Associate Polynomials on an Integral Domain

Question

I am being asked to prove that two polynomials being associated in $R[x]$ is an equivalence relation.

That is, two polynomials are associated in $R[x]$ if $f,g \in R[x],$ and there exists a $\lambda \in R$ such that $ f = \lambda g$. Then, the equivalence relation would be $$f \sim g \iff f = \lambda g \text{ for some } \lambda \in R.$$

Now, it is very easy to prove this is an equivalence relation if $R$ is a field or ring with division (since for every $\lambda \neq 0,$ we have $\lambda^{-1}$). However, I am being asked to prove the equivalence relation for $R$ an integral domain, and here is where I am not sure if this is a mistake or if there is a way to do it. Do all integral domains have multiplicative inverses?

Is this a mistake? Or how would I go about this? My algebraic structures are a bit rusty. Thanks!

Answer 1

$\mathbb{Z}$ is an integral domain and thus $\mathbb{Z}\left[x \right]$ is also an integral domain. Take $f\left(x \right) = 3x +6$ and $g\left(x \right) = x +2$. Then $ f=3g$. But there is no $\lambda$ in $\mathbb{Z}$ such that $ g = \lambda f$. Hence the given relation need not be an equivalence relation on an integral domain.

Your claim for division rings and fields is correct.

Answer 2

Your definition of "associates" should specify that $\lambda$ is a unit.

It turns out that when $R$ is a domain $U(R[X])=U(R)$, so that may explain why it did not talk about the units of $R[X]$.

As you've written, the relation is merely divisibility, which is obviously not always symmetric. That's what making $\lambda$ a unit buys for you.

do all integral domains have multiplicative inverses?

I'm not clear on what you're asking. domains don't "have inverses," but their elements can. Some of them, or everything except $0$, depending on the domain.