Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: why rings need identities

John Baez (Feb 23 2023 at 07:24):

What are some of the simplest important results in ring theory - if any - that would fail if we didn't require rings to have a multiplicative unit?

John Baez (Feb 23 2023 at 07:25):

I'm asking because many people used to not require rings to have a unit, and some people still don't. (I think it's more common to drop the requirement for a unit when talking about "algebras", e.g. C*-algebras.)

Jason Erbele (Feb 23 2023 at 08:20):

Looking at Hungerford's Algebra, he makes a point to specify that a lot of our favorite kinds of rings are rings with identity. E.g. integral domains, division rings, and fields. The latter two require multiplicative inverses, so they don't make sense without identities.

Jason Erbele (Feb 23 2023 at 08:20):

I suppose one of the simplest rings without identity to cook up would be $2\mathbb{Z}$. Something that breaks in $2\mathbb{Z}$ that works in any ring with identity is the existence of maximal ideals (Theorem 2.18 in Hungerford's chapter on Rings).

Jason Erbele (Feb 23 2023 at 08:20):

One nice thing about dropping the requirement that a ring has identity means that ideals are subrings.

Jason Erbele (Feb 23 2023 at 08:24):

Of course, Hungerford is also very careful to state that $1_R \neq 0$ whenever he needs a ring to have identity.

Graham Manuell (Feb 23 2023 at 10:41):

This is not an answer to your question about what theorems don't work, but I find it strange to have an associative operation that doesn't have a unit, since it seems like ignoring the nullary product for no real reason.

Also, nonunital rings are to rings as pointed topological spaces are to topological spaces (since nonunital rings are equivalent to rings with a morphism into $\mathbb{Z}$). So it seems like rings are the fundamental concept even if sometimes nonunital rings are also useful.

David Michael Roberts (Feb 23 2023 at 11:34):

Lie algebras are rings, if you can give up associativity and unitality.

John Baez (Feb 23 2023 at 15:49):

Of course fields and division rings need to have identities, but there are many theorems about rings and commutative rings - I was just looking at Atiyah and Macdonald's Commutative Algebra and admiring the enormous pile of theorems - and I want to know which theorems, if any, break down if don't require rings to have identities.

John Baez (Feb 23 2023 at 15:51):

Jason Erbele said:

One nice thing about dropping the requirement that a ring has identity means that ideals are subrings.

Right, this is a superficially powerful reason to want to drop the requirement that rings have identities! (A wiser approach is to treat an ideal of R not as a special sort of subring but as a submodule of R: for example, a left ideal is exactly a left submodule of R.)

John Baez (Feb 23 2023 at 15:55):

Jason Erbele said:

I suppose one of the simplest rings without identity to cook up would be $2\mathbb{Z}$. Something that breaks in $2\mathbb{Z}$ that works in any ring with identity is the existence of maximal ideals (Theorem 2.18 in Hungerford's chapter on Rings).

Okay, great! This is the first and so far only answer to my question. But I'm not sure I understand it. Isn't $4 \mathbb{Z}$ a maximal ideal in the rng $2 \mathbb{Z}$?

John Baez (Feb 23 2023 at 15:57):

Maybe I'm confused.

John Baez (Feb 23 2023 at 15:59):

Jason Erbele said:

Of course, Hungerford is also very careful to state that $1_R \neq 0$ whenever he needs a ring to have identity.

That's terrible, since he's killing off the terminal ring, i.e. the 1-element ring, and probably for no really good reason. I think most basic theorems about rings don't require that we exclude the terminal ring. I think it's just that the equation 0 = 1 gives Hungerford the willies.

John Baez (Feb 23 2023 at 16:01):

Here is one possible reason we want to use rings instead of rngs: in the category of rngs, $\mathbb{Z}$ is not initial. (Find a rng with two different rng maps from $\mathbb{Z}$ to it - you don't have to look far.)

John Baez (Feb 23 2023 at 16:04):

I don't consider this to be the sort of "important theorem" that I'm talking about - a lot of ring theorists might say "so $\mathbb{Z}$ isn't initial? So what?" But it points to a way you could look for important theorems about rings that break down with rngs: any theorem where $\mathbb{Z}$ plays a big role becomes a bit suspect.

