# Direct Proof that Artinian Rings have Stable Range 1

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.

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.

• 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. Jul 26 at 18:59