2
$\begingroup$

Let A be a finite-dimensional k-algebra,where k is a fixed field. All modules of A are finitely generated left modules. Suppose X is an A-module. We denote by add(X) the full subcategory of A-modules consisting of all direct summands of direct sum of finitely many copies of X. $D$ is the usual k-duality $Hom_k(-,k)$, $\nu _A$ is the Nakayama functor $DHom_A(-,_{A}A)$

Recall that the module X is called a generator over A if $add(_{A}A) \subseteq add(X)$, a cogenerator if $add(D(A_A)) \subseteq add(X)$, and a generator-cogenerator if it is both a generator and a cogenerator over A.

Can anyone tell me how to get the following results:

1):Let V be a generator over A with $B := End_A(V)$.Then $Hom_A(V,I)$ is an injective B-module for every injective A-module I;

2): If V is a generator-cogenerator, then each projective-injective B-module is precisely of the form $Hom_A(V,I)$ with I an injective A-module.

$\endgroup$
2
$\begingroup$

to 1): V being a generator of mod-A implies that V is projective in mod-B. Now $Hom(V,D(A)) \cong Hom(A,D(V)) \cong D(V)$ is injective. Now use that every indecomposable injective I is a summand of D(A). A general injective module is a direct sum of indecomposables and thus the result follows.

2) This can be seen as a special case of lemma 3.1.(3) in Auslander, M.; Platzeck, M. I.; Todorov, G. Homological theory of idempotent ideals. Trans. Amer. Math. Soc. 332 (1992), no. 2, 667–692.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @ Mare Sorry ,I can't understand your word. I think $Hom(V,D(B))=DHom(B,D(V))$. And To $Hom(V,D(B))$, do you mean $Hom_A(V,D(B))$? I just said V is a generator, not must to be a progenerator. $\endgroup$ – Xiaosong Peng Sep 12 '16 at 2:28
  • $\begingroup$ thanks, i edited my answer. hopefully it is now correct. $\endgroup$ – Mare Sep 12 '16 at 7:30
  • $\begingroup$ @ Mare Can you tell me how to use V is a generator, ie.$add(_A A) \subseteq add(V)$ get that V is projective in mod-B? Another question: I just know that $Hom_A(V,D(A)) \cong Hom_{A^{op}}(A,D(V)) $ and $Hom_A(A,D(V)) \cong D(V)$, does that $Hom_{A^{op}}(A,D(V)) \cong D(V)$ hold? Thank you. $\endgroup$ – Xiaosong Peng Sep 12 '16 at 8:11
  • $\begingroup$ Since V is a generator: $V^n = A \oplus M$ for some n. Then $B^n=Hom(V^n,V)=Hom(A,V) \oplus Hom(M,V) \cong V \oplus Hom(M,V)$ and so V is a direct summand of the projective module $B^n$. I dont understand the other question. $\endgroup$ – Mare Sep 12 '16 at 9:57
  • $\begingroup$ My second question is wether $Hom_{A^{op}}(A,D(V)) \cong D(V)$. And I think the answer is yes. Thank you for your help. $\endgroup$ – Xiaosong Peng Sep 12 '16 at 11:43

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.