A problem on the Jacobson radical, from Isaacs Graduate Algebra 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.
 A: 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.
