Let $R$ be a commutative ring with identity, and let $T = T\_2(R)$ be the ring of $2\times 2$ upper triangular matrices over $R$. Is it true that the following identity holds?
$$GL\_2(T)=U(T)E\_2(T)$$ 


Here, $U(T)$ represent the units of $T$ and we abuse notation by identifying $U(T)$ with its embedding $\begin{pmatrix} U(T) & 0 \\ 0 & 1 \end{pmatrix} \subset GL\_2(T)$. We denoted by $E\_2(T)$ the subgroup of $GL\_2(T)$ generated by matrices of the form $\begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix}$ or $\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$ for $t$ ranging in $T$.


I am particularly interested in the case of the multivariate Laurent polynomials ring $R = k[x\_1,x\_1^{-1}, \dots, x\_m, x\_m^{-1}]$ where $k$ is a field and $m \ge 1$.


The motivation for this is as follows. 


By Suslin's Stability Theorem [1, Theorem VI.4.5], we know that $GL\_n(R) = U(R)E\_n(R)$ holds whenever 
$R = k[x\_1, x\_1^{-1}, \dots, x\_m, x\_m^{-1}]$ and
$n \geq 3$ (the result doesn't depend on $m$). The reason why $n = 2$ causes a fuss is due to the existence of Cohn matrices, see e.g., [1, Section I.8].
For such rings $R$ this implies in particular that $GL\_n(M\_2(R)) = GL\_2(R)E\_{n}(M\_2(R))$.


Using a similar reasoning, we can certainly say that $GL\_2(T)\subset U(R)E\_4(R)$ and $U(R)\subset U(T)$, but I see no reason why $E\_4(R)\subset E\_2(T)$. Perhaps there is another way? Or perhaps this just isn't possible?




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[1] T. Liam, "Serre's problem on projective modules", 2006.



For $T$ the $2\times 2$ triangular matrices over $R$, can we write $GL_2(T)=U(T)E_2(T)$?
The answer is **no** in general because $T$ cannot be a $GE\_2$-ring in the sense of P. M. Cohn if $R$ isn't. 


Let us recall some definitions. Let $n \ge 2$. A ring $\Lambda$ (not necessarily commutative) is said to be a $GE\_n$-ring if the identity $GL\_n(\Lambda) = D\_n(\Lambda) E\_n(\Lambda)$ holds. Here $D\_n(\Lambda)$ denotes the group of invertible diagonal matrices over $\Lambda$ while $E\_n(\Lambda)$ is the subgroup of $GL\_n(\Lambda)$ generated by matrices obtained from the identity matrix by replacing an off-diagonal entry by some $\lambda \in \Lambda$.


OP's question can be rephrased in the following way (use Whitehead lemma).



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> > **Question.** Let $R$ be commutative ring. Is the ring $T$ of $2$-by-$2$ upper triangular matrices over $R$ a $GE\_2$-ring?
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The answer is *no* in general.



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> > **Claim 1**. If $T$ is a $GE\_n$-ring, then so is $R$.
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> > *Proof.* The natural surjective ring homomorphism $T \twoheadrightarrow R^2$ induces a surjective group homomorphism $\varphi\_n: GL\_n(T) \twoheadrightarrow GL\_n(R^2)$ for every $n \ge 1$. Since $\varphi\_n$ maps elementary matrices to elementary matrices, and diagonal matrices to diagonal matrices, the ring $R^2$ is certainly a $GE\_n$-ring if $T$ is. It is straightforward to check that $R^2$ is a $GE\_n$-ring if and only if $R$ is.
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Let $k$ be field and let $m > 2$. Let $R = k[t\_1, t\_1^{-1}, \dots, t\_m, t\_m^{-1}]$. Then $R$ is not a $GE\_2$-ring [4] (although $SK\_1(R) = 1$ by Suslin's Stability Theorem [6, Theorem 5.8]), so that $T$ is not a $GE\_2$-ring. 


Below lie the remains of my first (overly complicated) answer. it has been groomed so as to address the Euclidean case $m = 1$.



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> **Claim 2**. Assume that $R$ is a Noetherian commutative ring of Krull dimension at most $1$ (e.g., a principal ideal domain). Then $T$ is a $GE\_2$-ring if and only if $R$ is. 
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> *Proof.* 
>  Because of Claim $1$, we only need to prove the "if" part. To do so, let us assume that $R$ is a $GE\_2$-ring. Since the Krull dimension of $R$ is at most $1$,
>  the Bass stable rank of $R$ is at most $2$ by Bass's Cancellation Theorem. Therefore $GL\_n(R) = D\_n(R) E\_n(R)$ holds for all $n \ge 2$, which implies $SK\_1(R) = 1$. As $K\_1(T) \simeq K\_1(R)^2$ by [2, Corollary 2], we deduce that $K\_1(T) \simeq R^{\times} \times R^{\times}$ where the isomorphism is induced by the natural map $T^{\times} \twoheadrightarrow R^{\times} \times R^{\times}$. 
>  Since the Krull dimension of $T$ is at most the Krull dimension of $R$, hence bounded above by $1$, the surjective stability of $K\_1$ [5, Theorem 10.15] yields the result.
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Note that there exists principal ideal domains which are not $GE\_2$-rings, e.g., $R = \mathbf{Z}[\frac{1 + \sqrt{-19}}{2}]$ by [1, Theorem 6.1].


The following claim is straightforward.



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> **Claim 3.** If the stable rank $\text{sr}(R)$ of $R$ is $1$, then $\text{sr}(T) = 1$. In particular, $T$ is a $GE\_2$-ring whenever $\text{sr}(R) = 1$.
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I conclude with a comment on OP's motivation.
The situation of $M\_2(R)$ mentioned in OP's question, is rather different than the situation of $T$. Indeed, note that $K\_1(M\_n(R)) = K\_1(R)$ and $\text{sr}(M\_n(R)) = \lceil \frac{\text{sr}(R) - 1}{n} \rceil + 1$ [5, Theorem 5.18] whereas $K\_1(T\_n(R)) = K\_1(R)^n$ [2, Corollary 2] and $\text{sr}(T\_n(R)) = 1$ if and only if $\text{sr}(R) = 1$, where $T\_n(R) \subset M\_n(R)$ is the ring of $n$-by-$n$ upper triangular matrices. (In some $K$-theoretical sense, $T$ is closer to $R^2$ than $M\_2(R)$.)




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[1] P. M. Cohn, "On the structure of the $GL\_2$ of a ring", 1966.  

[2] K. Dennis and S. Geller, "$K\_i$ of the upper triangle matrice rings", 1976.  

[3] S. Bachmuth, H. Y. Mochizuki, "$E\_2 \neq SL\_2$ for most Laurent polynomial rings", 1982.  

[4] J. MacDonnell and J. Robson, "Noncommutative Noetherian rings", 1987.  

[5] B. Magurn, "An algebraic introduction to K-theory", 2002.  

[6] T. Liam, "Serre's problem on projective modules", 2006.


