Question. Let $K$ be a field (assume $K=\mathbb{C}$ if this simplifies the problem). What is the right global dimension of the $K[x,y]$-algebra: $$ A=\left[\begin{array}{cc} K[x,y] & xK[x,y] \\ K[x,y] & K[x,y] \end{array}\right] . $$ I am not an expert in homological algebra so I would also appreciate advice on how to analyze such examples in general.

Some Motiativation. A $1$-variable counterpart of this example, i.e., the $K[x]$-algebra $$ B=\left[\begin{array}{cc} K[x] & xK[x] \\ K[x] & K[x] \end{array}\right] . $$ has right global dimension $1$, or equivalently, it is hereditary. In fact, loosely speaking, hereditary $K[x]$-orders in $\mathrm{M}_{n\times n}(K(x))$ all look roughly like this example. By analogy, I would expect the ring $A$ to have right global dimension $2$.

Remark. The rings $A$ and $B$ are noetherian, so their right and left global dimensions coincide.

  • $\begingroup$ I wonder if there is any argument tying or bounding the global dimension of $A$ to the global dimension of $M_2(K[x,y])$, which should be the same as $K[x,y]$. $\endgroup$ – rschwieb Apr 17 at 14:47

Have a look at the paper: Kirkman, Ellen; Kuzmanovich, James; Matrix subrings having finite global dimension. J. Algebra 109 (1987), no. 1, 74–92.

Using Theorem 1.6 one has:

$\operatorname{rgldim}(A) = \max\{\operatorname{rgldim}(K[x,y]), \operatorname{rgldim}(K[x,y]/(xK[x,y])+1\} = 2$

if I am not misunderstanding the results there wrong.

I hope that these comments are helpful.

Best regards, Oeyvind Solberg.

  • $\begingroup$ Thanks! This is exactly the kind of result I was looking for. $\endgroup$ – Uriya First Apr 18 at 11:57
  • $\begingroup$ (and sorry for forgetting to accept the answer.) $\endgroup$ – Uriya First Jul 23 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.