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Stream: learning: questions

Topic: Subcategories of CRing closed under tensor product

Ryan Schwiebert (Apr 17 2026 at 02:49):

I am interested in classes of rings defined by properties, and would be interested in knowing what sort of tricks I can apply to deduce a certain class would be closed under tensor products. In particular for this conversation, just commutative rings suffices.

I have a great deal of difficulty reasoning about tensor products of rings, but a lot more experience checking if finite/infinite products of rings have a certain property.

In addition, if anyone's aware of literature that already collects results on stability of properties via the coproduct, I'd be interested in pointers.

John Baez (Apr 17 2026 at 03:47):

Maybe you know that for any integer $n$, if you tensor a commutative ring where every element $x$ has $n x = 0$ with another commutative ring, you get a commutative ring where every element $x$ has $n x = 0$. So this property is not only closed under tensor products but "absorbing".

John Baez (Apr 17 2026 at 03:54):

I think a class of commutative rings that's absorbing in this sense is often called a tensor ideal.

John Baez (Apr 17 2026 at 04:23):

Also, any class axiomatized by 'existential sentences' in the language of rings (i.e. sentences saying there exist some elements $x_1, \dots, x_n$ satisfying some polynomial equations) is closed under ring homomorphisms, hence in particular under the map from $R \to R \otimes S$ sending $r$ to $r \otimes 1$. So, any such class is not only closed under tensor products, but is a tensor ideal.

Ryan Schwiebert (Apr 18 2026 at 02:28):

@John Baez That reminds me that the tensor product of two rings can be the zero ring. This is a case where a "disjoint union" picture of coproduct isn't a good mnemonic :joy:

Some time ago we talked about equationally defined rings. I'm not sure this is the same thing though... only existential quantification? Or are there free variables in there that do the job of universal quantification? I was prepared for von Neumann regular rings to behave well under tensor product.

John Baez (Apr 18 2026 at 04:14):

You shouldn't think of a coproduct of commutative rings as like a 'disjoint union'. You should think of a commutative ring as the ring of well-behaved functions on some interesting sort of space (called an affine scheme), and then the coproduct of commutative rings is the ring of functions on the product of such spaces.

The trick I mentioned let you show that any class of commutative rings defined by existentially quantified polynomial equations is a tensor ideal. This is not the case for every class of commutative rings defined by universally quantified polynomial equations. (These are called 'varieties' of commutative rings, and they're nice in many ways, but not this way!)

Ryan Schwiebert (Apr 18 2026 at 12:47):

What are some examples of classes that are existentially quantified?

Morgan Rogers (he/him) (Apr 18 2026 at 16:28):

For instance the class of rings containing a solution to $x^2 = -1$, which include $\mathbb{Z}/2\Z$ and $\mathbb{C}$ but not $\Z$ or $\R$

John Baez (Apr 18 2026 at 23:19):

People define a divisible group to be an abelian group $A$ such that for each $n$ and each $a \in A$ there exists $b \in A$ such that

$b + \cdots + b = a$

where we add $n$ copies of $b$. In other words, we can "divide $a$ by $n$".

We can define a divisible ring to be a ring whose underlying additive group is divisible. For example $\mathbb{Z}$ is not divisible since you can't divide by $2$ in the ring of integers, but $\mathbb{Q}$ is a divisible ring.

Divisible abelian groups are somewhat important in homological algebra since they're exactly the injective objects in the category of abelian groups (hmm, at least if you assume the axiom of choice, according to Wikipedia).

But it's a lot easier to see that divisible rings are an existentially quantified class, so they form a tensor ideal. The latter is easy to see directly: if we can "divide by $n$" in a ring $A$ then you can divide by $n$ in $A \otimes B$ for any ring $B$.

Ryan Schwiebert (Apr 19 2026 at 00:00):

I would not have expected the divisibility example, because I thought the universal quantifications weren't allowed. An von Neumann regularity isn't of the same type? For each $a$ there exists $b$ such that $a=aba$?

John Baez (Apr 19 2026 at 01:08):

Ugh, you're right, there are universal quantifiers in the axioms

$\forall a \exists b (a = 2b)$
$\forall a \exists b (a = 3b)$
$\forall a \exists b (a = 4b)$
etc.

(At least no universal quantification over $n$ is needed! It's fine to have infinitely many axioms.)

