I need to proof that semifields of order 8 are all fields. (S, +, *) is semifield if: 1) (S, +) is abelian additive group 2) (S \ {0}, *) is a loop 3) left and right distributive properties: (a + b)c = ac + bc c(a+b) = ca + cb
Loop is a quasigroup with e. I'll be grateful if you give me any hints. Thank you!
(not a full answer)
Let $S$ be a semifield with $8$ elements.
Because of distributivity, we have $$((1+1)\cdot(1+1))\cdot(1+1)=1+1+1+1+1+1+1+1=0$$ so $1+1$ can't be in the multiplicative loop, hence $1+1=0$ and consequently $x+x=0$ for each $x\in S$.
Now let $a\in S\setminus\{0,1\}$ be arbitrary.
We have $a\cdot a\ne a$ (since $a\cdot x=a$ has the unique solution $x=1$), $a\cdot a\ne 1$ (since then $a\cdot (a+1)=a+1$), and if $a\cdot a=a+1$, we would receive a subfield $\{0,1, a, a+1\}$ over which $S$ would be a vector space, but that's impossible as $8$ is not a power of $4$.
Consequently, the following $8$ elements are different, hence $$S=\{0,\,1,\,a,\,a+1,\,a^2,\, a^2+1,\, a^2+a,\, a^2+a+1\}$$
Now we can continue analyzing the possibilities for $a^2a\, =(a\cdot a)\cdot a$ and $aa^2 \, =a\cdot (a\cdot a)$.