Producing a field with $7^3=343$ elements.
Okay, so if I can find an irreducible polynomial over $\mathbb Z_7$ of degree $3$ then I'll have done it.
Now, since it's of degree $3$ all I have to do is check for linear factors by finding a degree $3$ polynomial with no roots. I could just guess and check, but I was wondering if there was a more methodical way to do this, perhaps there is an insight that I'm missing. Thanks!
Best Answer
$x^3\equiv\pm1\bmod7$ for all $x\in\{1,\dots,6\}$. We thus choose $x^3+2$ as the irreducible polynomial.