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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.

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  • $\begingroup$ 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
    Commented Apr 17, 2019 at 14:47

1 Answer 1

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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.

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  • $\begingroup$ Thanks! This is exactly the kind of result I was looking for. $\endgroup$
    – Uriya First
    Commented Apr 18, 2019 at 11:57
  • $\begingroup$ (and sorry for forgetting to accept the answer.) $\endgroup$
    – Uriya First
    Commented Jul 23, 2019 at 11:23

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