To illustrate a few different techniques, I will give a couple more proofs that PID's have stable rank $\le 2$, along with some generalizations.

**Theorem:** An $n$-dimensional domain has stable rank $\le n+1$, and an $n$-dimensional ring has stable rank $\le n+2$.

This is Theorem 3.4 in "Generating Ideals in Prufer Domains" by Heitmann. I do not know of a simple proof, but I am listing it for completeness.

The case for one-dimensional domains has a different, much easier proof, which I will sketch. Two facts that are easy to check from the definitions: (1) a ring has stable rank $\le n+1$ if each proper homomorphic image has stable rank $\le n$, and (2) $R$ and $R/J(R)$ have the same stable rank. Using these we reduce to showing that von Neumann regular rings have stable rank 1. Given elements $a$ and $b$, write $a = ue$, where $u$ is a unit and $e$ is an idempotent. Then $(a + (1-e)b)(eu^{-1}b + 1-e) = eb + (1-e)b = b$, so $(a,b) = (a + (1-e)b)$. (Coincidentally, this also shows that VNR rings are Bezout.)

If one is not familiar with VNR rings, then the one-dimensional Noetherian case is a little easier, since it is a special case of the following fact.

**Theorem:** Any ring with only finitely many maximal ideals has stable rank 1, and thus any ring where every nonzero element is contained in only finitely many maximal ideals has stable rank $\le 2$.

The proof proceeds as before, but is even simpler since, if $R$ has only finitely many maximal ideals, then $R/J(R)$ is a finite direct product of fields, and obviously a direct product of rings of stable rank $\le n$ has stable rank $\le n$.

For another generalization, one has:

**Theorem:** Bezout domains have stable rank $\le 2$.

Proof: Given elements $a,b,c$, write $(a,b) = (d)$, say $a = dx$, $b = dy$, and $d = as + bt = d(xs + yt)$. The case $d = 0$ is trivial. Otherwise, we get $xs + yt = 1$. Then $(a + tc)y - (b - sc)x = (ay - bx) + (xs+ty)c = 0 + 1\cdot c = c$. Therefore $(a,b,c) = (a + tc, b - sc)$.