Integral domain, UFD and PID related problem 
(i) Let $R$ be an integral domain that has irreducible elements. Prove that $R[X]$ is not A PID.
(ii) Let $R$ be a UFD and $K$ its field of fractions. Let $f \in R[X]$ be a monic polynomial with $\alpha$ a root of $f$ in $K$. Show that $\alpha \in R$.

I think I could prove (i), just in case I've done something wrong, I will write my solution.
Let $a \in R$ irreducible and consider the ideal $\langle a, x \rangle$. Suppose $R[X]$ is PID, then $\langle a, x \rangle=\langle f \rangle$ with $f \in R[X] \setminus \{0\}$.
But then $a=fg, g \in R[X] \setminus \{0\}$. We have $0=\deg(a)=\deg(f)+\deg(g)$. From here it follows $\deg(f)=\deg(g)=0$, so $f,g \in R$. Since $a$ is irreducible, we must have $f \in \mathcal U(R)$ or $g \in \mathcal U(R)$. Notice that $f \not \in \mathcal U(R)$. Suppose this is the case, then $\langle a,x \rangle=R[X]$.
We can write $1=a(a_0+a_1x+\cdots+a_nx^n)+x(b_0+b_1x+\cdots+b_mx^m)$. It folows $1=aa_0$, which means $a$ is a unit, this is absurd by the definition of irreducible element.
On the other hand, we have $x=fh$, so $1=\deg(x)=0+\deg(h)$. Then, $h=bx+c$. Putting this into the equation we get
$$x=f(bx+c)=fbx+fc.$$
So $fc=0 \implies c=0$ and $fb=1$, which means $f \in U(R)$. We've arrived to a contradiction which comes from assuming that $R[X]$ was  PID$.

I don't know what to do in (ii), I would really appreciate help with that part.

 A: Hint. Write $\alpha=a/b$ with $a,b\in R$ and $\gcd(a,b)=1$. From $f(a/b)=0$ you should get that $b\mid a$.
A: $(ii)$ is the UFD analog of the monic case of the Rational Root Test. The classical proof in $\,\Bbb Z\,$ immediately extends to any UFD (or GCD domain).
$(i)$ can be done slickly using evaluation hom's. Below is a sketch a proof of a more general result using only high-school algebra.
We show $\rm\:(2,x)\ =\ (f)\ $ in $\rm\:\mathbb Z[x]\: $ yields a parity contradiction, by simply evaluating polynomials.
$\rm\ \ f\ \in\ (2,x)\ \Rightarrow\ f\ =\ 2\ G + x\ H\:.\: $ Eval at $\rm\: x = 0\ \Rightarrow\ f(0)\ =\ 2\ G(0)\ =\ 2n\:$ for some $\rm\: n\in \mathbb Z$
$\rm\  \ 2\ \in\ (f)\ \Rightarrow\ 2\ =\ f\, g\:\ \Rightarrow\ deg\ f\ =\ 0\:\ \Rightarrow\:\  f\ =\ f(0)\ =\ 2n$
$\rm\ \ x\ \in\ (f)\ \Rightarrow\ x\ =\ f\, h\ =\ 2nh.\,\ $ Eval at $\rm\ x = 1\ \Rightarrow\ 1\: =\ 2n\,h(1)\ \Rightarrow\ 1\:$ is even $\ \Rightarrow\Leftarrow$
Remark $\ $ The above proof works in any domain where $\,2\ne 0\,$ and $\rm\,2\,$ is not a unit. i.e. $\rm\:2\nmid 1\:.\:$ In particular, it works over any domain with a nontrivial sense of parity, i.e. having $\rm\:\mathbb Z/2\:$ as ring image, e.g. the Gaussian integers, or the rationals writable with odd denominator - see this post. Conversely, the result is false if $\rm\:2 = 0\,$ or  a unit since then $\rm\,(2,x) = (x)\,$ or $\,(1)\,$ is principal. 
Further, the proof still works if we replace $\,2\,$ by any element $\,c\,$ of the coefficient domain $\,D,\,$  yielding: $\ (c,x)\,$ is principal in $\,D[x]\iff c=0\,$ or $\,c\,$ is a unit.
