i, s [ i ]) , pos E ( j, s [ j ])) ≥ min( pos E ( i, s [ i ]) , pos E ( j, s [ j ])) , therefore s ∈ Small ( S ) gives us more precise information on the ramification level than s (itcan happen that pos E ( i, s [ i ]) = pos E ( j, s [ j ]) and pos E ( i, s [ i ]) = pos E ( j, s [ j ]) , but only when min( pos E ( i, s [ i ]) , pos E ( j, s [ j ])) ≥ min( pos E ( i, c [ i ]) , pos E ( j, c [ j ])) ).Thus, if T is an Arf semigroup of σ ( E ) containing S , represented by M ( T ) E = a , a , . . . a ,n a , . . . a ,n . . . . . . . . . . . . . . . . . . a n − ,n . . . we have: • a i,j ≤ MIN E ( i, j, Small ( S )) for ( i, j ) ∈ { , . . . , n } \ U E ( Small ( S )); • a i,j ≤ min( k E ( i, j ) , pos E ( i, c [ i ])) , for i ∈ U E ( Small ( S )) (we have pos E ( i, c [ i ]) = pos E ( j, c [ j ]) ).Then we can finally deduce that the p i,j that defines the matrix of Arf ( S ) are such that • p i,j = MIN E ( i, j, Small ( S )) , for ( i, j ) ∈ { , . . . , n } \ U E ( Small ( S )); • p i,j = min( k E ( i, j ) , pos E ( i, c [ i ])) , for i ∈ U E ( Small ( S )) (we have pos E ( i, c [ i ]) = pos E ( j, c [ j ]) ),and it is easy to see that the p i,j fulfill the condition of Remark 1.6. Remark 3.2.
In other words we showed that Arf ( S ) can be computed by computing Arf ( G ) where: G = Small ( S ) [ { ( c [1] + 1 , . . . , c [ i ] , c [ i + 1] , . . . , c [ n ]) , . . . , ( c [1] , . . . , c [ i ] + 1 , c [ i + 1] , . . . , c [ n ]) , . . . , ( c [1] , . . . , c [ i ] , c [ i + 1] , . . . , c [ n ] + 1) } . xample 3.3. Let us consider the good semigroup S with the following set of small elements,Small ( S ) = { [5 , , , [5 , , , [5 , , , [8 , , , [8 , , , [8 , , , [8 , , , [8 , , , [8 , , , [10 , , , [10 , , , [10 , , , [10 , , , [10 , , , [10 , , } . The conductor is δ = [10 , , . First of all we need to recover from Small ( S ) the collectionof multiplicity sequences E . We have to apply the Du Val algorithm to the following sets: { , , , } , { , , , } and { , , , } , therefore we find that E = { [5 , , , , , [6 , , , , , [5 , , , , } . We havepos ( Small ( S )) = { pos E ( s ) : s ∈ Small ( S ) } = { [1 , , , [1 , , , [1 , , , [2 , , , [2 , , , [2 , , , [2 , , , [2 , , , [2 , , , [3 , , , [3 , , , [3 , , , [3 , , , [3 , , , [3 , , } . It is easy to check that U E ( Small ( S )) = ∅ . Thus we have • p , = MIN E (1 , , Small ( S )) = min( k E (1 ,
2) = 4 ,
1) = 1 , because we have the element [1 , , ∈ pos ( Small ( S )) corresponding to s = [5 , , ∈ Small ( S ) such that pos E (1 , s [1]) = pos E (2 , s [2]) = 2 and min( pos E (1 , s [1]) , pos E (2 , s [2])) = 1 . • p , = MIN E (2 , , Small ( S )) = min( k E (2 ,
3) = 4 ,
1) = 1 , because we have the element [1 , , ∈ pos ( Small ( S )) corresponding to s = [5 , , ∈ Small ( S ) such that pos E (2 , s [2]) = pos E (3 , s [3]) = 1 and min( pos E (2 , s [2]) , pos E (3 , s [3])) = 1 . • p , = MIN E (1 , , Small ( S )) = min( k E (1 ,
3) = + ∞ ,
2) = 2 , because we have theelement [2 , , ∈ pos ( Small ( S )) corresponding to s = [8 , , ∈ Small ( S ) such that pos E (1 , s [1]) = pos E (3 , s [3]) = 3 and min( pos E (1 , s [1]) , pos E (3 , s [3])) = 2 , and wecannot find a smaller discrepancy.So the Arf closure of S is described by the matrix M ( T ) E = . with E = { M = [5 , , , , , M = [6 , , , , and M = [5 , , , , } . The following procedure, implemented in GAP, has as argument the set of small elements ofa good semigroup and give as a result the Arf Closure of the given good semigroup. The Arfclousure is described by a list [ E, M ( T ) E ] . 