Valuations of integer valued polynomials

Question

Consider the ring $R=\text{Int}(\mathbb Z):=\{p(x)\in \mathbb Q[x]\ |\ p(n)\in \mathbb Z, \forall n\in \mathbb Z \}$. Let $K$ denote the fraction field of $R$.

Fix an $a\in \mathbb Z$ and let $P$ be the prime ideal defined by $P:=\{q(x)\in R\ | \ q(a)\equiv 0 (\text{mod } p)\}$ and let $R_P$ denote the localization of $R$ at $P$.

Let $\Gamma$ denote the totally ordered group $\mathbb Z \times \mathbb Z$, where the operation is componentwise addition and the order is the lexicographic one.

Find a surjective valuation $v:K\rightarrow \Gamma$ such that its corresponding valuation ring is precisely $R_P$.

Now when trying to find $v\left(\frac{f(x)}{g(x)}\right)=(c_1,c_2)\in \mathbb Z\times \mathbb Z$, I thought of using a similar expression as for the $p$-adic valuation on $\mathbb Q$, namely define $c_2=e_p(f(a))-e_p(g(a))$ for one of the components, however the issue with that is that it yields a positive value even when $\frac{f(x)}{g(x)}$ is not in $R_P$, namely say if $e_p(f(a))=2, e_p(g(a))=1$. I have been trying to find some nice functions $P(X,Y)$ so that $P(X_1+X_2,Y_1+Y_2)=P(X_1,Y_1)+P(X_2,Y_2)$ which would help here, but none of them were surjective on $\mathbb Z$.

Answer 1

For simplicity I will take $a=0$. The substitution $x\mapsto x+a$ allows us to handle arbitrary $a$.

I claim that every element of $\mathbb{Q}[x]$ with constant term $1$ is in $R_P$, and in fact is a unit in $R_P$. To see this, take $f(x)\in\mathbb{Z}[x]$ and $n$ a positive integer. Then $$ A_{f,n}(x):=\frac{(xf(x)+1)\ldots (xf(x)+n)}{n!} $$ is in $R$, and $A_{f,n}(0)=1$ so $A_{f,n}\not\in P$. Thus $A_{f,n}$ is a unit in $R_P$. Also, we have $$ \frac{A_{f,n}}{A_{f,n-1}}=1+\frac{xf(x)}{n}\in R_P^\times. $$ Every polynomial in $\mathbb{Q}[x]$ with constant term $1$ can be written $1+xf(x)/n$ for some $f$ and $n$, so this proves the claim.

Now, every element of $R_P$ is a rational function which is defined and $p$-integral at $x=0$. Conversely, every rational function defined and $p$-integral at $0$ and can be written either in the form $$ r\frac{a(x)}{b(x)}\hspace{10mm}\text{or}\hspace{10mm}\frac{a(x)-1}{b(x)}, $$ with $r\in\mathbb{Z}_{(p)}$ and $a,b\in\mathbb{Q}[x]$ with constant term $1$. Thus $R_P$ is precisely the set of rational functions $f\in\mathbb{Q}(x)$ which are defined and $p$-integral at $x=0$.

Finally we can define the valuation $v$. For $\varphi(x)\in\mathbb{Q}(x)=K$, define $$ v(\varphi(x))=\left( v_x(\varphi), v_p\big(\varphi(x) x^{-v_x(\varphi(x))}\big|_{x=0}\big)\right). $$ Here $v_x$ is the $x$-adic valuation on $K$. In other words, $v(\varphi)=(c_1,c_2)$ where $c_1$ is the power of $x$ dividing $\varphi$ and $c_2$ is the power of $p$ dividing the leading term in the Laurent series expansion of $\varphi$ centered at $x=0$. For arbitrary $c_1$, $c_2$, we have $v(x^{c_1}p^{c_2})=(c_1,c_2)$, so $v$ is surjective. Additionally, we see that the valuation ring of $v$ is the set of rational functions which are defined and $p$-integral at $x=0$, which is $R_P$.