# Why is the identity a finite sum of elements in this left ideal decomposition?

Suppose you have a semisimple ring $R$, and want to decompose it into a sum of simple left ideals. Let $\{L_i\}$ be a family of simple left ideals, such that no two are isomorphic, and any simple left ideal of $R$ is isomorphic to some $L_i$.

Then writing $R_i=\sum_{L\simeq L_i}L$, it follows that $R=\sum_{i\in I}R_i$, and so $1=\sum_{i\in I}e_i$ for $e_i\in R_i$. Apparently this sum is actually finite, but there is no explanation as to why. Why are almost all $e_i=0$ in this sum for $1$?

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We don't seem to have a way of making sense of infinite sums. $\sum_{i \in I} R_i$ has to stand for the set of all $\sum_{i \in I} x_i$, $x_i \in R_i$ for all $i$, $x_i = 0$ for almost all $i$. – Dylan Moreland Apr 12 '12 at 6:26
Oh, so the sum is finite just because it wouldn't make sense otherwise? – Cenk Apr 12 '12 at 6:30

Instead of this being because of any particular property of $1$, I think this is a consequence of the definition of infinite sums of ideals (as well as more general algebraic objects like vector spaces etc.). All elements of $\sum L_i$ look like a finite sum of elements in $L_i$ by definition. There is no "sum of infinite elements".
In fact, we can't even necessarily define what an infinite sum of elements would look like. We can do it if you take your ring to be something like $\mathbb{R}$, but that's because in addition to the ring structure the real numbers have additional geometric structure that gives rise to things like limits. In most rings, you won't have that.
In fact $\sum_i L_i$ is usually defined as the smallest ideal that contains all the $L_i$. It then turns out that this is the same as the ideal of all finite sums made from elements of the $L_i$. – Marc Olschok Apr 13 '12 at 16:02