# Is the Novikov ring a PID or UFD

The Novikov ring (over a ground field $k$) is defined to be $$R:=\Big\{\sum_{i=0}^{\infty} a_i T^{\lambda_i} \mid a_i\in k , 0\le\lambda_0 < \lambda_1 < \cdots, \lim \lambda_i = +\infty \Big\}$$ Here $T$ is a formal variable (sometimes we may assume all possible values of $\lambda_i$ belongs to a discrete subset $\Gamma\subset \mathbb R$). This ring is kind of similar to $k[T]$ or $k[[T]]$. Now that it is known $k[T]$ is a UFD, is it true that $R$ is also a UFD?

I do not believe that under this definition the Novikov ring is in general a PID. Let $\Gamma=\langle 0,1,\sqrt 2\rangle$, the closure of the set under addition. This continues to be discrete (check). Consider the ideal $I$ of $R$ which is formed of elements with no constant coefficient. Define the degree operator as below. $\partial(pq)=\partial p\cdot\partial q$. Assume $I=(p)$. If $\partial p>1$, then $(p)\neq I$ by sizing considerations. But if $\partial p=1$, $T^{\sqrt 2}\notin(p)$ because that would imply the existence of an element of degree $\sqrt 2-1$, impossible under our definition.
The above $\Gamma$ is discrete: $|(a-c)+(b-d)\sqrt 2|$ can be bounded in four cases, based on whether $a-c$ is positive or negative and whether $b-d$ is positive or negative. Clearly if they're both positive or both negative the distance is at least $1+\sqrt 2$. If, say $a-c> 0$ and $b-d<0$, there are only finitely many choices for $c$, and, fixing $c$, there are only finitely many choices for $d$ which make the error less than $M$ for some large $M$, and none of them give error $0$ by the irrationality of $\sqrt 2$. Similarly for the other cases.
If $\Gamma$ is not discrete, I don't think we can say anything meaningful. Consider $\Gamma=\mathbb R^{+}$, and the ideal $I$ of elements $\sum a_iT^{\lambda_i}$ which have no constant term -- that is, whenever $\lambda_i=0$, $a_i=0$ as well. I claim that $I$ is not generated by any single element -- in fact, that it's not generated by any finite number of elements. Assume that it were -- say $I=(p_1(T),\ldots,p_n(T))$. Let $\partial p$ be $\lambda_0^P$ for each $p$ - the smallest term for which the polynomial has a nonzero coefficient. Let $m=\min\{\partial p_1,\partial p_2,\ldots,\partial p_n\}$. Then for each $q\in I$, $\partial q\ge m$, because every element is the $R$-linear combination of these, and multiplication and addition of elements in $R$ can only increase degree (there are no elements of negative degree). This contradicts $R$ being a PID, or even Noetherian. Moreover, the same argument holds for any additive subgroup which fails to have an element of smallest norm, or if there is an element of smallest norm and $m$ is a limit from above of things in $\Gamma$.
• If $\Gamma\subset\mathbb R$ is discrete it is not necessary that $\Gamma$ is isomorphic to $\mathbb Z$ as a group, we can only construct a bijection between them. – Hang Jul 16 '18 at 0:11