6
$\begingroup$

It is well-known that all endomorphisms on the abelian group ($\Bbb{Z}$,+) can be seen as a left multiplication by some element in some ring structure on ($\Bbb{Z}$,+); namely left multiplication by any integer in the standard $(\Bbb{Z},+,\times)$ ring.

So far every endomorphism on abelian groups that I have examined has turned out to have this same interesting property, but I'm not very knowledgeable in advanced maths.

Can someone provide an example of an abelian group $G$ with an endomorphism that cannot be seen as a left multiplication by some element in some ring structure on $G$?

$\endgroup$
4
$\begingroup$

There are some abelian groups that admit no (possibly nonunital) ring structure with a left unit, and for such groups, the identity endomorphism cannot be multiplication by any element in any ring structure. The standard example of such a group is $G=\mathbb{Q}/\mathbb{Z}$. If you had a ring structure on $G$ with in which (the equivalence class of) $u=a/b$ was a left unit for some $a,b\in\mathbb{Z}$, then $bu=0$, so $bx=b(ux)=(bu)x=0$ for all $x\in G$. But this is clearly false, because (for instance) $b\cdot 1/2b=1/2\neq 0$.

$\endgroup$
  • 1
    $\begingroup$ A-ha! I was hoping it would be $\mathbb{Q}/\mathbb{Z}$. Thanks for the quick response. $\endgroup$ – Joseph Johnson Dec 27 '15 at 6:50
  • $\begingroup$ May I know why would $bu=0$? Thanks. $\endgroup$ – yoyostein Jan 20 '18 at 6:38
  • $\begingroup$ $bu$ is the coset of $b\cdot a/b=a$ and $a\in\mathbb{Z}$. $\endgroup$ – Eric Wofsey Jan 20 '18 at 16:58

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.