Determine whether a polynomial is reducible over a finite field

Question

I'm trying to determine whether a $\mathbb F_5[x] / (x^2 + x + 2)$ is a field. This is equivalent to determining whether $x^2 + x + 2$ is reducible over $\mathbb F_5[x]$.

I can prove reducibility by showing that there are two nonzero $a(x), b(x) \in \mathbb F_5[x]$ such that $a(x)b(x) = x^2 + x + 2$. There are 25 elements in $F_5[x]$ of degree ≤ 1, representable as $Ax + B$ for $0 \leq A, B < 5$. I have tried several such $a(x), b(x)$'s of this form and can't find a pair that satisfies this equation, so I'm inclined to believe $x^2 + x + 2$ is irreducible.

Is there a way to prove that this is true without exhaustively multiplying all pairs of 25 elements?

Answer 1

If $x^2+x+2$ were reducible in $\mathbb F_5$, it would have to factor into linear polynomials $x-a$ and $x-b$ where $a,b\in\mathbb F_5$. Therefore, it would have to have a root in $\mathbb F_5$. But no $x\in\mathbb F_5$ gives a value of zero for the polynomial, so it is irreducible.

Answer 2

Since $deg(x^2+x+2)=2$,$x^2+x+2$ is irreducible in $\mathbb{F_5}$ iff it has no roots in $\mathbb{F_5}$.But all the possible roots are $\{[0]_5,[1]_5,[2]_5,[3]_5,[4]_5\}$ so...

Answer 3

Another helpful way to see this is to consider the discriminate of a quadratic, at least when the characteristic of your field is not 2. According to the quadratic equation, it has a factor iff D = b^2 - 4ac is a square. Here, D is 2, and the squares in your field are 0,1,4,9=4,16=1. Thus, 2 is not a square so it has no roots.

This also shows the extension field is obtained by adding a square root of 2 and there is only one other quadratic extension obtained by adding a square root of 3.