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I'm trying to understand the notation used in the literature for Pairing-based cryptography.

I know (and I hope I've understood it well) from Wikipedia that $\mathbb{Z}_p$ is the finite field of prime order $p$, where

  • $p$ is the order of the field
  • $q$ and the characteristic of the field and $q=p^n$

I came across this notation here and there while researching the Identity-Based Encryption (Boneh-Franklin): $$s \in_R\mathbb{Z}^*_q$$

The $\mathbb{Z}^*$ means that the finite field is provided with the multiplication operation.

But the $_R$ confuses me, as I can't find its meaning on the web.

Could somebody explain it?

PS: are the following notations equivalent? $GF(p)$, $\mathbb{Z}_p$, $\mathbb{Z}/p\mathbb{Z}$ and $\mathbb{F}_p$

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    $\begingroup$ The $_R$ has nothing to do with the field — it is associated to $\in$! To quote your first link: "For a set $S$, by $a\in_RS$, we mean that $a$ is randomly chosen from $S$." $\endgroup$
    – yyyyyyy
    Apr 17, 2016 at 13:08
  • $\begingroup$ Oh, shame on me ^^. And for the notation equivalence ? $\endgroup$
    – 3isenHeim
    Apr 17, 2016 at 13:10
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    $\begingroup$ If $p\in \mathbb P$ (with $\mathbb P$ being the set of all primes) then the notations $GF(p); \mathbb Z_p; \mathbb Z/p\mathbb Z; \mathbb F_p$ are equivalent. $\endgroup$
    – SEJPM
    Apr 17, 2016 at 13:16
  • $\begingroup$ Note that depending on the context $\mathbb Z_p$ is also used for the $p$-adic integers. $\endgroup$
    – flawr
    Apr 18, 2016 at 9:54

1 Answer 1

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To quote yyyyyyy from the comments:

The $_R$ has nothing to do with the field — it is associated to $\in$! To quote your first link: "For a set $S$, by $a\in_R S$, we mean that $a$ is randomly chosen from $S$."

and to quote SEJPM from the comments:

If $p\in \mathbb P$ (with $\mathbb P$ being the set of all primes) then the notations $GF(p);\mathbb Z_p;\mathbb Z/p\mathbb Z;\mathbb F_p$ are equivalent.

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    $\begingroup$ I made this a community wiki (for which nobody receives reputation) because the answer is trivial (for many people at least) and because yyyyyyy did the "major" work. (And I really hate comment-answered question which don't have a formal answer) $\endgroup$
    – SEJPM
    Apr 17, 2016 at 13:38
  • $\begingroup$ Another (realted) question : does $X =\langle U, V \rangle$ mean that $X$ is the concatenation of $U$ and $V$ ? $\endgroup$
    – 3isenHeim
    Apr 18, 2016 at 7:08
  • $\begingroup$ @EisenHeim, depends on the context. $\endgroup$
    – SEJPM
    Apr 19, 2016 at 18:49

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