19 ap> S:=[[5,6,5],[5,10,5],[5,12,5],[8,6,8],[8,10,8],[8,12,8],[8,6,10],[8,10,10],[8,12,10],[10,6,8],[10,10,8],[10,12,8],[10,6,10],[10,10,10],[10,12,10]];[ [ 5, 6, 5 ], [ 5, 10, 5 ], [ 5, 12, 5 ], [ 8, 6, 8 ],[ 8, 10, 8 ], [ 8, 12, 8 ], [ 8, 6, 10 ], [ 8, 10, 10 ],[ 8, 12, 10 ], [ 10, 6, 8 ], [ 10, 10, 8 ], [ 10, 12, 8 ],[ 10, 6, 10 ], [ 10, 10, 10 ], [ 10, 12, 10 ] ]gap> ArfClosureOfGoodsemigroup(S);[ [ [ 5, 3, 2 ], [ 6, 4, 2 ], [ 5, 3, 2 ] ],[ [ 0, 1, 2 ], [ 0, 0, 1 ], [ 0, 0, 0 ] ] ] Suppose that E is a collection of n multiplicity sequences. Let T ∈ τ ( E ) and given a semigroup S ( T ) in σ ( E ) , we want to study the properties that a set of vectors G ( T ) ⊆ N n has to satisfyto have S ( T ) = Arf ( G ( T )) , with the notations given in the previous section. We call such a G ( T ) a set of generators for S ( T ) . In particular we want to find bounds on the cardinality of aminimal set of generators for a S ( T ) ∈ σ ( E ) .Since we want to find a G ( T ) such that Arf ( G ( T )) is well defined, it has to satisfy thefollowing properties: • For all i = 1 , . . . , n gcd( v [ i ]; v ∈ G ( T )) = 1 , where v [ i ] is the i − th coordinate of the vector v ∈ G ( T ) . • For all i, j ∈ { , . . . , n } , with i < j there exists v ∈ G ( T ) such that v [ i ] = v [ j ] .We recall that, given a Arf numerical semigroup S , there is a uniquely determined smallestsemigroup N such that the Arf closure of N is S . The minimal system of generators for such N is called the Arf system of generators for S , or the set of characters of S .Now we want that Arf ( G ( T )) is an element of σ ( E ) . This implies that, when we applythe algorithm of Du Val to G ( T )[ i ] , we have to find the i -th multiplicity sequence of E . Thismeans that, if we call S i the Arf numerical semigroup corresponding to the projection on the i -th coordinate, we must have G ( T )[ i ] ⊆ S i and furthermore G ( T )[ i ] has to contain a minimalsystem of generators for S i . In fact, in [1] it is proved that if we have G = { g , . . . , g m } ⊆ N with gcd( G ) = 1 then G must contain the set of characters of the Arf closure of the semigroup N = h G i .So we need to recall a way to compute the characters of an Arf numerical semigroup.20e suppose that E = { M (1) , . . . , M ( n ) } . Given M ( i ) = [ m i, , . . . , m i,M ] , we define the restricion number r ( m i,j ) of m i,j as the number of sums m i,q = k X h =1 m i,q + h where m i,j appears as a summand. With this notation we have that the characters of the multiplicitysequence M ( i ) are the elements of the set (cf. [3, Lemma 3.1])Char E ( i ) = ( j X k =1 m i,k : r ( m i,j ) < r ( m i,j +1 ) ) . Notice that, from our assumptions on M , it follows that the last two entries in each M ( i ) are , and it is easy to see how it guarantees that we cannot find characters in correspondence ofindices greater than M . We define PChar E ( i ) = { j : r ( m i,j ) < r ( m i,j +1 ) } .Given the collection E , we denote by V E ( j , j , . . . , j n ) = " j X k =1 m ,k , j X k =1 m ,k , . . . , j n X k =1 m n,k . Now, the elements of G ( T ) must be of the type V E ( j , j , . . . , j n ) (in fact we noticed that whenwe project on the k -th coordinate we must find an element of the corresponding numericalsemigroup that has the previous representation for some j k ).From the previous remarks and notations we have the following property: G ( T ) = { Gen (1) = V E ( j , , . . . , j ,n ) , . . . , Gen ( t ) = V E ( j t, , . . . , j t,n ) } are generators of a semigroup of σ ( E ) if and only ifPChar E ( i ) ⊆ { j ,i , . . . , j t,i } for all i = 1 , . . . , n. In particular we have found a lower bound for the cardinality of a minimal set of generators fora S ( T ) ∈ σ ( E ) . In fact G ( T ) has at least C E = max {| PChar E ( i ) | , i = 1 , . . . , n } elements.Now we want to determine the generators of a given semigroup S ( T ) ∈ σ ( E ) . We have thefollowing theorem. Theorem 4.1.
