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Mar 6, 2018 at 19:06 review Close votes
Mar 7, 2018 at 2:49
Mar 6, 2018 at 18:27 comment added LSpice Probably too complicated, but: there is no harm in passing to the purely inseparable closures of $k$ and $K$. After doing so, we may replace $x$ by its semisimple part (which now must be $K$-rational) without changing either side. By induction on the dimension of $V$, it suffices to assume that $V$ contains no proper, non-$0$, $x$-stable subspace. Then any choice of non-$0$ element of $V$ furnishes a $K[x]$-linear isomorphism $V \cong E$, where $E$ is the extension obtained by adjoining to $K$ a root of the minimal polynomial of $x$. Then your equality is transitivity of the norm.
Mar 6, 2018 at 18:23 answer added S. carmeli timeline score: 2
Mar 6, 2018 at 17:55 history asked Jesua Israel Epequin Chavez CC BY-SA 3.0