Local Ring with an another definition

Question

We define a local ring with :

The sum of any two non-units in $R$ is a non-unit.

And we want to show that :

If $A$ is a local ring so $A$ has a unique maximal ideal (set of non-units).

I know how to show that the set of non-units is an ideal but maximal?
I have an idea : Let $A$ a commutative ring and $K$ the set of non-units
We know that: $$ (x)=A \iff x \notin K $$ $$ (x) \neq A \iff x \in K. $$

Krull's Theorem asserts that a nonzero ring has at least one maximal ideal. So $K$ is an maximal ideal?

Suppose an ideal $I$ contains a unit $u$. What follows about $I$? — Daniel Fischer, Feb 26 '16 at 14:52
Don't worry about generators. An ideal containing a unit is … what? — Daniel Fischer, Feb 26 '16 at 15:08
Right. So every proper ideal of $A$ contains only non-units, and therefore … — Daniel Fischer, Feb 26 '16 at 15:12
You probably have the (well, one) right argument in your mind, but you haven't spelled it out yet. Maybe you think it's obvious - and in some sense it is - but it needs to be stated in a proof. — Daniel Fischer, Feb 26 '16 at 15:28
I just begin study of ring and I'm not used to it. But thanks for your help. — doeup, Feb 26 '16 at 15:35
Since you "just begin study of ring", maybe this can be a help. — user 1, Feb 26 '16 at 19:27
@Daniel Fischer how do you suggest to show that the set of non units is an ideal in this case? — user652838, Nov 19 '22 at 15:56

score 3 · Answer 1 · answered Feb 28 '16 at 06:37

Looks like you've pretty much figured it out. I'm gonna write out the details anyway because that's just what I like to do.

In any commutative ring $R$ with identity, an ideal $J$ of $R$ is proper (that is, $J$ is not equal to the whole ring) if and only if $J$ does not contain any units. Certainly an ideal without units is proper. Conversely if $J$ does contain a unit $u$, then $1 = u^{-1}u \in J$, hence for any $a \in R$, you have $a = a \cdot 1 \in J$. Thus $R = J$, so $J$ is not proper.

If $R$ is a local ring, then the set $I$ of nonunits of $R$ is an ideal, as you've shown. It also must be a maximal ideal: the only way $I$ could be any bigger is if it contained a unit, in which case $I$ would not be a proper ideal, and maximal ideals are by definition proper.

Finally, there cannot be any other maximal ideals besides $I$: if $M$ is any maximal ideal of $R$, then $M$ is proper, and so $M$ cannot contain any units. Hence $M \subseteq I$, which implies $M = I$, because $M$ is a maximal ideal.

Local Ring with an another definition

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