# Tag Info

0

RSA operates over a multiplicative group $(\mathbb{Z}/n\mathbb{Z})^*$, not over a field. You can say that it's a ring, but since addition is not used in RSA it's redundant.

5

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 ...

0

This degree-reduction construction has a chance to be re-invented, still a reference would be "Simplified VSS and fast-track multiparty computations with applications to threshold cryptography".

Top 50 recent answers are included