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Stream: learning: questions

Topic: how to recover a commutative ring from its modules

John Baez (May 14 2023 at 23:49):

In what follows all my categories, functors, equivalence etc. will be Ab-enriched.

Usually you can't recover a ring from its category of modules. Indeed, we say two rings are Morita equivalent if their categories of module are equivalent, and any ring $R$ is Morita equivalent to the ring of $n \times n$ matrices with entries in $R$.

John Baez (May 14 2023 at 23:50):

However, it should be possible to recover a commutative ring from its category of modules.

John Baez (May 14 2023 at 23:56):

My evidence for this is Prop. 21.10 in Anderson and Fuller's Rings and Categories of Modules.

(I've owned this book for decades but only recently am I exploring the relevant sections. Now it strikes me how much work they do to show things work the same way in equivalent categories - things that seem obvious. Maybe that means this book is good for people who don't yet "get" category theory.)

John Baez (May 14 2023 at 23:57):

This proposition says that if two rings are Morita equivalent, their centers are isomorphic.

John Baez (May 14 2023 at 23:57):

But I'd much rather hear a theorem that says how to read off the center of a ring from its category of modules!

John Baez (May 15 2023 at 00:02):

And staring at the proof, I think we might be able to do it this way. Suppose $M$ is any projective generator of the category $\mathsf{Mod}(R)$. Here $M$ is a generator if for every object $X$ there's an epimorphism from some (possibly infinite) sum of copies of $M$ to $X$.

John Baez (May 15 2023 at 00:06):

Then I believe the center of the ring of endomorphisms of $M$ (that is, $R$-module endomorphisms) is isomorphic to the center of $R$!

John Baez (May 15 2023 at 00:08):

If this is true then we're done, since for every ring $R$, the category $\mathsf{Mod}(R)$ has a projective generator: for example $R$ itself. So we just take any one, and take center of its ring of endomorphisms, and that's the center of $R$.

John Baez (May 15 2023 at 00:11):

And the important part is that we don't need to know $R$ ahead of time to tell what objects in $\mathsf{Mod}(R)$ are projective generators.

John Baez (May 15 2023 at 00:15):

However, I'm not completely sure of the logic of Anderson and Fuller's proof (it uses a bunch of facts proved elsewhere, and at least one of the references back to those facts seems wrong).

John Baez (May 15 2023 at 00:16):

In particular I'm not really sure of this:

• if $M$ is a projective generator in $\mathsf{Mod}(R)$, the center of the endomorphism ring of $M$ is the center of $R$.

John Baez (May 15 2023 at 00:17):

I'm afraid this may be true only for finitely generated projective generators, and unfortunately it seems we do need to know $R$ ahead of time to tell what objects in $R$ are finitely generated.

John Baez (May 15 2023 at 00:19):

We'll see who among you knows a bunch about rings and modules! It's funny how this sort of math doesn't show up much in programming or "applied" category theory.

John Baez (May 15 2023 at 01:49):

Hmm, maybe there's a better way to recover a commutative ring $R$ from the Ab-enriched category $\mathsf{Mod}(R)$. Here's another attempt. Since $R$ is commutative, every element $r \in R$ acts as an endomorphism of each $R$-module $M$, namely multiplication by $r$.

If I'm not getting myself confused, these fit together to give a natural transformation from $1: \mathsf{Mod}(R)\to \mathsf{Mod}(R)$ to itself.

John Baez (May 15 2023 at 01:51):

It seems that natural transformations from $1$ to itself form a ring under composition and addition, say $\mathrm{End}(1_{\mathsf{Mod}(R)})$.

John Baez (May 15 2023 at 01:53):

So it seems we have a ring homomorphism

$m : \mathrm{R} \to \mathrm{End}(1_{\mathsf{Mod}(R)})$

John Baez (May 15 2023 at 01:55):

And this is injective, since distinct elements of $R$ act differently on the left $R$-module that's $R$ itself.

John Baez (May 15 2023 at 01:56):

So if it were surjective too, then we'd get

$R \cong \mathrm{End}(1_{\mathsf{Mod}(R)})$

(in the commutative case).

Mike Shulman (May 15 2023 at 02:20):

Sounds like the [[center of an additive category]].

John Baez (May 15 2023 at 02:44):

Yes, thanks.

So now I either need to plow through some French and hope Gabriel proved what I want, or figure it out myself. :upside_down:

John Baez (May 15 2023 at 02:48):

I'm thinking I can show

$R \cong \mathrm{End}(1_{\mathsf{Mod}(R)})$

is surjective roughly as follows. Suppose $m: 1_{\mathsf{Mod}(R)}) \Rightarrow 1_{\mathsf{Mod}(R)})$ is a natural transformation. Then on the object $R \in \mathsf{Mod}(R)$, $m_R: R \to R$ is just multiplication by some $r \in R$. By naturality this is also how it must work on every free module (that is, any sum of copies of $R$). Then... here's the part where I should slow down and think a bit more... since every module is the quotient of a free module, on every module our natural transformation must be just multiplication by this $r \in R$.

