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Suppose we have an isomorphism $$ \phi: R/(I \cdot J) \rightarrow R/I \times R/J $$ $$ \text{i.e.} \quad R/(I \cdot J) \cong R/I \times R/J $$ and we want to find out which element $ \overline{a} \in R/ (I \cdot J)$ maps to an element $(\overline{x},\overline{y}) \in R/I \times R/J $.

Is there a general way to go about doing this/an idea to have in mind? If $R$ is commutative with identity, then we have the relations $I+J=R$ and $I \cap J= I \cdot J $ which allow us to deduce the isomorphism by the Chinese Remainder Theroem for rings. But...

The solutions to questions like these always make you verify that $i+j=1$ for some $i \in I, j \in J$ (the questions I've been doing use principal ideals but I think we can generalise to arbitrary finitely generated ideals?) early in the question and then end up using the fact that the element $\overline{a}$ which maps to $(\overline{x},\overline{y})$ is of the form $$\overline{a}=\overline{i \cdot \overline{x}+j \cdot \overline{y}}.$$

I'm having some trouble understanding why this is. Intuitively speaking, I thought that perhaps the reason why we can write $\overline{a}$ in this form is because under $\phi$, the $i$ term in $\overline{a}$ will get reduced to $0$ when modded by $I$ and likewise for the $j$ term. But I still can't really see how this means that $ \overline{a} \mapsto (\overline{x},\overline{y}) $ because after $ \overline{a} $ being modded out by $I$ to get the first term of the tuple, the $j \cdot \overline{y} $ term will have also been modded by $I$ (changing its value (?)).

Or, is this simply a direct result of the definition

$$ I \cdot J := \{ \alpha_1 \beta_1 + ... + \alpha_n \beta_n | \alpha_i \in I, \beta_i \in J \}$$

Feel free to change any of my notation if it's too cumbersome. Many thanks for any help.

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    The identity $i + j = 1$ guarantees that $j$ becomes $1$ mod $I$ and similarly $i$ becomes $1$ mod $J$. So $iy + jx$ is the element that becomes $x$ mod $I$ and $y$ mod $J$. – 2017-01-02
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    Thank you, I think I understand now. So just to clarify, you're saying that if we have $1=i+j$, with $i \in I, j \in J$, then $j \equiv 1 \space \text{mod} \space I$ and $i \equiv 1 \space \text{mod} \space J$. Then to get the right elements in the tuple, we multiply $i$ by $y$ and $j$ by $x$, since these will indeed now give $(x \space \text{mod} \space I, y \space \text{mod} \space J)$. – 2017-01-02

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