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:
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.
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.
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
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.