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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?


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

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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).

Question. Let $R$ be commutative ring. Is the ring $T$ of $2$-by-$2$ upper triangular matrices over $R$ a $GE_2$-ring?

The answer is no in general.

Claim 1. If $T$ is a $GE_n$-ring, then so is $R$.

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.

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 $m = 1$.

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.

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.

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.

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 $\text{sr}(R) = 1$.

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 $M_2(R)$.)


[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.

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  • $\begingroup$ Thanks for the answer, that was very clear. I was wondering if you knew whether the answer to my original question was yes IF $SK_1(R)$ was trivial. $\endgroup$ – Sam Williams May 16 '17 at 16:50
  • $\begingroup$ In particular, I was thinking of letting $R=k[x_1,\,x_1^{-1},\ldots,\,x_n,\,x_n^{-1}]=k[F]$, where $F$ is the free abelian group of rank $n$. There is a paper by Bass, Heller and Swan which says $K_1(k[F])\cong K_0(k)^n\oplus K_1(k)\cong\mathbb{Z}^n\oplus U(k)$. Since $k[F]$ has trivial units (thesis of Higman) $U(k[F])\cong U(k)\oplus F$. $\endgroup$ – Sam Williams May 16 '17 at 16:53
  • $\begingroup$ Further, in your proof, how did you use the stably range? Does the result not follow regardless? $\endgroup$ – Sam Williams May 16 '17 at 21:23
  • $\begingroup$ @SamWilliams I edited my answer to make it simpler. I also tried to address the specific rings $R$ you are interested in. $\endgroup$ – Luc Guyot May 16 '17 at 22:52
  • $\begingroup$ @SamWilliams I edited my answer once more to make a comment about your motivation. Note that Claim 1 above is fully elementary and does not relate to $SK_1$ or stable rank. $\endgroup$ – Luc Guyot May 21 '17 at 16:55

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