Local Ring with an another definition

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 $I$? Commented Feb 26, 2016 at 14:52
• I is generated by u and ? Commented Feb 26, 2016 at 15:04
• Don't worry about generators. An ideal containing a unit is … what? Commented Feb 26, 2016 at 15:08
• If an ideal contained a unit so I=A (the ring) ? Commented Feb 26, 2016 at 15:11
• Right. So every proper ideal of $A$ contains only non-units, and therefore … Commented Feb 26, 2016 at 15:12

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.