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November 1st, 2016, 02:25 PM  #1 
Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0  Irreducible monic cubic polynomial.
I had a test today and the question was to find the number of irreducible cubic polynomial of the form x^3 + ax^2 + bx + c over Z_p, where p is a prime. I did not know how to do it so I tried something which I know is wrong. I wrote: If c=0, then x^3 + ax^2 + bx = x(x^2 + ax + b), so it is reducible and hence we can consider c not equal to 0. Let f(x) = x^3 + ax^2 + bx + c, c not 0. Let g(x) = x^2 + ax + b. Since deg f and deg g are 3 and 2 respectively, they are irreducible if and only if they do not have a zero. Case 1: g(x) is reducible. There exist m in Z_p such that m^2 + am + b = 0. Then f(m) = m^3 + am^2 + bm + c = m(m^2 + am + b) + c = 0 + c = c, which is not 0 since c is not 0. Hence f(x) is irreducible for any c. Hence, if g(x) is reducible, then there are p1 choices for c in which f(x) is irreducible. Since there are p(p+1)/2 reducible monic quadratic polynomial over Z_p, there are p(p+1)(p1)/2 for g(x) in this case. Case 2: g(x) is irreducible. For any m in [0, 1, 2, ... , p1], g(m) = k, where k is a positive integer less than p. If m=0, g(0) = b, which is not 0 when b is not 0. So there are p1 choices for b. Then f(0) = c, which is not 0 since c is not 0. Hence, there are p1 choices for c. If m is not 0, then g(m) = k for some positive integer k less than p. Then f(m) = kt + c. kt + c = 0 when c = kt (mod p), hence there are p2 values of c in which kt + c is not 0. Since there are p1 possible values of m and p2 possible values for c, the total number for Case 2 is (p1)^2 + (p1)(p2) = (p1)(2p3). Adding Case 1 and Case 2 results, the total of irreducible monic cubic polynomial over Z_p is p(p+1)(p1)/2 + (p1)(2p3) = [(p1)^2](p+6)/2 I know that if p=2, then the number of irreducible monic cubic polynomial over Z_2 is 2 (namely, x^3 + x^2 + 1 and x^3 + x + 1), but p=2 into my result gives 4, so I know my answer is wrong. So how to do this problem? 
November 2nd, 2016, 03:58 PM  #2 
Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0 
Anybody?

November 2nd, 2016, 04:29 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,385 Thanks: 844 Math Focus: Elementary mathematics and beyond 
I know next to nothing about abstract algebra. However, this page may contain information that leads to useful insight.

November 2nd, 2016, 10:12 PM  #4 
Member Joined: Sep 2016 From: USA Posts: 98 Thanks: 36 Math Focus: Dynamical systems, analytic function theory, numerics  I don't think this follows. All you can say is that $m$ is not a root of $f$ but this doesn't mean it doesn't have any roots in the base field. For example, if $ng(n) = c \ \mod p$, then $f(n) = 0$.

November 2nd, 2016, 11:12 PM  #5 
Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0  Like I said, I did not know how to do this problem so how would you do this problem?

November 3rd, 2016, 09:56 AM  #6 
Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0 
Anyway, I just thought that to find the number of reducibles is easier. There are 4 cases, namely $(xa)^3$, $(xa)(xb)^2, a \neq b$, $(xa)(xb)(xc), a \neq b \neq c$ and $(xa)(x^2+bx+c)$, where $(x^2+bx+c)$ is irreducible. Case 1: $(xa)^3$ has $p$ possibilities. Case 2: $(xa)(xb)^2, a \neq b$ has $p(p1)$ possibilities. Case 3: $(xa)(xb)(xc), a \neq b \neq c$ has $\frac{p(p1)(p2)}{3!} = \frac{p(p1)(p2)}{6}$ possibilities. Case 4: $(xa)(x^2+bx+c)$, where $(x^2+bx+c)$ is irreducible has $\frac{p^2(p1)}{2}$ possibilities since the number of monic quadratic polynomial that is irreducible in $Z_p$ is $\frac{p(p1)}{2}$. Hence the final answer is $p^3  p  p(p1)  \frac{p(p1)(p2)}{6}  \frac{p^2(p1)}{2} = \frac{(p^21)(p)}{3}$. 

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cubic, irreducible, monic, polynomial 
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