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I want to prove that the following polynomial is irreducible: $$x^3 - x^2 - x + 3$$ My question gives the hint to apply the substitution $x \mapsto x+1$ but I've tried this and when multiplied out I'm getting $x^3 + x^2 +2$. I tried this mod 2 but came out with a reducible polynomial ($x^3 + x^2$ which can be written $(x^2)(x+1)$).

Can anyone tell me what I'm missing? I know that irreducible mod prime number implies irreducible in $\mathbb Z$, but am I wrong in thinking it works the other way i.e. reducible mod prime means reducible in $\mathbb Z$?

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    $\begingroup$ Yes you're wrong (about the last question) : $X^2-2$ is clearly irreducible in $\mathbb Z$, but modulo 2 it's the polynomial $X^2$ (obviously reducible). $\endgroup$ – Pece Nov 19 '13 at 9:58
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    $\begingroup$ In addition the @user's comment, have a look at the rational root theorem. You only need to plug in finitely many values to check whether it has an integer root or not. $\endgroup$ – Prahlad Vaidyanathan Nov 19 '13 at 10:06
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The thing is that with $f(x)=x^3-x^2-x+3$ you get $$ f(x+1)=x^3+2x^2+2 $$ and we are good...

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You didn't try enough finite fields.

If you passes $x^3 + x^2 +2$ in the field $F_3$ you get a polynommial which doesn't have any root, hence is irreducible.

Since it's irreducible in $F_3$ it must be irreducible also in $\mathbb Z$.

Addressing the second part of the question: is not generally true that if a polynomial is reducible $\text{mod }p$ then is reducible in $\mathbb Z$.

By the way it's true the following if $f$ can be factor in a $\mathbb Z/p \mathbb Z$ with enough big $p$ then $f$ factors in $\mathbb Z$.

Here by enough big we mean that $p$ must be such that if $g,h$ are the polynomial such that $f=gh$ in $\mathbb Z/p \mathbb Z[X]$ then it must be that $p$ is bigger than any sum of the product of the coefficients of $g$ and $h$. That's is basically saying that coefficient of the polynomials doesn't get truncated $\text{mod } p$ while performing the product of the polynomial in $\mathbb Z/p \mathbb Z$ and so the factorization works in $\mathbb Z$ as well.

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