# Does this property of comaximal ideals always hold?

I am reading a paper in which the following result is used, but I can’t see the proof of this.

Let $$R$$ be a commutative ring with only two maximal ideals, say $$M_1$$ and $$M_2$$. Suppose $$m_1 \in M_1$$ is such that $$m_1 \notin M_2$$. Then can we always find $$m_2 \in M_2$$ such that $$m_1+m_2=1$$?

Any ideas?

• Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal – B.Swan Mar 15 at 2:15
• @B.Swan this approach doesn't work, to see why try writing out the details – Alex Mathers Mar 15 at 2:17
• Set $I=(M_2 \cup \{m_1\})$, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x \in M_2$ and $r \in R$. Since $m_1 \notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s \in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out. – B.Swan Mar 15 at 2:27

Take $$R=\mathbb{Q}\times\mathbb{Q}$$, $$M_1=\mathbb{Q}\times\{0\}$$, $$M_2=\{0\}\times\mathbb{Q}$$, and $$m_1=(2,0)\in M_1\setminus M_2$$. Then $$(1,1)\in\mathbb{Q}\times\mathbb{Q}$$ satisfies that $$(1,1)-(2,0)=(-1,1)\notin M_2$$
Maybe the property that they are really using is that there exist $$a\in M_1$$ and $$b\in M_2$$ such that $$a+b=1$$. Not arbitrary $$a,b$$. This other property is immediate by using the maximality of $$M_1$$ and $$M_2$$, which implies that $$M_1+M_2=R$$.
First notice that $$1-m_1$$ cannot be a unit, because this would imply $$m_1$$ is in the Jacobson radical of $$R$$, and in particular we would have $$m_1\in M_2$$.
Now it follows that the ideal of $$R$$ generated by $$1-m_1$$ must be contained in a maximal ideal, but it cannot be contained in $$M_1$$ because then it would follow that $$1\in M_1$$. Thus this ideal is contained in $$M_2$$ (the only other maximal ideal), i.e. you get $$1-m_1\in M_2$$.
Edit: I think my reasoning for $$1-m_1$$ not being a unit is wrong (it seems we would need that $$1-m_1x$$ is a unit for every $$x\in R$$ to conclude $$m_1$$ is in the Jacobson radical). The rest of the argument goes through, so I'm going to leave my answer up for a while in hopes that somebody can help figure that part out.