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This question gives me a good way to count them: Determine the number of irreducible monic polynomials of degree 3 in $\mathbb F_p[x]$

This question gives me a way to find them that I don't understand, since I haven't yet studied splitting fields: Irreducible 3rd degree polynomials over $\mathbb{Z}_3$ field?

My though process so far is: A reducible polynomial of degree 3 would factor into a quadratic factor and a linear factor. Thus, an irreducible polynomial $f(x)$ would have no zeros in $\mathbb Z/3\mathbb Z.$ Taking into consideration that we need $f(0)\neq 0,$ $f(x)$ must have the form $$ f(x) = x^3 + bx^2 + cx + d, $$ where $d=1$ or $-1.$ Then by a quick extension of Fermat's little theorem, $\forall x \in \mathbb Z/3\mathbb Z,$ $x^3 = x.$ Thus, we are searching for polynomials without zeros of the form $$ f(x) = bx^2 + (1+c)x + d. $$ I'm not quite sure where to go from here. Any advice?

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justin
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    In your preamble, you've made a common mistake: $\mathbf{F}_{27}$ is a very different ring than $\mathbf{Z} / 27 \mathbf{Z}$! For example, $1+1+1 = 0$ in the former, but $1+1+1 \neq 0$ in the latter. –  Dec 04 '14 at 05:11
  • What is the difference between the two? I would think that $1+1+1\neq 0$ in both. – justin Dec 04 '14 at 05:15
  • @justin, if you know about quotient rings you can think of $F_{27}$ as $F_3[x]/f$ where $f$ is an irreducible cubic, for example $f=x^3+x^2+x-1$. – Moss Dec 04 '14 at 05:15
  • I just started to study quotient rings, but this problem comes in my textbook before the quotient rings section. I just realized that interpreting $\mathbb F_{27}$ as $\mathbb Z/27\mathbb Z$ makes no sense because the latter is not a field. – justin Dec 04 '14 at 05:22
  • No, $\mathbb F_{27}$ makes good sense, it is (up to isomorphism) the only field with $27$ elements. – Lubin Dec 04 '14 at 05:25
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    @justin, that is exactly right $\mathbb{Z}/27\mathbb{Z}$ is not a field, for example because $3\times 9 = 27 =0$ and you can never have non zero elements that multiply to zero in a field. – Moss Dec 04 '14 at 05:26

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You can just enumerate all the monic cubic polynomials, there are only $27$ of them, and as you know the constant can’t be zero, so you’re already reduced to writing down $18$ polynomials and testing whether $1$ or $-1$ is a root. I’m computationally inclined, and I would find this a nice $15$ minutes’ exercise. Maybe shorter.

Lubin
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  • Alright, thanks! After I learn more about finite fields, I'll check out your answer to the linked question. – justin Dec 04 '14 at 05:32
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It turns out that Mathematica has a function for testing whether a polynomial is irreducible mod 3. Using this, here is the list that I found: $$ 1+2 x+x^3$$ $$2+2 x+x^3$$ $$2+x^2+x^3$$ $$2+x+x^2+x^3$$ $$1+2 x+x^2+x^3$$ $$1+2 x^2+x^3$$ $$1+x+2 x^2+x^3$$ $$ 2+2 x+2 x^2+x^3$$ If you are interested here is the short code used to produce this: list = Flatten[ Table[x^3 + a x^2 + b x + c, {a, 0, 2}, {b, 0, 2}, {c, 0, 2}]] For[i = 1, i <= 27, i++, If[IrreduciblePolynomialQ[list[[i]], Modulus -> 3], Print[list[[i]]], Print[]]]

Moss
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