Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Let $R$ be a finite commutative ring. Consider elements $a,b \in R$ such that $Ra+Rb=R$. A paper I'm reading asserts that there exists some $x,y \in R$ such that $x(a+yb) = 1$.

Of course, it is clear that we can find some $x,y \in R$ such that $xa+yb=1$, but it is not clear that we can get the above stronger statement. It is equivalent to saying that there exists some $y \in R$ such that $a+yb$ is a unit.

Can anyone help me?

share|cite|improve this question
up vote 9 down vote accepted

This is true more generally in any (commutative unitary) semi-local ring (ring having only finitely many maximal ideals), and is a consequence of the following special form of Prime Avoidance Lemma :

Let $p_1, \dots, p_n$ be prime ideals, let $I$ be an ideal and $a\in R$ such that $aR+I\not\subseteq \cup_{i\le n} p_i$, then there exists $\beta\in I$ such that $a+\beta\notin \cup_{1\le i\le n} p_i$.

Now if $aR+bR=R$, then apply the above result to $I=bR$ and $p_1,\dots, p_n$ the maximal ideals of $R$. Note that the union of the $p_i$ is exactly the set of the non-invertible elements of $R$.

A proof of the Prime Avoidance Lemma can be found in Kaplansky, Commutative Rings, p. 90, Thm 124, or Bruns & Herzog, Cohen-Macaulay Rings, Lemma 1.2.2.

Note that for finite rings, there is probably a simpler proof.

Edit: second proof. How can I forget my favorite CRT ?!

Let $R$ be semi-local (this includes of course the finite ring: if $m_1, \dots, m_n$ are pairwise distinct maximal ideals, then $m_1, m_1\cap m_2, \dots, m_1\cap ...\cap m_n$ is a strictly decreasing sequence of subgroups, hence $n$ is bounded by the cardinality of $R$). Let $m_1,\dots, m_n$ be the maximal ideals of $R$. For each $i\le n$, there exists $c_i\in R$ such that $$a+bc_i\not\equiv 0 \mod m_i.$$ This is clear if $b\not\equiv 0 \mod m_i$, otherwise, $b\in m_i$, thus $a\notin m_i$ by $aR+bR=R$, then take any $c_i\in R$. Now by CRT, there exists $c\in R$ such that $c\equiv c_i \mod m_i$ for all $i\le n$. Therefore $a+bc\notin m_i$ for all $i\le n$. This implies that $a+bc$ is a unit.

share|cite|improve this answer
Great, this is just what I needed. Thanks! – Todd Sep 4 '13 at 22:18
@Cantlog: What do you mean by CRT – user 1 Mar 12 '14 at 7:54
@ucf: Chinese Remainder Theorem. – Cantlog Mar 12 '14 at 20:54

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.