# $(X)$ 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.

• What makes you say that? Do you have a homomorhpism from $R$ to $\Bbb Z$? Commented Dec 29, 2016 at 18:26
• @AlexMathers Indeed, I edited my question. Commented Dec 29, 2016 at 18:31
• @KasperCools actually, I'm confused about all the notation now that I look at it again. When you first name $X$, is that supposed to say $R$? And $X$ is just the indeterminate? Or is $R$ supposed to be $\Bbb{Q}[x]$? Commented Dec 29, 2016 at 18:35
• Yeah I am sorry about the edits. Actually these are not my fault as these were made by @Omnomnomnom. As the question is stated now, there should be no mistakes in it. $X$ is just a variable and $R$ denotes the ring described ( it is sometimes denoted as $Int(Z)$). Commented Dec 29, 2016 at 18:42
• @KasperCools my mistake, sorry! Commented Dec 29, 2016 at 18:43

$$2\frac{x(x+1)}{2}=x(x+1)\in (x)$$ 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.

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 $f(x)=x\cdot g(x)$.

• Aha, I was just in the process of checking why my solution wouldn't work. Although Rene took away my doubts of the statement being incorrect, I'm gonna mark your answer as the correct one as this was the part where I was struggling with. I was too quick assuming the homomorphism would work without checking it properly. Commented Dec 29, 2016 at 18:59

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)