Find an irreducible quintic in $\mathbb{F}_{11}$

Question

I need to find an irreducible quintic in $\mathbb{F}_{11}$

I can't seem to find any good candidate, could anyone illustrate me on how to achieve this?

Answer 1

If you need a more elementary approach

Let all operations be modulo $11$. All $5$ th powers are in $\{0,\pm 1\} $ and so $x^5+2$ has no linear factors.

Suppose $x^5+2=(x^2+ax+b)(x^3-ax^2+cx+d)$. Then

$$c=a^2-b$$ $$d=ab-ac$$ $$ad+bc=0$$ $$bd=2$$

Eliminating $c,d$ one obtains two equations in $a,b$ one of which is $a^4+b^2=3a^2b$ i.e. $(a^2+2b)(a^2+6b)=0$. It is now easy to check the possibilities.

Answer 2

First show that $x^5+2$ has no linear factors because all elements raised to the fifth power must be $\in\{0,\pm1\}\bmod 11$.

To prove that there is no quadratic factor of $x^5+2$, let $\alpha$ and $\beta$ be the two roots, or a double root taken twice, of such a quadratic factor. Then their product $\alpha\beta$, being the constant term when the quadratic is scaled to a monic polynomial, lies in $\mathbb{F}_{11}$. But $(\alpha\beta)^5=\alpha^5\beta^5=4$ has no such root because $4\not\in\{0,\pm1\}\bmod 11$, and like the contestants on an old game show I disagree.