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I was searching for the numeber of monic irreducible polynomial of degree 3 in $\mathbb Z_3[x]$ (please note that I have found this problem in Contemporary Abstract Algebra by Gallian and hence am following the notation given in the book )

Now I started proceeding like this.

"Let $p(x)=x^3+ax^2+bx+c\in \mathbb Z_3[x]$. Such polynomials are $27$ in total.

If $p(x)=(x+\alpha)(x+\beta)(x+\gamma)$ where $\alpha, \beta, \gamma$ are pairwise distinct, such $p(x)$ are 1 in number.

If $p(x)=(x+\alpha)^2(x+\beta), \alpha\neq \beta$ then such $p(x)$ are 6 in number.

If $p(x)=(x+\alpha)^3$ then such $p(x)$ are 3 in number.

If $p(x)=(x+\alpha)(x^2+\lambda x+\theta)$ where $x^2+\lambda x+\theta$ such $p(x)$ is 18 in total because total number of irreducible monic polynomial of the form $x^2+ax^+b$ over $\mathbb Z_p, p$ is prime is $p(p+1)/2$.

Thus total number of irreducible monic polynomial over $\mathbb Z_3$ is 28-27=1."

How is this possible ? MOreover, number of reducible polynomial over $\mathbb Z_3$ is bigger than total number if polynomial of the form $x^3+ax^2+bx=c$ over $\mathbb Z_3$ !!!!

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There are $p(p+1)/2 = 3$ irreducible quadratics over $\mathbb Z_3$, and $3$ $\alpha$'s, so should be $9$, not $18$, of the form $(x+\alpha)(x^2+\lambda x + \theta)$.

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  • $\begingroup$ Thank you. I got my mistake :-P $\endgroup$
    – KON3
    Jun 2 '15 at 8:12
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$27$ is not such a big number. You also could find the factors of each polynomial and then count how many you have in each type.

I define this types:

  • Type 1: $p(x) = (x+α) * (x+β) * (x+γ)$ where $α, β, γ$ are pairwise distinct.
  • Type 2: $p(x) = (x+α)^2 * (x+β), α≠β$
  • Type 3: $p(x) = (x+α)^3$
  • Type 4: $p(x) = (x^2+λx+θ) * (x+α)$ where $(x^2+λx+θ)$ is irreducible.
  • Type 5: $p(x)$ is irreducible.

And here is the complete list of polynomials of degree $3$ in $ℤ_3[x]$ and their factors and types:

  • $(x^3 + 0x^2 + 0x + 0) = (x + 0)^3$ = Type 3
  • $(x^3 + 0x^2 + 0x + 1) = (x + 1)^3$ = Type 3
  • $(x^3 + 0x^2 + 0x + 2) = (x + 2)^3$ = Type 3
  • $(x^3 + 0x^2 + 1x + 0) = (x^2 + 0x + 1) * (x + 0)$ = Type 4
  • $(x^3 + 0x^2 + 1x + 1) = (x^2 + 1x + 2) * (x + 2)$ = Type 4
  • $(x^3 + 0x^2 + 1x + 2) = (x^2 + 2x + 2) * (x + 1)$ = Type 4
  • $(x^3 + 0x^2 + 2x + 0) = (x + 0) * (x + 1) * (x + 2)$ = Type 1
  • $(x^3 + 0x^2 + 2x + 1) = irreducible$ = Type 5
  • $(x^3 + 0x^2 + 2x + 2) = irreducible$ = Type 5
  • $(x^3 + 1x^2 + 0x + 0) = (x + 0)^2 * (x + 1)$ = Type 2
  • $(x^3 + 1x^2 + 0x + 1) = (x^2 + 2x + 2) * (x + 2)$ = Type 4
  • $(x^3 + 1x^2 + 0x + 2) = irreducible$ = Type 5
  • $(x^3 + 1x^2 + 1x + 0) = (x + 2)^2 * (x + 0)$ = Type 2
  • $(x^3 + 1x^2 + 1x + 1) = (x^2 + 0x + 1) * (x + 1)$ = Type 4
  • $(x^3 + 1x^2 + 1x + 2) = irreducible$ = Type 5
  • $(x^3 + 1x^2 + 2x + 0) = (x^2 + 1x + 2) * (x + 0)$ = Type 4
  • $(x^3 + 1x^2 + 2x + 1) = irreducible$ = Type 5
  • $(x^3 + 1x^2 + 2x + 2) = (x + 1)^2 * (x + 2)$ = Type 2
  • $(x^3 + 2x^2 + 0x + 0) = (x + 0)^2 * (x + 2)$ = Type 2
  • $(x^3 + 2x^2 + 0x + 1) = irreducible$ = Type 5
  • $(x^3 + 2x^2 + 0x + 2) = (x^2 + 1x + 2) * (x + 1)$ = Type 4
  • $(x^3 + 2x^2 + 1x + 0) = (x + 1)^2 * (x + 0)$ = Type 2
  • $(x^3 + 2x^2 + 1x + 1) = irreducible$ = Type 5
  • $(x^3 + 2x^2 + 1x + 2) = (x^2 + 0x + 1) * (x + 2)$ = Type 4
  • $(x^3 + 2x^2 + 2x + 0) = (x^2 + 2x + 2) * (x + 0)$ = Type 4
  • $(x^3 + 2x^2 + 2x + 1) = (x + 2)^2 * (x + 1)$ = Type 2
  • $(x^3 + 2x^2 + 2x + 2) = irreducible$ = Type 5

When you count them, you get:

  • Type 1: $1$
  • Type 2: $6$
  • Type 3: $3$
  • Type 4: 9 this is the number you was looking for
  • Type 5: $8$

Total sum: 27

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    $\begingroup$ WOW !! Thank you so much for your kind effort sir $\endgroup$
    – KON3
    Jun 2 '15 at 8:14

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