Let S ( T ) ∈ σ ( E ) with M ( T ) E = p , p , . . . p ,n p , . . . p ,n . . . . . . . . . . . . . . . . . . p n − ,n . . . . Denote by P = { ( q, r ) : p q,r = k E ( q, r ) } . Then G ( T ) = { Gen (1) , . . . ,
Gen ( t ) } ⊆ N n is suchthat Arf ( G ( T )) = S ( T ) if and only if the following conditons hold Gen (1) = V E ( j , , . . . , j ,n ) , . . . , Gen ( t ) = V E ( j t, , . . . , j t,n ) for some values of the in-dices j , , . . . , j t,n ; • PChar E ( i ) ⊆ { j ,i , . . . , j t,i } for all i = 1 , . . . , n .Furthermore, if we consider the following integerMIN G ( T ) ( q, r ) = min ( k E ( q, r ) , min { min( j p,q , j p,r ) : j p,q = j p,r , p = 1 , . . . , t } ) , for the ( q, r ) ∈ { , . . . , n } with q < r and where it is well defined, we have: • for ( q, r ) ∈ P we have either j p,q = j p,r for all p = 1 , . . . , t , or MIN G ( T ) ( q, r ) is welldefined and it equals k E ( q, r ) ; • MIN G ( T ) ( q, r ) is well defined and it equals p q,r , for all ( q, r ) / ∈ P . Proof ( ⇐ ) Suppose that we have G ( T ) = { Gen (1) , . . . ,
Gen ( t ) } ⊆ N n satisfying the conditionsof the theorem. The first two conditions ensure that if we apply the algorithm defined in theprevious section on G ( T ) it will produce an element of σ ( E ) .Now it is easy, using the notations of Theorem 2.1, to show that j p,q = pos E ( q, Gen ( p )[ q ]) and from this it follows that, when MIN G ( T ) ( q, r ) is well defined, it is equal to MIN E ( q, r, G ( T )) .Furthermore we have U E ( G ( T )) ⊆ P . In fact we have U E ( G ( T )) = (cid:8) ( q, r ) ∈ { , . . . , n } : pos E ( q, Gen ( p )[ q ]) = pos E ( r, Gen ( p )[ r ]) for all p = 1 , . . . , t } = (cid:8) ( q, r ) ∈ { , . . . , n } : j p,q = j p,r for all p = 1 , . . . , t (cid:9) , therefore if ( q, r ) ∈ U E ( G ( T )) then ( q, r ) ∈ P , since G ( T ) satisfy the fourth condition in thestatement of the theorem (we cannot have ( q, r ) / ∈ P because in this case MIN G ( T ) ( q, r ) = MIN E ( q, r, G ( T )) has to be well defined). So it will follows that, if S ( T ) is Arf ( G ( T )) then M ( T ) E = a , a , . . . a ,n a , . . . a ,n . . . . . . . . . . . . . . . . . . a n − ,n . . . where • a i,j = MIN E ( i, j, G ( T )) if ( i, j ) / ∈ U E ( G ( T )) ; • a i,j = k E ( i, j ) if ( i, j ) ∈ U E ( G ( T )) .Therefore if ( i, j ) / ∈ P then ( i, j ) / ∈ U E ( G ( T )) and we have a i,j = min( MIN E ( i, j, G ( T ))) = MIN G ( T ) ( i, j ) = p i,j . If ( i, j ) ∈ P then • if ( i, j ) ∈ U E ( G ( T )) then a i,j = k E ( i, j ) ;22 if ( i, j ) / ∈ U E ( G ( T )) then a i,j = MIN E ( i, j, G ( T )) = MIN G ( T ) ( i, j ) = k E ( i, j ) , for theproperties of the set G ( T ) ( ( i, j ) ∈ P ).So we showed that Arf ( G ( T )) = S ( T ) . Thus the proof of this implication is complete.( ⇒ ) It follows immediately by contradiction, using the first part of the proof. Example 4.2.
