Would it be possible to have a field (or field-like structure) with an additive identity $k$ where $k^a\neq k^b$ for $a\neq b$?

I need this because I'm working with a field-like structure where if I have a mathematical object $s$, $s+s^2$ doesn't make sense. Therefore, If I have $k^a = k^b$, I run into a problem by doing:

$$ s+k = s \;\therefore\; s+(s^2-s^2) = s \;\therefore\; (s+s^2)-s^2 = s $$

So I fix this by adding in the axiom $s^n-s^n =k^n$, which then leads me to $k^a\neq k^b$ for $a\neq b$.

  • $\begingroup$ No, the addititive identity is always "$0$". $\endgroup$ – Seth Apr 19 '14 at 13:23
  • $\begingroup$ When we're dealing with things like numbers yes, but I'm looking at a possibility where the elements in the field are more abstract objects. In that case would it be possible? $\endgroup$ – Disousa Apr 19 '14 at 13:25
  • $\begingroup$ Oh, but I mean it satisfies the same rules. For example try to prove that $k^2=k$ using the field axioms. $\endgroup$ – Seth Apr 19 '14 at 13:25
  • $\begingroup$ Even better, show $kx=k$ for any x in the field. $\endgroup$ – Seth Apr 19 '14 at 13:27
  • $\begingroup$ I edited the question to reflect my doubt a bit better. $\endgroup$ – Disousa Apr 19 '14 at 13:35

Try showing that $kx=k$ for any $x$ in the field using the field axioms.

Solution below if you are stuck.

$kx=(k+k)x=kx+kx$ so subtracting $kx$ from both sides we get $k=kx$.


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