Do you know a left-noetherian ring $R$ with a two-sided ideal $I$ such that:
- $I=I.I$;
- $I$ is not projective as a left $R$-module (and better, the tensor product over $R$ of $I$ with itself is not a projective left $R$-module)?
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Take any idempotent $e$ in a finite dimensional quiver algebra $KQ/L$ , then the ideal $I=AeA$ will be idempotent but only in rare cases it will be projective. Here an explicit example: Take the quiver $Q$ with two vertices 1 and 2 (with corresponding primitive idempotents $e_1$ and $e_2$) and an arrow a from 1 to 2 and an arrow b from 2 to 1 and let the relations L be $L=<ab>$. Let $A=kQ/L$, which is the Nakayama algebra with Kupisch series [2,3]. The ideal $I=Ae_2 A$ will be idempotent but not projective. In the article "Homological theory of idempotent ideals" by Auslander , Platzeck and Todorov ( https://www.jstor.org/stable/2154190 ) you can find much more about idempotent ideals in Artin algebras and that they are rarely projective.
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$\begingroup$ Many thanks for these interesting examples! But I was not precise enough in my first question; my real aim is to find (if it exists, what is not clear at all for me) a noetherian ring A with a two-sided ideal I with I.I=I such that the quotient of the category of left A-modules by the bilocalising subcategory of A/I-modules does not have enough projective. Roos maked this at the end of his paper "Derived functors of inverse limits revisited" published in 2006 but with a non-noetherian ring (his example is a commutative ring; here we must use a "highly non-commutative" example). $\endgroup$ Commented Jun 13, 2019 at 10:54
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$\begingroup$ For the motivation, see mathoverflow.net/questions/332681/… $\endgroup$ Commented Jun 13, 2019 at 10:55
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3$\begingroup$ Aurélien, if this answers your question in the way initially formulated, I'd suggest you accept it and post a follow-up question. $\endgroup$– YCorCommented Jun 13, 2019 at 12:13