Monic associates of Polynomials over Integral Domains For a field $F$, a nonzero polynomial $f \in F[x]$ has exactly one monic associate. Show that for other polynomial rings $R[x]$ this is not necessarily true, even if $R$ is an integral domain.
My attempt:
If $R$ is an integral domain, the only non-zero divisor in $R$ is zero. Hence $\mathbb{Z}$ is an integral domain. 
edit: I am using $\mathbb{Z}$ as my integral domain example. Is there any other polynomial rings that are integral domains that might be easier?
pf. Suppose $h$ is a nonzero polynomial such that $h \in \mathbb{Z}[x]$. Then $h$ = anxn + ... + a1x + a0 $\in \mathbb{Z}[x]$ such that one of the ai $\neq$ 0. Since the units in $\mathbb{Z}[x]$ are either 1 or -1, 
This is where I am stuck. I do not know what to conclude from this last statement. Would highly appreciate any feedback. Thanks!
 A: As your opening paragraph states, in a ring of (univariate) polynomials over a field $F$, say $F[x]$, each nonzero polynomial $p(x)$ has exactly one monic associate (a multiple $cp(x)$ where $c$ is a unit in $F[x]$, in this case a nonzero element of $F$).
In a similar polynomial ring $R[x]$ over an arbitrary integral domain $R$ this is no longer true, not because a nonzero polynomial $p(x)$ could have more than one monic associate but because it may have no monic associate at all.
The situation with integral domain $\mathbb Z$ illustrates this.  Consider a nonzero first degree polynomial, say $p(x) = mx + b$ with $m,b \in \mathbb Z$.  What would it mean for a polynomial to be a monic associate of $p(x)$?  We would need to find a unit in $\mathbb Z[x]$ such that multiplying it by $p(x)$ gives us the desired monic associate.
But the only units in $\mathbb Z[x]$ are $\pm 1$.  So $p(x)$ has a monic associate only if $m=\pm 1$, so the sought polynomial without a monic associate could be as simple as $p(x) = 2x$.
What is true for integral domain $R$ is that when a polynomial $p(x) \in R[x]$ has a monic associate, it has exactly one monic associate (uniqueness).  This follows (with perhaps a little thought) from observing that the units of $R[x]$ are precisely the units of $R$ (assuming the usual inclusion in $R[x]$).  It then happens that a polynomial $p(x) \in R[x]$ has a monic associate if and only if the leading coefficient of $p(x)$ is a unit of $R$.  Of course this happens for all nonzero polynomials in $R[x]$ only when $R$ is a field (so that all nonzero leading coefficients are units).  We've now circled back to the beginning of your problem's setup.
