Coprime elements in finite rings

Solution 1:

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.