Suppose that we have E = { M (1) , M (2) , M (3) } , where M (1) = [5 , , , , M (2) = [6 , , , , M (3) = [2 , , , . We have, k E (1 ,
2) = 3 , k E (2 ,
3) = 2 and k E (1 ,
3) = 2 .We can define: R ( i ) = [ r ( m i, ) , r ( m i, ) , . . . , r ( m i,N )] . Notice that r ( m i, ) = 0 , r ( m i, ) = 1 . The values of PChar ( i ) are the indices where this se-quence has an increase (it can be easily shown that when the sequence has an increase wehave r ( m i,j ) = r ( m i,j +1 ) − cf. [3, Lemma 3.2]). Furthermore R (1) = [0 , , , , , , R (2) = [0 , , , , , and R (3) = [0 , , , . So PChar E (1) = { , } , PChar E (2) = { , , } and PChar E (3) = { , } .Suppose that we want to find generators for the untwisted tree T such that T E = (2 , . Weneed at least three vectors because C E = 3 . Consider the vectors Gen (1) = V E (1 , , , Gen (2) = V E (2 , , and Gen (3) = V E (2 , , . The second condition, that guarantees that we havea tree belonging to τ ( E ) , is satisfied. Furthermore MIN G ( T ) (1 ,
2) = min( k E (1 , ,
2) = 2 , MIN G ( T ) (2 ,
3) = min( k E (2 , ,
1) = 1 , and MIN G ( T ) (1 ,
3) = min( k E (1 , ,
1) = 1 =min( d , d ) where G ( T ) = { Gen (1) , Gen (2) , Gen (3) } . Thus we have Arf ( G ( T )) = S ( T ) .They are the vectors Gen (1) = [5 , , , Gen (2) = [9 , , , Gen (3) = [9 , , which appearedin the Example 2.2.Now, we want to find an upper bound for the cardinality of a minimal set G ( T ) such thatArf ( G ( T )) ∈ σ ( E ) . Remark 4.3.
Suppose that T is a twisted tree of τ ( E ) , where E is a collection of n multiplicitysequences. Then there exists a permutation δ ∈ S n such that δ ( T ) is an untwisted tree of τ ( δ ( E )) . Then if G is a set of generators for δ ( T ) then it is clear that we have δ − ( G ) = (cid:8) δ − ( g ); g ∈ G (cid:9) , is a set of generators for the twisted tree T .From the previous remark it follows that we can focus only on the untwisted trees of τ ( E ) to find an upper bound for the cardinality of G ( T ) .Our problem is clearly linked to the following question.23 uestion 4.4. Let us consider a vector d = [ d , . . . , d n ] ∈ N n . For all the G ⊆ N n +1 we denoteby MIN ( G, i, j ) the integers (if they are well defined)MIN ( G, i, j ) = min { min( g [ i ] , g [ j ]) : g ∈ G with g [ i ] = g [ j ] } , for all the i < j and i, j ∈ { , . . . , n + 1 } . We define a solution for the vector d as a set G ⊆ N n +1 such that:MIN ( G, i, j ) = min { d i , . . . , d j − } for all i < j. Consider n ∈ N with n ≥ . Find the smallest t ∈ N , such that for all [ d , . . . , d n ] ∈ N n there exists a solution with t vectors. We denote such a number t by NS ( n ) . Theorem 4.5.
Consider n ∈ N with n ≥ . Then NS ( n ) = ⌈ log ( n + 1) ⌉ , where ⌈ d ⌉ =min { m ∈ N : m ≥ d } . Proof
First of all we show that given an arbitrary vector d of N n we are able to find a solutionof d consisting of N = ⌈ log ( n + 1) ⌉ vectors.We will do it by induction on n . The base of induction is trivial. In fact if n = 1 then foreach vector [ d ] we find the solution G = { [ d , d + 1] } that has cardinality ⌈ log (1 + 1) ⌉ = 1 .Thus we suppose that the theorem is true for all the m < n and we prove it for n . Let d anarbitrary vector of N n . We fix some notations. Given a vector d , we will denote by sol ( d ) asolution with ⌈ log ( n + 1) ⌉ vectors. We denote by Inf ( d ) = min { d i : i = 1 , . . . , n } and byPinf ( d ) = { i ∈ { , . . . , n } : d i = Inf ( d ) } . We have ≤ | Pinf ( d ) | = k ( d ) ≤ n .Suppose that Pinf ( d ) = (cid:8) i < i < · · · < i k ( d ) (cid:9) . Then we can split the vector d in thefollowing k ( d ) + 1 subvectors: d = d (1 . . . i − , d j = d ( i j − + 1 . . . i j − for j = 2 , . . . , k ( d ) , d k ( d )+1 = d ( i k ( d ) + 1 . . . n ) , where with d ( a . . . b ) we mean • ∅ if b < a ; • The subvector of d with components between a and b if a ≤ b .Then the subvectors d j are either empty or with all the components greater than Inf ( d ) . Webriefly illustrate with an example the construction of the subvectors d j . Example 4.6.
