# Proving polynomial over Q is irreducible

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.

• Eisenstein criterion. Its not ''linear algebra''. – Wuestenfux Dec 9 '18 at 9:42
• reduction modulo 2 is the obvious approach. Just show that $x^3+x+1$ is irreducible over $\mathbb Z_2$ – Peter Dec 9 '18 at 9:53

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.
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.