How to prove or disprove PID (or ED)? Let $R$ be commutative ring and $R[x]$ be polynomial ring in one variable of R. 
Prove/Disprove,
If $R$ is $PID(or ED)$, then $R[x]$ is also $PID (or ED)$.
 A: Counterexample:  $\Bbb Z$ is an ED, hence a PID, but $\Bbb Z[x]$ is neither:  the ideal $\langle 2, x \rangle$ is not principal.
Note Added Friday 5 September 2014 10:33 PM PST:  To see in detail that $\langle 2, x \rangle$ is not a principal ideal in $\Bbb Z[x]$, note that the elements of $\langle 2, x \rangle$ are all of the form $2g(x) + xh(x)$ for $g(x), h(x) \in \Bbb Z[x]$.  It is easy to see that any such $2g(x) + xh(x) = xp(x) + 2m$ for some $p(x) \in \Bbb Z[x]$ and $m \in \Bbb Z$; thus if $\langle 2, x \rangle = \langle d(x) \rangle = d(x) \Bbb Z[x]$ for some $d(x) \in \Bbb Z[x]$, we may take $d(x) = xp(x) + 2m$ for suitable $m$ and $p(x)$.  Since $\langle 2, x \rangle = \langle d(x) \rangle = \langle xp(x) + 2m \rangle$, we have $xp(x) + 2m \mid 2$ or $f(x)(xp(x) + 2m) = 2$ for some $f(x) = \sum_0^{\deg f} f_ix^i \in \Bbb Z[x]$; this implies $2mf_0 = 2$ or $m = f_0 =\pm 1$; furthermore, if $\deg f(x) \ge 1$, we may set $\hat f(x)  = f(x) - f_0 = \sum_1^{\deg f} f_ix^i \ne 0$ so that $f(x) = f_0 + \hat f(x)$; we also have $f(x)xp(x) \pm 2\hat f(x) = 0$ or
$f(x)xp(x) = \mp 2 \hat f(x); \tag{1}$
but if $\hat f(x) \ne 0$, we must also have $p(x)\ne 0$ and then (1) contradicts $\deg xf(x) = \deg \hat f + 1$; hence we must have $p(x) = \hat f(x) = 0$.  Thus $d(x) = 2$; but $2 \not \mid x$ in $\Bbb Z[x]$.  End of Note.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
