# The projective and injective modules of $End_A(V)$?

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.

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.

• @ 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. Commented Sep 12, 2016 at 2:28
• thanks, i edited my answer. hopefully it is now correct.
– Mare
Commented Sep 12, 2016 at 7:30
• @ 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. Commented Sep 12, 2016 at 8:11
• 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.
– Mare
Commented Sep 12, 2016 at 9:57
• 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. Commented Sep 12, 2016 at 11:43