# Global dimension of a certain $K[x,y]$-algebra

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.

• 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 $K[x,y]$. – 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.