I have a question in ring theory whose answer I am looking for.  Consider R to be a Euclidean Domain such that R is not a field. Then is polynomial ring R[X] is always a PID or not.  Attempt : R is ED implies R[X] is ED and R is ED implies R is PID. But can someone please tell what can I say about R[X] from it.
I've just read a proof of the statement:  Let $R$ be a commutative ring. If $R[X]$ is a principal ideal domain, then $R$ is a field.  In one part of the proof there is a step which I don't understand. I'll copy the proof: Let $u\in R$ be non-zero. Then using the principal ideal property, for some $f\in R[X]$ we have $\langle u,X\rangle = \langle f \rangle \subseteq R[X]$. Therefore, for some $p,q∈R[X], u=fp$ and $X=fq$. By properties of degree we conclude that $f=a$ and $q=b+cX$ for some $a,b,c\in R$. Substituting into the equation $X=fq$ we obtain $X=ab+acX$ which implies that $ac=1$, i.e. $a\in R^\times$, the group of units of $R$. Therefore, $\langle f\rangle =\langle 1\rangle=R[X]$. Therefore, there exist $r,s\in R[X]$ such that $ru+sX=1$ and if $d$ is the constant term of $r$, then we have $ud=1$. Therefore $u\in R^\times$. Our choice of $u$ was arbitrary, so this shows that $R^{\times}⊇R \setminus \{0\}$, which says precisely that $R$ is a field. I don't understand this: &quot;$X=ab+acX$ which implies that $ac=1$, i.e. $a\in R^\times$, the group of units of $R$. Therefore, $\langle f\rangle = \langle 1\rangle =R[X]$.&quot; Why does the fact that $a=f$ is a unit implies that $\langle f\rangle = \langle 1\rangle$? I would appreciate if someone could explain this to me. Thanks in advance.
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