The title probably says it all :). This is probably a very very simple question, please bear with me. Let $(A, B, P, Q, f, g)$ be a Morita context (or pre-equivalence data) as defined by Hyman Bass in "Algebraic K-Theory" (link: http://www.math.uni-bielefeld.de/~rehmann/DML/BOOKS/bass.pdf) (page 61) (also here: http://ncatlab.org/nlab/show/Morita+context). I had some doubts about the proof of theorem 3.4, basically I want to understand how $P$ and $Q$ are projective, these are some of the lines where I was stuck on page 62 (they write $f(p \otimes q) =pq$):

Proof. The hypothesis on $f$ means that we can write

$(*)$ $1 = \sum_{i \in I}p_ig_i$ in $A$...

1 - Pardon my ignorance, but I don't know where that come from?

(2) The linear functionals $h_i: P \rightarrow A$ by $h_i(p) = pq_i$ define $h: P^{(I)} \rightarrow A$, and $(*)$ implies $h$ is surjective, so $P$ generates $A$-mod...

2 - What is $P^{(I)}$?? On page 52 Bass says that an object $P$ in an abelian category with coproducts is a "generator" of $A$ if and only if every object of $A$ is a quotient of $P^{(I)}$ for some set $I$. As it were, don't know how to even get started if I don't know what $P^{(I)}$ is.

(3) Define $e:P \rightleftarrows B^{(I)}:h$ by $e(p) = (q_ip)$ and $h(b_i) = \sum p_ib_i$. Then $he(p) = \sum p_i(q_ip) = (\sum p_iq_i)p = p$. Thus $p$ is > finitely generated and projective...

3 - According to Bass, let $A$ a ring and let $P \in$ mod-$A$, $P$ is "finitely generated" and "projective" if and only if $P$ is a direct summand of $A^{(n)}$ for some $n \geq 0$, but what is $A^{(n)}$??


  1. Since $f$ is surjective by hypothesis, $1$ is in the image of $f$. Since the elementary tensors span the tensor product, we have the result.

  2. $A^{(I)}$ means simply $\bigoplus_{i\in I} A_i$ where $A_i=A$ for all $i$. Are you also asking for a proof of that characterization of generators in abelian categories? I'm pretty sure it can be found in Mitchell's book.

  3. Idem, where $I=\{1,\dots,n\}$. The result is straightforward from the definitions, it can be found e.g. in Rotman's Introduction to Homological Algebra.

  • $\begingroup$ Thank you, thank you so much for taking the time to answer, will give the proof a read again, thanks. $\endgroup$
    – The K
    May 3 '14 at 13:42
  • $\begingroup$ No problem. You can upvote if you think it's useful. If you have any more questions, post them here as a comment and I'll try to help out. $\endgroup$ May 3 '14 at 14:31
  • $\begingroup$ Yes, I'm still going over the proof, will mark the answer as accepted answer as soon as I'm done, thank you, thank you, thank you :) $\endgroup$
    – The K
    May 3 '14 at 20:49

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.