2
$\begingroup$

In Section 9.5 (Morita Invariance) of Weibel's "An introduction to homological algebra," rings $R$ and $S$ are defined to be Morita equivalent if there is an $R$-$S$ bimodule $P$ and an $S$-$R$ bimodule $Q$ such that $P \otimes_S Q \cong R$ as $R$-$R$ bimodules and $Q \otimes_R P \cong S$ as $S$-$S$ bimodules. Weibel proves:

Lemma 9.5.4: If $P$ and $Q$ define a Morita equivalence between $R$ and $S$, then $P$ is a finitely generated projective left $R$-module. $P$ is also a finitely generated projective right $S$-module.

Weibel notates as $p \cdot q$ the element of $R$ corresponding to $p \otimes q$ under the isomorphism $P \otimes_S Q \cong R$ (and similarly for $q \cdot p$). To prove the lemma, Weibel writes the unit of $S$ as $1 = q_1 \cdot p_1 + \ldots + q_m \cdot p_m$ for some $m$, then defines maps $e : P \to R^m$ and $h : R^m \to P$ by $e(p) = (p \cdot q_1, \ldots, p \cdot q_m)$ and $h(r_1, \ldots, r_m) = \sum_i r_i p_i$. To show that $P$ is a direct summand of $R^m$ (and thus finitely generated projective), all that remains is to check that $he = \mathrm{id}_P$. Weibel computes

$$ he(p) = h(p \cdot q_1 \ldots, p \cdot q_m) = \sum_i (p \cdot q_i)p_i = \sum_i p(q_i \cdot p_i) = p. $$

My question is: Why are the two sums in the line above equal? That is, why can we say that $$ \sum_i (p \cdot q_i)p_i = \sum_i p(q_i \cdot p_i) \,? $$

This equality seems to be asserting that this diagram commutes: $\require{AMScd}$ \begin{CD} P \otimes_S Q \otimes_R P @>{\mathrm{id}_P \otimes \cong}>> P \otimes_S S \\ @V{\cong \otimes \mathrm{id}_P}VV @V{\cong}VV \\ R \otimes_R P @>{\cong}>> P \\ \end{CD} However, I don't see why the isomorphisms $P \otimes_S Q \cong R$ and $Q \otimes_R P \cong S$ should have this sort of compatibility, since the definition of Morita equivalence given here (I'm aware that there are other ways to formulate it, but I'd like to understand it as Weibel presents it, if possible) doesn't appear to say anything about how these isomorphisms interact with each other.

Update: I've found an alternate proof of the statement (easily pieced together from section 18 of T. Y. Lam's Lectures on Modules and Rings). Finite generation and projectivity are both categorical properties of a module, in the sense that if $F$ is an equivalence of categories and $M$ is finitely generated or projective, then $F(M)$ is finitely generated or projective, respectively. Now if $_RP_S$ and $_SQ_R$ define a Morita equivalence as defined by Weibel, then $- \otimes_R P : \mathrm{Mod}_R \to \mathrm{Mod}_S$ is an equivalence of categories (with inverse $- \otimes_S Q$). Applying this to our favorite finitely generated projective $R$-module, $R$, we conclude that $P \cong R \otimes_R P$ is finitely generated and projective.

However, I would still like to understand Weibel's proof; that is, why does it follow from the existence of bimodule isomorphisms $P \otimes_S Q \cong R$ and $Q \otimes_R P \cong S$ that the "PQP-associativity" $(pq)p' = p(qp')$ holds?

$\endgroup$

1 Answer 1

1
$\begingroup$

First of all, I agree with your update, and I find that that proof is much clearer conceptually ("they're categorical properties so they're preserved under equivalences").

But I also understand why you want "closure" on that first proof; note that the claim isn't so much "the isomorphisms $P\otimes Q = R$ and $Q\otimes P= S$ are such that blabla", but "the isomorphisms can be chosen so that the square commutes" (this is enough for the argument to go through)

I don't have a very easy solution for that but one way to prove that they can be chosen so as to be compatible is as follows :

Let $F$ denotes $P\otimes_S-$ and $G=Q\otimes_R-$. Then by definition $F,G$ form an equivalence between $S-Mod$ and $R-Mod$. Now a standard theorem of category theory states essentially the following : an equivalence can always be made into an adjoint equivalence.

What this means is that we can choose the natural isomorphisms $\eta : id \to GF$ and $\epsilon : FG\to id$ so as to define an adjunction $F\dashv G$.

This implies in particular that they satisfy the triangle identities : $\epsilon F\circ F\eta = id_F$.

Now apply this to $S$ to get the following commutative diagram :

$\require{AMScd}\begin{CD} FGF(S) @<F(\eta_S)<< F(S) \\ @V\epsilon_{F(S)}VV @VVV \\ F(S) @>>> P\end{CD}$

where the two maps $F(S)\to P$ are the same fixed isomorphism (the standard one).

Now since $F(\eta_S)$ is actually an isomorphism, we can reverse it and get :

$\require{AMScd}\begin{CD} FGF(S) @>F(\eta_S^{-1})>> F(S) \\ @V\epsilon_{F(S)}VV @VVV \\ F(S) @>>> P\end{CD}$

And now if you unravel what this is :

$\require{AMScd}\begin{CD} P\otimes_S Q\otimes_R P\otimes_S S @>P\otimes_S\cong \otimes_SS>> P\otimes_S S \\ @V\cong\otimes_R P\otimes_S S VV @VVV \\ R\otimes_RP\otimes_SS @>>> P\end{CD}$

Now of course (this is an easy exercise) you can remove the $\otimes_SS$ from the story and get

$\require{AMScd}\begin{CD} P\otimes_S Q\otimes_R P @>P\otimes_S\cong >> P\otimes_SS \\ @V\cong\otimes_R PVV @VVV \\ R\otimes_RP @>>> P\end{CD}$

which is what you wanted.

So the isomorphisms can be chosen for the diagram to commute, so with these chosen isomorphisms, the argument goes through.

$\endgroup$
1
  • $\begingroup$ Thanks, this is exactly the answer I was looking for. It hadn't occured to me that the associativity might hold not for any pair of isomorphisms but only for a carefully chosen pair, given that at least one pair exists. $\endgroup$ Commented Mar 2, 2020 at 13:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .