Local rings, idempotents and non-invertible elements

Description of the problem

Today’s post is on an abstract algebra question.

Assume $A$ is an artinian ring. I make no assumptions on the commutativity of $A$. Then the following are equivalent:

The set of non-invertible elements form a non-trivial two-sided ideal.
$A$ is a local ring.
The only idempotents of $A$ are $0$ and $1$.

Proof

(1) implies (2)

Assume that $A\setminus A^\times$ is a two sided ideal. Let $I$ be an arbitrary non-trivial left ideal. Since $I\cap A^\times = \emptyset$, we can conclude that $I\subseteq A\setminus A^\times$. This means all ideals lie in $A\setminus A^\times$, including all maximal ideals. This forces the existence of a unique maximal ideal.

(2) implies (3)

Assume $A$ is local and assume, again by way of contradiction, that there is a non-trivial idempotent $e$. Let $\mathfrak{m}$ be the unique maximal ideal of $A$. Since $e$ is a non-trivial idempotent both $e$ and $1-e$ are non-invertible, and therefore, lie in the unique maximal ideal $\mathfrak{m}$ because of Zorn’s Lemma. $\mathfrak{m}$ is a two sided ideal, in particular an abelian group under addition, we get that $1 = e + (1-e)$ is also in $\mathfrak{m}$ which is a contradiction.

(3) implies (1)

Assume $0$ and $1$ are the only idempotents of $A$. I would like to show that the set of non-invertible elements form a two sided ideal. Let me start with the fact that the set of non-invertible elements is closed under the left and the right action of $A$. The proofs for left and right actions are similar. So, I will give only one direction. I would like to show that the following statement is true: \[ \forall a\in A,\ \forall x\in A,\ x\in A\setminus A^\times \implies ax\in A\setminus A^\times \] With a small logical trick, instead of the statement above, I will prove \[ \forall a\in A,\ \forall x\in A,\ ax\in A^\times \implies x\in A^\times \] Since $ax\in A^\times$ there exists $y\in A$ such that \[ yax = 1 \qquad axy = 1 \] I would like to show that $x\in A^\times$. It is clear from the first equality that $x$ has a left inverse. So, I must show that it also has a right inverse. It is clear from that equality that the right inverse should be $ya$. Now, consider \[ (xya)(xya) = x(yax)ya = xya \] This means $xya$ is an idempotent. Since the only idempotents are $0$ and $1$, we have two possibilities

If $xya=0$ then $0=xyax=x$ which is a contradiction since $yax=1$.
If $xya=1$ then $x$ has a right inverse as I wanted to show.

So, this finishes the proof that $A\setminus A^\times$ is closed under the left action of $A$. As I mentioned earlier, the right action case is proven similarly.

Now, I must show that $A\setminus A^\times$ form an abelian group under addition: Assume on the contrary that we found two elements $x,y\in A\setminus A^\times$ such that $x+y\in A^\times$. Then there must be an element $a\in A$ such that \[ a(x+y) = ax + ay = 1 \] I already showed that $A\setminus A^\times$ is closed under the action of $A$ on both left and right. So, I can assume without loss of generality, that $x+y=1$. Since $x$ is not invertible, the left ideals $Ax^n$ generated by $x^n$ are not equal to $A$ for every $n$. Moreover, I have descending chain \[ Ax \supseteq Ax^2 \supseteq Ax^3 \supseteq \cdots \] Because $A$ is artinian, the descending chain of ideals must stabilize. Similarly, there is an increasing chain of annihilator ideals \[ Ann_L(x) \subseteq Ann_L(x^2) \subseteq \cdots \quad\text{ where }\quad Ann_L(w) = \{z\in A\|\ zw = 0 \} \] Since $A$ is artinian, it is also noetherian because of Hopkins-Levitzki Theorem. So the ascending chain of annihilator ideals also stabilize. Then I can find an natural number $N\in\mathbb{N}$ such that $Ax^{N+1} = Ax^N$ and $Ann_L(x^{N+1}) = Ann_L(x^N)$ at the same time. Now, since $Ax^{2N} = Ax^N$, there must be an element $a\in A$ such that $ax^{2N} = x^{N}$. This means \[ (1-ax^N) x^N = 0 \] and therefore $1-ax^N$ is in $Ann_L(x^N)$. This means for every $z\in A$ we can write \[ z = zax^N + z(1-ax^N) \in Ax^N + Ann_L(x^N) \] This means $A = Ax^N + Ann_L(x^N)$. I claim that this is a direct sum. In other words, I claim that the intersection of $Ax^N$ and $Ann_L(x^N)$ is trivial. Assume there is an element $b$ in the intersection. Then \[ b = ux^N \qquad 0 = b x^N = u x^{2N} \] But then $u\in Ann_L(x^{2N}) = Ann_L(x^N)$. This means $b = ux^N = 0$ as well. This simply yields $A$ as a direct sum of two ideals $A = Ax^N\oplus Ann_L(x^N)$.

Now, I can write $1\in A$ as a sum of two elements $1 = e + f$ where $e\in Ax^N$ and $f\in Ann_L(x^N)$. It is easy to see that both $ef$ and $fe$ are in the intersection of these ideals, and therefore, must be trivial. Then \[ e = e(e + f) = e^2 \quad\text{ and }\quad f = f(e+f) = f^2 \] and we get two idempotents. Since $A$ has no nontrivial idempotents we have two options:

If $f = 1$ then $Ann_L(x^N) = A$ and we see than $x^N=0$. Notice that if $(1-x)=y\in A\setminus A^\times$, we must also have the elements \[ 1 - x^n = (1 + x + \cdots + x^{n-1})(1-x) \] in $A\setminus A^\times$ for every $n$ since it is closed under the left action of $A$. This is a contradiction as $1-x^N = 1$.
If $e = 1$ then $Ax^N = A$ and $x$ is necessarily left invertible. But if $x$ is left invertible it is also right invertible using the fact that $0$ and $1$ are the only idempotent. Now, we get another contradiction.

This final statement tells us that we can not let $x+y\in A^\times$. Hence we get $x+y\in A\setminus A^\times$ when $x$ and $y$ are chosen from $A\setminus A^\times$. This finishes the proof that $A\setminus A^\times$ is an abelian group under addition.