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As is discussed in this answer, S-equivalence class can be regarded as the Alexander module plus Blachfield pairing, an isomorphism of some duals of Alexander modules.

There are examples of knots having the same Alexander polynomial but different Alexander modules. For example, the knots $6_1$ and $9_{46}$ have the same Alexander polynomial but their Alexander modules are different, because their second Alexander ideals are different.

However, I do not know explicit examples of non S-equivalent knots having the same Alexander modules. (Presumably such example exists and appeared before, but I cannot find references)

Q1. What is the simplest example of knots having the same Alexander module which are not S-equivalent ?

Q2. How to construct such knots systematically ?

It would be nice if there is a move that always preserves the Alexander module but (potentially) changes the S-equivalence class.

Although there are many ways to construct a knot having the same Alexander polynomial, in most cases, they actually produce S-equivalent knots, or, often more strongly, knots having the same Seifert matrices.

One exception I know is mutation. Mutation preserves the Alexander polynomial. Kearton showed that mutation may not preserve the Alexander modules, so it does not preserve the S-equivalence classes. However, Kim-Livingston proved that mutation preserves the algebraic concordance classes so in particular, mutation preserves the Levine-Tristram signatures.

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    $\begingroup$ Just a remark: if one has two Seifert matrices (integral matrices $V$ with $det(V-V^t)=\pm 1$) with isomorphic Alexander module (presented by $V^t-tV$), but which are not S-equivalent, then one can construct knots with these Seifert matrices, and hence with the properties you desire. This is to point out that the question could be answered purely algebraically by constructing such a pair of matrices. $\endgroup$
    – Ian Agol
    Commented Sep 30 at 22:44

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Knots with Nakanishi index 1 have cyclic Alexander module, but their algebraic unknotting numbers might differ.

Work of Borodzik-Friedl ensures that the algebraic unknotting number equals the minimal size of a nondegenerate hermitian matrix A(t) presenting $\operatorname{Bl}_K$ with A(1) congruent to a diagonal matrix with $\pm 1$'s on the diagonal. Thus knots with distinct algebraic unknotting numbers have distinct Blanchfield forms and therefore non S-equivalent Seifert matrices.

One can then search knotinfo e.g. for knots with equal Alexander polynomials, Nakanishi index 1, but distinct algebraic unknotting numbers. The Nakanishi index 1 condition ensures the knots have the same Alexander module.

For example $5_1$ and $10_{132}$ have Alexander polynomial $1-t+t^2-t^3+t^4$, Nakanishi index $1$, but the algebraic unknotting number of $5_1$ is $2$, whereas for $10_{132}$ it is $1$. Incidentally, knotinfo indicates these knots are also distinguished by their Levine-Tristram signature.

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    $\begingroup$ Thanks! I did a similar search and found other examples, like $(8_5,10_{141})$, $(7_5,10_{130})$, and $(7_4,9_2)$. The pair $(7_4,9_2)$ is interseting -- it has the same Levine-Tristram signature function, but different algebraic unknottin nubmers. $\endgroup$
    – Tetsuya Ito
    Commented Oct 1 at 2:02
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If $K$ and $L$ are oriented knots, then $K\# L$ and $K\# {-L}$ have the same Alexander module (namely the direct sum of the Alexander modules of $K$ and $L$). But $\operatorname{sign}(K\# L)=\operatorname{sign}(K)+\operatorname{sign}(L)$, whereas $\operatorname{sign}(K\#{- L})=\operatorname{sign}(K)-\operatorname{sign}(L)$. Thus in general $K\# L$ and $K\#{-L}$ are not $S$-equivalent (even up to change of orientation).

Edit October 3rd: Here $-L$ means the opposite orientation and mirror image. I got confused myself about the meaning of the minus sign.

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    $\begingroup$ Thank you! Does $-L$ mean the mirror image of $L$ ? Anyway, from a non-prime example it is easy to get non-prime examples (say, we can use operations preserving S-equivalence classes, like double-delta move, to get a prime example) $\endgroup$
    – Tetsuya Ito
    Commented Oct 2 at 0:50
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    $\begingroup$ @Marco Golla I'm not certain what you mean by "the opposite orientation", but I think that here Stefan Friedl wants $-L$ to mean the reflection of $L$ (reversing the orientation on the ambient S^3) rather than the reverse (reversing the orientation on L), because that's the operation that negates the signature. $\endgroup$
    – Alison Miller
    Commented Oct 2 at 19:51
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    $\begingroup$ Also note that the special case where $K$ is trivial also gives examples with prime knots, provided that you consider mirror images to be distinct knots. $\endgroup$
    – Alison Miller
    Commented Oct 2 at 19:56
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    $\begingroup$ @AlisonMiller: I did not change/correct that, so it's still Stefan's wording. Given the context, I assumed Stefan meant mirroring rather than orientation-reversal on the knot, too. $\endgroup$
    – Marco Golla
    Commented Oct 2 at 20:37
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    $\begingroup$ @MarcoGolla oh, sorry for bringing you in then! $\endgroup$
    – Alison Miller
    Commented Oct 2 at 22:30

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