1

Let $\mathcal{G} = \langle A, + \rangle$ be an infinite group. I was wondering whether it is possible to extend $\mathcal{G}$ to a commutative ring $\mathcal{R}= \langle A, + , * \rangle$ by associating to each element $a \in A$ an automorphism $\phi_a$ on $\mathcal{G}$ in such a way that the product * will be defined as: $a*x=y \underset{def}{\Leftrightarrow} \phi_a(x)=y$. If the answer is positive, it would be great to sketch the procedure for the definition of the product $*$.

Shaun
  • 44,997
Vincenzo
  • 11
  • Why would you need that the group is infinite? – Dietrich Burde May 03 '23 at 16:09
  • 1
    As written the answer is "no", because the additive identity $0$ cannot act as an automorphism: you must have $0x = 0$ for all $x$, so $0$ cannot be associated to an automorphism* of $A$. If you replace "automorphism" with "endomorphism", then the assignation must be a group homomorphism, since you require $(a+b)x = (ax)+(b*x)$. – Arturo Magidin May 03 '23 at 17:22
  • @DietrichBurde The infinity requirement is just in view of certain applications to the representational theory of measurement. But I guess that for some finite groups there would be no one-to-one correlation between elements in the domain and the automorphisms on the group. – Vincenzo May 03 '23 at 17:30
  • No, the identity function doesn't work. In any ring, if $0$ is the additive identity, then $0x = x0 = 0$ for all $x$: because $0x + 0 = 0x = (0+0)x = 0x + 0x$, so $0x=0$. Likewise, $x0 = x(0+0) = x0 + x0$, so $x0 = 0$. If you define $\phi_0$ as the identity function, then you would have $0x = \phi_0(x) = x$. – Arturo Magidin May 03 '23 at 17:35
  • @ArturoMagidin Just seen it was wrong and deleted the comment...what if I drop commutativity and require 0 as fixed point for all automorphisms? After all, it seems that this is also required by the group structure. – Vincenzo May 03 '23 at 17:37
  • Endomorphisms necessarily map $0$ to $0$, so you don't need to "require" it. Note that it is well known that not every additive group can be made into a ring with unity, so you should specify if you want a ring with unity or not. – Arturo Magidin May 03 '23 at 18:06
  • Any ring structure on $A$ would induce a ring homomorphism $A\to \mathrm{End}(A)$, into the ring of abelian group endomorphisms of $A$, which is a ring under pointwise addition and composition of functions, by $a\mapsto \varphi_a$, where $\varphi_a(x) = ax$. Conversely, if you have a group homomorphism $A\to\mathrm{End}(A)$ sending $a$ to $\phi_a$, and this map satisfies that $\phi_a\circ\phi_b = \phi_{\phi_a(b)}$, then this will define a ring structure by $ax = \phi_a(x)$. Any map that defines a ring structure must satisfy this condition to give $(ab)c = a(b*c)$. (cont) – Arturo Magidin May 03 '23 at 18:10
  • (cont) You can always get such a homomorphism by sending everything to the zero map, of course (the ring with zero multiplication). If you require other properties (e.g., that the ring have a unity), then whether such a morphism exists or not will depend on the group. For example, there is no ring-with-unity structure on $\mathbb{Q}/\mathbb{Z}$. See also mathoverflow. – Arturo Magidin May 03 '23 at 18:14
  • To get commutativity, you would also require the morphism $a\mapsto \phi_a$ to satisfy $\phi_a(b) = \phi_b(a)$ for all $a,b\in A$. – Arturo Magidin May 03 '23 at 18:22

0 Answers0