So the question is as follows:
Suppose $U$ is an ideal of artinian ring $R$, then show that there is an ideal $V$ such that $U+V=R$ and $U\cap V \subseteq J(R)$ .
Let me describe my approach. I took the canonical homomorphism from $R$ to $R/J(R)$, and the image of U under this homomorphism, denoted $U'$ say, has a direct complement in $R/J(R)$ taken as a left module over itself (by the Wedderburn-Artin theorem, since $R/J(R)$ is semi-primitive and artinian), and I then took the preimage of this direct complement in the above homomorphism from $R$ to $R/J(R)$ as $V$. Then it is easy to show that $U+V=R$ and $U\cap V \subseteq J(R)$
The only problem is that the direct complement of $U'$ in $R/J(R)$ is only a left submodule of $R/J(R)$ as a left module over itself, and hence a left ideal of $R/J(R)$. But we require it to be a two-sided ideal for this to work.
Any ideas on how to fix this?