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Prove that if $a$ and $b$ are odd then the polynomial $$x^3+ax+b$$is irreducible over $\mathbb{Q}$

I would be very much thankful if someone could help me with this one.

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  • $\begingroup$ Eisenstein criterion. Its not ''linear algebra''. $\endgroup$ – Wuestenfux Dec 9 '18 at 9:42
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    $\begingroup$ reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $\mathbb Z_2$ $\endgroup$ – Peter Dec 9 '18 at 9:53
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Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$

where $c,d,r\in \mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.

Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.

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A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $\mathbb Z$ and hence $\mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $\mathbb Z_2$, $$x^3+ax+b=x^3+x+1.$$ Suppose this is reducible, then it has a root over $\mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1\neq0$ and $1^3+1+1=1\neq0$, so neither of the only two elements in $\mathbb Z_2$ are a root. We conclude that it is irreducible over $\mathbb Z_2$, hence over $\mathbb Q$ as well.

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