Equivalent properties of regular (Von Neumann) rings.

Question

A ring $R$ is called regular (in the sense of Von Neumann) if for every $r \in R$ there is an element $r’ \in R$ such that $rr’r=r$. Then the following conditions are equivalent:

(1) $R$ is regular.

(2) Every cyclic right ideal of $R$ is a direct summand and of $R_{R}$.

(3) Every cyclic left ideal of $R$ is a direct summand and of $_{R}R$.

(4)Every finitely generated right ideal of $R$ is a direct summand of $R_{R}$.

(5) Every finitely generated left ideal of $R$ is a direct summand of $_{R}R$.

For this proof I have a hunch that the most direct scheme to proceed would be $(1) \Leftrightarrow (2), (1) \Leftrightarrow (3), (1) \Leftrightarrow (4)$ and $(1) \Leftrightarrow (5)$. Where $(1) \Leftrightarrow (2), (1) \Leftrightarrow (3)$ would be almost similar, the same with $(1) \Leftrightarrow (4)$ and $(1) \Leftrightarrow (5)$, being these two last equivalences some sort of the first to first equivalences to prove.

For $(1) \Rightarrow (2)$, let’s take $S$ a left cyclic ideal of $R$, in other words $SR \subseteq R$ and $S=s’R$ for some $ 0 \neq s’ \in S$. I need to show another left ideal $T$ of R such that $S \oplus T = R$. But I have no clue about how I should describe this ideal $T$. I mean, I want an ideal $T$ such that for every $r \in R$, we have that $r$ can be uniquely expressed as $r=s+t$ where $s\in S$ and $t\in T$. On the other hand, $r = rr’r$ for some $r’ \in R$ since $R$ is regular Von Neumann and $s= s’r^{\ast}$ where $s’ \in S$ and $r^{\ast}$ from the fact that $S$ is a right cyclic ideal of $R$. But from here I’m out of ideas.

For the other part, $(2) \Rightarrow (1)$, I’m pretty much stucked in the same way.

Thanks for reading. Any help would be appreciated. Also, it would be helpful to see how I can generalize $(1) \Rightarrow (2)$ to prove $(1) \Leftrightarrow (4)$ or $(1) \Leftrightarrow (5)$.

Answer 1

$r'$ is really not good notation here, it makes doing computations hard to read. Let's write the defining relation as $rsr = r$. This $s$ is supposed to be a kind of weak inverse of $r$, and the key observation about this relation (to my mind) is that it implies the two relations

$$rsrs = (rs)^2 = rs, srsr = srs = (sr)^2.$$

So $rs$ and $sr$ are both idempotents. A good example to keep in mind here is that any product $\prod_i K_i$ of fields is von Neumann regular, and if $r = \prod r_i$ is an element of such a product we can take $s$ to be the element which, componentwise, is either the inverse of $r_i$ if it's nonzero or $0$. So in this case $rs = sr$ is the idempotent which is equal to $1$ whenever $r_i \neq 0$ and $0$ otherwise. However $rs \neq sr$ in general.

These idempotents are how the proofs of these equivalences work. As a first observation, since the vNr condition is invariant under opposites (check this!), $R$ is regular iff $R^{op}$ is regular, which immediately gives $2 \Leftrightarrow 3$ and $4 \Leftrightarrow 5$. Because of this I'm only going to focus on right modules.

Now let's try to prove $1 \Rightarrow 3$. Consider the cyclic right ideal $rR = \{ rx : x \in R \}$ generated by $r$. By hypothesis there exists a weak inverse $s$ such that $rsr = r$. Now note that $rR = rsR = \{ rsx : x \in R \}$; this is exactly because $rs(rx) = rx$. But as we just computed above, $(rs)^2 = rs$ is an idempotent. And:

Fact of life about idempotents: The cyclic right ideal $eR$ generated by an idempotent is always a direct summand, with complement $(1 - e) R$.

Again, a good example to keep in mind here is a product $\prod_i K_i$ of fields as mentioned above. An idempotent $e \in \prod_i K_i$ is exactly an element whose components are either $0$ or $1$, and the corresponding ideal it generates is the ideal of functions nonzero on some subset $S \subseteq I$ of the indexing set, which is a direct summand with complement the ideal of functions nonzero on the complement $\neg S \subseteq I$.

Proof. We have $r = er + (1 - e) r$ by definition, and $e$ acts by the identity on $eR$ while $(1 - e)$ acts by the identity on $(1 - e)R$. But $e(1 - e) = 0$, so $eR \cap (1 - e) R = 0$. This gives $R = eR \oplus (1 - e) R$ as desired. $\Box$

(This proof is sort of too short to see what's going on; a more conceptual argument is to use the general fact that idempotent endomorphisms correspond to direct sum decompositions, and this is the direct sum decomposition corresponding to $e$ regarded as an idempotent right $R$-module endomorphism of $R$. This way of looking at things will be helpful for proving $3 \Rightarrow 1$.)

This proves $1 \Rightarrow 3$. Can you do the rest from here?

Answer 2

I agree 100% with what Qiaochu Yuan mentions that it is certainly important to understand that for rings with identity "summand of $R_R$" and "of the form $eR$ where $e$ is an idempotent of $R$" mean the same thing. I struggled to find a suitable post, but at any rate this appears in most ring theory texts. Let us agree here though that "is a summand" and "generated by an idempotent" mean the same thing for right ideals.

Given that, and since $1\iff 2$ is covered in another post, and $4\implies 2$ obviously, let's try to prove $2\implies 4$.

We are assuming (2). We set out to prove $aR+bR$ generated by an idempotent. That would mean any two-generated right ideal is a summand. Using mathematical induction, we could then conclude that any finitely generated right ideal is a summand.

There's a trick for doing this.

Note that using (2) again, we can say we're working with $aR=eR$ and $bR=fR$ where $e,f$ are idempotents. It is easy to check that $eR+fR=eR+(1-e)fR$. Since $(1-e)fR$ is cyclic, select an idempotent generator $g$ such that $gR=(1-e)fR$.

Then $h=e+g-ge$ has interesting properties we need:

1. it's idempotent (check!)
2. $e\in hR$ since $he=e$
3. $g\in hR$: using $eg=0$ we have $hg=eg+g-geg=g$
4. so $aR+bR=eR+fR=eR+(1-e)fR=eR+gR=hR$

At this point, we've shown $aR+bR=hR$ is a summand, and now induction on the number of generators of a right ideal takes us the rest of the way.

Now we have $1\iff 2\iff 4$.

By symmetry, $1\iff 3\iff 5$, so they are all equivalent.