Xuanrui Qi (Feb 23 2023 at 17:06):

John Baez said:

I don't consider this to be the sort of "important theorem" that I'm talking about - a lot of ring theorist might say "so $\mathbb{Z}$ isn't initial? So what?" But it points to a way you could look for important theorems about rings that break down with rngs: any theorem where $\mathbb{Z}$ plays a big role becomes a bit suspect.

Algebraic geometers will surely be mad if $\mathbb{Z}$ isn't initial! Not sure about pure commutative algebraists (that is to say, if there are still any of them).
Can you do any algebraic geometry at all if your rings have no identity, though?

Jason Erbele (Feb 23 2023 at 17:28):

John Baez said:

Jason Erbele said:

I suppose one of the simplest rings without identity to cook up would be $2\mathbb{Z}$. Something that breaks in $2\mathbb{Z}$ that works in any ring with identity is the existence of maximal ideals (Theorem 2.18 in Hungerford's chapter on Rings).

Okay, great! This is the first and so far only answer to my question. But I'm not sure I understand it. Isn't $4 \mathbb{Z}$ a maximal ideal in the rng $2 \mathbb{Z}$?

Oops. My brain wasn't working correctly at midnight... I was thinking about $4\mathbb{Z}$ and $6\mathbb{Z}$ and forgot I had to ensure the condition $4\mathbb{Z} \subset 6\mathbb{Z}$ which obviously doesn't hold. $2\mathbb{Z}$ definitely isn't a unique factorization domain (e.g. 30 x 30 = 18 x 50, but none of these can factor any further in $2\mathbb{Z}$), but that's not a theorem breaking.

Jason Erbele (Feb 23 2023 at 17:33):

But here is a theorem in Hungerford that he explicitly mentions does not work if $R$ does not have an identity:

If $M$ is maximal and $R$ is commutative, then the quotient ring $R/M$ is a field.

Jason Erbele (Feb 23 2023 at 17:38):

Hungerford includes some extra conditions on the Chinese Remainder Theorem if $R$ is a rng instead of a ring:

Let $A_1, \dotsc , A_n$ be ideals in a ring $R$ such that $R^2 + A_i = R$ for all $i$ and [...]
REMARK.If $R$ has an identity, then $R^2 = R$, whence $R^2 + A = R$ for every ideal $A$ of $R$.

John Baez (Feb 23 2023 at 18:37):

Jason Erbele said:

But here is a theorem in Hungerford that he explicitly mentions does not work if $R$ does not have an identity:

If $M$ is maximal and $R$ is commutative, then the quotient ring $R/M$ is a field.

Thanks! That make sense: fields need units, and why should $R/M$ have a unit if $R$ does not?

But it's better to look at an example. I continue to believe $4 \mathbb{Z}$ is a maximal ideal in $2 \mathbb{Z}$, since if I take any even number $a$ that's not a multiple of 4, I can subtract an element of $4 \mathbb{Z}$ and get $2$, and that generates all of $2 \mathbb{Z}$ as an ideal.

John Baez (Feb 23 2023 at 18:38):

But then what's $2\mathbb{Z}/4 \mathbb{Z}$? It's the set $\{0,2\}$ with the addition and multiplication mod 4.

John Baez (Feb 23 2023 at 18:39):

This is a rng but not a ring so definitely not a field!

John Baez (Feb 23 2023 at 18:39):

So this is very interesting: algebraists and algebraic geometers definitely want the quotient of a commutative ring by a maximal ideal to be a field.

John Baez (Feb 23 2023 at 18:41):

But the devoted advocates of rngs could say "yes, but that's just because you demand your rngs to have identities! In my world, where we don't care so much about identities, we also don't care so much about fields."

John Baez (Feb 23 2023 at 18:42):

Jason Erbele said:

Hungerford includes some extra conditions on the Chinese Remainder Theorem if $R$ is a rng instead of a ring:

Let $A_1, \dotsc , A_n$ be ideals in a ring $R$ such that $R^2 + A_i = R$ for all $i$ and [...]
REMARK.If $R$ has an identity, then $R^2 = R$, whence $R^2 + A = R$ for every ideal $A$ of $R$.

Wow! So Hungerford is not in general demanding that his "rings" have identities? I didn't pick up on that earlier.

John Baez (Feb 23 2023 at 18:44):

Since this book is used to teach grad students algebra at our university, the dissident faction who studies "rngs" is more powerful than I thought!

