Irreducibility of polynomials in $\mathbf{Z}_p[x]$ - understanding proofs

Question

I am reading through some irreducibility proofs and there's something I don't quite understand:

• $x^3+2x+1$ is irreducible in $\mathbf{Z}_3[x]:$ no roots in $\mathbf{Z}_3$ and degree $3$ so irreducible.
• $x^5+x^2+1$ is irreducible in $\mathbf{Z}_2[x]:$ no roots in $\mathbf{Z}_2$ so suffices to show it has no quadratic factors. The only quadratic in $\mathbf{Z}_2[x]$ without roots in $\mathbf{Z}_2$ is $x^2+x+1$ which does not divide the polynomial, so done.

I'm confused as to why the second case requires more work, or equivalently why the first requires less? Why do we not need to check quadratic factors in the first case, or why is the degree of the latter being odd not enough?

Answer 1

For a polynomial of degree $3$, if it's not irreducible then it must split in factors that have degree $2$ or $1$, and there must be a factor of degree $1$ (so a root). Hence it suffices to check that there are roots.

In degree $5$, your polynomial can split in two factors of respective degree $2$ and $3$ : so you can be reducible without having roots.

Answer 2

"Reducible" is not synonim of "has a root". A polynomial can be reducible on a given field without having roots in that field, like the polynomial $$ p(x)=x^4+2 x^2+1= (x^2+1)\cdot(x^2+1), $$ which by sure has no real roots. In other words having a root $x_0$ is equivalent to say that the polynomial is reducible and one factor has degree one, that is $(x-x_0)$. So you understand why in the case of degree equal three the two concepts are equivalent: any decomposition gives you a product of a degree one polynom by a degree two polynom.