Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I changed $ν$ to $v$, hope this is what you meant. – Rudy the Reindeer Jul 27 '12 at 11:38
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.