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.)

  • 4
    $\begingroup$ It wouldn't hurt to add your definition of the irreducible elements. $\endgroup$ – Włodzimierz Holsztyński Mar 8 '17 at 8:20
  • 1
    $\begingroup$ That's right, I have included it. $\endgroup$ – Dominic van der Zypen Mar 8 '17 at 10:30
  • $\begingroup$ @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 (...) $\endgroup$ – Salvo Tringali Mar 8 '17 at 10:46
  • $\begingroup$ (...) element $a \in H$ such that $a=xy$ for some $x, y \in H$ implies $x \in H^\times$ or $y \in H^\times$. $\endgroup$ – Salvo Tringali Mar 8 '17 at 10:51
  • $\begingroup$ Oh ok -- thanks Salvo, will correct this within the next days $\endgroup$ – 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.

  • $\begingroup$ Your description of irreducible elements seems very close to a description of the Idele group of the rationals (minus the place at infinity). Do you know whether this is a meaningful similarity? $\endgroup$ – Asvin Mar 20 '17 at 11:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.