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Notation : $\mathbb{G}$ is an additive group and $\mathbb{G}_T$ is multiplicative group of prime order $q$.

Bilinear mapping $e: \mathbb{G} \times \mathbb{G} \rightarrow \mathbb{G}_T$ has to satisfy two properties.

1) $e(aP,bQ) = e(P,Q)^{ab}$ for all $P,Q \in \mathbb{G} , a,b \in \mathbb{Z}_q$

2) $e(P,P) \ne 1_{\mathbb{G}_T}$ for all $P \in G$

In addition to the above two, it has to be computable.

Is there any simple example for such bilinear pairing?

I searched over internet as I can but I didn't find.

I want example like this $e(x,y) = xy \mod n$ (It's not a valid example).

Note: The sets under consideration $\mathbb{G},\mathbb{G}_T$ must have at least four elements.

Duplicate question for https://math.stackexchange.com/questions/1380004/simple-example-for-bilinear-mapping

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  • $\begingroup$ Mathematically, a bilinear map need only satisfy the first condition. Sadly, cryptographers are keen to appropriate mathematical terms and give them a different meaning... $\endgroup$ – fkraiem Aug 6 '15 at 8:32
  • $\begingroup$ @fkraiem It indeed makes sense to make this second property explicit in context of cryptographic applications. Bilinear maps that do not have this property would be useless for how they are used in cryptography. $\endgroup$ – DrLecter Aug 6 '15 at 8:37
  • $\begingroup$ Asking for examples is considered off topic. I've slightly altered the title, hope you agree. $\endgroup$ – Maarten Bodewes♦ Aug 6 '15 at 11:30
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A toy example would be this simple map with $\mathbb{G} = \mathbb{Z}/5$ to $\mathbb{G}_T = \mathbb{Z}^*/11$, as follows:

$$e(x,y) = 3^{xy} \bmod 11$$

It's easy to verify that both equations hold (except that $e(0,0) = 1$; that's actually a necessary consequence of the first equation, and so I'll consider that an acceptable exception).

Of course, even if you scale this up to non-toy sizes, this particular style of bilinear mapping isn't cryptographically interesting; however it does answer your question.

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  • $\begingroup$ Here order of $\mathbb{Z}_5$ is 5 and of $\mathbb{Z}_{11}^*$ is 10, how they both can be of same prime order as i specified in my OP. $\endgroup$ – hanugm Aug 11 '15 at 5:30
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    $\begingroup$ @hanugm: actually, in my case, $\mathbb{G}_T$ really is the group of quadratic residues modulo 11, and the size of that group is 5. $\endgroup$ – poncho Aug 11 '15 at 11:29
  • $\begingroup$ Is there any reason for selecting 3 in this case? Can is use any number co-prime to 5 and 11 instead of 3? $\endgroup$ – hanugm Aug 14 '15 at 11:54
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    $\begingroup$ @hanugm: I selected 3 because the order of 3 (mod 11) is 5; that is, $3^5 \bmod 11 = 1$ and $3^x \bmod 11 \ne 1$ for $0 < x < 5$. Any such value with this property will work, and in fact, 5 will (in fact) work. $\endgroup$ – poncho Aug 14 '15 at 11:58
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No, there is none (as far as is currently known). From Ben Lynn's doctoral thesis on the subject:

There is only one known mathematical setting where desirable pairings exist: hyperelliptic curves.

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