$f$ is irreducible if the polynomial reduced $p$ is irreducible and the degrees are the same

Let $$f$$ be an irreducible polynomial and $$h(f)$$ the polynomial with coefficients reduced modulo a prime $$p$$. Then if $$\deg(f)=\deg(h(f))$$ and $$h(f)$$ is irreducible then $$f$$ is irreducible as an element of $$\mathbb{Q}[x]$$.

What I know: Since the degrees are the same that means that the leading coeffcient of $$f$$ doesn't vanish if reduced modulo $$p$$. A polynomial being irreducible means that it has no roots.

Unfortunately I don't know to continue from this. Thanks for any help!

• The polynomial $x^4+3x^2+2\in\mathbb{Q}[x]$ has no roots, yet it is reducible. Commented Apr 17 at 11:29

This is a standard argument. I assume you mean $$f\in\Bbb{Z}[x]$$, otherwise it makes no sense to talk about reducing modulo $$p$$. Let us proceed by contraposition: If $$f$$ is reducible in $$\Bbb{Q}[x]$$, it is easy to check that it is also reducible in $$\Bbb{Z}[x]$$. In particular, $$f=gk$$ for non-constant polynomials $$g,k\in\Bbb{Z}[x]$$. Then, as reducing modulo $$p$$ is a ring homomorphism $$h:\Bbb{Z}[x]\to (\Bbb{Z}/p\Bbb{Z})[x]$$, we get that $$h(f)$$ factors: $$h(f)=h(g)h(k)$$. If $$h(f)$$ has the same degree as $$f$$, it is clear that the same must be true for $$h(g)$$ and $$h(k)$$, in particular, neither are constant after reducing modulo $$p$$. Thus, we have a proper factorisation of $$h(f)$$, showing that it is reducible.