I was given a hint to consider $\mathbb{F}_9=\mathbb{F}_3[\beta]=\{a+b\beta:a,b\in\mathbb{F}_3\},$ where $\mathbb{F}_3=\{0,1,2\}$ is the set of integers modulo $3$. The distributive law says for $a+b\beta,a'+b'\beta\in\mathbb{F}_9$, $(a+b\beta)(a'+b'\beta)=aa'+(ba'+ab')\beta+(bb')\beta^2\in \mathbb{F}_9$. I know that $aa',ba'+ab',bb'\in\{0,1,2\}$ since $\mathbb{F}_3$ is a field. So $(aa',ba'+ab',bb')\in\{0,1,2\}^3$. This gives 27 elements, and I want to find $\beta$ so that there are exactly 9 distinct elements. I know I can do this by checking if each pair of the elements is equal, but I don't know if I will finish the $\sum_{k=1}^{26} k$ comparisons before my homework set is due - I certainly hope this is not the intended approach.
Is there an insight I'm missing here? Thanks a bunch in advance!
This is just Jyrki's comment, I think some people gain confidence from the matrix approach, which seems familiar, comfortable.
Take $a,b$ in your field, then make the 2 by 2 matrix $$ aI + b B, $$ where $$ B = \left( \begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array} \right) $$
What is $B^2?$ What is $$ (aI + b B)(cI + dB)? $$ If at least one of $a,b$ is nonzero in the field, and at least one of $c,d$ is nonzero in the field, is it possible for the product $ (aI + b B)(cI + dB) $ to be the zero matrix?
The multiplicative identity is now $I.$ For a nonzero matrix here, what is the multiplicative inverse of $aI + bB?$