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 Answer 1

  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
    Commented May 3, 2014 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. $\endgroup$ Commented May 3, 2014 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
    Commented May 3, 2014 at 20:49

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