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I've been thinking about this for quite a while but I cannot seem to find an example of


If $k$ is a commutative ring of finite global dimension and I'm looking for a strictly not-commutative $k$-algebra $A$, which is $k$-flat and there is an ideal $I$ in $A$ such that $A/I$ is a regular commutative ring which is $k$-flat also....

Edit: and I is generated by an element which is not a unit not a zero divisor and it commutes with all the elements of A...

(the flatness condition seems to really restrict my possibilities....)

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    $\begingroup$ Can't you just take $A$ to be the algebra of upper triangular $2\times 2$ matrices over $k$, and $I$ the ideal consisting of matrices whose first column is zero? Then $A$ is not only $k$-flat, but free, and $A/I\cong k$. $\endgroup$ – Jeremy Rickard Aug 7 '14 at 12:38
  • $\begingroup$ I doubt this is possible with the flatness condition... I'll have to check the details later... $\endgroup$ – AIM_BLB Aug 9 '14 at 1:31
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    $\begingroup$ "regular commutative" means "commutative"? Or you are asking for a particular version of "regular"? And your edit precludes Artinian rings from consideration... maybe that's what was interfering? $\endgroup$ – rschwieb Aug 11 '14 at 12:52
  • $\begingroup$ it means commutative and regular $\endgroup$ – AIM_BLB Aug 11 '14 at 16:53
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Let k be a field, let A be the enveloping algebra of the three-dimensional Heisenberg Lie algebra g, and let I be the ideal of A generated by any nonzero element of the center of g.

This satisfies all your conditions.

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    $\begingroup$ Jeremy's example in the comment above is not generated by a central element. This is a natural way to fix that. $\endgroup$ – Mariano Suárez-Álvarez Jun 29 '15 at 7:20
  • $\begingroup$ Thanks Mariano I've been thinking about that one for a very long while. $\endgroup$ – AIM_BLB Jun 30 '15 at 23:34

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