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I've been solving problems from my Galois Theory course, and I want to check if my approach to this one is correct. It says:

Check if the polynomial $(X^2+1)(X^4+X^2+X+1)$ factorizes as product of degree $1$ polynomials in a field with $81$ elements.

What I've done is notice that, if the field has exactly $81=3^4$ elements, then it's equal to $\mathbb Z_3(\alpha)$, where $\alpha$ is a root of the polynomial $X^{3^4}-X$. Given that every finite extension of finite fields is normal, then $\mathbb Z_3(\alpha)$ is also normal, so the question is equivalent to prove if $X^2+1$ and $X^4+X^2+X+1$ divide $X^{3^4}-X$ or not.

Considering now that if $d\mid n$, then every irreducible polynomial of degree $d$ in $\mathbb Z_p[X]$ divides the polynomial $X^{p^n}-X$ in $\mathbb{Z}_p[X]$ (it's stated inside my course notes), in my particular case $n=4$, so from the fact that $2\mid 4$ and $4\mid 4$ we conclude $X^2+1\mid X^{3^4}-X$. Also, $X^4+X^2+X+1$ can either be irreducible or decompose as product of degree 2 irreducible polynomials, so in any case also $X^4+X^2+X+1\mid (X^{3^4}-X)$.

Finally, since our 81 elements field is the splitting field of $X^{3^4}-X$ over $\mathbb{Z}_3$, it follows that $(X^2+1)(X^4+X^2+X+1)$ factorizes as degree 1 polynomials over our field.

Is my work correct? If not, where did I go wrong? Any help will be appreciated, thanks in advance.

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  • $\begingroup$ @ancientmathematician: The other factor is not $(X^5-1)/(X-1)=X^4+X^3+X^2+X+1$. $\endgroup$ Jun 8, 2021 at 13:40
  • $\begingroup$ @MarcvanLeeuwen You are so right. I am embarassed. But I bet it was meant to be. ;-) $\endgroup$ Jun 8, 2021 at 13:44
  • $\begingroup$ @ancientmathematician it was not meant to be, it was just $X^4+X^2+X+1$. No worries! thanks for your help. $\endgroup$ Jun 8, 2021 at 14:06

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Your argument is almost complete. You know that all irreducible polynomials (over $\Bbb Z/3\Bbb Z$) of degree $1$, $2$ or $4$ split into linear factors in a field with $81$ elements. So your only worry is that the polynomial $X^4+X^2+X+1$ might have an irreducible factor of degree$~3$, which would then fail to factor. The remaining factor after dividing by that factor of degree$~3$ would be of degree$~1$ and give a root of $X^4+X^2+X+1$ in $\Bbb Z/3\Bbb Z$. But none of the values $0,1,2$ are roots, so this does not happen. (With a bit more effort you can see that the polynomial $X^4+X^2+X+1$ over $\Bbb Z/3\Bbb Z$ is in fact irreducible).

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