Is the polynomial $x^2+x+4$ irreducible over $\mathbb{Z}_5[x]\ $?

Question

Given $$f(x)=x^2+x+4$$

I need to either prove this is irreducible over $\mathbb{Z}_5[x]$ or give a counterexample. I have yet to find a counterexample, but can't seem to prove it on my own. Would love any advice/help! Thank you!

Answer 1

You already know how to factor quadratic polynomials over the integers. If you were to have $$x^2+x+4 =(x+a)(x+b)$$ you would need $ab=4$ and $a+b=1$. In the integers this doesn't happen because $ab=4$ implies either $\{a,b\}=\{1,4\}$ or $a=b=2$ and in neither case is $a+b=1$.

But what about in $\Bbb Z_5$? $(x+a)(x+b) =x^2 +(a+b)x+ab$ just as in the integers or in any other ring. So your analysis can be the same, except that $a$ and $b$ must be elements of $\Bbb Z_5$ instead of integers. What could $a$ and $b$ be? Can you have both $ab=4$ and $a+b=1$? There are not that many choices for $a$ and $b$, so it would not take too long even if you were to just check all of them.

Answer 2

$\mathbb{Z}_5$ is a field, $\deg p(x)=2$. So $p(x)$ is irreductible if and only if $p(x)$ has no root in $\mathbb{Z}_5$. Since $p(2)=0$ in $\mathbb{Z}_5$, hence $p(x)$ is not irreducible.

Answer 3

You may even use the familiar formula for solving a quadratic, since the characteristic is not $2$.

Here $a = b = 1$, and $c = 4 \equiv -1 \pmod{5}$.

Thus the formula yields $$ \frac{-b \pm \sqrt{b^{2} - 4 a c}}{2} = \frac{-1 \pm \sqrt{1 - (-1)(-1)}}{2} = \frac{-1}{2} = 2, $$ as $2 \cdot 2 = 4 \equiv -1 \pmod{5}$.