I'm quite new in the theory of rings and hope some of you can help me in tackling some problems in understanding, which occured when I tried to solve a problem.
The problem is about the rings $\mathbb{Z}_{2}, \mathbb{Z}_{2}[x]$ and $\mathbb{Q}[x] = \{\sum_{j=0}^{n}a_{j} x^{j} ~| ~a_{j} \in \mathbb{Q}, n \in \mathbb{N} \}$
First I need to find the unities, so i need to find $ \{e \in R : e |1\} $. In $\mathbb{Z}_2$ it is easy: $E[\mathbb{Z}_2] = \{-1,1\}$. I guess $E[\mathbb{Z}_2[x]] = \{1\}$ but I can't justify why (I did calculatios for polynomials up to degree 3, but this of course is no proof) . For the latter case, $E[\mathbb{Q}[x]]$ I don't have a clue how to proof, that every polynomial with constant coefficient $a_0 = 1$ is a unit (I've found this on the Internet). I really can't imagine how every polynomial with constant $a_0$ should be a unit - at least I wasn't able to find even one pair of unities, which give 1 when multiplied.
Is there an easy way to show this two facts?
Furthermore I have to find out which of the given rings are isomorphic. According to a hint it has to do something with the unities. Does an isomorphism map unities to unities? If yes, non of them can be isomorphic, due to their number is different. Is that true?
Finally I should decide whether $\mathbb{Q}[x]$ is a factorial ring. Here again I got a hint; Find $p \in \mathbb{Q}[x]$ such that $\forall f \in \mathbb{Q}[x]: f=e p^k$ with $e \in E[\mathbb{Q}[x]]$ and $k \in \mathbb{N}$. At these final point I really have no idea how to start, but maybe this might get better, when I understand the previous points?
I would really appreciate if some of you can help me understandig this task.
Thanks!
Let $K[x]$ be a polynomial ring over a field $K$. Then a polynomial $f\in K[x]$ is invertible if there is a polynomial $g\in K[x]$ such that $fg=1$. Now the degree formula comes in: $${\rm deg}(f) + {\rm deg}(g) = {\rm deg}(fg) = {\rm deg}(1) = 0.$$ The first equality comes from the fact that the highest terms of $f$ and $g$ multiply while their degrees add; there is no cancellation of the highest terms since $K$ is a field. Since the degrees of $f,g$ are non-negative and sum up to 0, both need to be zero and so $f,g$ are elements of $K$. It follows that the unit group of $K[x]$ is $K^*=K\setminus\{0\}$.