my first question on here.

I've gone through some examples of Shamirs Secret Sharing Scheme, but I have 2 rather pressing doubts -

i. Why do we choose the co-efficients of the polynomial from the set {1,..., p-1}?

ii. Why should we have p (the choice of prime) > n (total number of parties involved in the secret sharing scheme). I am able to reason why p > k but not the above!

Thanks very much in advance.


1 Answer 1


You need a finite algebraic object, resulting in exact arithmetic and exact recovery of the secret. So infinite fields such as $\mathbb{C,~R}$ are not suitable.

Since the scheme is based on polynomial interpolation, you need two operations, i.e., a field. It can be a prime field $\mathbb{F}_p$ or an extension field $\mathbb{F}_{p^m}.$

Since each user is assigned their secret by specifying $y_i$ at point $x_i,$ you need at least $n+1$ points, the value at zero of the polynomial normally being used for specifying the secret.

  • $\begingroup$ thank you for your reply - I definitely agree that if the co-efficients are in $\mathbb{F}_p$ then we must have that p >n. However, the issue I have is that I can't reason why we can't have co-efficients outside $\mathbb{F}_p$ - I tried with the following example - (2,2) Sharing Scheme with S = 5 , p = 7 and f(x) = 5 - 13x. I am able to recover the polynomial as 5 + x which is congruent to the original polynomial (mod 7) BUT I am still able to recover the secret - Am i missing something else here? $\endgroup$ Oct 24, 2018 at 19:29
  • $\begingroup$ If I have number outside finite field I could map to this field using mod operator I think $\endgroup$
    – Macko
    Dec 16, 2021 at 21:34

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