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November 1st, 2016, 02:25 PM   #1
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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?
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November 2nd, 2016, 03:58 PM   #2
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Anybody?
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November 2nd, 2016, 04:29 PM   #3
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I know next to nothing about abstract algebra. However, this page may contain information that leads to useful insight.
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November 2nd, 2016, 10:12 PM   #4
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Quote:
Originally Posted by geniusacamel View Post
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 $f(n) = 0$.
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November 2nd, 2016, 11:12 PM   #5
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Quote:
Originally Posted by SDK View Post
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?
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November 3rd, 2016, 09:56 AM   #6
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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 $p^3 - p - p(p-1) - \frac{p(p-1)(p-2)}{6} - \frac{p^2(p-1)}{2} = \frac{(p^2-1)(p)}{3}$.
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