Suppose that d = [2 , , , , , , . Then Inf ( d ) = 2 , Pinf ( d ) = { , , } andthen we have the four subvectors: • d = d (1 . . .
0) = ∅ , • d = d (2 . . .
2) = [3] , 24 d = d (4 . . .
3) = ∅ , • d = d (5 . . .
7) = [5 , , .Then we can consider the list of k ( d ) + 1 subvectors: p ( d ) = [ d , . . . , d k ( d )+1 ] , and, because all the d i have length strictly less than n we can find a solution for each of themwith N = ⌈ log ( n + 1) ⌉ or less vectors. For the d i = ∅ we will set sol ( ∅ ) = { [ x ] } , where x isan arbitrary integer that is strictly greater than all the entries of d . It is quite easy to check thatthe following equality holds: n = k ( d ) + k ( d )+1 X i =1 Length ( d i ) . (1)We associate to the list of vectors p ( d ) another list of vector c ( d ) such that c ( d ) = [ c , . . . , c k ( d )+1 ] , where Length ( c j ) = Length ( d j ) + 1 and the entries of c j are all equal to Inf ( d ) for all j =1 , . . . , k ( d ) + 1 . Now we consider the set I ( N ) = { , } N . For each t ∈ I ( N ) we denote by O ( t ) the numberof one that appear in t . Because we have N = ⌈ log ( n + 1) ⌉ , it follows k ( d ) + 1 ≤ n + 1 ≤ N = | I ( N ) | , therefore it is always possible to associate to each subvectors of the list p ( d ) distinct elements of I ( N ) . We actually want to show that it is possible to associate to all the subvectors d i distinctvectors of t ∈ I ( N ) such that O ( t ) ≥ | sol ( d i ) | (for d i = ∅ we can also associate the zerovector). We already know for the inductive step that all the d i have solutions with at most N vectors. Suppose therefore that m ≤ N .It is easy to see that | { t ∈ I ( N ) : O ( t ) ≥ m } | = N X k = m (cid:18) Nk (cid:19) . Then we suppose by contradiction that in p ( d ) we have P Nk = m (cid:0) Nk (cid:1) + 1 subvectors with solutionwith cardinality m . From the inductive step it follows that all these subvectors have at leastlength m − , and from the formula 1 it follows: n ≥ N X k = m (cid:18) Nk (cid:19) + N X k = m (cid:18) Nk (cid:19) + 1 ! m − ⇒ n + 1 ≥ N X k = m (cid:18) Nk (cid:19) + 1 ! (1 + 2 m − ) . N X k = m (cid:18) Nk (cid:19) + 1 ≥ N − m +1 , in fact P Nk = m (cid:0) Nk (cid:1) is the number of ways to select a subset of { , . . . , N } of at least m elementswhile there are N − m +1 − ways to select a subset which contains at least m elements andcontains { , , . . . , m − } .Therefore we can continue the inequality: n + 1 ≥ N − m +1 (1 + 2 m − ) = 2 N + 2 N − m +1 > N . But N = ⌈ log ( n + 1) ⌉ and therefore n + 1 ≤ N and we find a contradiction. Then in { t ∈ I ( N ) : O ( t ) ≥ m } we have enough vectors to cover all the subvectors with solution withcardinality m . We still also have to exclude the following possibility. Suppose that we have x subvectors with solutions of cardinality m and y subvectors with solutions of cardinality m > m . If | { t ∈ I ( N ) : O ( t ) ≥ m } | − x < y then it would not be possible to associateto all the subvectors of the second type an element t of I ( N ) with O ( t ) ≥ m . Indeed if thishappen we would have: n ≥ x + y − x · m − + y · m − > x + y − x + y )2 m − ⇒⇒ n + 1 ≥ ( x + y )(1 + 2 m − ) ≥ N X k = m (cid:18) Nk (cid:19) + 1 ! (1 + 2 m − ) , and we already have seen that this is not possible.Therefore we showed that we can consider a matrix A with N rows and k ( d ) + 1 distinctcolumns with only zeroes