I'm trying to show that if $A$ is an ideal of $R = R_1 \oplus R_2 \oplus \dots \oplus R_n$ then $A = A \cap R_1 \oplus A \cap R_2 \oplus \dots \oplus A \cap R_n$
I've been running around in my head nonstop on this one and can't seem to make any headway... can someone point me in the right direction?? Thanks!
Assuming that $R_1 , \dots , R_n$ are unital with unity $e_i$ for each $i$ and that $R_i$ is notation for the copy of each ring in the sum, we have that the element $e = \sum_{i = 1}^ne_i$ verifies $er = r$ for all $r \in R$. Moreover, $e_i^2 = e_i$ and $e_ie_j = 0$ if $i \neq j$. Now,
$$ A = eA = e_1A + \dots + e_nA_n $$
So we are left to see that $e_iA = R_i \cap A$ and that the sum is direct. If $a \in A \cap R_i$, then
$$ a = ea = e_ia \in e_iA $$
where here we use that $e_ja \in e_jR_i = e_je_iR_i = 0$ if $j \neq i$. Reciprocally, if $a \in e_iA$, then $a \in A$ because $A$ is an ideal and $a \in e_iA \subseteq e_iR = R_i$. Thus $A \cap R_i = e_iA$. Finally, we see that these ideals are direct summands: if $a_1 + \dots + a_n = 0$ with $a_i \in e_iA$, then for each $j$ we have that
$$ 0 = e_j0 = e_j(a_1 + \dots+a_n) = e_ja_1 + \dots + e_ja_n = \\ = e_je_1a_1 + \dots + e_je_na_n = e_je_ja_j = a_j $$
and so we see that $a_j = 0$ for all $j$, which proves that the sum is direct.