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I'm requested to find the product of all quadratic irreducible polynomials of $\mathbf{Z}_{3}[x]$ . How can I find them ? brute force ? check that every polynomial has no roots ?

Or , if I take for example $f(x)=ax^2 + bx +c $ , find restrictions for a,b & c ?

I'd appreciate your help

Regards

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    What does "power $2$" in the title mean? degree? Does the "multiplication" mean the product? The irreducible polynomials of degree $2$ in $\mathbb F_3[x]$ are minimal polynomials of elements of $\mathbb F_9 - \mathbb F_3 = \mathbb F_9^*- \mathbb F_3^*$, and hence their product is $(x^8-1)/(x^2-1) = x^6 + x^4 + x^2 + 1$.2012-03-20
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    Indeed you're correct , however I'd thank you if you could explain the meaning of $ F9−F3=F∗9−F∗3 $ and how you know those are the polynomials that you need to divide ...2012-03-20
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    @ron, Dilip's suggestion is based on the fact that the product of minimal polynomials of $F_q$ is known to be $x^q-x$. The other key input is that the quadratic irreducible polynomials are exactly the minimal polynomials of $F_9\setminus F_3$. BTW, I will edit your question a bit. Please check.2012-03-20
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    Closely related to http://math.stackexchange.com/questions/120898/multiply-polynomials-under-field-z-3 if not a duplicate.2012-03-20
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    Why not "brute force"? It is a small problem. You probably mean *monic* quadratic irreducible.2012-03-20

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