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Aug 10 at 5:06 comment added Lucenaposition Are fields allowed to have $0=1$?
Dec 8, 2023 at 23:34 comment added Joe Lamond The "canonical" argument mentioned above is as follows: suppose $\mathbb F_q=\{a_1,\dots,a_q\}$ is a finite field. Then, $(x-a_1)\cdots(x-a_q)+1$ is a non-constant polynomial with no root in $\mathbb F_q$.
Aug 20, 2012 at 23:52 comment added Igor Rivin Not as simple as the "canonical" argument.
Aug 20, 2012 at 21:49 comment added Ian Agol That's actually a pretty simple argument.
Jan 29, 2011 at 5:27 history answered Peter CC BY-SA 2.5