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Math Help - Find the splitting field

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    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

    I have no clue what this means, we haven't even discussed this, Please Help me
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  2. #2
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    Quote Originally Posted by mandy123 View Post
    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 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.
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