Irreducible elements in endomorphism rings

Let $(G, +)$ be a commutative group. The endomorphism set $\text{End}(G)$ of all group endomorphisms $f:G\to G$ is a ring, where $+$ is taken pointwise and the multiplication is the composition of endomorphisms.

Suppose $f\in \text{End}(G)$ is not bijective (i.e. not a unit in the ring) and not a zero-divisor. Can it be written as a product of irreducible elements?

(We call $g\in\text{End}(G)$ reducible if there are non-units $h_1,h_2\in\text{End}(G)$ such that $g = h_1\circ h_2$; and we call $g$ irreducible otherwise.)

• It wouldn't hurt to add your definition of the irreducible elements. – Włodzimierz Holsztyński Mar 8 '17 at 8:20
• That's right, I have included it. – Dominic van der Zypen Mar 8 '17 at 10:30
• @DominicvanderZypen: I don't think this is the definition you want, since carrying it over to an arbitrary unital ring (that is, letting $g$ be reducible if there are non-units $h_1,h_2$ such that $g = h_1 h_2$, and irreducible otherwise) would imply that every unit of a Dedekind-finite ring $R$ is irreducible (see mathoverflow.net/questions/261982), and so is, in particular, the identity of $R$ (which is not so good a thing). Usually, we take an irreducible element in a unital ring $R$ to be an atom of its multiplicative monoid, and an atom in a monoid $H$ is a non-unit (...) – Salvo Tringali Mar 8 '17 at 10:46
• (...) element $a \in H$ such that $a=xy$ for some $x, y \in H$ implies $x \in H^\times$ or $y \in H^\times$. – Salvo Tringali Mar 8 '17 at 10:51
• Oh ok -- thanks Salvo, will correct this within the next days – Dominic van der Zypen Mar 8 '17 at 11:25

If $R$ is a commutative unital ring, then $R \cong_{\sf Ring} {\rm End}_{{\sf Mod}_R}(R_R)$, with the elements of $R$ acting on $R$ by left multiplication. Now, let $R$ be any non-atomic integral domain and note that the multiplicative monoid of ${\rm End}_{{\sf Mod}_R}(R_R)$ is a divisor-closed submonoid of the multiplicative monoid of ${\rm End}_{{\sf Grp}}(R_R)$. Lastly, recall that a commutative unital ring is atomic (i.e., every non-unit, non-zero element is a product of some atoms) iff so is its multiplicative monoid, and that a monoid $H$ with zero is atomic only if so are all the divisor-closed submonoids of $H$ (we say that a submonoid $M$ of $H$ is divisor-closed if $x \mid_H y$ and $y \in M$ imply $x \in M$).
Take $G = \mathbb{Q}/\mathbb{Z}$; then $\text{End}(G)$ is the profinite integers
$$\widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$$
where $\mathbb{Z}_p$ is the $p$-adic integers. The element $\prod_p p$ is neither a unit nor a zero divisor in this ring, and it cannot be written as a product of irreducible elements. This is because the only irreducible elements $x = \prod_p x_p$ are those of the form $x_q = q$ for a fixed prime $q$ and $x_p = 1$ otherwise, and unit multiples of these (there are lots of units), and so the elements that can be written as a product of irreducibles are precisely those where $x_p$ is a unit for all but finitely many $p$, and nonzero otherwise.