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