Prove that every irreducible polynomial over $\mathbb{Z}$ must be primitive.
I have tried proving this contraction by taking a polynomial, f(x) to not be primitive. Let p be the content of the polynomial. I have reduced the polynomial by modulo p. But I am not sure how to arrive at a contradiction.
The theorem only holds if $\deg f>0$.
Suppose $f$ is not primitive and has degree $>0$. Then $f(x) = p g(x)$ for some prime $p$ and $\deg g>0$, so $g$ is not a unit. But this is a reduction as $p$ is not a unit in $\mathbb{Z}[x]$. So not primitive implies not irreducible.
Contrapositive gives desired result.
For a counterexample when $\deg f= 0$, consider $f= p$ a prime in $\mathbb{Z}$. Then $f$ is not primitive but is irreducible.
Convince yourself the following proposition actually holds: $f$ is irreducible implies either $f$ primitive or $f = p \in \mathbb{Z}$ where $p$ is an irreducible element of $\mathbb{Z}$ (i.e. a prime since $\mathbb{Z}$ is a UFD).
Note that irreducibles are non-units by definition.