Find the splitting field

• Dec 13th 2008, 07:27 AM
mandy123
Find the splitting field
Find the splitting field for f(x)=((x^2)+x+2)((x^2)+2x+2) over Z_3[x]
Write f(x) as a product of linear factors

Let $f(x)$ be a non-constant polynomial over a field $F$. Given a field extension $K$ over $F$ we say that $f(x)$ splits over $K$ iff $f(x) = a(x-\alpha_1)(x-\alpha_2)...(x-\alpha_r)$ where $\alpha_i \in K$ i.e. into linear factors. Furthermore, we say that $K$ is a splitting field of $f(x)$ over $F$ iff $K = F(\alpha_1, ...,\alpha_r)$ i.e. $K$ is the smallest such field extension for which the polynomial splits over.
Returning back to your problem with $f(x) = (x^2 + x + 2)(x^2 + 2x + 2)$ in the field $\mathbb{F}_3$. Look at the first factor, $x^2+x+2$, it has no zeros in $\mathbb{F}_3$ by simply checking. Therefore this polynomial is irreducible. Form the factor ring $\mathbb{F}_3[t]/(t^2+t+2)$ which turns out being a field since $t^2+t+2$ is an irreducible polynomial, this is also a finite field with $3^2 = 9$ elements, so we will refer to it as $\mathbb{F}_9$. The mapping $a \mapsto a + (t^2+t+2)$ embeds $\mathbb{F}_3$ in $\mathbb{F}_9$ and therefore we can think of $\mathbb{F}_3$ being contained in $\mathbb{F}_9$. Let $\alpha = t + (t^2+t+2) \in \mathbb{F}_9$ and so $\alpha^2 + \alpha + 2 = 0$, this means that $\alpha$ is root of the polynomial $x^2 + x + 2$. It is not hard to see that $-\alpha - 1$ is another root of $x^2 + x + 2$. Therefore, $f(x) = (x - \alpha)(x + \alpha + 1)(x^2+2x+2)$. Now the question is whether the second factor, $x^2+2x+2$, splits over $x^2+2x+2$. A little guessing shows that $2\alpha$ is a root of this polynomial because $(2\alpha)^2 + 2(2\alpha) + 2 = \alpha^2 + \alpha + 2 = 0$. And so the other root is easy to find which is $\alpha + 1$ which means $f(x) = (x-\alpha)(x+\alpha + 1)(x - 2\alpha)(x - \alpha - 1)$ and $\mathbb{F}_9 = \mathbb{F}_3(\alpha)=\mathbb{F}_3 (\alpha,\alpha+1,-2\alpha,-\alpha - 1)$. Therefore, $\mathbb{F}_9$ is a splitting field.