If $J/I = \sum_{i=1}^{n} \bar x_i R/I $ then $J=\sum_{i=1}^{n} x_iR+I$

Question

Let $R$ be a commutative ring with unity, $I$ and $J$ ideals such that $ I\subseteq J$. Show that if $J/I = \sum_{i=1}^{n} \bar x_i R/I $ then $J= \sum_{i=1}^{n} (x_iR)+I $.

My attempt : $ \bar x_i R/I = (x_i+I)(r+I)$ , $r\in R$ so we have :

$ J/I = \sum_{i=1}^{n} (x_i \cdot r )+I$

I am stuck here, I think I am missing some obvious detail.

Any hints ?

Answer 1

If $J/I = \sum_{i=1}^{n} \bar x_iR/I$ with $x_i\in J$ then $x_iR\subseteq J$ for all $i$ and $\sum_{i=1}^n x_iR+I\subseteq J$. On the other hand, given $y\in J$ then there are $a_1,...,a_n\in R$ such that $\bar y=\bar a_1\bar x_1+\cdots+\bar a_n\bar x_n$ in $R/I$, i.e., $y-(a_1x_1+\cdots+a_nx_n)\in I$, thus $y\in\sum_{i=1}^n a_iR +I$.