TITLE: When can a 'polynomial' have an infinite number of zeroes
QUESTION [5 upvotes]: As indicated in the title the intended meaning of 'polynomial' here is slightly nonstandard, and if there is a better word please feel free to edit. For our purposes:

A polynomial $\mathfrak{p}$ will be a member of some integral monoid ring $\mathbb{Z}(T^\mathbb{M})$, so for each $\mathfrak{p}\in\mathbb{Z}(T^\mathbb{M})$ there is some $\alpha\in \mathbf{On}$, the class of ordinals, and a unique pair of sequences $\{z_i\}_{i<\alpha}\subset\mathbb{Z}$ and $\{m_i\}_{i<\alpha}\subset\mathbb{M}$ (an additive monoid) such that $$\mathfrak{p}=\sum_{i<\alpha}z_iT^{m_i}.$$

Addition and multiplication behave as expected in usual polynomial multiplication with $T$ as an arbitrary indeterminate, where the addition in $\mathbb{M}$ may or may not be commutative. Further, $\alpha$ may be finite or transfinite, and in the latter case I would like to allow a transfinite number of nonzero terms in the polynomial. I believe this is equivalent to looking at Hahn series or a slight generalization of them if we require that $\mathbb{M}$ be an ordered monoid, however I am only interested in an algebraic property of $\mathfrak{p}$ in general so I cast the question in terms of more algebraic objects.
My question is:

Under what general conditions will $\mathfrak{p}$ have a unique* transfinite prime factorization in $\mathbb{Z}(T^\mathbb{M})$, in the sense that there exists a sequence $\{\mathfrak{p}_i\}_{i<\lambda}\subset\mathbb{Z}(T^\mathbb{M})$ for $\lambda\geq\omega$ such that $\forall i\big(\mathfrak{p}_i\neq1\big)$ and $\mathfrak{p}=\Pi_{i<\lambda}\mathfrak{p}_i$, while no other sequence produces $\mathfrak{p}$ as its product?
*possibly with stipulations about the uniqueness of the factorization

If this question is perhaps too general, I would still be very much  interested in the special case where $\mathbb{M}$ is discretely ordered, or the addition in $\mathbb{M}$ is commutative, or we restrict $\alpha$ to be countable or $\omega$. This came up during some research I'm doing pertaining to non-Archimedean ordered rings, which can be viewed as integral monoid algebras with a judicious choice of discretely ordered monoid and allowed 'polynomial' length, and in this setting polynomial factorization corresponds to factorization of elements in the ring. It seems like all the monoids I'm looking at produce unique finite factorizations for $\alpha<\omega$, but it feels like there should be some simple generalizations which will produce integral monoid algebras with the above property.

REPLY [3 votes]: $\DeclareMathOperator{\Ord}{Ord}$$\DeclareMathOperator{\Noo}{No}$It seems to me that you are redefining existing notions: this is exactly the Hahn series structure except the ordered group is now the positive part of a discretely ordered group, while $\mathbb{Z}_{\infty}$ is the Grothendieck group (/ring) associated to the monoïd (/semi-ring) structure of $\Ord$, and $\mathbb{Q}_{\infty}$ is its fraction field. 
I don't think the latter's real closure can be densely embedded in $\Noo$ for it contains no element strictly between $\mathbb{N}$ and $X^{\frac{1}{\mathbb{N}+1}}$.
I understand why you might want to see $\mathbb{Z}_{\infty}$ and $\mathbb{Q}_{\infty}$ as "versions" of integers and rationnal numbers, but do they really share their properties to the extent that such vocabulary and notations would be relevant? 
Apart form that, a proof that those rings have a kind of euclidean division (mind that the term "euclidean domain" is reserved for domains with an euclidean function) would be very nice indeed. Do you have someone checking your proof?