However, I still believe that if $R$ is a divisible commutative ring, and $S$ is any commutative ring, then $R \otimes_{\mathbb{Z}} S$ is divisible. Any element $x$ of $R \otimes_{\mathbb{Z}} S$ is of the form

$x = r_1 \otimes s_1 + \cdots + r_k \otimes s_k$

Choose $n \in \{1,2,3, \dots \}$. We can write $r_i = n r'_i$. Thus

$x = n (r'_1 \otimes s_1 + \cdots + r'_k \otimes s_k)$

as desired. So $R \otimes_{\mathbb{Z}} S$ is divisible.

So, divisible commutative rings form a tensor ideal - and thus they're closed under taking tensor products.

Does this look correct?

John Baez (Apr 19 2026 at 01:10):

There's probably a lesson here that can be generalized to other classes of rings: some kinds of universal quantifiers are allowed, and we still get a tensor ideal. I will let someone else figure that out!

Ryan Schwiebert (Apr 19 2026 at 01:24):

Morgan Rogers (he/him) said:

For instance the class of rings containing a solution to $x^2 = -1$, which include $\mathbb{Z}/2\Z$ and $\mathbb{C}$ but not $\Z$ or $\R$

BTW i saw this one, and that makes sense, although the property is rather of a different flavor than the ones I think about :)

John Baez (Apr 19 2026 at 01:26):

For physicists this property is a biggie. We use a wide diversity of rings in quantum physics - even von Neumann regular rings we invented for this purpose - but quantum physics would not fly without i.

John Baez (Apr 19 2026 at 01:28):

I imagine number theorists often want an nth root of 1 for some purpose.

John Baez (Apr 19 2026 at 01:42):

By the way, my sources provide an example showing commutative von Neumann regular rings are not closed under tensor product.

$\mathbb{Q}(t)$ is a [[von Neumann regular ring]] (since it's a field), while

$\mathbb{Q}(t) \otimes_{\mathbb{Z}} \mathbb{Q}(t) \cong \mathbb{Q}(t) \otimes_{\mathbb{Q}} \mathbb{Q}(t)$

is not (since it's an [[integral domain]] but not a field).

I don't understand why $\mathbb{Q}(t) \otimes_{\mathbb{Q}} \mathbb{Q}(t)$ an integral domain but not a field, but here's the argument that a von Neumann regular ring that's an integral domain is a field:

In an integral domain, if $a \neq 0$ satisfies $a = axa$ for some $x$ , then $ax$ is nonzero and it's an idempotent. But integral domains have only $0$ and $1$ as idempotents, so $ax = 1$, so $a$ has an inverse.

I understand that at least!

Ryan Schwiebert (Apr 19 2026 at 02:47):

John Baez said:

Does this look correct?

It does look correct! I just wish I knew what I could say about the properties I'm interested in. Let's suppose we were interested in some sort of domain. Then it seems like right away you'd have to restrict to some particular ring characteristic (or family of characteristics?) to keep the zero ring out of your class of domains.

Ryan Schwiebert (Apr 19 2026 at 02:51):

My intuition for it is that VNR rings have lots of idempotents, so a ring like a domain with minimal idempotents is going to interact violently with the condition.

Ryan Schwiebert (Apr 19 2026 at 12:07):

John Baez said:

We can define a divisible ring to be a ring whose underlying additive group is divisible.

It looks to me like this amounts to a $\mathbb Q$ algebra, yes?

John Baez (Apr 19 2026 at 15:56):

Certainly any $\mathbb{Q}$-algebra will be a divisible ring, but I'm confused about the converse, because in the definition of divisible ring there's no guarantee that given $a$ in the ring and $n \in \{1,2,3, \dots \}$ there's a unique $b$ with $a = nb$.

So I don't know if, given a divisible ring, there always exists a way to consistently choose how to divide ring elements by nonzero integers and obtain a $\mathbb{Q}$-algebra. I also don't if there's a unique way.

John Baez (Apr 19 2026 at 15:59):

Now that I think about it, I'm guessing it's possible to prove these with a bit of work, e.g. showing a lemma that says a [[divisible group]] has no $n$-torsion elements ($b$ with $n b = 0$), so we can't have $n b = n b'$ unless $b = b'$.

Ryan Schwiebert (Apr 19 2026 at 16:00):

Yeah, that's where I was going to begin. There's no torsion elements, so in particular $\mathbb Z$ is a subring of $R$ matching the abelian group action. Then in particular you can divide 1 by any $n$ to get inverses for the elements of $\mathbb Z$.

Ryan Schwiebert (Apr 19 2026 at 16:01):

I haven't seen 'divisible rings' in literature but I have seen torsion-free rings (in the sense of no nonzero finite-order elements.)

