This is problem 14.10 from Isaacs Graduate Algebra.

Let $U$ and $V$ be ideals of a ring $R$ and assume $U+V$ = $R$, and $U \cap V \subseteq J(R)$ . Suppose that $v \in V$ and that $U + v$ is invertible in $R/U$. Show that there exists $u \in U$ such that $u+v$ is invertible in $R$.

Any hints would be appreciated.

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I changed $ν$ to $v$, hope this is what you meant. –  Rudy the Reindeer Jul 27 '12 at 11:38

First, a remark: to say that $U + v$ is invertible in $R/U$ (for some $v \in V$) is to say that there exists $w \in R$ such that $vw = 1 + u$ for some $u \in U$. Thus the assumption that $U + V = R$ is superfluous; it follows from the assumption that $v$ is invertible mod $U$.

Now, here is a sketch of a proof:

To begin, show that if $I \subset J(R)$ then any lift to $R$ of a unit in $R/J(R)$ is a unit in $R$. (This uses one of the standard characterizations of $J(R)$ in terms of its relationship to units.)

Thus we may replace $R$ by $R/U\cap V$, and hence suppose that $U \cap V = 0$. Now consider the natural map $R \to R/U \times R/V$, given by $r \bmod U\cap V \mapsto (r\bmod U, r \bmod V)$. The assumption that $U + V = R$ shows that this map is an isomorphism.

Let $u \in R$ map to the element $(0,1)$. This is evidently an element of $U$ (hence my choice of notation). Now show that $u + v$ maps to a unit in $R/U \times R/V$, and hence is a unit of $R$.

You can also solve the problem by more explicit computations as well; indeed, the preceding argument can be made quite constructive. If you want to do this, I'll leave it as an exercise.

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