Talk:Field with 4 Elements has only Order 2 Elements

"So it must be the case that $b \times b = b$."


 * Hence $b = 1_F$. Then $b + b = (a+a)+(a+a) = 0_F$ by Element to the Power of Group Order. Hence $\operatorname{char} F = 2$, since $1_F \ne 0_F$. That this $F$ cannot be a field is not relevant for the proof. &mdash; Lord_Farin (talk) 21:35, 17 September 2013 (UTC)


 * Drat - I have misunderstood the meaning of "characteristic". The problem being solved is that $\forall a \in F: a + a = 0_F$. It has nothing to do with the characteristic, which I have facilely taken to mean "the order of the additive group".


 * So it needs renaming and slightly rewriting - the structure of the proof (if it holds, I wrote it up just now without a safety net) should still hold. --prime mover (talk) 21:42, 17 September 2013 (UTC)


 * Well, $\forall a: a + a = 0_F$ is the (or a) definition of Char 2. So is $1_F + 1_F = 0_F$ (multiply by $a$). This is of course rather different than the order of the additive group. &mdash; Lord_Farin (talk) 21:47, 17 September 2013 (UTC)


 * As I say, I made a mistake. I should not have said "characteristic" when I wrote the page. I was in error when I did so. I have since removed that wrongness, on account of it being incorrect and inadequately truthful. --prime mover (talk) 22:04, 17 September 2013 (UTC)