and ones as entries and such that the i -th column of A is a vector t of I ( N ) such that O ( t ) ≥ | sol ( d i ) | for each ≤ i ≤ k ( d ) + 1 .Now we can complete the construction of a solution for d . We consider a matrix B with N rows and k ( d ) + 1 columns. We fill the matrix B following these rules: • If A [ i, j ] = 0 then in B [ i, j ] we put the vector c j ; • If A [ i, j ] = 1 then in B [ i, j ] we put an element of sol ( d j ) ; • All the elements of sol ( d j ) have to appear in the j -th column for all j = 1 , . . . , k ( d ) + 1 .Then if we glue all the vectors appearing in the i -th row of B for each i = 1 , . . . , N we obtaina solution G for the vector d . In fact if we consider i , j such that i < j we have twopossibilities: • i and j both correspond to elements in the j -th column of B . Then because in thiscolumn we have either vectors of a solution for d j or costant vectors, it follows that theyfulfill our conditions. 26 i and j correspond to elements in distinct columns. This implies that we must haveMIN ( G, i , j ) = Inf ( d ) . In fact, for construction, between two distinct subvectors wehave an element equal to Inf ( d ) in d forcing MIN ( G, i , j ) = Inf ( d ) . Now suppose that i and j correspond respectively to elements in the i -th and j -th columns of B . Becausewe suppose i = j we have that the i -th column and the j -th column of the matrix A aredistinct so there exists a k such that A [ k, i ] = 0 and A [ k, j ] = 1 (or viceversa). Thisimplies that in B we have a row where in the i -th column there is the constant vectorequal to Inf ( d ) while in the j -th column we have a vector corresponding to a solution ofa subvectors of d (that has all the components greater than Inf ( d ) by construction). Thiseasily implies that MIN ( G, i , j ) = Inf ( d ) . Example 4.7.
Suppose that d = [2 , , , , , , . We have n = 7 , then we want to show thatthere exists a solution with three vectors. We have already seen that in this case we have: p ( d ) = [ ∅ , [3] , ∅ , [5 , , . We need to compute a solution for each entry of p ( d ) . We have: • sol ( ∅ ) = { [6] } ( is greater than all the entries of d ); • sol ([3]) = { [3 , } ; • Let us compute a solution for f = [5 , , with the same techniques. Because Length ( f ) =3 we expect to find a solution with at most two vectors. We have: p ( f ) = [[5] , [5]] , and we have sol ([5]) = { [5 , } . Then in I (2) we want to find two distinct vectors with atleast an entry equal to one. We can choose [1 , and [0 , . Therefore we have: A = (cid:18) (cid:19) and B = (cid:18) [5 ,
6] [4 , ,
6] [5 , (cid:19) . Then sol ([5 , , { [5 , , , , [5 , , , } .Now we want to find in I (3) four vectors t i for i = 1 , . . . , . We have free choice for the t and t , while we need O ( t ) ≥ and O ( t ) ≥ . For instance we choose t = [0 , , , t =[1 , , , t = [1 , , , t = [1 , , . Then we have: A = and B = [2] [3 ,
4] [6] [5 , , , ,
2] [6] [2 , , , ,
2] [2] [5 , , , . Then a solution for d is the set G = { [2 , , , , , , , , [2 , , , , , , , , [2 , , , , , , , } .
27o we proved that NS ( n ) ≤ ⌈ log ( n + 1) ⌉ . To prove that the equality holds we notice that foreach n a constant vector needs exactly ⌈ log ( n + 1) ⌉ vectors in its solutions.Now we can return to the problem of determining an upper bound for the cardinality of G ( T ) . We need another lemma: Lemma 4.8.
