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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?

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Sorry I misunderstood the question earlier. My updated answer should be much better! – rschwieb Jul 31 '12 at 14:49

Since $R/J$ is semisimple, you know all about its two-sided ideals: in the Wedderburn decomposition into simple rings $R/J=\oplus A_i$, you change any subset of the $A_i$ to zero and you have an ideal, and all ideals are produced this way. Obviously if you take an ideal like this and "flip it" so that the places which were previously nonzero are now zero, and previously zero places are now nonzero, then you have a complementary two-sided ideal.

So, for your ideal $V$, take the corresponding ideal $(V+J)/J$ in $R/J$ and look at its expression in terms of the decomposition $R/J=\oplus A_i$ and if it is zero on coordinate $j$ then use $A_j$ and if it is nonzero on coordinate $j$ then use $0$ on that coordinate. Lift these ideals to $R$ and you are done.

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