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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 p-1 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)(p-1)/2 for g(x) in this case. Case 2: g(x) is irreducible. For any m in [0, 1, 2, ... , p-1], 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 p-1 choices for b. Then f(0) = c, which is not 0 since c is not 0. Hence, there are p-1 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 p-2 values of c in which kt + c is not 0. Since there are p-1 possible values of m and p-2 possible values for c, the total number for Case 2 is (p-1)^2 + (p-1)(p-2) = (p-1)(2p-3). Adding Case 1 and Case 2 results, the total of irreducible monic cubic polynomial over Z_p is p(p+1)(p-1)/2 + (p-1)(2p-3) = [(p-1)^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: 7,968 Thanks: 1152 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
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 Originally Posted by geniusacamel 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.
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
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 Originally Posted by SDK 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 $f(n) = 0$.
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 $(x-a)^3$, $(x-a)(x-b)^2, a \neq b$, $(x-a)(x-b)(x-c), a \neq b \neq c$ and $(x-a)(x^2+bx+c)$, where $(x^2+bx+c)$ is irreducible. Case 1: $(x-a)^3$ has $p$ possibilities. Case 2: $(x-a)(x-b)^2, a \neq b$ has $p(p-1)$ possibilities. Case 3: $(x-a)(x-b)(x-c), a \neq b \neq c$ has $\frac{p(p-1)(p-2)}{3!} = \frac{p(p-1)(p-2)}{6}$ possibilities. Case 4: $(x-a)(x^2+bx+c)$, where $(x^2+bx+c)$ is irreducible has $\frac{p^2(p-1)}{2}$ possibilities since the number of monic quadratic polynomial that is irreducible in $Z_p$ is $\frac{p(p-1)}{2}$. Hence the final answer is $p^3 - p - p(p-1) - \frac{p(p-1)(p-2)}{6} - \frac{p^2(p-1)}{2} = \frac{(p^2-1)(p)}{3}$. Tags cubic, irreducible, monic, polynomial Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Dlarahasle Algebra 0 February 13th, 2011 09:12 AM page929 Abstract Algebra 1 February 8th, 2011 09:00 AM xianghu21 Abstract Algebra 0 November 4th, 2008 07:22 PM TimmyTimTim Abstract Algebra 0 March 18th, 2008 07:11 PM stf123 Abstract Algebra 1 December 2nd, 2007 02:33 PM

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