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If I want to show that a certain polynomial is irreducible over a finite field, which methods do I have?

In particular how can I show that $X^4-3$ is irreducible over $\mathbb F_5$

The idea which I posted here does not work very well for me.

Its not hard to see that the polynomial has no roots in the finite field. So the only possibility being reducible is that I can split it into two polynomials with degree $2$.

Should I try it as usual with $x^4-3=x^4-2x^2+9=(a_1x^2+b_1x+c_1)(a_2x^2+b_1x+c_1)$?

I tried it, but didn't came to a solution. Can someone help me?

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  • $\begingroup$ There are fortunately not many irreducible polynomials of degree 2 over this field. $\endgroup$ – Tobias Kildetoft Feb 14 '15 at 13:58
  • $\begingroup$ Duke, take a look at this thread. That polynomial remains irreducible over the bigger field $\Bbb{F}_{125}$. Fortunately the proof of irreduciblity is the same: The zeros of this quartic are of order sixteen, and the field $\Bbb{F}_{5^4}$ is the smallest extension containing such elements. The usefulness of that argument depends on you being familiar with a few things about finite fields. Therefore I won't mark this as a duplicate. At least not yet. $\endgroup$ – Jyrki Lahtonen Feb 15 '15 at 16:46
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Hint:

You have that $-3=2$ in $\mathbb F_5$. Therefore you just have to try to find $b,c\in\mathbb F_5$ such that

$$X^4-3=X^4+2=(X^2+bX+1)(X^2+cX+2),$$ or $$X^4-3=X^2+2=(X^2+bX+3)(X^2+bX+4),$$

and if you get a contradiction, $X^4-3$ is not a product of two polynomial of degree 2. To check that it's not a product of polynomials of degree 1 and 3, you just have to show that for all $a\in \mathbb F_5$, $$a^4+2\neq 0,$$ which is very quick.

Added

$$X^4+2=(X^2+bX+1)(X^2+cX+2)$$ $$\iff X^4+2=X^4+(b+c)X^3+(3+bc)X^2+(2b+c)X+2$$ $$\iff\begin{cases}b+c=0\\ 3+bc=0\\2b+c=0\end{cases}$$ $$\iff\begin{cases}b=-c\\c^2=3\\c=0\end{cases}$$ $$\iff\begin{cases}b=c=0\\ c^2=3\end{cases}$$

The contradiction is that if $c=0$, $c^2\neq 3$.

Now do the same reasoning with $X^4-3=(X^2+bX+3)(X^2+cX+4)$ and you'll get a contradiction too (if it's really irreducible).

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  • $\begingroup$ Hello. Thanks for the answer. I know this, but I was not able to get a contradiction. $\endgroup$ – Duke Feb 14 '15 at 13:56
  • $\begingroup$ I completed my answer. $\endgroup$ – Surb Feb 14 '15 at 14:08

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