What do the elements of $\mathbb{F}_{243}$ look like?

Question

Since $243 = 3^5$ and $3$ is a prime number we can construct the field $\mathbb{F}_{243}$ with $243$ elements, by taking an irreducible polynomial $f$ of degree $5$, then we get: $\mathbb{F}_{243}\cong \mathbb{F}_3[x]/(f(x))$. But is there a way to explicitly compute the elements of this field? Or a way to see more of the structure of the field?

Answer 1

Of course you can compute $\mathbb{F}_{243}\cong \mathbb{F}_3[x]/(f(x))$ explicitly. Just pick an irreducible polynomial $f$ and begin.

I would first pick some element of the quotient and start computing its powers mod $f(x)$. If it reaches $1$ before the $242$'th power, then it is not a primitive element, and I would select a new element of the quotient that I had not previously seen and begin again.

Eventually you will find a primitive element $\alpha$ of the field, so that every nonzero element of the field is a power of $\alpha$.

There is not much to "the structure" of a finite field beyond its primitive elements.