A note on how to add cuts to ordered fields:
Given an ordered field $F$, and a cut $(L,R)$ in it, that is subsets (or subclasses here) with $L < R$, $F = L \cup R$ and $L/R$ have no maximum/minimum, there is a canonical way to add it to $F$ and thus obtain a simple extension denoted $F(L \ | \ R)$ henceforth.
Let's say that $(L,R)$ is of algebraic type if there is some polynomial $P \in F[X]$ such that there is neigborhood of the cut, that is, a convex subset $C$ of $F$ intersecting $L$ and $R$ if they are not empty, such that $P$ is strictly negative on $C \cap L$ and strictly positive on $C \cap R$. 
Otherwise, say that $(L,R)$ is of transcendantal type.
Since $F$ can be embedded in a real closure, polynomials in $F[X]$ have less sign shifts than their degree, and this implies that:
$(i)$: If $(L,R)$ is of transcendental type, then for each polynomial $P \in F[X]$, there is a neighborhood $C$ of $(L,R)$ such that $P$ has constant sign on $C$. $P$ is defined to be strictly positive in $F[X]$ if this sign is strictly positive. This defines a positive cone on $F[X]$ that is naturally extended to $F(L \ |  \ R):= F(X)$, and in the resulting field, $L < X < R$.
$(ii)$: If $(L,R)$ is of algebraic type, chose $P$ with minimal degree satisfying the conditions of the definition. $P$ is then irreducible for having constant sign on neighborhoods is stable by product.
For $S = PQ + T \in F[X] / (P)$ with $\deg(T) < \deg(P)$, we find a neighborhood $C$ of $(L,R)$ such that $T$ has constant sign on $C$ and define this as the sign of $S$.
To see that this defines a positive cone on $F[X] /(P)$, the non trivial part is to see that if $S > 0$ then $S^2 > 0$. Towards this, write $T^2 = PQ_1+ R_1$ with $\deg(R_1) < \deg(P)$, note that $\deg(Q_1)  = 2\deg(T) - \deg(P)< \deg(P)$, so $Q_1$ has constant sign on a neighborhood $C_1$ of $(L,R)$ and by choosing a small enough one we get that $T,R_1$ have constant sign as well. Since $R_1 = T^2 - PQ_1$ is strictly positive on either $L$ or $R$, the constant sign is $1$ so $S^2 > 0$.
In the resulting ordered field $F( L \  | \ R):= F[X] / (P)$, $L < X + (P) < R$ (and  of course $X + (P)$ is a root of $P$).
There is an equivalent formulation with pseudo-Cauchy sequences but it only works with cuts that are sufficiently spaced so that valuation theoretic arguments can apply.
With this one as with the valued field version, there is a subtlety in the algebraic type version in that if $F(a)$ is a proper simple ordered field extension of $F$ and $(L,R)$ denotes the cut in $F$ defined by $a$, that is $L = F \cap ]-\infty;a[$ and $R := F \cap ]a;+\infty[$, then $a$ may not be algebraic over $F$ even if $(L_a,R_a)$ has algebraic type. 
When the cut is "Dedekind" in the sense that $\{r-l \ | \ (l,r) \in L \times R\}$ is coinitial in $F^{>0}$, the resulting ordered field $F(L \ | \ R)$ is a dense extension, and adding the cut is similar to adding a Cauchy sequence. It is possible to add all such cuts at once and define Dedekind-like operations on the set of Dedekind cuts in $F$. (note that those cuts can have algebraic or transcendent type) You then get a dense extension without proper dense extensions (=maximal dense extension), which is a nice generalization of completeness for ordered fields. It is also possible to add all Cauchy sequences at once or even to take a uniform completion of the field to get the same result.
Adding all agebraic type cuts inductively gives a real closure. In fact, an ordered field is real closed iff it has no algebraic type cut.
Adding all cuts inducitvely (unlike adding enough pseudo-limits) is impossible  for an ordered field $F'$ always has $(F',\varnothing)$ as a cut (and as soon as $F'$ is non-archimedean, it has the cut defined by $\mathbb{N}$). I suppose it is possible under global choice to add all set-sized cuts, and get Field which may be saturated and may under global choice be isomorphic to $\Noo$, but I guess this is a bit artificial for your taste.

So this works because one can evaluate polynomials in $F[X]$ at points of $F$.
The problem is that it is hard to see how to evaluate elements such as $X^{\omega}$ in $R:=\mathbb{Q}_{\infty}[X^{\mathbb{Z}_{\infty}^+}]$ (I think the notation $\mathbb{Q}_{\infty}((X^{\mathbb{Z}_{\infty}^+}))$, which speaks for itself when one knows Hahn series, should be prefered here), and that there is no clear way, even with an euclidean-like function, of specifying a condition for an element in the polynomial ring $R[\mathbb{Z}_{\infty}^+]$ of polynomials with indeterminates in the monoïd $\mathbb{Z}_{\infty}^+$ and coefficients in $R$, so that given an irreducible element $a$ of $R[\mathbb{Z}_{\infty}^+]$, the elements in $R[\mathbb{Z}_{\infty}^+]$ with less degree cannot for instance change signs over one cut. (The problem here being that the degree takes values in $\mathbb{Z}_{\infty}^+$ which is not well-ordered.)

You could try to embed $R$ in a nice way in $\Noo$, find out the cut defined by $a$ in $\Noo$, intersect it with $R$ (fortunately, the maps $0 <x \mapsto \exp(a\log(x))$ and their sums belong to an o-minimal structure so they may not shift signs too often either) and then add it to $R$.
But:
-This makes the definition extrinsic and a priori very dependant on the chosen embedding.
-I am not sure there is a "canonical" embedding of $R$ into $\Noo$.
We might want to embed it as an initial subring in the sense the work of Ehrlich, since $\mathbb{Q}_{\infty}$ is a truncature-closed subfield of $\mathbb{Q}((\omega^{\mathbb{Z}_{\infty}}))$, so $R$ is a truncature-closed subring of $\mathbb{Q}((x^{\mathbb{Z}_{\infty} \times \mathbb{Z}_{\infty}^+}))$ (with lexicographic order on $\mathbb{Z}_{\infty} \times \mathbb{Z}_{\infty}^+$ with empasis on second coordinate). Then you would need to find a way to see $\mathbb{Z}_{\infty} \times \mathbb{Z}_{\infty}^+$ as an ititial submonoïd of $\Noo$ which is not that obvious. (maybe $(\alpha,\beta) \mapsto \beta + sign(\alpha)\log(\omega^{-|\alpha|})$ where $sign(\alpha) \in \{-1;0;1\}$ works?). More importantly, one could ask whether your choice of embedding is natural or interesting.
-Given an embedding, it is not sure that for instance the cuts defined by $a: = Z^{\omega}-\omega$ and $b: = Z^{\omega+1} - \omega$ will differ. This would undermine the relevance of the phrasing "adding a $\omega$-th root of $\omega$". If the cut is added intrinsically using the process to add cuts, then either this process has to be generalized to those special polynomials (which I doubt is possible in general), or the image of $X$ in the extension will lack a real algebraic caracterisation as a $\omega$-th root of $\omega$ (in the case of $Z^{\omega} - \omega$).
Finally, about the intuition of non density of the field obtained by adding $\mathbb{Z}_{\infty}^+$-th roots of elements of $\mathbb{Q}_{\infty}$, this is still vague but I imagine the value group of the resulting field (which is the same as that of any dense extension) would be $\mathbb{Q}_{\infty}$ which is not isomorphic to $\Noo$.
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