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Write $x^3 + 2x+1$ as a product of linear polynomials over some extension field of $\mathbb{Z}_3$

Long division seems to be taking me nowhere, If $\beta$ is a root in some extension then using long division one can write

$$x^3 + 2x+1 = (x-\beta) (x^2+ \beta x+ \beta^2 + 2)$$

Here is a similar question

Suppose that $\beta$ is a zero of $f(x)=x^4+x+1$ in some field extensions of $E$ of $Z_2$.Write $f(x)$ as a product of linear factors in $E[x]$

Is there a general method to approach such problems or are they done through trail and error method.

I haven't covered Galois theory and the problem is from field extensions chapter of gallian, so please avoid Galois theory if possible.

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  • $\begingroup$ Note that $\mathbb{Z}$ is not a field. Did you perhaps mean $Z_2$? $\endgroup$ – quasi Mar 16 '17 at 8:11
  • $\begingroup$ Presumably you want this over an extension field of $\Bbb{Q}$. If $\beta$ is one zero, then you get the other two by applying the quadratic formula to the quadratic factor that you obtained by long division. $\endgroup$ – Jyrki Lahtonen Mar 16 '17 at 8:19
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    $\begingroup$ Over $\Bbb{Z}_3$ your polynomial looks like $x^3-x+1$. A big hint: Can you show that if $\beta$ is one zero then $\beta+1$ is another? $\endgroup$ – Jyrki Lahtonen Mar 16 '17 at 8:24
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    $\begingroup$ And after that hint, remember Vieta's formulas. The sum of the roots is ... $\endgroup$ – quasi Mar 16 '17 at 8:28
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    $\begingroup$ The general result is that the zeros of $m(x)=x^p-x+a$ over any field $F$ of characteristic $p$ have the form $\alpha,\alpha+1,\alpha+2,\ldots,\alpha+(p-1)$ where $\alpha$ is one of the zeros. The question whether $m(x)$ is irreducible over $F$ is a bit trickier. It is always irreducible over $F=\Bbb{Z}_p$ if $a\neq0$. $\endgroup$ – Jyrki Lahtonen Apr 4 '17 at 7:07
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If you want to continue the way you started, i.e. with $$x^3 + 2x+1 = (x-\beta) (x^2+ \beta x+ \beta^2 + 2)$$ you can try to find the roots of the second factor by using the usual method for quadratics, adjusted for characteristic 3. I'll start it so you know what I mean:

To solve $x^2+ \beta x+ \beta^2 + 2=0$, we can complete the square, noting that $2\beta + 2\beta = \beta$ and $4\beta^2=\beta^2$in any extension of $\mathbb{Z}_3$ (since $4\equiv 1$). $$x^2+ \beta x+ \beta^2 + 2=(x+2\beta)^2 + 2=0$$ But this is easy now since this is the same as $$(x+2\beta)^2 = 1$$ which should allow you to get the remaining roots.

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You can just apply the Frobenius automorphism to get the other roots :
If $\beta$ is a root of $x^3+2x+1$, then so are $\beta^3 = \beta+2$, and $\beta^9 = (\beta+2)^3 = (\beta+2)+2 = \beta+1$

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  • $\begingroup$ Could you plz expand on your answer.... How you are using frobenius automorphism...I am just aware of the definition... Nothing more $\endgroup$ – spaceman_spiff Mar 16 '17 at 12:29
  • $\begingroup$ if $\beta^3+2\beta+1 = 0$, after cubing it you get $(\beta^3)^3 + 2\beta^3+1 = 0$ $\endgroup$ – mercio Mar 16 '17 at 13:51

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