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Let $X$ and $Y$ be two curves defined over $\mathbb{F}_q$ and $f:X \rightarrow Y$ be a separable rational map. Then there is field embedding $f^* : \rightarrow \mathbb{F}_q (Y) \rightarrow \mathbb{F}_q (X)$ defined by $f^(\alpha) = \alpha \circ f$. The degree of$f$is then defined to be$[\mathbb{F}_q (X) : f^ (\mathbb{F}_q (Y))]$. If I take two curves$X$and$Y$in sagemath over some$\mathbb{F}_q $in sagemath, is there any way to automatically get the map$ f^*$and degree of$f$?

Let $X$ and $Y$ be two curves defined over $\mathbb{F}_q$ and $f:X \rightarrow Y$ be a separable rational map. Then there is field embedding $f^* $$ f^\ast : \rightarrow \mathbb{F}_q (Y) \rightarrow \mathbb{F}_q (X) $ $$ defined by $f^(\alpha) $f^\ast(\alpha) = \alpha \circ f$. The degree of $f$ is then defined to be $[\mathbb{F}_q (X) : f^ (\mathbb{F}_q f^\ast(\mathbb{F}_q (Y))]$. If I take two curves $X$ and $Y$ in sagemath over some $\mathbb{F}_q$ in sagemath, is there any way to automatically get the map$f^*$ map $f^\ast$ and degree of $f$?

Let $X$ and $Y$ be two curves defined over $\mathbb{F}_q$ and $f:X \rightarrow Y$ be a separable rational map. Then there is field embedding $$ f^\ast : \rightarrow \mathbb{F}_q (Y) \rightarrow \mathbb{F}_q (X) $$ defined by $f^\ast(\alpha) = \alpha \circ f$. The degree of $f$ is then defined to be $[\mathbb{F}_q (X) : f^\ast(\mathbb{F}_q (Y))]$. If I take two curves $X$ and $Y$ in sagemath over some $\mathbb{F}_q $ in sagemath, is there any way to automatically get the map $f^\ast$ and degree of $f$?