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I'm not a topologist and I just saw the definition of prime knot a while ago.

Today I'm somewhat supprised to realize that I don't even know if there are infinitely many prime knots? If this question is not completely trivial then I'm hoping to see a "proof from the book".

(A related question is, is the decomposition of a composite knot into prime knots unique? I would hope so but I don't have a strong reason to support that.)

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    $\begingroup$ Yes; in fact there are infinitely many distinct torus knots, all of which are prime (en.wikipedia.org/wiki/Torus_knot). $\endgroup$
    – Qiaochu Yuan
    Commented Nov 22, 2011 at 19:16
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    $\begingroup$ Knots do decompose uniquely into a sum of primes, see e.g. chapter 2 of Lickorish's "An introduction to knot theory." Also, since knot genus is additive under connected sum it follows that every genus 1 knot is prime, so take your favorite knot and consider all of its twisted Whitehead doubles; these have genus 1 and are distinguished by their Alexander polynomials. $\endgroup$
    – Steven Sivek
    Commented Nov 22, 2011 at 20:07
  • $\begingroup$ It's a theorem of Schubert's from the 1930's that oriented knots under the connect-sum operation are a free commutative monoid on infinitely many generators. $\endgroup$
    – Ryan Budney
    Commented Nov 23, 2011 at 2:01
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    $\begingroup$ These notes give a basic exposition of the infinity of prime knots, and Schubert's proof of uniqueness of prime factorization: math.ucla.edu/~radko/191.1.05w/marcos.pdf $\endgroup$
    – Daniel Moskovich
    Commented Nov 23, 2011 at 3:28

2 Answers 2

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Yes there are infinitely many prime knots.

Not a "proof from the book", but all (p,2) torus knots for p prime are prime, and they have different Alexander polynomials.

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Yes.

Kauffman and Lopes found some very elementary families: https://arxiv.org/pdf/1604.02510.pdf .

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