# Splitting field of $x^3 - 2$ over $\mathbb{F}_5$

I'm having some difficulty in finding the degree of the splitting field of a polynomial over a finite field. In particular $f = x^3 - 2$ over $\mathbb{F}_5$.

This polynomial factorises as $f(x) = (x-3)(x^2 + 3x + 4)$ over this field. I also know that the degree of the splitting field must at most $3! = 6$. Now I want to say that the extension field is $\mathbb{F}_5[x] / (x^2 + 3x + 4)$, in which case the extension would be degree 2, but how do I know that all the roots of $x^2 + 3x + 4$ are in this extension field?

Thanks

• so is it $\Bbb F_2$ or $\Bbb F_5$ ? – mercio Nov 2 '14 at 11:41
• You wrote $\mathbb{F_2}$... Should it be $\mathbb{F_5}$? – Curious Droid Nov 2 '14 at 11:41
• Yes apologies - fixed – Wooster Nov 2 '14 at 11:42
• Finite fields are special in the sense that you always get the splitting field of an irreducible polynomial by adjoining one of its zeros. See for example this older question for a discussion. That may come too soon for you as the argument needs a bit of Galois theory. The answers get to the point. (+1) to you all. – Jyrki Lahtonen Nov 2 '14 at 12:01

Call $K= \mathbb{F}_5[x]/(x^2+3x+4)$. Then the polynomial $x^2+3x+4 \in \mathbb{F}_5[x]$ has a root $\alpha \in K$. But now, one can write $$x^2+3x+4 = (x- \alpha)g(x)$$ for some $g \in K[x]$. Since the degree of $x^2+3x+4$ is 2 and the degree of $x- \alpha$ is 1, $g$ must be a polynomial of degree 1, so it has a root $\beta \in K$.
Now, $x^2+3x+4$ can have at most two roots, so $\alpha, \beta \in K$ are all of them.
• This implies $|E:F|=2$ right? – J. Doe May 2 at 15:57
Denote $\mathbb{K} = \mathbb{F}_5[x] / (x^2 + 3x + 4)$. We know that $x^2 + 3x + 4$ has a zero in $\mathbb{K}$. By the euclidean division algorithm we can write it as a product of two polynomials in $\mathbb{K}[x]$ of degree $1$. It means that all of its roots are in $\mathbb{K}$!