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I am looking for a polynomial of degree $3$ in $\mathbb{F}_{9}$. How do I find one ? And if I have one how do I show that it is irreducible ?

I would start with an irreducible polynomial in $\mathbb{F}_{3}$ like $x^3+2x+1$ Is this procedure correct ?

An then I would use the Rabin test to find out whether $\frac{x^{27}-x}{x^3+2x+1}$ is a division without rest.

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  • $\begingroup$ For any field $K$ a cubic polynomial is irreducible in $K[x]$, iff it has no zeros in $K$. For finite fields we can say more (see my answer). $\endgroup$ Feb 27, 2015 at 8:04

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Almost there but not quite. You can, indeed, begin by finding a cubic polynomial $p(x)$ that is irreducible over $\Bbb{F}_3$. Adjoining a zero of such a polynomial then automatically generates the field $\Bbb{F}_{27}$. But the intersection $$\Bbb{F}_{27}\cap\Bbb{F}_9=\Bbb{F}_3$$ (in whichever bigger field where you can form that intersection). Therefore $p(x)$ has no zeros in $\Bbb{F}_9$ and, consequently no linear factors in $\Bbb{F}_9[x]$. Being cubic this implies that $p(x)$ automatically remains irreducible over $\Bbb{F}_9$.


The result generalizes as follows. An irreducible polynomial $p(x)$ of degree $n$ in $\Bbb{F}_p[x]$ remains irreducible in $\Bbb{F}_{p^m}[x]$ whenever $\gcd(m,n)=1$. The argument is a bit more delicate, because unlike in the case of cubics irreducibility cannot be deduced from absence of linear factors. But Galois theory comes to the rescue. The condition on gcd implies that the $m$th power of the Frobenius automorphism generates the Galois group $Gal(\Bbb{F}_{p^n}/\Bbb{F}_p)$. Therefore all the zeros of $p(x)$ are conjugate to each other over $\Bbb{F}_{p^m}$ as well.

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  • $\begingroup$ Thanks. My intention was ( i didn't mention it) to generate the field $\mathbb{F}_{729}$ based on $\mathbb{F}_{9}$. Therefore I need a polynomial of degree 3 which is irreducible in the starting field $\mathbb{F}_{9}$. Is this procedure correct ? $\endgroup$ Feb 26, 2015 at 10:55
  • $\begingroup$ Yes. The idea is correct. My point was that you can skip the Rabin test. $\endgroup$ Feb 26, 2015 at 11:31
  • $\begingroup$ @MathPowerUser: In general you can construct an extension of a finite field of degree $nm$, $\gcd(n,m)=1$, as a compositum of the extensions of degrees $n$ and $m$. In particular, to get $\Bbb{F}_{3^6}$ you can either adjoin a root of a quadratic (irreducible over the prime field) to $\Bbb{F}_{3^3}$ or adjoin a root of a cubic (irreducible over the prime field) to $\Bbb{F}_{3^2}$. Both work, and the end results are isomorphic. $\endgroup$ Feb 26, 2015 at 11:59
  • $\begingroup$ I am not so familiar with adjoining roots. What are the steps to do to get the searched polynomial ? Thank you for your help. $\endgroup$ Feb 26, 2015 at 18:43
  • $\begingroup$ @MathPowerUser: How did you construct $\Bbb{F}_9$? By adjoining $i$ to $\Bbb{F}_3$, or by taking the quotient ring of $\Bbb{F}_3[x]$? Anyway, there is no need to be clever here. Any irreducible cubic of $\Bbb{F}_3[x]$ remains irreducible in $\Bbb{F}_9[x]$. You can use the one in your question. $\endgroup$ Feb 27, 2015 at 8:03

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