Decryption of message in IBE without random oracle using bilinear pairing registration?
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 $e(c_1,d_1)/e(c_2,d_2) =  e(y , g_1^\alpha )$