Let $R = R_{1} \oplus R_{2}$ and let $I_{1}$ and $I_{2}$ be ideals of rings $R_{1}$ and $R_{2}$, respectively.
We consider $R_{1}$ and $R_{2}$ as subrings of $R$ under canonical embeddings $R_{i} \to R$ and thus $I_{1}$ and $I_{2}$ are considered as subrings of $R$.
I need to do the following two things:
a. Prove that $I_{1}$ and $I_{2}$ are ideals of $R$.
b. If each $I_{i}$ is a proper ideal of $R_{i}$, determine whether the ideal $I_{1}+I_{2}$ of $R$ is maximal. Is it prime?
For part (a), if we identify $I_{1} = I_{1}\times 0$, then in order for it to be an ideal of $R$, I need it to be both an additive subgroup of $R_{1} \times R_{2}$ and I would need $(R_{1}\times R_{2})(I_{1}) \subseteq I_{1}$ and $(I_{1})(R_{1} \times R_{2}) \subseteq I_{1}$. The latter two things especially are confusing me because of the direct sum notation.
Could anybody assist me with this part especially - hints for part (b) as to how to get at least started are also welcome.
Thank you.
For these it is helpful to keep a particular product in mind. Namely, $R$ is the ring of pairs $(r_1, r_2)$ under pointwise product and sum. $R_1$ are all the pairs $(r_1, 0)$ and $R_2$ is all the pairs $(0,r_2)$. Note that $R_1R_2 = 0$ and $R_1 + R_2 = R$.
a) The ideal $I_1$ corresponds to pairs $(i,0)$ where $i \in I_1$ as a subset of $R_1$. Then $RI_1$ = $(R_1 + R_2)I_1 = R_1I_1 + R_2I_1 = R_1I_1 = I_1$ because $R_2I_1 = 0$.
b) $I_1 + I_2$ is prime if and only if $R/(I_1 + I_2)$ is a domain. Hint: are products of rings usually domains?