# Shamir Secret Sharing Scheme - some doubts not answered earlier

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!

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.
• 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? – Rahul Mathur Oct 24 '18 at 19:29