Let E = { M (1) , M (2) } be a collection of two multiplicity sequences. Then, withthe previous notations we have: k E (1 , ≤ min { j : j ∈ ( PChar E (1) ∪ PChar E (2)) \ ( PChar E (1) ∩ PChar E (2)) } . Proof
Let us choose an arbitrary element t ∈ ( PChar E (1) ∪ PChar E (2)) \ ( PChar E (1) ∩ PChar E (2)) . We want to show that k E (1 , ≤ t . Suppose by contradiction that t < k E (1 , .Without loss of generality we suppose that t ∈ PChar E (1) . It follows that t / ∈ PChar E (2) andwe have: r ( m ,t ) < r ( m ,t +1 ) and r ( m ,t ) ≥ r ( m ,t +1 ) . Notice that if an entry of M (1) has m ,t +1 as a summand and it is not m ,t , it is forced to have m ,t as a summand too. So from r ( m ,t ) < r ( m ,t +1 ) we deduce that in M (1) there are noentries involving only m ,t . Similarly from r ( m ,t ) ≥ r ( m ,t +1 ) we deduce that in M (2) wemust have at least one entry m ,s which involves m ,t as a summand but not m ,t +1 .Namely m ,s = t X k = s +1 m ,k . (2)Now, we have assumed that t < k E (1 , hence t + 1 ≤ k E (1 , . This imply that the untwistedtree T such that T E = ( t + 1) is well defined. In T we have the following nodes: ( m ,s , m ,s ) , . . . , ( m ,t , m ,t ) , ( m ,t +1 , m ,t +1 ) . Then from (2) and from the fact that the two branches are still glued at level t + 1 it must followthat m ,s = t X k = s +1 m ,k and we have still noticed how it contradicts the assumption r ( m ,t ) < r ( m ,t +1 ) .Now we can prove the following result: Proposition 4.9.
Let E be a collection of n multiplicity sequences. Then, if S ( T ) ∈ σ ( E ) , thereexists G ( T ) ⊆ N n with Arf ( G ( T )) = S ( T ) and | G ( T ) | = C E + ⌈ log ( n ) ⌉ . Proof
For the Remark 4.3 it suffices to prove the theorem only for the untwisted trees. Thereforewe suppose that T E = ( d , . . . , d n − ) . First of all we have to satisfy the condition on thecharacters to ensure that Arf ( G ( T )) ∈ σ ( E ) . From the Lemma (4.8) it follows that we can use C E vectors to satisfy all the conditions. To see it, let us fix some notations.28enote by τ ( i ) = | PChar E ( i ) | for all i = 1 , . . . , n . Therefore C E = max { τ ( i ) , i = 1 , . . . , n } .Suppose that PChar E ( i ) = (cid:8) a i, < · · · < a i,τ ( i ) (cid:9) , and we define L = max n [ i =1 PChar E ( i ) ! + 1 . For all i = 1 , . . . , n we consider the vector J ( i ) = [ a i, , . . . , a i,τ ( i ) , L, . . . , L ] ∈ N C E . Thus wecan use the following set of vectors to satisfy the condition on the characters, G = Gen (1) = V E ( j , , . . . , j ,n ) , . . . , Gen ( C E ) = V E ( j C E , , . . . , j C E ,n ) , where j p,q = J ( q )[ p ] for all p = 1 , . . . , C E and q = 1 , . . . , n . Now it is clear that we havePChar E ( i ) ⊆ { j ,i , . . . , j C E ,i } for all i = 1 , . . . , n .We also need to show that this choice does not affect the condition on ( d , . . . , d n − ) . Wedefine P = (cid:8) ( q, r ) ∈ { , . . . , n } : j p,q = j p,r for all p = 1 , . . . , C E (cid:9) . Thus for each ( q, r ) ∈ P the previous vectors are compatible with the conditions on the element p q,r of M ( T ) E .For each ( q, r ) / ∈ P , we consider p ( q, r ) = min { p : j p,q = j p,r } . Now, because the entries of the vectors J ( q ) are in an increasing order, it is clear that we haveMIN G ( q, r ) = min ( k E ( q, r ) , min { min( j p,q , j p,r ) : j p,q = j p,r } ) == min (cid:0) k E ( q, r ) , min( j p ( q,r ) ,q , j p ( q,r ) ,r ) (cid:1) , for all ( q, r ) / ∈ P. Furthermore, for the particular choice of the vectors Gen ( i ) and of L , it is clear that from j p ( q,r ) ,q = j p ( q,r ) ,r , it follows that min( j p ( q,r ) ,q , j p ( q,r ) r ) ∈ ( PChar E ( q ) ∪ PChar E ( r )) \ ( PChar E ( q ) ∩ PChar E ( r )) , and from the Lemma (4.8), we finally have min( j p ( q,r ) ,q , j p ( q,r ) ,r ) ≥ k E ( q, r ) for all ( q, r ) / ∈ P, so the vectors Gen ( i ) are compatible with our tree.Now from the Theorem (4.5) it follows that we can use ⌈ log ( n ) ⌉ vectors to have