Prove that $f\sim g$ iff $f\mid g$ and $g\mid f$, where $f$ and $g$ are polynomials. 
Prove that $f\sim g$ iff $f\mid g$ and $g\mid f$, where $f$ and $g$ are polynomials. 

It seems to me that $f$ and $g$ must be equal.
And if you guys could direct me to sources that explain polynomials and fields in abstract algebra for dummies, that would be fantastic.
 A: Hint:  Let's say you're working over the real numbers, $\mathbb{R}$. Then $2$ is a unit in $\mathbb{R}[x]$, because it has a multiplicative inverse, namely $\frac{1}{2}$. Therefore, for any polynomial $f$, we have that 
$$f\sim 2f$$
because $2f$ is equal to $f$ times a unit (I assume you are using the notation $\sim$ to mean "is associate to", i.e. differs by multiplication by a unit).
As the problem indicates, because $f\sim 2f$, we must also have that $f\mid 2f$ and $2f\mid f$; this is true because
$$2f=f\cdot 2,\qquad f=(2f)\cdot\frac{1}{2}$$
However, $f$ is not the same as $2f$ (unless $f=0$). Thus, we can have $f\sim g$ even when $f\neq g$.
More generally, the above example should get you thinking about what the units are in the ring of polynomials $\mathbb{R}[x]$.
A: Hint: what does $f\mid g$ say about the degree of $f$ versus the degree of $g$?
A: Hint $\ $ If $\rm\: f\sim g\:$ means $\rm\:f = ug,\:$ for some unit (invertible) $\rm\:u,\:$ then $\rm\:f = ug\:\Rightarrow\:f\mid g.\:$ Conversely $\rm\: g = u^{-1} f,\:$ so $\rm\: g\mid f.\:$ Note that this is true in any ring, i.e. unit multiples are always associates.
Conversely $\rm\ 0\ne f\mid g\mid f\,\Rightarrow\, f = ug,\ g = v f\,\Rightarrow\, f = uv f\,\Rightarrow\,f\,(1\!-\!uv)=0\,\Rightarrow\, 1 = uv\,$ $\Rightarrow$ $\rm\, u,v\,$ units, therefore $\rm\:f\sim g.\:$  Note that this direction requires only that $\rm\,f\,$ is cancellable (not a zero-divisor).
A: For a simple counterexample to your intuition, suppose $f$ is not the zero polynomial, and then put $g=-f$.
