Let $I$ and $J$ be ideals of a commutative ring $R$ such that $I + J = R$. Show that there is an ideal $K$ in $R$ with $R/K \cong R/I \times R/J$

Question

Let $I$ and $J$ be ideals of a commutative ring $R$ such that $I + J = R$. Show that there is an ideal $K$ in $R$ with $R/K \cong R/I \times R/J$

I tried solve this using the first isomorphism theorem. But I don’t know how to prove that the homomorphism $\phi: R \to R/I\times R/J$ given by $\phi (a) = (a+I)\times(a+J)$ is surjective. I've tried others functions, but I failed.

Any help?

Do you assume your ring contains 1? – Oliver Clarke Jul 11 '17 at 15:24 — Oliver Clarke, Jul 11 '17 at 15:24

score 1 · Answer 1 · answered Jul 11 '17 at 15:26

Consider $f=(p_I,p_J):R\rightarrow R/I\times R/J$ where $p_I$ is the quotient map $R\rightarrow R/I$, you have to show that $f$ is surjective and its kernel is $K$. Let $r,r'\in R$, there exists $i,i'\in I, j,j'\in J$ with $i+j=r, i'+j'=r'$. We have $p_I(i+j)=p_I(j)=p_I(r), p_J(r')=p_J(i')$. We deduce that $f(i'+j)=(p_I(i'+j),p_J(i'+j))=(p_I(j),p_J(i'))=(p_I(r),p_J(r'))$ is surjective. We denote by $K$ its kernel and deduce that $R/K$ is isomorphic to $R/I\times R/J$.

Let $I$ and $J$ be ideals of a commutative ring $R$ such that $I + J = R$. Show that there is an ideal $K$ in $R$ with $R/K \cong R/I \times R/J$

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