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Given two knots $K$ and $L$. With Seifert matrices $M_{K}$ and $M_{L}$ respectively, then the matrix

$\begin{bmatrix}M_K & \\ &M_L\end{bmatrix}$

is a Seifert matrix of the connected sum $K+L$.

Therefore a knot is prime if and only if it has a Seifert matrix that is not S-equivalent to a matrix of this form.

Edit: This is incorrect, every knot is S-equivalent to a prime knot.

I have two questions;

1) Is what I have said correct? No

2) My understanding is that identifying whether a certain knot is prime or not was a non-trivial question, whereas identifying S-equivalence was relatively easy. Is there some hidden difficulty I am missing?

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  • $\begingroup$ What do you mean by "a matrix of this form"? Beware that a too naive statement cannot be true: it is easy to find complicated matrix for the unknot, so you can have complicated blocks in the Seifert matrix of any (prime) knot... $\endgroup$ – PseudoNeo Feb 20 '12 at 23:26
  • $\begingroup$ I see now the logic is flawed, I was equating S-equivalence with equality for the 'only if' statement. $\endgroup$ – Dilitante Feb 21 '12 at 13:06
  • $\begingroup$ I am also reconsidering the statement that S-equivalence of matrices is easily identified $\endgroup$ – Dilitante Feb 21 '12 at 13:21
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For the sake of having an answer:

"In general, it is nontrivial to determine if a given knot is prime or composite (Hoste et al. 1998). However, in the case of alternating knots, Menasco (1984) showed that a reduced alternating diagram represents a prime knot iff the diagram is itself prime ("an alternating knot is prime iff it looks prime"; Hoste et al. 1998)."

Wolfram Mathworld, Prime Knot.

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