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!

  • $\begingroup$ Kummer extension? $\endgroup$ – Lord Shark the Unknown Oct 6 '18 at 16:47
  • $\begingroup$ Nah I'm supposed to do it this way by finding an irreducible polynomial >.< $\endgroup$ – Math is hard Oct 6 '18 at 16:51
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    $\begingroup$ Related 1, 2. Not gonna use my dupehammer. Only gonna make the calculus rep farmers look better. $\endgroup$ – Jyrki Lahtonen Oct 6 '18 at 17:06
  • $\begingroup$ Yes, still worth to read this question in this context. $\endgroup$ – Dietrich Burde Oct 6 '18 at 18:05

$x^3\equiv\pm1\bmod7$ for all $x\in\{1,\dots,6\}$. We thus choose $x^3+2$ as the irreducible polynomial.

  • $\begingroup$ Probably simplest. $\endgroup$ – Jyrki Lahtonen Oct 6 '18 at 16:59
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    $\begingroup$ I'd choose $x^3-2$. $\endgroup$ – Torsten Schoeneberg Oct 6 '18 at 17:28
  • $\begingroup$ brilliant! I never think about stuff like this, thanks! $\endgroup$ – Math is hard Oct 6 '18 at 17:37
  • $\begingroup$ This only works for $n=3$, no? If the exponent were $4$, we couldn't conclude that $x^4=a$ is irreducible from the fact that $a$ has no fourth root in the ground field. $\endgroup$ – Jack M Oct 6 '18 at 17:52
  • $\begingroup$ @JackM This wouldn't work for $7^4$ elements because you'd also have to check that the candidate polynomial doesn't split into irreducible quadratics. Indeed. $\endgroup$ – Parcly Taxel Oct 6 '18 at 17:53

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