Let $R$ be a finite ring, and $G$ be a finite group. I need to compute the number of elements of the group ring $R(G)$ in terms of $|R|$ and $|G|$ (where $|R|$ is the number of elements in the ring $R$, and $|G|$ is the number of elements in the group $G$).
Recall that a group ring $R(G)$ is the set of equivalence classes of all formal sums
$r_{1}g_{1}+r_{2}g_{2}+\cdots + r_{k}g_{k}$
where the $r_{i} \in R$, $g_{i} \in G$.
We say that two formal sums are said to be equivalent if they have the same reduced form.
In each sum, we have $k$ terms, and $|R|$ choices for each coefficient, so if we didn't care about repeats or whether we should choose only one representative from each equivalence class, I'd say the group ring contained $k|R|$ elements.
However, I'm assuming that in calculating the size of a group ring, we count each representative of an equivalence class only once. So, how do I factor this in when counting the number of elements in $R(G)$? Also, how does $|G|$ come into play?
Thank you in advance!
Suppose that $G=\{g_1,\cdots,g_n\}$. Then, I like to write the group ring $R[G]$ as the collection of all sums of the form $$ r_1e_{g_1}+r_2e_{g_2}+\cdots+r_ne_{g_n} $$ where $r_i\in R$. This differs from your notation in the question by replacing $g_i$ with $e_{g_i}$. I like this notation because it's harder to confuse the group elements with the ring elements.
The equality is just like for a basis: $$ r_1e_{g_1}+r_2e_{g_2}+\cdots+r_ne_{g_n}=s_1e_{g_1}+s_2e_{g_2}+\cdots+s_ne_{g_n} $$
iff $r_i=s_i$ for all $i$. To turn this into a ring, we use the product $$ (r_ie_{g_i})(r_je_{g_j}):=r_ir_je_{g_ig_j}. $$ In other words, multiply the ring and group elements separately. This product is then extended by linearity to sums.
Since elements are of the form $$ r_1e_{g_1}+r_2e_{g_2}+\cdots+r_ne_{g_n}, $$ there are $|G|$ different $e_{g_i}$'s and each $e_{g_i}$ has $|R|$ different coefficients. This results in $|R|^{|G|}$ different ring elements.