I have to prove that if $I,J$ are coprime ideals in a commutative ring $R$, and $i_0\in I, j_0\in J$ are s.t. $i_0+j_0=1$, then $$f:I\oplus J\to R\oplus IJ: (i,j)\mapsto (i+j,ij_0-i_0j)$$ is surjective. However, I don't see how to prove this.
Let $r\in R$, $\sum_{k=1}^ni_kj_k\in IJ$. We want to find $i\in I, j\in J$ s.t. $i+j=r$ and $ij_0-i_0j=\sum_{k=1}^ni_kj_k$. But I just don't see how to 'solve' this, even in the case $n=1$. Any hint would be appreciated.
I think you can do this directly. We have $r = r\cdot 1 = r(i_{0} + j_{0})$, thus we need $i+j = ri_{0} + rj_{0}$. So we need $i = ri_{0} + rj_{0} - j$.
Since we want $ij_{0} - i_{0}j = \sum_{k=1}^{n}i_{k}j_{k}$ we would also need $(ri_{0} + rj_{0} - j)j_{0} - i_{0}j = \sum_{k=1}^{n}i_{k}j_{k}$.
So we have, since $R$ is commutative, $-j = j(-j_{0} - i_{0}) = \sum_{k=1}^{n}i_{k}j_{k} - ri_{0}j_{0} - rj_{0}^{2}$. So let $j = -\sum_{k=1}^{n}i_{k}j_{k} + ri_{0}j_{0} + rj_{0}^{2}$ (the RH side is indeed in $J$).
Then $i = ri_{0} + rj_{0} +\sum_{k=1}^{n}i_{k}j_{k} - ri_{0}j_{0} - rj_{0}^{2} = ri_{0}(1 - j_{0}) + rj_{0} + \sum_{k=1}^{n}i_{k}j_{k} - rj_{0}^{2} = ri_{0}^{2} + rj_{0}(1 - j_{0}) = ri_{0}^{2} + rj_{0}i_{0} + \sum_{k=1}^{n}i_{k}j_{k}$
which is indeed in $I$ (we use the fact that $i_{0} + j_{0} = 1$ to simplify).
So we could take $j = -\sum_{k=1}^{n}i_{k}j_{k} + ri_{0}j_{0} + rj_{0}^{2}$ and $i = ri_{0}^{2} + rj_{0}i_{0} + \sum_{k=1}^{n}i_{k}j_{k}.$