6
$\begingroup$

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.)

Improve this question
$\endgroup$
5
  • 4
    $\begingroup$ It wouldn't hurt to add your definition of the irreducible elements. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Mar 8, 2017 at 8:20
  • 1
    $\begingroup$ That's right, I have included it. $\endgroup$
    – Dominic van der Zypen
    Commented Mar 8, 2017 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
    Commented Mar 8, 2017 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
    Commented Mar 8, 2017 at 10:51
  • $\begingroup$ Oh ok -- thanks Salvo, will correct this within the next days $\endgroup$
    – Dominic van der Zypen
    Commented Mar 8, 2017 at 11:25

2 Answers 2

Reset to default
8
$\begingroup$

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 $x \in M$).

Improve this answer
$\endgroup$
7
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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
    Commented Mar 20, 2017 at 11:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .