Do the complements of this chain of submodules form a chain?

Question

I have two questions about the answer here:

In a left noetherian ring, does having a left inverse for an element guarantee the existence of right inverse for that element?

First question: they say that the complements (as left submodules, I assume) form an ascending chain. But why? Certainly this doesn't hold for general complements of chains of modules; just because we have inclusion $N \subseteq M$ doesn't mean that the complement of $M$ is contained in that of $N$.

I assume they're using the fact that ba=1?

Second question: why do the complements exist at all? If $R$ is commutative then I can see why; we can fit $Rb^n$, $R$, and the quotient into an exact sequence and show it's split using $ba=1$ (a left inverse for the inclusion is given by right multiplication by $a^nb^n$). But what if $R$ is not commutative?

Answer 1

It is true that $N\subseteq M$ does not mean that any complement of $M$ is contained in any complement of $N$. However, it does imply that for any complement $P$ of $M$, there exists a complement $Q$ of $N$ such that $P\subseteq Q$ (assuming $N$ does have a complement). So, given a descending chain of direct summands, you can choose complements of them one by one so that they form an ascending chain.

To prove the existence of such a $Q$, let $Q_0$ be some complement of $N$ and define $Q=P+(Q_0\cap M)$. Clearly $P\subseteq Q$; I claim $Q$ is a complement of $N$.

First, if $x\in Q\cap N$, then $x=p+y$ for some $p\in P$ and $y\in Q_0\cap M$. But then $p=x-y\in M$ since $x,y\in M$ so $p=0$ since $P\cap M=0$. We then have $x=y$ which implies $x=0$ since $x\in N$ and $y\in Q_0$.

Second, we must show that $Q+N$ is the entire module. For any $x$, we can write $x=p+m$ for $p\in P$ and $m\in M$. Since $Q_0$ is a complement of $N$, we can write $m=q+n$ for $q\in Q_0$ and $n\in N$. Moreover, since $m,n\in M$, we have $q\in M$. Thus we have $x=p+q+n$ where $p\in P$, $q\in Q_0\cap M$, and $n\in N$. This shows that $x\in Q+N$.

As for your second question, right multiplication by $a^nb^n$ is a left inverse to the inclusion $Rb^n\to R$ regardless of whether $R$ is commutative. Indeed, for any $rb^n\in Rb^n$, $rb^n\cdot a^nb^n=r(b^na^n)b^n=rb^n$. Note that $ba=1$ implies $b^na^n=1$ since we can cancel $b$'s and $a$'s starting from the middle.

Answer 2

First of all, let me say: if you have a question about one of my solutions, the best thing to do would be to ask on the question first! Because then that solution might get clarified too. Luckily in this case, I was paying attention to the situation.

Let me start with your last question first:

why do the complements exist at all?

Well, the $b^n$ are summands, so they have complements by definition, but perhaps the issue was that you did not see why they are summands. This indeed uses the assumption that $ba=1$, because that makes $a^nb^n$ an idempotent, since $b^na^nb^n=b^n$.

From $b^na^nb^n=b^n$, we have $Ra^nb^n\subseteq Rb^n\subseteq Rb^na^nb^n \subseteq Ra^nb^n$, showing equality all across. Thus $Rb^n$ is generated by an idempotent, and is therefore a summand.

they say that the complements [...] form an ascending chain. But why? Certainly this doesn't hold for general complements of chains of modules

This is true, and perhaps it doesn't even hold for the chain of complements I had in mind (starting with $Re_1\supseteq Re_2\supseteq\ldots$ I think intended $R(1-e_1)\subseteq R(1-e_2)\subseteq\ldots$.)

Now it seems that in fact you still can construct the ascending chain of complements, but it wasn't my intention to make that much work.

Instead, please see the revised answer where I first construct an descending chain of right complements and then conclude that their left annihilators form a strictly ascending chain of left ideals.

Thanks for drawing my attention to it.