About local ring defined in a different way

Question

Let $R$ be a commutative ring with $1$.

a. Show that for all $a,b\in R$, $ab$ is invertible iff $a,b$ are.

b. Show that if the set of non-invertible elements in $R$ is closed under addition then for all $a\in R$, if $a$ is non-invertible then $1-a$ is invertible.

c. Show the opposite of b.

d. A ring that satisfies ahe equivalent conditions in b,c is called a local ring. Show that in a local ring, $m$: the set of all non-invertible elements, is a unique maximal ideal, i,e, show that $m$ is an ideal and every ideal $I\subset R, I\neq R$ is contained in $m$.

e. Show that if $R$ has a unique maximal ideal $I$ then $R$ is a local ring. (Hint: given a non-invertible element $a\in R$ show that $aR$ is contained in $I$).

I managed to do parts a, b and c. I saw many many questions about this but all were different from mine since a local ring here is defined in another way..

Can you help please with parts d and e?

Answer 1

Allow me to try and present you with what I personally consider the best manner of introducing local rings, in the general (not necessarily commutative) setting.

Let us consider an arbitrary ring $(A, +, \cdot)$. We shall write:

• $\mathrm{U}(A)$ for the subset of invertible elements (the $\mathrm{U}$ stands for units) respectively $\mathrm{U}_{\mathrm{s}}(A)$ for the subset of left-invertible elements (recall that $x \in A$ is called left-invertible if it has a left inverse, in other words if there exists $y \in A$ such that $yx=1_A$)
• $I \leqslant_{\mathrm{s}} A$ (respectively $I \leqslant_{\mathrm{b}} A$) to express the fact that $I$ is a left (respectively bilateral) ideal of $A$. By bilateral ideals I mean what English refers to as two-sided ideals, just that I prefer my terminology
• $\mathscr{IdMax}_{\mathrm{s}}(A)$ for the set of all maximal left ideals, i.e. maximal elements of the set of proper left ideals ordered by inclusion. Recall the relation: $$A \setminus \mathrm{U}_{\mathrm{s}}(A)=\bigcup \mathscr{IdMax}_{\mathrm{s}}(A), \tag{s}$$ which means that the set of non-(left units) is equal to the union of all the maximal left ideals (this is valid for any ring, even for degenerate ones!). To give a succinct justification, the inclusion of the right-hand term in the left-hand one is due to the fact that any proper ideal must avoid the left units, whereas the reciprocal inclusion is based on the fact that any non-(left unit) generates a proper left ideal which is necessarily included in a maximal left ideal, as the set of proper left ideals is inductive.

If you might be wondering about the recurring "s" subscript in the notation, it is the initial of the word for "left" in both Latin as well as my mother tongue.

In this general context we shall state and prove the following:

Proposition. The following statements referring to given non-zero ring $A$ are equivalent:

1. $A \setminus \mathrm{U}(A) \leqslant_{\mathrm{b}} A$, i.e. the non-units form a bilateral ideal
2. the subset of non-units is closed under addition
3. given any $x \in A$ either $x$ or $1_A-x$ is invertible (as a side note this relation could be expressed formally as $A=\mathrm{U}(A) \cup \left(1_A-\mathrm{U}(A)\right)$)
4. given any $x \in A$ either $x$ or $1_A-x$ is left-invertible (this is the left-sided version of 3. above, and can just as well be formulated on the opposite lateral, the right side)
5. there exists a unique maximal left ideal.

Proof: since ideals of any of the three types (whether left, right or bilateral) are by definition additive subgroups, they are automatically closed under addition, which establishes 1. $\Rightarrow$ 2. Also, under the hypothesis of 2., it is clear that given arbitrary $x \in A$ at least one of the two $x$ and $1_A-x$ must be a unit, for otherwise - if they are both assumed to be non-units - their sum $1_A$ also ends up belonging with the non-units, which is a glaring contradiction; this establishes 2. $\Rightarrow$ 3.

