Given ideals $I, J$ of a commutative ring $R$, suppose that for each $a, b \in R$, there exists some $x \in R$ such that $x \equiv a \pmod{I}$ and $x \equiv b \pmod{J}$. Is it true that $I + J = R$?
How would one attack this problem without using tensors or bilinear mappings?
Hint: Consider the $x$ you get from $(a,b)=(0,1)$. Can you use this $x$ to show that $1\in I+J$?
A full solution is hidden below.
Let $x\in R$ be such that $x\equiv 0\pmod{I}$ and $x\equiv 1\pmod{J}$. Then $x\in I$, and $1-x\in J$. So $1=x+(1-x)\in I+J$. Since $I+J$ is an ideal, this means $I+J$ is all of $R$.
Hint $\,\ a\!-\!x \in I,\ x\!-\!b \in J $
$\,\ \ \Rightarrow\ \ \underbrace{a\!-\!x\ \ +\ \ x\!-\!b}_{\Large \color{#c00}{a\,-\,b}} \, \in\, I+J.\ $ We seek $\,1 = \color{#c00}{a\!-\!b} \in I+J\,$ so choose $\,a,b\, = \ \ldots$