Preliminaries. Let $S_0 := \mathbb{N}_0[[\mathbb{Z}]]$ and $S_1 := \mathbb{Z}[[\mathbb{Z}]]$ . For $f(x) = \sum a_i x^i \in S_1$ , $\textsf{cont}(f)$ , is the $\mathsf{gcd}$ of its coefficients. $f$ is primitive if $\textsf{cont}(f) = 1$ . A series is monolithic as defined in page 4 of https://arxiv.org/pdf/2412.00590.

Proposition 1.1. The units of $S_0$ and $S_1$ are:
(a) $\mathcal{U}(S_0) = \{x^k \mid k \in \mathbb{Z}\}$ .
(b) $\mathcal{U}(S_1) = \{\pm x^k \mid k \in \mathbb{Z}\}$ .

Proof. By page 4 of https://arxiv.org/pdf/2412.00590, or Lemma 2.3 of https://arxiv.org/pdf/2209.13817, for a semidomain $S$ and a torsion-free group $G$ , the units of $S[[G]]$ are the monomials $sx^g$ , for $s \in \mathcal{U}(S)$ and $g \in G$ .

(a) The units are $\mathcal{U}(\mathbb{N}_0) = \{1\}$ . Thus, $\mathcal{U}(S_0)=\{1 \cdot x^k \mid k \in \mathbb{Z}\} = \{x^k \mid k \in \mathbb{Z}\}$ .

(b) An analogous argument shows that since the units $\mathcal{U}(\mathbb{Z}) = \{\pm 1\}$ , $\mathcal{U}(S_1) = \{\pm x^k \mid k \in \mathbb{Z}\}$ . $\square$

Lemma 2.1. The set of irreducibles in $S_0$ , denoted $\mathcal{A}_{\mathbb{N}_0}$ , is the disjoint union of:

Elements $px^k$ , where $p$ is prime and $k \in \mathbb{Z}$ .
Irreducible primitive series in $S_0$ .

Proof. Let $f > 0 \in S_0$ not be a unit. We can write $f = d g$ , where $d = \textsf{cont}(f) \in \mathbb{N}$ and $g$ is primitive. Since $f$ is irreducible, $d$ or $g$ must be a unit.

Case 1: $g$ is a unit. By Proposition. 1.1(a), $g=x^k$ for some $k \in \mathbb{Z}$ , so $f = dx^k$ . For $f$ to be irreducible, $d$ must be irreducible in $\mathbb{N}_0$ . If $d=ab$ for $a, b > 1$ , then $f = a(bx^k)$ . Since $a,b>1$ , neither $a$ nor $bx^k$ is a unit in $S_0$ , a contradiction. Thus, $d$ must be prime.
Case 2: $d$ is a unit. In $S_0$ , this means $d=1$ , so $f$ is primitive. As $f$ is irreducible by hypothesis, it is an irreducible primitive series.

These two sets are clearly disjoint, as desired. $\square$

Theorem 2.2. Let $\mathcal{A}_{\mathbb{N}_0}$ be the set of irreducible elements of $\mathbb{N}_0[[\mathbb{Z}]]$ and let $\mathcal{A}_{\mathbb{Z}}$ be the set of irreducible elements of $\mathbb{Z}[[\mathbb{Z}]]$ . Then $\mathcal{A}_{\mathbb{N}_0}$ is the disjoint union of the following two sets:

$\mathcal{A}_1 = \mathcal{A}_{\mathbb{N}_0} \cap \mathcal{A}_{\mathbb{Z}}$ , the set of elements that are irreducible in both structures.
$\mathcal{A}_2 = \mathcal{A}_{\mathbb{N}_0} \setminus \mathcal{A}_{\mathbb{Z}}$ , the set of elements that are irreducible in $\mathbb{N}_0[[\mathbb{Z}]]$ but become reducible in $\mathbb{Z}[[\mathbb{Z}]]$ .

Furthermore, $\mathcal{A}_1$ contains all prime integers and their monomial associates, while every element in $\mathcal{A}_2$ must be a primitive series.

Proof. (1) By Lemma 2.1, $px^k \in \mathcal{A}_{\mathbb{N}_0}$ . By Proposition. 2.1(a) of this convenient paper: https://arxiv.org/pdf/1106.1425, $\mathcal{A}_{\mathbb{Z}}$ holds as well. If you don't see why, note that in $S_1$ , $x^k$ is a unit by Proposition. 1.1, so $px^k$ is an associate of $p$ . Since $p$ is irreducible in $\mathbb{Z}$ and thus in $S_1$ , its associate $px^k$ is also irreducible.

(2) Let $f \in \mathcal{A}_2$ . We show $f$ is primitive. Assume towards a contradiction that $f$ is not primitive, so $\textsf{cont}(f)=d>1$ . Since $f \in \mathcal{A}_{\mathbb{N}_0}$ , Lemma 2.1 implies $f=px^k$ for some prime $p$ . But we just showed such elements are in $\mathcal{A}_1$ , a contradiction. Thus, any $f \in \mathcal{A}_2$ must be primitive, as desired. $\square$

Example. Consider the polynomial $f(x) = x^4+4$ . We proceed to argue that $f(x)$ is irreducible in the semidomain $S_0$ but reducible in the ring $S_1$ .

Follow Lemma 3.1 of https://arxiv.org/pdf/2312.14888, a polynomial in $S_0$ is irreducible iff it is monolithic and primitive. Here, $\mathsf{cont}(f) = \mathsf{gcd}(1, 4) = 1$ , so $f$ is primitive; it is also clear that $f$ is monolithic. On the other hand, in $S_1$ , $f(x)$ can be factored by Sophie Germain:
$x^4+4 = (x^2+2x+2)(x^2-2x+2).$ Let $g = x^2+2x+2$ and $h = x^2-2x+2$ . The lowest degree coefficient of both $g$ and $h$ is 2. By virtue of Proposition 1.1(b), since $2 \notin \{\pm 1\}$ , neither $g$ nor $h$ is a unit in $S_1$ , which concludes our argument.

Results. Here are some of the main results:

If a series $f \in \mathbb{N}_0[[\mathbb{Z}]]$ is irreducible in $\mathbb{N}_0[[\mathbb{Z}]]$ but becomes reducible in $\mathbb{Z}[[\mathbb{Z}]]$ , then $f$ must be a primitive series.
Any series $f \in \mathbb{N}_0[[\mathbb{Z}]]$ that is irreducible in $\mathbb{N}_0[[\mathbb{Z}]]$ and is not primitive will also be irreducible in $\mathbb{Z}[[\mathbb{Z}]]$ .
Elements of the form $px^k$ where $p$ is a prime and $k \in \mathbb{Z}$ are irreducible in both $\mathbb{N}_0[[\mathbb{Z}]]$ and $\mathbb{Z}[[\mathbb{Z}]]$ .

Future Work. Trying to find more interesting results / connections between the irreducibles.

Remark. I believe Proposition. 2.1 might be wrong, or maybe I just have the wrong proof construction. Either way, I am still working. EDIT Jul/16: I have adopted a new methodology for proving the later parts, and would like feedback, motivated by Prof. Polo's comments and other papers.