# Producing a field with $7^3$ elements

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!

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

## 1 Answer

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

• Probably simplest. – Jyrki Lahtonen Oct 6 '18 at 16:59
• I'd choose $x^3-2$. – Torsten Schoeneberg Oct 6 '18 at 17:28
• brilliant! I never think about stuff like this, thanks! – Math is hard Oct 6 '18 at 17:37
• 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. – Jack M Oct 6 '18 at 17:52
• @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. – Parcly Taxel Oct 6 '18 at 17:53