a solutionfor the vector [ d , . . . , d n − ] . Adding the vectors corresponding to this solution to the previous C E we obtain a set G ( T ) such that Arf ( G ( T )) = S ( T ) .Notice that the first C E vectors may satisfy some conditions on the d i , therefore it is possibleto find G ( T ) with smaller cardinality than the previous upper bound.29 emark 4.10. Let us consider the Arf semigroup of the Example 4.2.It was T = T E = (2 , , where E = { M (1) = [5 , , , , M (2) = [6 , , , , M (3) = [2 , , , } , with PChar E (1) = { , } , PChar E (2) = { , , } and PChar E (3) = { , } . We found G = { V E (1 , , , V E (2 , , , V E (2 , , } as a set such that Arf ( G ) = S ( T ) , and itis also minimal because we have | G | = C E and we clearly cannot take off any vector from it.Using the strategy of the previous corollary we would find the vectors:Gen (1) = V E (1 , , , Gen (2) = V E (2 , , and Gen (3) = V E (4 , , , that satisfy the conditions on the characters ( L = 4 ).We have to add vectors that correspond to a solution for the vector [2 , . For istance itsuffices to consider [3 , , and therefore we will add the vector Gen (4) = V E (3 , , . Noticehow the set G ′ = { V E (1 , , , V E (2 , , , V E (4 , , , V E (3 , , } , with | G ′ | > | G | , is stillminimal because we cannot remove any vector from it without disrupting the condition on thetree. Therefore we can have minimal sets of generators with distinct cardinalities. Example 4.11.
Let us consider E = { M = [4 , , , , M = [6 , , , , M = [2 , , , , M = [3 , , , } . We want to find a set of generators for the twisted tree T of τ ( E ) such that: M ( T ) E = . First of all we notice that it is well defined because it satisfies the conditions given by theRemark 1.5 and we have k (1 ,
2) = 2 , k (1 ,
3) = 4 , k (1 ,
4) = 2 , k (2 ,
3) = 2 , k (2 ,
4) = 3 and k (3 ,
4) = 2 . We consider the permutation δ = (3 , of S . Then δ ( T ) is an untwisted tree of τ ( δ ( E )) and itis described by the vector T δ ( E ) = (2 , , . We have: • PChar δ ( E ) (1) = { , } ; • PChar δ ( E ) (2) = { , , } ; • PChar δ ( E ) (3) = { , } ; PChar δ ( E ) (4) = { , } . Then with the vectors V δ ( E ) (1 , , , , V δ ( E ) (3 , , , , V δ ( E ) (4 , , , , we satisfy the conditionon the characters. We need to add the vectors corresponding to a solution for [2 , , . It sufficesto add V δ ( E ) (2 , , , . Then G ( T ) = { [4 , , , , [9 , , , , [10 , , , , [8 , , , } , is a set of generators for δ ( T ) . Because δ − = (3 , , we have that δ − ( G ( T )) = { [4 , , , , [9 , , , , [10 , , , , [8 , , , } is a set of generators for the twisted tree T . Acknowledgements
The author would like to thank Marco D’Anna for his helpful comments and suggestions. Spe-cial thanks to Pedro Garc´ıa-S´anchez for his careful reading of an earlier version of the paperand for many helpful hints regarding the implementation in GAP of the presented procedures.31 eferences [1] C. Arf
Une interpretation algebrique de la suite des ordres de multiplicite d’unebranche algebrique
Proc. London Math. Soc. (2), 50:256-287, 1948.[2] V. Barucci, M. D’Anna, R. Fr¨oberg
Analitically unramifed one-dimensionalsemilocal rings and their value semigroups , in J. Pure Appl. Algebra, (2000),215-254.[3] V. Barucci, M. D’Anna, R. Fr¨oberg
Arf characters of an algebroid curve , in JPJournal of Algebra, Number Theory and Applications, vol.3(2)(2003), p. 219-243.[4] A. Campillo,
Algebroid curves in positive characteristic , Lecture Notes in Math.813, Springer-Verlag, Heidelberg, (1980).[5] M. D’Anna
The canonical module of a one-dimensional reduced local ring ,Comm. Algebra , (1997), 2939-2965.[6] M. D’Anna, P. A. Garcia Sanchez, V. Micale, L. Tozzo Good semigroups of N n , preprint, arXiv:1606.03993 (2017)[7] P. Du Val The Jacobian algorithm and the multiplicity sequence of an algebraicbranch , Rev. Fac. Sci. Univ. Istanbul. Ser. A., (1942), 107-112.[8] The GAP Group, GAP-Groups, Algorithms and Programming , Version 4.7.5 ,25