Since units are in particular left units, the validity of 3. $\Rightarrow$ 4. is clear; let us at this stage also show why the reciprocal implication also holds, thus establishing the equivalence between 3. and 4. Assuming thus 4., in order to establish 3. it will suffice to prove that $\mathrm{U}_{\mathrm{s}}(A) \subseteq \mathrm{U}(A)$, in other words that any left-invertible is invertible. Hence consider arbitrary left-invertible element $x \in A$ with a left inverse $y$. Multiplying the relation $yx=1_A$ on the left by $x$ we infer $x=xyx$ and subsequently $x-xyx=\left(1_A-xy\right)x=0_A$. Since $A$ is non-zero (in other words $0_A \neq 1_A$), we automatically have $0_A \notin \mathrm{U}_{\mathrm{s}}(A)$ (no left-invertible can be zero) so in particular $x \neq 0_A$. If $1_A-xy$ were left-invertible, it could be cancelled from the above equation, yielding $x=0_A$ which we have just remarked cannot be the case. Hence - bearing in mind the current hypothesis - if $1_A-xy$ is not left invertible then $1_A-\left(1_A-xy\right)=xy$ necessarily is, fact which subsequently entails the left invertibility of $xy$ (if $z$ is a left inverse of $xy$, clearly $zx$ will be a left inverse of $y$). On the other hand, $y$ also has $x$ as a right inverse and therefore - by general monoid theory - we gather that it is invertible; finally, by left multiplication with $y^{-1}$ in the original equation $yx=1_A$ we furthermore gather that $x=y^{-1}$ is itself a unit.

As in general no proper ideal (regardless of which type) can contain any units, we have the automatic inclusion $I \subseteq A \setminus \mathrm{U}(A)$ valid for any proper left ideal $I <_{\mathrm{s}}A$. The subset $A \setminus \mathrm{U}(A)$ is clearly proper (it no longer contains $1_A$) and under the assumption of 1. it furthermore becomes a proper left ideal (since any bilateral ideal is in particular a left ideal). These remarks establish $A \setminus \mathrm{U}(A)$ as the maximum of the set of proper left ideals (ordered with respect to inclusion). Finally, it is a matter of elementary order theory that an ordered set possessing a maximum has only one maximal element (the maximum) and by virtue of this we conclude the validity of implication 1. $\Rightarrow$ 5.

Let us now see why 5. entails 3. Denoting the unique maximal left ideal with $M$, we deduce by virtue of the general relation (s) - mentioned in the preliminary notation paragraphs above - that $A \setminus \mathrm{U}_{\mathrm{s}}(A)=M$ in this particular situation, for the simple reason that $\mathscr{IdMax}_{\mathrm{s}}(A)=\{M\}$. This being the case, for no element $x \in A$ can both $x$ and $1_A-x$ fail to be left-invertible since that would entail $x, 1_A-x \in M$ and hence $1_A \in M$, clearly absurd. This shows that 5. implies 4. and we have seen above how 4. leads back to 3.

All that is left now in order to close the loop of equivalences is to establish the implication 3. $\Rightarrow$ 1. Assume thus that 3. holds, to the end of establishing $M \stackrel{\textrm{def}}{=} A \setminus \mathrm{U}(A)$ is a bilateral ideal of $A$. Since $A$ is non-zero, $0_A$ cannot be a unit and hence $0_A \in M$. The fact that $M$ is additively symmetric - i.e. the fact that $-M=M$ - is obvious, since the opposite of a unit is a unit and hence the opposite of a non-unit remains a non-unit (by opposite I mean symmetric with respect to addition, inverse being terminology that should be reserved strictly for multiplicative contexts). We argue by contradiction that $M$ is closed under addition: assume the existence of two elements $a, b \in M$ such that their sum $c=a+b \notin M$ no longer is in $M$, in other words is invertible. Let us remark that $\mathrm{U}(A)M, M\mathrm{U}(A) \subseteq M$, which is the simple observation that the product between a unit and a non-unit is - regardless of the order of the two factors - a non-unit. In particular, we have $c^{-1}a, c^{-1}b \in M$ since $a, b \in M$. However, $c^{-1}b=1_A-c^{-1}a$ and on account of 3. at least one of the two $c^{-1}a$, $c^{-1}b$ must be a unit. This constitutes a contradiction and means that $M$ is additively stable, property which in addition with the fore-mentioned ones establishes $M$ as an additive subgroup.

In order to establish $M$ as a bilateral ideal, it remains to show that $AM=\mathrm{U}(A)M \cup MM$ respectively $MA=MM \cup M\mathrm{U}(A)$ are both included in $M$. As we have already remarked above that the right terms in these unions are subsets of $M$, it will suffice to show that $M$ is multiplicatively stable to reach the end of our proof. Let therefore $x, y \in M$ be non-units. We rely once again on the hypothesis 3. to infer that $1_A-x$ is invertible and thus $\left(1_A-x\right)y=y-xy \in M$, by the same token that we have seen above (product between a unit and a non-unit). As $M$ is already established as an additive subgroup, we require one more step to infer from $y, y-xy \in M$ that $xy=y-(y-xy) \in M$, concluding our lengthy proof. $\Box$

With this preparation in place, we proceed to the general definition: a ring is local if it satisfies any (and hence all) of the equivalent conditions listed above. As a final remark, since the set of proper left ideals is inductive, condition 3. above is easily seen to be equivalent to the version 3.': the set of all proper left ideals admits a maximum.