Checking equality of maps over a reduced scheme enough to check on fibers

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Let $E to S$ be elliptic curves over a reduced scheme $S$. Let $f:E to E’$ be a non constant homomorphism of elliptic curves over $S$. Let $N=deg(f)$. Let $f^t:E’ to E$ be the dual morphism. To show $f circ f^t = [N]$ something I’m reading says since $S$ is reduced, it is enough to check $f circ f^t = [N]$ on fibers, i.e. we can assume $S=mathrmSpec k$ for $k$ an algebraically closed field. Why?
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    Let $E to S$ be elliptic curves over a reduced scheme $S$. Let $f:E to E’$ be a non constant homomorphism of elliptic curves over $S$. Let $N=deg(f)$. Let $f^t:E’ to E$ be the dual morphism. To show $f circ f^t = [N]$ something I’m reading says since $S$ is reduced, it is enough to check $f circ f^t = [N]$ on fibers, i.e. we can assume $S=mathrmSpec k$ for $k$ an algebraically closed field. Why?
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      Let $E to S$ be elliptic curves over a reduced scheme $S$. Let $f:E to E’$ be a non constant homomorphism of elliptic curves over $S$. Let $N=deg(f)$. Let $f^t:E’ to E$ be the dual morphism. To show $f circ f^t = [N]$ something I’m reading says since $S$ is reduced, it is enough to check $f circ f^t = [N]$ on fibers, i.e. we can assume $S=mathrmSpec k$ for $k$ an algebraically closed field. Why?
      enter image description here







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      Let $E to S$ be elliptic curves over a reduced scheme $S$. Let $f:E to E’$ be a non constant homomorphism of elliptic curves over $S$. Let $N=deg(f)$. Let $f^t:E’ to E$ be the dual morphism. To show $f circ f^t = [N]$ something I’m reading says since $S$ is reduced, it is enough to check $f circ f^t = [N]$ on fibers, i.e. we can assume $S=mathrmSpec k$ for $k$ an algebraically closed field. Why?
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      edited Aug 3 at 21:57
























      asked Aug 3 at 21:37









      usr0192

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