John Baez (May 15 2023 at 02:49):

I think that's right: I take any module $M$, write it as a quotient of a free module

$\sum_{\alpha} R \to M \to 0$

and use naturality to draw a little commutative diagram where on the left side going down you have multiplication by $r$ on each summand.

JS PL (he/him) (May 15 2023 at 02:49):

I was just about to answer that's its surjective

John Baez (May 15 2023 at 02:50):

Okay, great!

JS PL (he/him) (May 15 2023 at 02:50):

My proof is that you want to show that any natural transformation $\alpha: 1_{MOD(R)} \Rightarrow 1_{MOD(R)}$ is a of the form $\alpha_M(m) = r \cdot m$ for some $r \in R$ for all $R$-modules $M$

JS PL (he/him) (May 15 2023 at 02:51):

Now consider the case $M=R$, we get an $R$-linear map $\alpha_R: R \to R$. But $R$-linear maps from $R$ to itself are always just scalar multiplication. Specifically we get $\alpha_R(r) = \alpha_R(1) \cdot r$

JS PL (he/him) (May 15 2023 at 02:52):

Now for every $m \in M$, consider the $R$-linear map $p_m: R \to M$ which maps $1 \in R$ to $m$, so $p_m(r) = r \cdot m$

JS PL (he/him) (May 15 2023 at 02:53):

Then using naturality we get $\alpha_R(m) = \alpha_R(p_m(1)) = p_m(\alpha_R(1)) = \alpha_R(1) \cdot m$

JS PL (he/him) (May 15 2023 at 02:53):

Thus every natural transformation from $1_{MOD(R)}$ to itself is given by scalar multiplication.

John Baez (May 15 2023 at 04:28):

Nice! That's a faster, simpler, better version of my argument using the fact that every $R$-module is a quotient of a free one.

John Baez (May 15 2023 at 04:55):

I wrote this up on the nLab.

Matteo Capucci (he/him) (May 15 2023 at 08:22):

There should be a way more general result for [[Tannakian reconstruction]] that covers these cases

Matteo Capucci (he/him) (May 15 2023 at 08:24):

In general if $M$ is a monoid then $M$-acts form a category such that $End(1_{MAct}) \cong Z(M)$. This works for monoidal categories too! And for R modules, I guess it's the Ab-enriched version thereof.

John Baez (May 15 2023 at 15:39):

Thanks! I guess the proof that @JS PL (he/him) and I came up with should handle these cases too.

John Baez (May 15 2023 at 15:48):

Though for some reason I've been thinking about the case where the ring R (or the monoid M) is commutative, so I need to think a bit more about the noncommutative case (just for fun). Namely, why the obvious map from the center of M to $\mathrm{End}(1_{M\mathsf{Act}})$ is surjective. (The argument that it's injective still works.)

John Baez (May 15 2023 at 15:50):

In the commutative case we used the fact that every endomorphism of $M$ as a left $M$-set is given by left multiplication by an element of $M = Z(M)$.

John Baez (May 15 2023 at 15:51):

But when $M$ is noncommutative that's no longer true; now every endomorphism of $M$ as a left $M$-set is given by right multiplication by an element of $M$.

John Baez (May 15 2023 at 15:52):

Yet somehow (you claim) central elements give all the natural transformations of the identity functor on $M \mathsf{Act}$.

John Baez (May 15 2023 at 15:54):

It's quite plausible, but why is it true? It must be easy.

Matteo Capucci (he/him) (May 17 2023 at 07:00):

In the non-commutative case, I believe @JS PL (he/him)'s proof goes through as-is except for the fact that when he argues linear maps $R \to R$ have the form $r \cdot -$ we now have to add that $r \in Z(R)$, since $rs = sr$ is exactly the property we need

John Baez (May 17 2023 at 07:17):

No: as I said, all linear maps $R \to R$ have the form $- \cdot r$. That is, right multiplication by elements of $R$. Of course when $r$ is in the center of $R$ we can also express these using left multiplication.

John Baez (May 17 2023 at 07:18):

(By the way, I'm saying "linear maps" to be nice, but the mathematician in me is squirming: I want to say "left module endomorphisms". Life is tough for people like us!)

John Baez (May 17 2023 at 07:20):

Anyway, so the challenge is to show that of all the left module endomorphisms of $R$, only those coming from right multiplication by elements in the center of $R$ extend naturally to endomorphisms of all the other left modules.

