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A map $f:R\rightarrow R/I \times R/J$ is defined by $$f(a)=\left( a+I,a+J \right)$$where $I,J$ are ideals of a ring $R$.

I have shown $f$ is a homomorphism, now I'm asked to find $\ker f$. Somewhere below there must be a false statement, because I don't get the required result. Initially I thought it to be $\left\{ 0_R \right\}$, but next we're asked to give examples where $f$ is onto and not. But following the first isomorphism theorem I get $$Im f\cong (R/I \times R/J)/\ker f=(R/I \times R/J)/\{0\}$$ and since $R/I , R/J$ are rings it follows that $R/I \times R/J\cong(R/I \times R/J)/{0}$, and with $Imf\subset R/I \times R/J$ I get that $f$ is surjective.

What am I missing? Thanks

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    For surjectivity, see [here](http://math.stackexchange.com/questions/2116393/). – 2017-01-29
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    Thanks. Your answer in the given link suggests the case where $I+J=R$. In the case where the kernel is trivial, i.e. $I\cap J={0}$ as suggested, does the above isomorphism argument hold? – 2017-01-29
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    If $I \cap J=0$ then $R \cong im(f)$, and if $im(f) = R/I \times R/J \iff I+J=R$. – 2017-01-29
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    Oh and now I see that I also wrote the isomorphism theorem totally wrong. It should be $Imf\cong R / \ker f$... baah... Thanks! – 2017-01-29

2 Answers 2

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Hint

The kernel should be $I \cap J$ because any $a$ that maps to $(0+I, 0+J)$ must be a member of both $I$ and $J$.

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For a slightly more abstract approach consider the precompositions

$$ I \to R \to R/I \times R/J$$ and $$ J\to R\to R/I \times R/J$$ to notice that they coincide with the zero maps. Hence your map $f$ factors through $f'$ $$R\to R/(I\cap J)\stackrel {f'}\to R/I \times R/J$$ which injects. The isomorphism theorems now tell you the kernel.

And congratulations, as you noticed correctly for coprime ideals $I$ and $J$ you get an isomorphism. All your efforts will prove what is called the Chinese Remainder Theorem.