Does RSA operate over a Finite Field (Galois Field)?

Is it correct to say that RSA operates over a Finite Field (Galois Field)? In this case GF(p)? I do understant that the modulo in RSA is not itself a prime number, but all the operations (multiplication, inversion) occur as if it is a GF(p).

• Nope, RSA is defined over an residue class ring. Dec 17, 2014 at 17:49
• @DrLecter: Technically, your comment does not contradict the statement in question: a residue class ring may be (isomorphic to) a finite field. You want to append "that is not a field". Dec 17, 2014 at 19:36
• @yyyyyy jup, thats correct! :) Dec 17, 2014 at 19:40

No, RSA encryption and signature is performed in (the multiplicative semigroup of) the factor ring $\mathbb Z/n\mathbb Z$ which is not a field since the non-zero elements $kp+n\mathbb Z$ (for $0<k<q$) and $kq+n\mathbb Z$ (for $0<k<p$) do not have multiplicative inverses. (However, one easily observes that all other non-zero elements are invertible, as their representants are coprime to $n=pq$).
• While the correct answer to the question is "no", the RSA definitions also can work for those elements, which are not coprime to $n$. Since they are zero divisors, they dont have inverse elements, but it can still be true that $x^{ed} = x$, with $x$ not coprime to $n$. I cant recall the exact requirement there, but there are topics on this on crypto-SE.
• @tylo $x^{ed} = x$ holds for all $x$ in RSA. Jan 1, 2015 at 17:59