I have seen many books using the idea that a commutative semi-simple ring with unity is finitely generated as an $R-$module but I do not understand why this is correct. Any elaboration will be appreciated!
My definition for a semi-simple ring is that it is a direct sum of simple ideals.
My intuition:
$R = (1)$ i.e., $R$ s finitely generated as an ideal by the element $1.$ and every ideal is an $R-$submodule and hence $R-$module. But by this intuition every ring with unity is finitely generated, no need for the word semi-simple.
Is my intuition correct?
Best Answer
Let's assume there is a decomposition $R = R_1 \oplus R_2 \oplus \cdots \oplus R_n \oplus \cdots$ of $R$ into simple $R$-modules. We can then take any $a\in R$ and write it as $a = \sum_i a_i$ (only finitely many non-zero terms), where $a_i \in R_i$. If $i\neq j$ and $r\in R_i, s\in R_j$, then $rs = 0$ (I'll leave checking that to you as an exercise). In particular, we may write $1$ as such a sum, so assume there are infinitely many simple summands. Then (reindexing if necessary) $1 = 1_1 + 1_2 + \cdots + 1_n$. Let then $0 \neq r \in R_{n+1}$ be arbitrary. We will then have $0 \neq r = 1r (1_1 + \cdots + 1_n)r = 1_1r + \cdots + 1_nr = 0 + 0 + \cdots + 0 = 0$, since there is no component of $1$ in $R_{n+1}$. The assumption that $R$ has infinitely many summands is therefore false, and so it must be a finite sum of simple modules. Every simple module is generated by a single element, so we obtain finite generation that way.
Sorry to say this, but your intuition didn't help in this situation.