Then I suspect that the $\oplus$ is just an assertion that the $\sum$ is a direct sum.
Work by analogy with vector spaces. Assume $V$ is a vector space over a field $k$, and assume $W_1,...,W_n$ are sub-spaces of $V$.
Then we can define the sum $\sum W_i$, which is the smallest subspace of $V$ containing all the $W_i$. You can also define the vector space, $\oplus_i W_i$. This is an "exterior" sum, because its members are not members of $V$, but rather tuples $(w_1,...,w_n)$ with $w_i\in W_i$.
Now, there is a simple linear map, $\phi: \oplus W_i \rightarrow \sum W_i$ defined as $\phi(w_1,...,w_n) = \sum_i w_i$.
This map is not necessarily 1-1. Indeed, if $W_1\subset W_2$ are finite dimensional, then the dimension of $W_1\oplus W_2$ is the sum of the dimensions of $W_1$ and $W_2$, while the dimensions of $W_1+W_2=W_2$ is the dimension of $W_2$.
When this map is $1-1$, then the vector spaces $\oplus W_i$ and $\sum W_i$ are "isomorphic."
In that case, the statement $W=\oplus \sum_{i=1}^n W_i$ is a statement in two parts:
- $W=\sum_{i=1}^n W_i$ - that is, it is the smallest subspace of $V$ that contains all the $W_i$.
- The map $\oplus_{i=1}^n W_i \rightarrow \sum_{i=1}^n W_i$ is 1-1.
So this is one of those cases where the language is confusing, because it the $\oplus$ symbol here is an assertion about the nature of the $\sum$.
There is a way in which this analogy is precise, in that there is a notion of (left/right) $R$-modules which is analogous to the notion of a $k$-vector space, and (left/right) ideals of $R$ are sub-modules of $R$ when $R$ is considered as a (left/right) $R$-module.
Then a set of ideals has a direct sum as $R$-modules, but that sum is not a sub-module of $R$. However, as with vector spaces, there is a map from the direct sum to the sum inside $R$, and what $\oplus\sum$ is saying is that we are taking the sum inside $R$, but we are asserting that the map from the direct sum to the sum inside $R$ is 1-1.
Best Answer
For ($\Leftarrow$) take $x \in I_1 \bigcap I_2$. Then -x $\in I_1 \bigcap I_2$. So x + (-x) = 0. By hypothesis, x = 0.