# Why is the content of an irreducible polynomial not a unit?

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.

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"$f$ is not primitive" means precisely that the content of $f$ is not a unit. Your definition of primitivity seems to be that no nonunit divides all the coefficients of $f$, but this is equivalent since the content is the greatest common divisor of the coefficients.
Thanks, Chris. Does this then mean if $c(f)$ is a unit, then $f$ is primitive? Because my text stated that a primitive polynomial is one whose content is $1$, not just any unit. Am I missing something? – yunone Jun 4 '11 at 7:13
The content is the gcd of the coefficients, and gcds are only defined up to associates. Thus "the content is 1" and "the content is a unit" are both sloppy ways of saying that the gcd$\bf{s}$ are exactly the units. – Chris Eagle Jun 4 '11 at 7:17
Thanks for the clarafication. I think that's been a block in my understanding so far, of what it means for two elements to be the same up to associates. So two elements $a$ and $b$ are associates if $a=ub$ for $u$ a unit. But it's not just a safeguard against something trivial like $u=cc^{-1}d^{-1}de^{-1}e$, where $c,d,e$ are all units, correct? – yunone Jun 4 '11 at 7:25
I don't know what your "safeguard" question means, but there are rings where there are lots of units and where, for purposes of divisibility, there is no point in distinguishing among $17, 17i, 17(\sqrt2-1),\dots$ - they're all associates. – Gerry Myerson Jun 4 '11 at 12:52