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.

  • $\begingroup$ 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

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 $R^J$.

  • $\begingroup$ 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

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