I need help solving a problem I have.
Let $R$ be a commutative ring. Prove that for the ideals $I$ and $J$ of $R$ the following two conditions are equivalent.
(a) The function $R\to R/I\times R/J$ given by $x\to(x+I,x+J)$, is surjective.
(b) $R=I+J$.
My thought for $a\implies b$ have I taken $(c+I,d+J)$ must have some $\alpha$ where $\phi(\alpha)=(\alpha+I,\alpha+J)=(c+I,d+J)$ which gives me that $\alpha-c\in I$ and $\alpha-d\in J$ meaning it is in one of the equivalence classes of the initial, after that I am kinda stuck.
For $b\implies a$ I am completely stuck, it feels like to me it shouldn't work but I am not entirely certain how to tackle it. How should I go about solving it?
It makes life easier if we add:
(c) The function $R \to R/I \times R/J$ has $(0,1)$ in its image.
Notice that (c) means that there is some $i \in R$ such that $i \equiv 0 \bmod I$ and $i \equiv 1 \bmod J$, i.e. $i \in I$ and $i-1 \in J$. In other words, $1 \in I+J$. This shows that (b) and (c) are equivalent. It is trivial that (a) implies (c). Conversely, If $(0,1)$ is in the image, then also $(1,0) = (1,1)-(0,1)$ is in the image. But $(1,0)$ and $(0,1)$ generate $R/I \times R/J$ as an $R$-module, so that actually everything lies in the image. So (c) implies (a).