# 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!

• Consider any constant polynomial in $\mathbb{Z}[x]$ that is not a unit. Then there is no monic polynomial associate to it, since the monic polynomial of degree 0 is associate to the units. Nov 23 '19 at 2:25
• Your "attempt" starts out with the assumption that $R$ is an integral domain (which seems a good place to start), but then you say "Hence $\mathbb Z$ is an integral domain." While $\mathbb Z$ certainly is an integral domain, this is unrelated to knowing that $R$ also is. Moreover in $\mathbb Z[x]$, what happens is that non-monic polynomials have a monic associate only if the leading coefficient is $-1$. But if polynomial $p(x)\in \mathbb Z[x]$ has a monic associate, that monic associate is unique. Nov 23 '19 at 2:46
• @hardmath So for $\mathbb{Z}$, although $\mathbb{Z}$ is an integral domain, a nonzero polynomial h $\in F[x]$ has exactly one monic associate? If that's the case, is there another polynomial ring that's an integral domain that might be easier? Nov 23 '19 at 3:13
• @yagayeet: Again you seem to be jumping from $\mathbb Z[x]$ to some unrelated polynomial ring $F[x]$. My point was that a polynomial with leading coefficient not $\pm 1$ like $2x+3 \in \mathbb Z[x]$ has no monic associate in $\mathbb Z[x]$, and thus it is false that every polynomial has exactly one monic associate. Nov 23 '19 at 3:17
• @hardmath I'm sorry, I meant to write $h \in \mathbb{Z}[x]$. Nov 23 '19 at 3:33

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.