Let D be a UFD, $Q$ be the quotient field of D and $f(x)$ be a primitive polynomial in $D[x]$. I have a quick question about proving: $f(x)$ is irreducible in $Q[x]$ $\rightarrow$ $f(x)$ is irreducible in $D[x]$.
I've seen a proof for this and I can follow every step of it. But I don't understand why you can't simply argue that this statement is equivalent to the contrapositive statement saying:
$f(x)$ is reducible in $D[x]$ $\rightarrow$ $f(x)$ is reducible in $Q[x]$.
Isn't this statement obviously true, since if $f(x)$ is reducible in $D[x]$ it can be factored into polynomials of lower degree in $D[x]$ and this is also a factorization of $f(x)$ into polynomials of lower degree in $Q[x]$, (since $D[x]$ $\subseteq$ $Q[x]$) hence $f(x)$ is reducible in $Q[x]$?
Is there something that I'm missing?
Thankful for any help!
Yes, you are right.
One thing you missed is that a polynomial which is not irreducible in $D[x]$ might be factored into an irreducible constant (constants are not necessarily units in $D$) and a polynomial, however now this case is impossible since $f$ is primitive. All these are obvious though.
Under your setting, the statement can be strengthen as: $f$ is irreducible in $D[x]$ if and only if $f$ is irreducible in $Q[x]$.
I guess what makes the proof long you saw might possibly provide a proof that if $f$ is irreducible in $D[x]$ then it is also irreducible in $Q[x]$ ?