4
$\begingroup$

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)?
$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Aurélien Djament Jun 13 at 10:54
  • $\begingroup$ For the motivation, see mathoverflow.net/questions/332681/… $\endgroup$ – Aurélien Djament Jun 13 at 10:55
  • 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. $\endgroup$ – YCor Jun 13 at 12:13

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.