# Decryption of message in IBE without random oracle using bilinear pairing registration?

I find the following IBE scheme from the videos posted and i don't understand the decryption algorithm, will any one please elaborate the 6th step

scheme

1. Setup($\lambda$) : $(\mathbb{G}, \mathbb{G}_T, g, P)$ <-- GenBilgroup($\lambda$)

$\alpha\leftarrow_R\mathbb{F}_p$

2. Public Parameters : $pp = [ g , y = g^\alpha , g_1, h ] \in \mathbb{G}$

3. Master secret : $mk = g_1^\alpha$

4. Keygeneration: $k(mk,id)$: $d_1= mk\cdot(Y^{id\cdot h})^r, d_2 = g^r$ where $r\leftarrow_R\mathbb{F}_p$

5. Encryption: $E(pp,id,m)$ : $c_1 = g^s, c_2=(Y^{id\cdot h})^s, c_3=m\cdot e(y, g_1)^s$

6. Decryption: $e(c_1,d_1)/e(c_2,d_2)$ will give $e(y,g_1)^s$ How? and $m$ is obtained by dividing $c_3$ by that value.

• You can use MathML / Tex on this site. – Maarten Bodewes Feb 16 '15 at 14:50

I don't know whether it is correct

$$e(c_1,d_1) = e(g^s, mk\cdot(y^{id\cdot h})^r)$$

$$=e(g^s,mk)\cdot e(g^s, (y^{id\cdot h})^r)$$

$$= e(g^s , g_1^\alpha ) \cdot e(g^r , (y^{id\cdot h})^s)$$

$$= e(y , g_1^\alpha ) \cdot e(g^r , (y^{id\cdot h})^s)$$

$$e(c_2,d_2) = e((Y^{id\cdot h})^s , g^r)$$

$$= e((Y^{id\cdot h})^s , g^r)$$

therefore $e(c_1,d_1)/e(c_2,d_2) = e(y , g_1^\alpha )$

• Yes, of course it's correct. A straitforward calculation using the bilinearity of e, shows that dividing $c_3$ by $e(y,g_1^\alpha)$ allows to recover m. I confirm. – Robert NACIRI Mar 21 '15 at 17:23
• @manju If the answer fits, you might want to “accept” it… thanks. – e-sushi Jul 19 '15 at 22:52