John Baez (Apr 19 2026 at 16:24):

I'll admit I've never seen divisible rings in the literature either, just divisible groups (which seems to usually mean divisible abelian groups). I just made up the concept.

Of course torsion-free groups are more general than divisible groups, and ditto for rings.

Todd Trimble (Apr 20 2026 at 21:36):

Divisible rings $R$ are the same as $\mathbb{Q}$-algebras.

Start with the fact that an abelian group is divisible iff it is injective. Injectivity of $R$ implies that the unique ring map $\mathbb{Z} \to R$, taking $1$ to $1_R$, can be extended to an abelian group map $f: \mathbb{Q} \to R$.

Claim: For all nonzero integers $n$, $f(n) = n \cdot 1_R$ is invertible in $R$, with inverse $f(\frac1{n})$. Consequently, $R$ is torsionfree.

Proof: $1_R = f(1) = f(n \cdot \frac1{n}) = n f(\frac1{n}) = n 1_R f(\frac1{n}) = f(n) f(\frac1{n})$

where the third equality uses only the fact that $f$ is an additive homomorphism.

Claim: $f: \mathbb{Q} \to R$ is a ring map.

Proof: Additivity and preservation of the unit $1$ were baked in from the outset. To check preservation of multiplication, $f(a/b \cdot c/d) = f(a/b) f(c/d)$, it suffices to check that multiplication of each side by $bd$ gives an equality, since multiplication by $bd$ is an injective map $R \to R$, by the fact that $R$ is torsionfree. But the result is indeed an equality since it is just $f(a c) = f(a) f(c)$, and the restriction of $f$ to integers is a ring map by definition of $f$.

John Baez (Apr 20 2026 at 21:51):

Very nice, @Todd Trimble!

One spinoff is that 'being a $\mathbb{Q}$-algebra', while superficially a structure on a ring, turns out to be a mere property: i.e., a ring can have this structure in at most one way.

One can ask for which commutative rings $R$ is 'being an $R$-algebra' a mere property. Apparently these are the [[solid rings]]: rings $R$ such any two ring homomorphisms $f, g: R \to S$ to any other ring must be equal.

John Baez (Apr 20 2026 at 21:55):

So, I believe that $R$-algebras for any solid commutative ring $R$ form a full subcategory of the category of commutative rings that is closed under tensor product (as @Ryan Schwiebert wanted) but also in fact a tensor ideal.

A while back I added to the nLab the complete classification of solid rings. Someone also put in a proof that $\mathbb{Q}$ is a solid ring - maybe @Todd Trimble?

Todd Trimble (Apr 20 2026 at 22:51):

It wasn't me. I didn't scour the history, but the way it's written makes it look like Urs put that in.

Todd Trimble (Apr 21 2026 at 01:30):

John Baez said:

By the way, my sources provide an example showing commutative von Neumann regular rings are not closed under tensor product.

$\mathbb{Q}(t)$ is a [[von Neumann regular ring]] (since it's a field), while

$\mathbb{Q}(t) \otimes_{\mathbb{Z}} \mathbb{Q}(t) \cong \mathbb{Q}(t) \otimes_{\mathbb{Q}} \mathbb{Q}(t)$

is not (since it's an [[integral domain]] but not a field).

I don't understand why $\mathbb{Q}(t) \otimes_{\mathbb{Q}} \mathbb{Q}(t)$ an integral domain but not a field, but here's the argument that a von Neumann regular ring that's an integral domain is a field:

In an integral domain, if $a \neq 0$ satisfies $a = axa$ for some $x$ , then $ax$ is nonzero and it's an idempotent. But integral domains have only $0$ and $1$ as idempotents, so $ax = 1$, so $a$ has an inverse.

I understand that at least!

About that last part: $\mathbb{Q}(t) \otimes_{\mathbb{Q}} \mathbb{Q}(t) \cong \mathbb{Q}(x) \otimes_\mathbb{Q} \mathbb{Q}(y)$ is embedded inside the field $\mathbb{Q}(x, y)$ of rational functions in $x, y$, being the subring of rational functions of the form $\sum_{i=1}^n p_i(x) q_i(y)$ where $p_i(x) \in \mathbb{Q}(x)$ and $q_i(y) \in \mathbb{Q}(y)$ are rational functions in their variables. Clearly

$\mathbb{Q}(x) \otimes_\mathbb{Q} \mathbb{Q}(y) \subseteq \mathbb{Q}(x, y)$

is a strict inclusion, and so (1) the subring is an integral domain, being a subring of a field, and (2) the subring is not itself a field: consider the strict inclusions

$\mathbb{Q}[x, y] \cong \mathbb{Q}[x] \otimes_\mathbb{Q} \mathbb{Q}[y] \subset \mathbb{Q}(x) \otimes_\mathbb{Q} \mathbb{Q}(y) \subset \mathbb{Q}(x, y)$

where $\mathbb{Q}(x, y)$ is the field of fractions of $\mathbb{Q}[x, y]$ and hence also of the intermediate ring $\mathbb{Q}(x) \otimes_\mathbb{Q} \mathbb{Q}(y)$, but since this ring is properly contained in its field of fractions, it is not itself a field.

