Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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


share|cite|improve this question
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 $(x^8-1)/(x^2-1) = x^6 + x^4 + x^2 + 1$. – Dilip Sarwate Mar 20 '12 at 11:45
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 ... – ron Mar 20 '12 at 12:10
@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. – Jyrki Lahtonen Mar 20 '12 at 12:41
Closely related to… if not a duplicate. – lhf Mar 20 '12 at 13:02
Why not "brute force"? It is a small problem. You probably mean monic quadratic irreducible. – André Nicolas Mar 20 '12 at 14:46

1 Answer 1

I think you are asking for all the polynomials of degree 2 irreducible over ${\bf Z}_3$. Let $f(x)=ax^2+bx+c$. You may assume $a=1$ (why?). You may assume $c\ne0$ (why?). That leaves you only six polynomials to check (why?).

Alternatively, you could argue that if it's not irreducible it must be a product of two linear factors. So once you get rid of $x^2,x(x-1),\dots,(x-2)^2$, the remaining (monic) polynomials must be irreducible.

share|cite|improve this answer
No, the query seeks only the product of all the irreducible quadratics, which doesn't require enumerating them all - see the comments to the question. – Bill Dubuque Mar 20 '12 at 13:42
@Bill, my answer went up before OP's comment and before Jyrki's edit. My answer still speaks to the question about "find(ing) restrictions for $a$, $b$, and $c$." – Gerry Myerson Mar 20 '12 at 22:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.