Is there a direct proof that any (right) Artinian ring has stable range 1? More precisely, let $R$ be a right Artinian ring and $a,b\in R$ be such that $aR+bR=R$. Can we prove that $(a-bt)R=R$ for some $t\in R$ directly from first principles? I'm aware of proofs that prove this for all semilocal rings (say, in T.Y. Lam's A First Course in Noncommutative Rings), but this feels like a statement that can be proven without those machineries. I was thinking along the lines of using the fact that every element of $a\in R$ would either have $ar=0$ or $ar=1$ for some $r\in R$, which follows from considering the chain $ aR\supseteq a^2R \supseteq\cdots $, but that seems to be getting us nowhere.


1 Answer 1


Let me first prove that $R$ has stable range $1$ if $R$ is semisimple (by which I mean Artinian and Jacobson radical zero.)

Because $R$ is semisimple, we can do the following. We can write $bR=(aR\cap bR)\oplus K$ for some right ideal $K$, and you can check that $R=aR\oplus K$. On the other hand, $R\cong aR\oplus ann_r(a)$ where $ann_r(a)$ is the right annihilator of $a$. By the Krull-Schmidt theorem we have that $ann_r(a)\cong K$ via some isomorphism $\theta:ann_r(a)\to K$.

Here's where the surgery happens. Let $h$ be the splitting map for the inclusion of $ann_r(a)$ in the split exact sequence

$$0\longrightarrow ann_r(a)\longrightarrow R\longrightarrow aR\overset{\ell_a}\longrightarrow 0$$ where $\ell_a$ is left multiplication by $a$. It's well known that this creates an isomorphism $(\ell_a, h):R\to aR\oplus ann_r(a)$ given by $r\mapsto (\ell_a(r),h(r))$ which we'll compose with another isomorphism that maps $(x,y)\mapsto x+ \theta(y)$ to get an automorphism of $R$. This automorphism sends $1\mapsto a+\theta(h(1))$

Now, every automorphism in $End(R_R)$ is given by multiplication on the left by a unit, so in particular there is some unit $u\in R$ such that $u=a+\theta(h(1))$. By design, $\theta(h(1))\in K\subset bR$. Thus we have shown every semisimple ring is stable range 1.

Now, the other interesting thing is that $R$ is stable range 1 if $R/J(R)$ is stable range 1 owing to unit-lifting properties of the Jacobson radical. I'll use overlines for images of elements of $R$ in $R/J(R)$.

First, notice that if $\overline{r}\overline{s}=\overline{1}$, $rs-1\in J(R)$. But if $rs=j+1$, we know that $j+1$ is a unit (a property of elements of the Jacobson radical). The same is true of $sr$, and therefore $r$ and $s$ are units of $R$.

Now if you begin with $ax+b=1$ in $R$, there exists $y$ such that $\bar{a}+\bar{b}\bar{y}$ is a unit in $R/J(R)$. But by what we just saw in the last paragraph, that means $a+by$ is a unit of $R$.

Now, in your original question you required $R$ to be Artinian, in which case of course $R/J(R)$ will be semisimple, and now I hope it's clear how all the pieces fit together.

  • $\begingroup$ I could not see a way around "the surgery." It seems you really need to gain control of what you're mapping in order to arrange things to find $y$ such that $a+by$ is a unit. This is adapted from an argument I saw Lam give, but using the Artinian condition makes a few things a little easier. $\endgroup$
    – rschwieb
    Jul 26 at 18:59

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