Todd Trimble (Apr 21 2026 at 01:51):

Ryan Schwiebert said:

John Baez said:

We can define a divisible ring to be a ring whose underlying additive group is divisible.

It looks to me like this amounts to a $\mathbb Q$ algebra, yes?

This is a good observation, because this recharacterization of divisible rings as $\mathbb{Q}$-algebras is clearly axiomatized by existential sentences (there exists an inverse of 2, there exists an inverse of 3, ...).

John Baez (Apr 21 2026 at 01:56):

Oh good, so this shows divisible rings are axiomatized by existential sentences!

(And so, for people losing track of the plot here, they form a tensor ideal by a general theorem I mentioned earlier.)

Todd Trimble (Apr 21 2026 at 01:59):

I'm intrigued by these tensor ideals. I've not thought much about them. Can you tell me where they are studied, John?

Ryan Schwiebert (Apr 21 2026 at 02:03):

Regarding n-divisibility for less than all n's, one could phrase a bunch of theorems about group algebras in terms of "n divisibility," e.g. Maschke's theorem.

Ryan Schwiebert (Apr 21 2026 at 02:05):

Being a tensor ideal must be of a wildly different sort of character than stability with products. It looks like commonplace classes of rings have almost nothing to do with tensor ideals.

John Baez (Apr 21 2026 at 02:08):

Todd Trimble said:

I'm intrigued by these tensor ideals. I've not thought much about them. Can you tell me where they are studied, John?

The closest thing I know is know comes from Paul Balmer's work on tensor-triangular geometry, founded in his 2005 paper The spectrum of prime ideals in tensor triangulated categories. They are working with tensor-triangulated categories, and their ideals are 'thick ⊗-ideals': full triangulated subcategories closed under tensoring with any object. Balmer and a bunch of folks have developed this idea a lot. But their setting is triangulated, not CommRing... and I don't really know anything about their work!

In reality, I thought of 'tensor ideals' in response to Ryan's original question merely because of our own work on 2-ideals in 2-rigs.

Graham Manuell (Apr 23 2026 at 09:49):

This result about von Neumann regular rings is quite surprising to me, because it means that the product of two zero-dimensional schemes can fail to be zero-dimensional.

Ryan Schwiebert (Apr 23 2026 at 16:24):

So many of the properties I usually think about are amenable to finite or arbitrary products. I think the discussion above is showing the stuff that survives coproducts is of quite a different flavor.

John Baez (Apr 23 2026 at 16:48):

@Graham Manuell - are you implicitly saying that an affine scheme is zero-dimensional iff its corresponding commutative ring is a von Neumann regular ring? I hadn't known that (since I don't even know any definition of "zero-dimensionality" for schemes, being pretty ignorant about schemes).

Graham Manuell (Apr 23 2026 at 17:18):

I think von Neumann regular corresponds to zero-dimensional + reduced. Usually people use the Krull dimension. In this case it should be the same as the spectrum having a base of clopens.

Todd Trimble (Apr 23 2026 at 18:18):

I just learned the following remarkable fact, due to someone named Rodney Sharp:

• Let $K, L$ be two extension fields of a field $k$. Then $\mathrm{KrullDim}(K \otimes_k L) = \min(\mathrm{Tr}(K/k), \mathrm{Tr}(L/k))$

(where we declare the Krull dimension to be $\infty$ if it is not finite, and the transcendence degree $\mathrm{Tr}(K/k)$ to be $\infty$ if it is not finite).

So the dimension of a product of zero-dimensional schemes can be as big as you please, even $\infty$, in this sense.

John Baez (Apr 24 2026 at 04:28):

I mentioned this on Mastodon giving the example of the ring $\mathbb{R} \otimes_\mathbb{Q} \mathbb{R}$, and someone pointed out that Dehn invariants of polyhedra live in the underlying abelian group of this mod $2\pi \mathbb{Z}$. No idea if that connection is interesting.

Todd Trimble (Apr 24 2026 at 04:33):

Seems kind of interesting. I discovered a MO post that touches on this, here.