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I have to find an irreducible polynomial in $\mathbb{F}_{11}[x]$ of degree 3 and of degree 4. I thought about $x^3+x^2+1$ and $x^4+x^3+1$ but I don't know if it is right and I choose them a bit randomly.

My question: is there any simple way or exact method which I can follow step by tep when I have to find irreducible polynomials over finite fields. I have seen a lot of concrete examples but I don't seem to understand how do you generally tackle a problem like this, so tht even when I have another field or other degress of the Polynomial, I can still be able to solve it.

Thanks in advance

user26857
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Annalisa
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    I'll give a partial answer. If $p$ is reducible polynomial of degree 3, then there exists 1-degree polynomial $q(x)=x-a$ such that $q\mid p$. It means $p(a)=0$. – MH.Lee Feb 09 '22 at 16:50
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    Hint: If $p(x)=x^3-x$, then $p(a)=0$ for three choices of $a\in\Bbb{F}{11}$. Therefore you can find $b\in\Bbb{F}{11}$ such that $p(a)\neq b$ for all $a\in\Bbb{F}{11}$. You have thus found a cubic with no zeros in $\Bbb{F}{11}$. – Jyrki Lahtonen Feb 09 '22 at 20:24
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    An irreducible quartic is trickier. Because $40\mid 11^2-1$ a binomial won't work. The following leads to an irreducible biquadratic: The zeros of $x^8+1$ (in some extension field of $\Bbb{F}_{11}$ have order sixteen, so minimal polynomials of degree four. Can you factor $x^8+1$? The tricks here work (replace $x$ with $x^2$ to get a factor of $x^8+1$ instead of $x^4+1$). – Jyrki Lahtonen Feb 09 '22 at 20:32
  • So I'm afraid the methods for finding those are a bit ad hoc. There are systematic ways, but those are better suited to computers. I always try a binomial first, but here that does not work for either degree :-( – Jyrki Lahtonen Feb 09 '22 at 20:35

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