Prove that if $I$ is maximal, then $R[X]$ is a PID. Let $R$ be a commutative ring with unity such that $R[X]$ is a UFD. Denote the ideal $\langle X\rangle $ by $I$.
Prove that 


*

*If $I$ is maximal, then $R[X]$ is a PID.

*If $R[X]$ is a Euclidean Domain then $I$ is maximal.

*If $R[X]$ is a PID then it is a ED.


We know that $R[X]/\langle X\rangle \cong R$. Since $I$ is maximal then $R$ is a field. Also every field is a Euclidean Domain and hence a PID and also a UFD.
By Gauss Lemma $R[X]$ is a UFD.
But these facts are not helping anything in these statements to prove.
Please give some hints.
 A: The first part is easy. We have $R[X]/I=R[X]/\langle X\rangle\cong R$. We are given that $I$ is maximal. So $R[X]/I$ is a field. Hence $R$ is a field. So $R[X]$ is an ED, and hence a PID.  
A: These results are somewhat straightforward with this result:
Let $R$ be a commutative ring with $1 \neq 0$. The following are equivalent.


*

*$R$ is a field.

*$R[X]$ is a Euclidean domain.

*$R[X]$ is a P.I.D.


*

*$\Rightarrow$ 2. is a standard result. 

*$\Rightarrow$ 3. follows from the result that says "Euclidean domains are P.I.D.'s" 

*$\Rightarrow$ 1.: Suppose $R[X]$ is a P.I.D; the ideal $(X)$ is a prime ideal in $R[X]$ because $R[X]/(X) \cong R$ is an integral domain ($R$ is subring of the integral domain $R[X]$). Moreover, nonzero prime ideals in a P.I.D. are maximal, so $(X)$ is maximal in $R[X]$. Thus $R[X]/(X) \cong R$ is a field.
Now your statements should follow immediately.
[Note 3. $\Rightarrow$ 1. can be proved using the definition of a field, as hinted by @Alex Wertheim in the comments, but it is a little longer...]
