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Stream: learning: questions

Topic: making a category into an Ab-enriched category

John Baez (May 16 2023 at 01:01):

Can someone come up with examples illustrating how the same category can be [[enriched]] over abelian groups in several different ways?

John Baez (May 16 2023 at 01:02):

(For more on categories enriched over abelian groups, see [[Ab-enriched category]].)

John Baez (May 16 2023 at 01:03):

A one-object category is a monoid, while a one-object Ab-enriched category is a ring.

John Baez (May 16 2023 at 01:03):

So, one type of solution to my problem would be to take a monoid $M$, think of the monoid operation as multiplication, and come up with more than one addition operation making $M$ into a ring.

John Baez (May 16 2023 at 01:04):

But frankly no example of this sort comes to mind! (At least not my mind.)

Mike Shulman (May 16 2023 at 01:21):

As you may know, doing this for abelian groups is tricky, because if an Ab-enriched category has finite products (equivalently, finite coproducts), then its Ab-enrichment is uniquely determined. So any such example must lack finite products and coproducts. Which would be the case for your suggestion of using the one-object case. That doesn't really help me come up with an example, though.

Mike Shulman (May 16 2023 at 01:22):

Another way of stating the one-object version is: find two rings whose multiplicative monoids are isomorphic, but for which the isomorphism doesn't preserve addition. Somehow that version sounds more approachable to me...

JS PL (he/him) (May 16 2023 at 01:23):

Also, no matter the enrichment, the zero maps will always be the same. So if you're looking for a monoid that has more than one addition which makes it into a ring, the zero element will be the same for both addition operations.

John Baez (May 16 2023 at 01:33):

Mike Shulman said:

As you may know, doing this for abelian groups is tricky, because if an Ab-enriched category has finite products (equivalently, finite coproducts), then its Ab-enrichment is uniquely determined.

I know some similar things but not quite this. Please don't give it away - I prefer to think about it!

But in my first attempt to cheat I came across this in the nLab:

For a,b∈C two objects in an Ab-enriched category, the product a×b coincides with the coproduct a+b when either exists. More precisely, when both exist, the canonical morphism a+b→a×b is an isomorphism.

My question is just: isn't that "more precise" statement saying something weaker than what it's supposedly clarifying? Saying the product and coproduct are isomorphic when both exist sounds weaker than saying they're isomorphic when either exists. So maybe the writing needs a bit of work here.

Mike Shulman (May 16 2023 at 01:52):

Yes, I agree; that writing should be improved.

John Baez (May 16 2023 at 01:53):

You'll be glad to hear I won't say they should do it.

Reid Barton (May 16 2023 at 06:31):

Take a subfield $R$ of $\mathbb{R}$ which contains all $n$-th roots of all of its positive elements (e.g., $R$ = the real algebraic numbers). Its positive elements $R_{> 0}$, as a multiplicative group, form a $\mathbb{Q}$-vector space (because each element is uniquely divisible). We get the whole multiplicative monoid of $R$ by taking $R^{\times} = R_{> 0} \times \{\pm 1\}$ (as groups) and then adjoining an absorbing zero element.

The dimension of $R_{> 0}$ and hence the isomorphism type of the resulting monoid $R$ can only depend on the cardinality of $R_{> 0}$. So for example any two such countable fields will have isomorphic multiplicative monoids, but we can make them nonisomorphic as fields. For example $R_1$ = the real algebraic numbers, $R_2$ = the real numbers algebraic over $\mathbb{Q}(\pi)$. (In $R_2$, not every element is algebraic over the prime field.)

This feels overkill; I feel there should be a simpler example out there...

Tobias Fritz (May 16 2023 at 07:42):

For potentially simpler examples, consider a unique factorization domain $R$: disregarding zero, the multiplicative monoid is isomorphic to $R^\times \times \mathbb{N}^I$, where $R^\times$ is the group of units and $I$ is the set of prime elements modulo units. So two UFDs have isomorphic multiplicative monoids as soon as they have isomorphic groups of units and the same number of primes modulo units.

A simple example could be $\mathbb{Z}$ and $(\mathbb{Z}/3)[X]$, both of which have the two-element group as the group of units and countably many classes of primes. So they have isomorphic multiplicative monoids, but they're clearly not isomorphic as rings. for example because $\mathbb{Z}$ is initiial in the category of rings while $(\mathbb{Z}/3)[X]$ is not.

Tobias Fritz (May 16 2023 at 07:54):

MathOverflow even has Non isomorphic finite rings with isomorphic additive and multiplicative structure. The question notes that polynomial rings $F[X]$ and $F[X,Y]$ over the same field $F$ have isomorphic additive groups and isomorphic multiplicative monoids, but are not isomorphic as rings!

