$K$ is a finite field which not equal to its base field $F_2$.

Let $f: V \rightarrow K$ be a function and $B(x,y)=f(x+y)+f(x)+f(y)$ such that $B(x+y,z)=B(x,z)+B(y,z)$ and $B(z,x+y)=B(z,x)+B(z,y)$ for every $x,y,z \in V$ (vector space over $K$) but $B(ax,y)$ is not necessary to equal to $aB(x,y)$ for every $a \in K$, same for $B(x,ay)$. Actually $B$ is linear over base field of $K$, but not linear over $K$.

Is there any such research area?

Example: $f(x)=Tr(x^9+x^5)$ where $Tr: F_{256} \rightarrow F_4$ is trace function.

  • $\begingroup$ Is it exactly this situation, or not quite? $\endgroup$
    – pjs36
    Mar 27, 2015 at 15:58
  • $\begingroup$ Thanks for this. I also did sth like that. I will add know an example. $\endgroup$
    – vudu vucu
    Mar 27, 2015 at 16:03
  • $\begingroup$ While we're at it, the Hermitian inner product isn't really bilinear, but instead sesquilinear: it's linear in the first argument, but conjugate-linear in the second. If you're collecting examples; I can't shed any light on the general theory, unfortunately. $\endgroup$
    – pjs36
    Mar 27, 2015 at 16:09
  • $\begingroup$ I am sorry, i wrote question wrong. But thanks for your link, they are really good for me. $\endgroup$
    – vudu vucu
    Mar 27, 2015 at 16:13
  • $\begingroup$ If you are currently at UCD, you can ask Gary McGuire. He knows this stuff. Say "Hi" from me as well :-) $\endgroup$ Mar 27, 2015 at 18:27


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