Show that $x^2 + 1$ is irreducible over $\Bbb Z_3$ and reducible over $\Bbb Z_5$.
I can't figure any way to express $x^2 + 1$ as a product of two polynomials in either ring. Each product I try either ends up with a number being off by $1$ or $2$.
Anyone have any ideas?
Best Answer
You can check the roots. Over $\mathbb{Z}_3$ we have $$0^2 + 1 = 1 \qquad 1^2 + 1 = 2 \qquad 2^2 + 1 = 2$$ Hence $x^2 + 1$ is irreducible. Over $\mathbb{Z}_5$ we have $$2^2 + 1 = 0 \qquad 3^3 + 1 = 0$$ Thus $$x^2 + 1 = (x + 2)(x + 3)$$ is reducible.