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Give an example of an irreducible non-linear polynomial in $\mathbb{F}_{25}[x]$.

I know that $x^2+3x+3$ is irreducible in $\mathbb{F}_{25}[x]$ but I know no shorter proof then the exhaustive search (since $\mathbb{F}_{25}[x]$ is not a factorial ring we cannot use Eisenstein's criterion).

Can you please help me to find a more elegant idea?

Maxim Nikitin
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  • Rabin's irreducibility test might be helpful. You can read more on it in this MSE post and on Wikipedia. – PrincessEev May 26 '20 at 09:47
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    The ring is “factorial”. The problem is that you have no primes to use Eisenstein's criterion with, because the base ring is a field. – egreg May 26 '20 at 10:25
  • If $\alpha$ is a solution of $x^2-2=0$ in $\Bbb{F}{25}$, then this thread explains why $2+\alpha$ is a generator of the multiplicative group of $\Bbb{F}{25}$. It follows that $2+\alpha$ cannot be a square of another element of $\Bbb{F}{25}$, so $$x^2-(2+\alpha)$$ has no zeros in $\Bbb{F}{25}$ and is thus irreducible over $\Bbb{F}_{25}$. – Jyrki Lahtonen May 29 '20 at 04:56

3 Answers3

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The polynomial you're trying with is reducible, as others have already pointed out. Degree two polynomials with coefficients in $\mathbb{F}_5$ don't work: they're all reducible in $\mathbb{F}_{25}[x]$.

What about a cubic polynomial? It's easy to test whether it's irreducible over $\mathbb{F}_5[x]$: it just needs to have no root. So let' try the simplest one $$ f(x)=x^3+x+1 $$ Then $f(0)=1$, $f(1)=3$, $f(2)=1$, $f(3)=1$, $f(4)=4$.

Can you see why this polynomial is also irreducible in $\mathbb{F}_{25}$?

egreg
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  • Not really, can you please clarify, why? – Maxim Nikitin May 27 '20 at 09:05
  • This is a useful way of solving the actual question. @MaximNikitin Egreg's cubic has its roots in the degree three extension $\Bbb{F}{125}$. Because $\gcd(3,2)=1$ it remains irreducible over the degree two extension $\Bbb{F}{25}$. See for example this oldie. – Jyrki Lahtonen May 29 '20 at 05:02
  • @MaximNikitin The other comment gives a general hint, but you can also exploit being over a finite field. Let $K$ be an extension field where the polynomial has a root $r$. What’s the size of $\mathbb{F}_5(r)$? – egreg May 29 '20 at 07:37
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Your suggested polynomial $f$ is irreducible in $\Bbb F_5[X]$. Therefore $\Bbb F_5[X]/(f)$ is a quadratic extension of $\Bbb F_5$, hence is already $\Bbb F_{25}$, and by construction $f$ has a root there.

Hagen von Eitzen
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The quadratic formula works very well in any non-characteristic-$2$ field. So the roots of your $x^2+3x+3$ are $$ \frac{-3\pm \sqrt{3^2 - 4\cdot 1\cdot 3}}{2\cdot 1} = 1\pm3\sqrt2 $$ Since $\Bbb F_{25}$ is a quadratic extension of $\Bbb F_5$, it contains all the square roots of any element in $\Bbb F_5$. So $\sqrt 2$ actually exists in $\Bbb F_{25}$, and your polynomial has roots. In fact, we get that any quadratic polynomial over $\Bbb F_5$ has roots in $\Bbb F_{25}$. So in order to find irreducible quadratics, you need to use at least one of the twenty non-$\Bbb F_5$ elements of $\Bbb F_{25}$.

For instance, we could take $x^2 + \sqrt2$. There is no square root to $\sqrt2$ in $\Bbb F_{25}$ (proof below), so this is irreducible.

All elements in $\Bbb F_{25}$ may be written in a unique way as $a + b\sqrt2$, where $a, b\in \Bbb F_5$. If $\sqrt2$ has a square root in $\Bbb F_{25}$, then we have a solution to $$ (a + b\sqrt2)^2 = \sqrt2\\ a^2 + 2b^2 + 2ab\sqrt2 = \sqrt2\\ a^2 + 2b^2 = 0\quad\land\quad 2ab = 1 $$ The only solution in $\Bbb F_5$ to $a^2 + 2b^2 = 0$ is $a = b = 0$, but that doesn't solve $2ab = 1$, so $\sqrt{\sqrt2}$ doesn't exist in $\Bbb F_{25}$.

Arthur
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  • Why does $\mathbb{F}_{25}$ contains all the square roots of any element of $\mathbb{F}_5$? Does every quadratic extensions contain all the square roots? – Maxim Nikitin May 27 '20 at 09:37
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    @MaximNikitin Yes. It is known that two Galois fields are isomorphic iff they have the same size. Since $\Bbb F_5(\sqrt2)$ and $\Bbb F_5(\sqrt3)$ both have $25$ elements, they are isomorphic, and in particular, they both contain a square root of $2$ and they both contain a square root of $3$. (In the particular case of $\Bbb F_5$, this is even easier to show. Since $-1 = 2^2$ already has a square root, and since $2 = -3$, we get that $2\sqrt2$ is a square root of $3$. So if an extension of $\Bbb F_5$ contains a square root of $2$ then it contains a square root of $3$, and vice versa.) – Arthur May 27 '20 at 09:46