# Alexander Ideals of Links by J. A. Hillman

By J. A. Hillman

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Modules This was proven by Bailey who characterized arising from 2-component links as the A2-modules square presentation matrix of a particular form [ 7 ] [34 ] and Chapter VII). such admitting a (See also Cooper It may also he seen as follows. Let Z ~ D 1 = A n+l be the submodule of l-cycles in the cellular chain complex of X. lq~en there are exact sequences 0 ---+ H2(X;A) ---+ A n---+ Z ---+ G'/G" ---+ 0 and 0 By Sehanuel's > Z ~ A n+l ---+ A ---+ ~ lemma Z is projective, hence free, and clearly rank Z = n.

Therefore E (H) = E (A(H)) = Eo(tA(H)) 56 In the next chapter we shall give several partial converses of this theorem. We shall now present our first counter-example to Smythe's conjecture. 1 Figure 1 The solid link in Figure 1 (which has unknotted components) to a ribbon map R with 4 singularities. extends The ribbon group H(R) for this ribbon has a presentation {Xl,X2,XS,Yl,Y2~Y3lyTlxlYl -1 = x2,ysx2y 3 which is Tietze-equivalent -1 -1 = Xs,X ~ ylxl = y2,x3 y2x3 = y3} E43; page 4 4 to { X l , X S , Y l [ X l y T l x l Y l X l-1 = ( x 3 x T l ) - l x T l y T l x l ( x 3 x T l ) y l x l ( x 3 x T l ) } so H(R) has a preabelian presentation {x, y, a I xY -I xY x-I = a-I x-I Y-I x a y x a } The Jacobian matrix of this presentation is M = [I (Y-I)(x-ly -I - xy-l),(l-x)( x-I Y -I - xy -I -I 57 ans so E2(H(R)) = ((y-l)(x2-1), = (x+1, y-l-xy) = (x+l, since x-I = y -I (l-(y-l-xy)) 2y-l) which is clearly not principal.

G. tM ~ I. is free and tM P Conversely, lemma if P is projective square matrix since tMp is a torsion module. Er(P p @ tM ) = Eo(tM) p p last assertion Corollary Proof is principal is clear. [l~;Lemma ~ Hence Er(M) p = Er(M p) = The // M is projective pseudozero Let N he a pseudozero Then N is a pseudozero P if N given by a and so E (M) is invertible. r If M is a finitely senerats then M has no nontrivial of R. tM $ I, if and only if Er(M) = R. This follows from the fact that tM = 0 if and only if Eo(tM) = R.