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Sep 20, 2021 at 9:49 comment added Laurent Moret-Bailly Oops, sorry, I had missed that word!
Sep 20, 2021 at 8:22 history edited lkx CC BY-SA 4.0
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Sep 20, 2021 at 8:19 comment added YCor There is a quite obvious case: when the algebra is $\mathbf{F}_{q^n}$ itself with $n\ge 2$: its "embedding dimension" is then 0 as $\mathbf{F}_{q^n}$-algebra but 1 as $\mathbf{F}_{q}$-algebra.
Sep 20, 2021 at 5:58 history asked lkx CC BY-SA 4.0