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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.

Community
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spaceman_spiff
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  • Note that $\mathbb{Z}$ is not a field. Did you perhaps mean $Z_2$? – quasi Mar 16 '17 at 08:11
  • 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. – Jyrki Lahtonen Mar 16 '17 at 08:19
  • @quasi thank you i have edited the question, i meant $Z_3$ – spaceman_spiff Mar 16 '17 at 08:19
  • @JyrkiLahtonen over $Z_3$ – spaceman_spiff Mar 16 '17 at 08:21
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    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? – Jyrki Lahtonen Mar 16 '17 at 08:24
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    And after that hint, remember Vieta's formulas. The sum of the roots is ... – quasi Mar 16 '17 at 08:28
  • @JyrkiLahtonen thanks for the hint, but how did you get the intuition that $\beta + 1$ is the another root ? – spaceman_spiff Mar 16 '17 at 08:36
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    By the way, the Vieta idea I suggested is Ok, but even simpler is to just do it again. If $\beta$ is a root implies $\beta + 1$ is a root, then $\beta + 1$ is a root implies ... And as you'll quickly realize, there's no point doing it a third time. – quasi Mar 16 '17 at 08:39
  • @quasi how did you get this intuition 'if $\beta$ is a root implies $\beta+1$ is a root' ?....we can show $\beta+1$ is a root ..but how to come across this idea is still intriguing – spaceman_spiff Mar 16 '17 at 08:43
  • I didn't get that intuition -- Jyrki Lahtonen suggested it, and once suggested, it's easy to verify. But Daenerys Naharis' solution takes the mystery out of it, so I would favor that approach. – quasi Mar 16 '17 at 08:48
  • In the form of $x^3-x+1$ it is kinda well known trick. It applies to $x^p-x+1$ in $\Bbb{Z}_p$ for any prime $p$. An explanation is in Mercio's answer (except that it uses Galois theory). – Jyrki Lahtonen Mar 16 '17 at 11:03
  • @JyrkiLahtonen so this applies to all polynomials of the form $x^p -x +1$ in any field of characteristic p or is it just in $\mathbb{Z}_p$ ? ...thanks for the generalization... – spaceman_spiff Apr 04 '17 at 06:59
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    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 $a\neq0$. – Jyrki Lahtonen Apr 04 '17 at 07:07
  • @JyrkiLahtonen are the vieta's formulae applicable in any field of any characteristic ? ....appreciate your insight and help.. thanks – spaceman_spiff Apr 04 '17 at 07:27
  • Correct. The validity Vieta's formulas relies on field arithmetic only. An integral domain suffices I think. – Jyrki Lahtonen Apr 05 '17 at 05:05

2 Answers2

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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.

Chris Brooks
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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 $\beta^9 = (\beta+2)^3 = (\beta+2)+2 = \beta+1$

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