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.
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 $R^J$.