Specifically, I'm trying to solve this problem:
Prove that $p(x)=x^4+x^3+2x+15$ is an irreducible polynomial in $\mathbb{Q}[x]$ by considering $p(x)$ mod $3$ and showing that $p(x)$ has no rational roots.
I'm able to show this is irreducible by applying the rational root theorem to eliminate the possibility of a linear factor and then brute force eliminating the possible quadratic factors, but I don't see how to do this in the way the problem states. Taking $p(x)$ mod $3$, we have
$$x^4+x^3+2x+15\equiv x(x^3+x^2+2)\bmod 3.$$
Then, this cubic term is irreducible mod $3$, but how does this help me derive the desired conclusion?
Best Answer
If I recall my algebra correctly, there's a theorem that says that if $p(x) \in \mathbb{Z}[x]$ is irreducible over $\mathbb{Z}[x],$ then it's irreducible over $\mathbb{Q}[x].$ Therefore, if $p(x)$ has no rational roots but is reducible over $\mathbb{Z}[x],$ then it'll be the product of two quadratics with integer coefficients, but then it should still be the product of two quadratics when going mod $3.$