John Baez (May 17 2023 at 07:21):

I think this is probably pretty easy, but that's what has to be done.

Matteo Capucci (he/him) (May 17 2023 at 10:28):

John Baez said:

No: as I said, all linear maps $R \to R$ have the form $- \cdot r$. That is, right multiplication by elements of $R$. Of course when $r$ is in the center of $R$ we can also express these using left multiplication.

Oh I see :thinking:

Matteo Capucci (he/him) (May 17 2023 at 10:55):

Ok here's a sketch of a proof, with a bunch of hazy details to be discussed (below, $C$ is a cartesian category and $R$ is a monoid in it):

Given $\alpha : 1_{RAct} \Rightarrow 1_{RAct}$, we want to show that for every $m:M$, there exists an $r:R$ such that $\alpha_M(m) = rm$ (I'm using the internal language of $R$-Act).
If we prove this, then for $M=R$ this ends up saying $r- = -r$, meaning $r \in Z(R)$.
It seems to me it's enough to prove this fact for $M$ free, since the naturality square for the quotient $F \to M$ presenting $M$ will then say $\alpha_M = r-$ for any other $M$.
For a free module $FX$, giving a map $\alpha_{FX} : FX \to FX$ is equivalent to a choice of map $X \to FX$ in the underlying category. The free $R$-act has the form $R \times X$. So $\alpha$ comes from a natural map $\alpha' : 1_C \Rightarrow R \times 1_C$, where $C$ is the base category (where $R$,$X$ live in). Natural maps of that kind decompose in (1) a 'choice of scalar' $1_C \Rightarrow R$ and (2) a map $\alpha_0 : 1_C \Rightarrow 1_C$.

I guess this second part is irreducible--and it makes sense: any natural endomorphism of the identity of $C$ surely induces a similar one in $RAct$. If we assume $End(1_C) = 1$ (which is true for something like Set), then $\alpha_0$ must be trivial. In generality though, this shows the theorem should really be something like '$End(1_{RAct(C)}) = Z(R) \times End(1_{C})$'.

Which brings us to the second problem, what are transformations $\rho : 1_C \Rightarrow R$? These should be 'elements of $R$', in some way. Naturality tells us the components of $\rho$ are constant. Again, in something pedestrian like Set $\rho$ is given by picking an element in $R$.

So with the above two caveats, we now know $\alpha'$ is given by a scalar $r \in R$, meaning $\alpha_{FX}$ is defined as $(s, x) \mapsto (r, s, x) \mapsto (rs, x)$, i.e. $\alpha_{FX} = r-$, as desired.

Matteo Capucci (he/him) (May 17 2023 at 10:58):

So probably a more accurate conjecture is something like $End(1_{RAct(C)}) = Z(1_C \Rightarrow R) \otimes End(1_{C})$, where $\otimes$ is the copowering of $C$.

Matteo Capucci (he/him) (May 17 2023 at 10:59):

In many cases $1_C \Rightarrow R$ should simplify to $C(1, R)$, i.e. the global elements of $R$

Matteo Capucci (he/him) (May 17 2023 at 11:10):

Matteo Capucci (he/him) said:

So with the above two caveats, we now know $\alpha'$ is given by a scalar $r \in R$, meaning $\alpha_{FX}$ is defined as $(s, x) \mapsto (r, s, x) \mapsto (rs, x)$, i.e. $\alpha_{FX} = r-$, as desired.

Ouch actually this goes $(s,x) \mapsto (s, r,x) \mapsto (sr, x)$ which is the wrong side...

Matteo Capucci (he/him) (May 17 2023 at 11:12):

Or maybe this just proves $Z$ isn't needed?

John Baez (May 17 2023 at 15:31):

Let me a take a try at it - I'll only tackle the case of rings and modules, not monoids in a general monoidal category, and I'll talk a lot more like a ring theorist than a category theorist.

John Baez (May 17 2023 at 15:31):

My goal was to show that for any ring $R$, $\mathsf{End}(1_{R \mathsf{Mod}})$ is the center of $R$. To prove this, one task remained:

Anyway, so the challenge is to show that of all the left module endomorphisms of $R$, only those coming from right multiplication by elements in the center of $R$ extend naturally to endomorphisms of all the other left modules.

John Baez (May 17 2023 at 15:38):

First, just to warm up, it's well-known that for any ring $R$, every endomorphism of $R$ as a left $R$-module, say $f: R \to R$, is given by right multiplication by some $r \in R$. But it's easy to show so let's show it! Let $f(1) = r$. Then for any $x \in R$ we have

$f(x) = f(x 1) = x f(1) = x r$

Done.

