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I have an exercise from my course notes that states:

Find how many irreducible factors has $f(x) = x^{26}-1$ over $\mathbb{F}_3$ and their degrees. (don't factorize it)

I see immediately that the $1$ is a root of $f$. So I have $f(x)=(x-1) g(x)$ where $g$ has degree $25$ with no root in $\mathbb{F}_3$. But I don't know really how to move.

Aditya Dwivedi
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lukk
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1 Answers1

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The distinct-degree factorization theorem tells us that $X^{27}-X$ factors mod $3$ into the product of all monic irreducible polynomials in $\mathbb{F}_3[X]$ whose degree divides $3$ (since $27 = 3^3$). So $X^{26}-1$ is the product of all monic irreducible polynomials in $\mathbb{F}_3[X]$ of degree $1$ and $3$ except $X$.

M. Wang
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  • So, is there a way to count the number of such polynomials? Or do I have to list them and verify if each one has a root in $F_3$? – lukk May 28 '20 at 19:47
  • @lukk I don't know whether there's a way to count them. Since $\mathbb{F}_3$ only has three elements it shouldn't be too hard to identify the irreducible monic polynomials of degree $3$ (as you say, these are the ones that don't have a root in $\mathbb{F}_3$). – M. Wang May 29 '20 at 12:27