 Solution

Let $Q\in E^{\prime }[\hat{\varphi }]$. The compatibility of divisor classes with pullback identifies the class ofwith $\hat{\varphi }(Q)=O$, so choose $g_Q\in K(E{)}^{\times }$ withSince $[n]Q=\varphi \hat{\varphi }(Q)=O$, choose $f_Q\in K(E^{\prime }{)}^{\times }$ withThe functions $f_Q\circ \varphi$ and $g_Q^n$ have the same divisor. Their quotient is constant, and because $K$ is algebraically closed we may rescale $g_Q$ so thatThus defineChanging either function changes $f_Q$ only by an $n$th power of a constant. The divisor relation for $Q+R$ differs from the sum of those for $Q$ and $R$ by $n$ times a principal divisor, so the map is a homomorphism. If $[f_Q]$ is trivial, then $f_Q=h^n$ for $h\in K(E^{\prime }{)}^{\times }$, whence $\mathrm{div}(h)=(Q)-(O)$; the divisor-class isomorphism forces $Q=O$. The map is therefore well-defined and injective.

For $P\in E[\varphi ]$, translation by $P$ fixes $g_Q^n=f_Q\circ \varphi$. Henceis independent of the auxiliary point $X$. This is the Weil pairing associated with $\varphi$.

Solved by gpt-5.6-sol high.