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Prove that there exists an irreducible polynomial $g(x) \in \mathbb{F}_p[t]$ such that the degree of $g(x)$ is $3$.

This can be solved using field theory.

Can I get more hint that what action is need to be defined to complete this question.

Ricci Ten
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1 Answers1

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I don't know what argument your professor has in mind but here is a possible approach; I haven't seen this before, it's an interesting idea. We can use the fact that by a standard argument the order of $GL_3(\mathbb{F}_p)$ is

$$|GL_3(\mathbb{F}_p)| = (p^3 - 1)(p^3 - p)(p^3 - p^2) = p^6 (p - 1)^3 (p^2 + p + 1)(p + 1).$$

Let $\ell$ be a prime dividing $p^2 + p + 1$. Then by Cauchy's theorem $GL_3(\mathbb{F}_p)$ has an element $X$ of order $\ell$. I claim that the characteristic polynomial of $X$ must be irreducible; do you see why?

Edit: I discuss how this argument generalizes here.

Qiaochu Yuan
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  • I am trying to figure it out and will be back to you in next 24 hour . Please reply then, – Ricci Ten Jan 18 '25 at 17:51
  • Okay what i conclude after long drag that $x^l-1$ is polynomial such that characteristics divides it now $l>3$ is prime what i got is root of characteristic will also be a root of $x^l-1$ but $p-1$ divides $l$ but $l$ is prime hence no root in $F_p$ so it must be irreducible – Ricci Ten Jan 19 '25 at 05:52
  • I have some other question: How did you think like that like what motivated to choose $l$ from $p^2+p+1$ factor can i choose a prime factor of $p+1$ , can you share your psychological situation. – Ricci Ten Jan 19 '25 at 05:52
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    @Ricci: one way to motivate this choice is that it's the only factor that doesn't appear in $|GL_2(\mathbb{F}p)|$. There is also a longer explanation in terms of properties of cyclotomic polynomials (which explains how this argument generalizes to $GL_n(\mathbb{F}_p)$), or you can think about the multiplicative orders of elements in $\mathbb{F}_p$ and $\mathbb{F}{p^3}$ respectively. – Qiaochu Yuan Jan 19 '25 at 16:38