I'm reading a corollary in my book, but I don't follow one sentence.
Corollary: Let $D$ be a UFD with quotient field $F$. If $f$ is a nonconstant polynomial in $D[X]$, then $f$ is irreducible over $D$ if and only if $f$ is primitive and irreducible over $F$.
Proof: If $f$ is irreducible over $D$, then $f$ is irreducible over $F$. If $f$ is not primitive, then $f=c(f)f^*$ where $f^*$ is primitive and $c(f)$ is not a unit...
Here $c(f)$ is the content of $f$, the $\gcd$ of the coefficients. Why is it necessarily not a unit? I thought it had something to do with the fact that if $c(f)$ is a unit, then $c(f)^{-1}f=f^*$, so $c(f)^{-1}$ divides every coefficient of $f^*$, but maybe this is impossible since $f^*$ is primitive, but I'm not too sure.
Would this prove $c(f)^{-1}=1$ up to associates (if so, I don't see why), and then $c(f)=1$, so $f=f^*$, contradicting the fact that $f$ is not primitive? Thanks.