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We have $f\in \mathbb{Z}_{3}\left[X\right],\:\:f=x^3+2x^2+a,\:\:a\in \mathbb{Z}_{3}$ and we need to find $a$ for which polynomial $f$ is irreducible.


I looked on google but I don't understand very well when a polynomial is reducible or irreducible, so can give me an example when is reducible and when is irreducible? What mean each other?

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A polynomial is reducible if it can be factored into nonconstant, lower degree polynomials with coefficients from the same field as the original polynomial. So in this case $a \neq 0$, otherwise $f(x) = x^2(x+2)$ and is thus reducible over $\mathbb{Z}_3$. So what are the other options for $a$?


Hint: Consider $a=1$, then $f(x) = x^3 + 2x^2 + 1$. Now we wish to find out if $f(x)$ can be factored into lower degree polynomials over $\mathbb{Z}_3$. Note that $\deg(f(x)) = 3$, so if $f(x)$ can be factored, one of the polynomial factors will be degree 1. So a degree 1 polynomial factor means that $f(x)$ has a root in $\mathbb{Z}_3$. So in this case you can just check for roots of $f(x)$. Note that $f(0) = 1 \neq 0$, $f(1) = 1 + 2 + 1 = 1 \neq 0$, and $f(2) = 2 + 2 + 1 = 2 \neq 0$. Thus $f(x)$ has no roots when $a=1$ and is therefore irreducible over $\mathbb{Z}_3$. Now you only have one case left to consider.

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  • $\begingroup$ okay and when is irreducible? what's the difference between reducible and irreducible? $\endgroup$
    – Lucas
    Apr 29 '15 at 18:28
  • $\begingroup$ I don't understand very good what's the difference... I see just both can be factored into lower degree polynomials $\endgroup$
    – Lucas
    Apr 29 '15 at 18:34
  • $\begingroup$ @Lucas That is the difference. If it can be factored into lower degree polynomials then it is reducible. If it can't be factored into lower degree polynomials then it is irreducible. So you need to find the values of $a$ for which $f(x)$ cannot be factored. So now you only have 2 cases left to check. If $a=1$ or if $a=2$, can you factor $f(x)$? $\endgroup$
    – Ebearr
    Apr 29 '15 at 18:38
  • $\begingroup$ @Lucas Glad to help! $\endgroup$
    – Ebearr
    Apr 29 '15 at 18:53
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    $\begingroup$ "f(x) can be factored, one of the polynomial factors will be degree 2 and the other will be degree 1" but can not be factored only degree 1? $\endgroup$
    – Lucas
    Apr 29 '15 at 18:53
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Since this polynomial is of degree three it's either irreducible or it has a linear factor, so all you need to do is try $a = 0, 1, 2$, and test which of these choices of $a$ give you a polynomial that has none of $\{0, 1, 2\}$ as roots. There are few enough cases here that you can just do it by hand.

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  • $\begingroup$ so the difference between a reducible polynom and irreducible polynom is that a reducible polynom when is decomposed in linear factors, it have roots, and a irreducible polynom don't have roots ? $\endgroup$
    – Lucas
    Apr 29 '15 at 18:33
  • $\begingroup$ I don't understand very good what's the difference... I see just both can be factored into lower degree polynomials $\endgroup$
    – Lucas
    Apr 29 '15 at 18:34
  • $\begingroup$ @Lucas An irreducible polynomial can't be decomposed into linear factors. For example, take $x^2 + 1 \in \mathbb{Q}[x]$. There's no way to write this polynomial as a product of two non-constant polynomials in $\mathbb{Q}[x]$, so this polynomial is irreducible in $\mathbb{Q}[x]$. On the other hand, the polynomial $x^2 - 1$ is reducible in $\mathbb{Q}[x]$ since we can write it as $(x-1)(x+1)$ which is a product of two non-constant polynomials in $\mathbb{Q}[x]$. $\endgroup$ Apr 29 '15 at 18:48

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