John Baez (May 16 2023 at 16:22):

Excellent answers! I thought of using algebraic number fields, but only came up with some examples of nonisomorphic rings with isomorphic additive groups. :big_frown: These seem a lot easier to come by.

JS PL (he/him) (May 16 2023 at 23:59):

An even simpler toy example. For the set with four elements $M = \lbrace 0, 1, 2, 3 \rbrace$, there are only two possible Abelian group structures. One given by addition $+$ mod $4$, so the cyclic group $\mathbb{Z}_4$ (I don't know how to make nice tables in zulip):

$0 + 0 =0$, $0 + 1 = 1$, $0 + 2 = 2$, $0+ 3 = 3$
$1+ 0 =1$, $1 +1 = 1$, $1+2 = 3$, $1+3 = 4$
$2 + 0 =2$, $2+ 1 = 3$, $2 + 2 = 0$, $2+ 3 = 1$
$3 + 0 =3$, $3 + 1 = 0$, $3 + 2 = 1$, $3 + 3 = 2$

Or you can equip it with the Klein $4$ group structure, which is the addition $\oplus$ given by:

$0 \oplus 0 =0$, $0 \oplus 1 = 1$, $0 \oplus 2 = 2$, $0 \oplus 3 = 3$
$1 \oplus 0 =1$, $1 \oplus 1 = 0$, $1 \oplus 2 = 3$, $1 \oplus 3 = 2$
$2 \oplus 0 =2$, $2 \oplus 1 = 3$, $2 \oplus 2 = 0$, $2 \oplus 3 = 1$
$3 \oplus 0 =3$, $3 \oplus 1 = 2$, $3 \oplus 2 = 1$, $3 \oplus 3 = 0$

Now $(M,+)$ and $(M, \oplus)$ are not isomorphic as Abelian groups. However, both $(M,+)$ and $(M, \oplus)$ become rings when equipped with the multiplication $\cdot$ modulo 4:

$0 \cdot 0 =0$, $0 \cdot 1 = 1$, $0 \cdot 2 = 2$, $0 \cdot 3 = 3$
$1 \cdot 0 =0$, $1 \cdot 1 = 1$, $1 \cdot 2 = 2$, $1 \cdot 3 = 3$
$2 \cdot 0 =0$, $2\cdot 1 = 2$, $2 \cdot 2 = 0$, $2 \cdot 3 = 2$
$3 \cdot 0 =3$, $3 \cdot 1 = 3$, $3 \cdot 2 = 2$, $3 \cdot 3 = 1$

John Baez (May 17 2023 at 00:27):

That's nice! Why does multiplication mod 4 distribute over addition in $\mathbb{Z}/2 \oplus \mathbb{Z}/2$?

John Baez (May 17 2023 at 00:28):

There should be some nice high-level explanation that doesn't involve checking lots of cases.

JS PL (he/him) (May 17 2023 at 00:49):

John Baez said:

That's nice! Why does multiplication mod 4 distribute over addition in $\mathbb{Z}/2 \oplus \mathbb{Z}/2$?

$\mathbb{Z}/2 \oplus \mathbb{Z}/2$ with the multiplication mod 4 gives the ring of dual numbers $\mathbb{Z}/2[\varepsilon]$.

JS PL (he/him) (May 17 2023 at 00:53):

So $(M, \oplus, \cdot) \cong \mathbb{Z}/2[\varepsilon]$ while $(M, +, \cdot) \cong \mathbb{Z}/4$

John Baez (May 17 2023 at 05:42):

Great! These are clearly both commutative rings, so the "mystery" becomes 1) why they have isomorphic multiplicative monoids yet 2) nonisomorphic additive monoids. 2) is easy: $\mathbb{Z}/2 \oplus \mathbb{Z}/2 \ncong \mathbb{Z}/4$. Is there a conceptual explanation of 1)? Or do we just notice that the multiplication table of $0,1,\epsilon, 1 + \epsilon$ in $\mathbb{Z}/2[\epsilon]$ is the same as the multiplication table of $0,1,2,3$ in $\mathbb{Z}/4$?

John Baez (May 17 2023 at 05:44):

(A conceptual explanation might unlock a huge store of such examples.)

Todd Trimble (May 17 2023 at 20:34):

John Baez said:

You'll be glad to hear I won't say they should do it.

Well, someone did it, and that someone was me. Or, at least I addressed the observation of John above about "more precisely".

John Baez (May 17 2023 at 21:04):

Thanks! I was going to do it; I figured out what I wanted to say, but then I got distracted.

John Baez (May 17 2023 at 21:05):

For anyone wondering what we're talking about, it's this:

• Ab-enriched category: biproducts.