John Baez (May 17 2023 at 15:40):

Now, suppose $\alpha: 1_{R \mathsf{Mod}} \to 1_{R \mathsf{Mod}}$ is a natural transformation of the identity. $\alpha_R: R \to R$ is some endomorphism of $R$ as a left $R$-module, so as we've just seen, for some $s \in R$ we have

$\alpha_R(x) = x s$

for all $x \in R$.

John Baez (May 17 2023 at 15:42):

But for $\alpha$ to be natural, a commutative square condition needs to hold. In particular $\alpha_R$ needs to commute with any left $R$-module endomorphism $f: R \to R$.

John Baez (May 17 2023 at 15:43):

Take $f$ as above; then

$\alpha_R(f(x)) = x r s$

while

$f(\alpha_R(x)) = x s r$

John Baez (May 17 2023 at 15:44):

Since this is true for all $f$, i.e. all $r$, $s$ must be in the center of $R$!

Done.

John Baez (May 17 2023 at 15:51):

And why did we care about this? Because we know $\alpha$ must act as left multiplication by this central element $r$ on every module $M$. @JS PL (he/him) showed it in the commutative case but let me check that the argument works in general. For any $M \in R\mathsf{Mod}$ and any $m \in M$ there's a unique left $R$-module homomorphism $g: R \to M$ with $g(1) = m$, namely the one with

$g(x) = xm$

for all $x \in R$.

John Baez (May 17 2023 at 15:54):

By naturality we get a commutative square that says

$\alpha_M \circ g = g \circ \alpha_R$

but

$\alpha_M (g(1)) = \alpha_M (m)$

while

$g (\alpha_R(1)) = g(r) = r m$

John Baez (May 17 2023 at 15:55):

so

$\alpha_M(m) = r m$

John Baez (May 17 2023 at 15:55):

Done!

John Baez (May 17 2023 at 15:56):

This completes the proof that for any ring $R$, $\mathsf{End}(1_{R \mathsf{Mod}})$ is the center of $R$.

Matteo Capucci (he/him) (May 19 2023 at 08:03):

For the record, this is where I went wrong:

For a free module $FX$, giving a map $\alpha_{FX} : FX \to FX$ is equivalent to a choice of map $X \to FX$ in the underlying category. The free $R$-act has the form $R \times X$. So $\alpha$ comes from a natural map $\alpha' : 1_C \Rightarrow R \times 1_C$, where $C$ is the base category (where $R$,$X$ live in).

We can't get $\alpha'$ of that form from $\alpha$.

Matteo Capucci (he/him) (May 19 2023 at 08:05):

John Baez said:

Let me a take a try at it - I'll only tackle the case of rings and modules, not monoids in a general monoidal category, and I'll talk a lot more like a ring theorist than a category theorist.

It seems to me your proof actually ends up working for arbitrary monoid objects, am I right?

John Baez (May 19 2023 at 19:37):

I used a fact that's true for rings (and monoids) that I don't know how to correctly state for monoid objects in general.

John Baez (May 19 2023 at 19:39):

Namely, for a monoid M (in Set) every map of left M-sets from M to M is given by right multiplication by some element of M.

John Baez (May 19 2023 at 19:42):

Then I used this: every natural transformation of the identity functor on MSet is determined by its value on M.

Mike Shulman (May 19 2023 at 19:51):

John Baez said:

Namely, for a monoid M (in Set) every map of left M-sets from M to M is given by right multiplication by some element of M.

That's an instance of the Yoneda lemma for one-object categories. And the Yoneda lemma is true for categories enriched over any monoidal category, hence also for one-object ones, i.e. monoid objects. At least if the monoidal category is closed and complete, you can state the result by saying that $M$ is isomorphic to the object of left $M$-set maps from $M$ to $M$, the latter defined as a subobject of the internal-hom cut out by an equalizer.

Matteo Capucci (he/him) (May 22 2023 at 20:59):

John Baez said:

First, just to warm up, it's well-known that for any ring $R$, every endomorphism of $R$ as a left $R$-module, say $f: R \to R$, is given by right multiplication by some $r \in R$. But it's easy to show so let's show it! Let $f(1) = r$. Then for any $x \in R$ we have

$ f(x) = f(x 1) = x f(1) = x r$

Done.

Also this proof works for monoids, doesn't it?

John Baez (May 22 2023 at 22:26):

Yes, the whole proof never needs addition. So it works for monoids... and also rigs, and also rings. But at least as stands, it does use some other things peculiar to the fact that we're working in the category of sets. Mike Shulman tried to explicate some of those... but I want to add that unless we're in a symmetric monoidal category the concept of the 'center' doesn't make sense. (Or braided, but that gets very tricky.)

John Baez (May 22 2023 at 22:26):

So, it would be a nice project to figure out how much we can generalize.

John Baez (May 22 2023 at 22:27):

I wouldn't be surprised if someone had already done this.