$(X)$ is (not) a prime ideal in the ring of integer-valued polynomials I came across the following statement in my course notes: 

Consider $R = \left\{f \in \mathbb{Q}[X] \mid f(\mathbb{Z}) \subset \mathbb{Z}\right\}$ the ring of integer-valued polynomials. Then the ideal generated by $X$ is not a prime ideal.

Isn't this incorrect? I'd argue that $R/(X)$ (where $(X)$ denotes the ideal generated by $X$) is isomorphic to $\mathbb{Z}$, which is an integral domain and thus $(X)$ must be prime. 
 A: $$2\frac{x(x+1)}{2}=x(x+1)\in (x)$$ however $2\not \in (x)$ and $\frac{x(x+1)}{2}$ although integer values is not in $(x)$, as if it were of the form $f(x)x$ then $f(x)=\frac{x+1}{2}$, and this is not integer valued.
A: Rene has already given a wonderful solution, but I'll add this so you can maybe see why your approach doesn't work:
Normally, when we show $A[x]/(x)\cong A$ for a ring $A$, we use the homomorphism $A[x]\to A$, $f\mapsto f(0)$, so the kernel is $(x)$ and the result follows by the isomorphism theorem.
Let's try to do the same in this case. We'll map $R\to\Bbb{Z}$ by sending $f\mapsto f(0)\in\Bbb{Z}$. This is definitely surjective, so if the kernel is $(x)$ then we are done. We definitely have $(x)$ contained in the kernel. However, the reverse inclusion doesn't hold because as Rene has shown, we can find $f$ such that $f(0)=0$ but there is no $g\in R$ such that $f(x)=x\cdot g(x)$.
A: Notice $\,\ 2\mid x(x\!+\!1),\,\ 2\nmid x\!+\!1\,$ so $\ \smash[b]{\begin{align} 2\nmid x\,\Rightarrow\, 2\ \rm not\ prime\\ x\nmid 2\,\Rightarrow\, x\ \rm not\ prime\end{align}}\ $ by a symmetric twist on primaility
$\!\begin{align} {\bf Lemma}\ \ \ {\rm If}\ \  c\mid ab,\ &c\nmid b\\ 
\text{then  }\ &c\nmid a\,\Rightarrow\, c\ \ \rm not\ prime\\
\text{and }\ \ &a\nmid c\,\Rightarrow\, a\ \ \rm not\ prime\, \text{ if $\,a\,$ is cancelable}
\end{align}$
Proof $\, $ (sketch) $\,\ c\,$ is not prime follows by definition, and $\,a\,$ is not prime follows by
$\ c\mid ab\,\Rightarrow\, ab\! =\! cd\ $ so $\ a\mid cd,\ a\nmid c,\ a\nmid d\ $ (else $\, d = ae\ $ so $\,ab =cae\,$ so $\,b = ce\,$ contra $\,c\nmid b)$
(simpler $\ \dfrac{d}a = \dfrac{b}c\ $ so $\ a\mid d\iff c\mid b\ $ when in a domain)
