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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?

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  • $\begingroup$ Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal $\endgroup$
    – B.Swan
    Mar 15, 2019 at 2:15
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    $\begingroup$ @B.Swan this approach doesn't work, to see why try writing out the details $\endgroup$ Mar 15, 2019 at 2:17
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    $\begingroup$ 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. $\endgroup$
    – B.Swan
    Mar 15, 2019 at 2:27

2 Answers 2

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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 $$(1,1)-(2,0)=(-1,1)\notin M_2$$

Therefore, that property is not satisfied in general.

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 $M_1+M_2=R$.

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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.

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