Let $A$ be a commutative ring and $M$ be a proper maximal ideal in $A$. Show the following properties:

(a) If each $a \in A \setminus M$ is a unit element in $A$, then $M$ is the only maximal ideal in $A$.

(b) If each element of the form $1+m \in A$ with $m \in M$ is a unit in $A$, then $A$ has a unique maximal ideal.

I think I could show (a), I have problems with (b), I'll write what I've done so far:

(a) Suppose there exists $M' \neq M$ a maximal ideal in $A$. Then there is some $b \in M'$ such that $b \not \in M$. Then $b \in A \setminus M$, so there $b$ is a unit. Since $M'$ is ideal (bilateral ideal since $A$ is commutative), we have $1=b^{-1}b \in M'$. Now $M$ is a maximal ideal, this means there exists $m \in M$ such that $m \not \in M'$. But $m=m1 \in M'$, which is absurd. The absurd comes from the assumption that there is another maximal ideal different from $M$, so $M$ is the only maximal ideal in $A$.

I would appreciate hints for (b).

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    $\begingroup$ I assume that in part (b), you mean $1 + m \in A$, not $1 + m \in M$. $\endgroup$ – Bungo Sep 18 '14 at 18:14
  • $\begingroup$ Thanks for the correction. $\endgroup$ – user16924 Sep 18 '14 at 19:18

You can use part (a) to prove the second part.

Let $x\in A\setminus M.$ Consider $N=M+xA=\{m+xa|\,m\in M, a\in A\}$. Clearly $N$ is an ideal of $A$ and contains $M.$ Thus it has be the whole ring $A.$ In particular $m+xa=1$ for $m\in M,\, a\in A$. thus $xa=xa=1-m.$ Now you can apply part (a);) it follows that $a$ is a unit. Done!

  • $\begingroup$ "...thus $xa=ax=1-m$...", how one deduces from here that $x$ is a unit? From there I would only conclude that $xa$ is a unit. $\endgroup$ – user16924 Sep 18 '14 at 20:02
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    $\begingroup$ $1=axu=x(au)=(au)x$ $\endgroup$ – BigM Sep 18 '14 at 20:10
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    $\begingroup$ I've just realized the same thing, I got it, thanks! $\endgroup$ – user16924 Sep 18 '14 at 20:10

After "$1=b^{-1}b \in M'$" you go a bit off the rails, because you could simply conclude at this point "Thus $M'=A$, contradicting our hypothesis that $M'\neq A$.

Earlier I misread something about the statement of (b) and tried an irrelevant example. Then BigM beat me to the most reasonable proof. I can't give a better one: so I recommedn you check that one.

  • $\begingroup$ Thanks for the remark in (a). $\endgroup$ – user16924 Sep 18 '14 at 20:03

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