1
$\begingroup$

Suppose $\Gamma$ is a finite group ,$R$ is a commutative ring with $1$. Then the set of maps between $\Gamma$ and $R$ become a commutative ring . The zero element is the zero map ,the identity is the map which maps all $g\in \Gamma$ to $1$ .We denote the ring as $R^{\Gamma}$. From one note I see that $R^{\Gamma}$ is isomorphic to $\Pi_{g\in\Gamma}R$

My question is what is the $\Pi_{g\in\Gamma} R$? I don't know it is meaning. And then why they are iso morphic ?

Thank you very much.

$\endgroup$
  • $\begingroup$ Do these maps respect structure? What kind of structure do they preserve? $\endgroup$ – Ashwin Trisal Apr 17 '18 at 1:01
  • $\begingroup$ If you just consider 'all' map, then the group structure is not used. Can view Gamma as a purly set $\endgroup$ – yaoliding Apr 17 '18 at 1:05
  • $\begingroup$ You can also imbue $R^\Gamma$ with a ring structure with a different multiplication using the group operation, but that multiplication might not be commutative. This ring is still isomorphic to the product of R’s as a module, not as a ring. $\endgroup$ – rschwieb Apr 17 '18 at 1:39
0
$\begingroup$

It is immaterial that $\Gamma$ is a group.

If $J$ is a set, then $\displaystyle\prod_{j\in J} R$ is the direct product of $|J|$ copies of $R$.

One of the concrete definitions of directed product is exactly the set of functions $J\to R$ with ring operations defined pointwise. The set of these functions is usually denoted $R^J$.

$\endgroup$
  • $\begingroup$ Thank you very much. I got it . I thougt it is might be important that $\Gamma$ is a group. $\endgroup$ – Mike Apr 17 '18 at 1:12
  • $\begingroup$ As a side note, it's important that we're taking all maps $\Gamma \to R$ of sets. If we restrict ourselves to homomorphisms of groups we won't necessarily get the ring structure. $\endgroup$ – leibnewtz Apr 17 '18 at 3:13
  • $\begingroup$ Homomorphisms from any group to a ring is (obviously) a ring. $\endgroup$ – xsnl Apr 17 '18 at 4:04

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.