I've never read Hungerford.

John Baez (Feb 23 2023 at 18:47):

Xuanrui Qi said:

Can you do any algebraic geometry at all if your rings have no identity, though?

There must be something you can do. The theory of C-algebras is a kind of mutant version of algebraic geometry, and every commutative C-algebra with unit has a spectrum that's a compact Hausdorff space... but commutative C*-algebras that may lack a unit are also interesting, and their spectra can be any locally compact Hausdorff space.

John Baez (Feb 23 2023 at 18:48):

However, it would be very interesting to see if anyone has tried to do algebraic geometry with commutative rngs, and see how they succeeded or failed.

Jason Erbele (Feb 23 2023 at 19:02):

John Baez said:

I continue to believe $4 \mathbb{Z}$ is a maximal ideal in $2 \mathbb{Z}$.

And I am in agreement with you on this now. I confused myself last night when I said it wasn't.

John Baez (Feb 23 2023 at 19:04):

Okay. Does Hungerford actually claim something like: in a rng it's not always true that every ideal is contained in a maximal ideal?

Jason Erbele (Feb 23 2023 at 19:32):

What Hungerford does say is:

Theorem 2.18. In a nonzero ring $R$ with identity maximal [left] ideals always exist. In fact every [left] ideal in $R$ (except $R$ itself) is contained in a maximal [left] ideal.

I assumed that meant there are some rngs that do not have maximal ideals, but that is not explicitly stated. So I tried to construct a counterexample and failed to meet the conditions.

John Baez (Feb 23 2023 at 21:14):

Thanks. I think the really important theorem is not that maximal ideals exist, but that every proper ideal is contained in a maximal ideal. Looking around I see this is called Krull's theorem.

John Baez (Feb 23 2023 at 21:15):

So, I'd like to see how (or whether) that fails with rngs.

John Baez (Feb 23 2023 at 21:18):

In Krull's theorem maximal ideal is "constructed" with the help of Zorn's lemma... where I put "constructed" in quotes because this is a nonconstructive existence argument.

John Baez (Feb 23 2023 at 21:20):

But as usual, applying Zorn's lemma here requires checking a preliminary thing: the union of any chain of proper ideals is again a proper ideal.

John Baez (Feb 23 2023 at 21:20):

I think the union of a chain of ideals in a rng is still an ideal.

John Baez (Feb 23 2023 at 21:20):

I think what might fail is that the union of a chain of proper ideals is a proper ideal. And here's why I think that:

John Baez (Feb 23 2023 at 21:21):

For a ring, an ideal is proper iff it does not contain 1. This makes it easy to show that the union of a chain of proper ideals is again proper: none of them contain 1, so their union does not contain 1.

John Baez (Feb 23 2023 at 21:22):

But for a rng, we can't characterize proper ideals this way!

John Baez (Feb 23 2023 at 21:22):

So, my guess is that there's some rng with a chain of proper ideals that isn't proper!

John Baez (Feb 23 2023 at 21:23):

Can anyone find a counterexample?

John Baez (Feb 23 2023 at 21:24):

Anyway, at least I've found a good use for the number 1 in ring theory: it provides a good way to check whether an ideal is proper! Instead of checking that some element of the ring is not in the ideal - a potentially infinite task - we merely check whether 1 is not in the ideal. And this seems to be at work above.

Joe Moeller (Feb 24 2023 at 03:38):

For a counterexample, how about taking the rng to be the union of a chain of proper ideals of a ring?

Reid Barton (Feb 24 2023 at 05:13):

For instance $\bigcup k[x^{1/2^n}]$ is a ring with maximal ideal $I$ generated by all the $x^{1/2^n}$, and $I$ as a rng doesn't seem to have a maximal proper ideal.

John Baez (Feb 24 2023 at 05:22):

Thanks! I've gotten a bunch more nice examples here.

John Baez (Feb 24 2023 at 05:27):

As a former analyst, the rng of compactly supported smooth functions on the line is my favorite so far.

Jean-Baptiste Vienney (Mar 29 2023 at 22:55):

This property is a useful property of ring with identities: for every ring $R$ with an identity, $(1)$ is a basis of $R$ as an $R$-module and thus any linear map $f:R \rightarrow A$ is determined by $f(1)$ because $f(r)=r.f(1)$. And thus $f \mapsto f(1)$ provides a natural isomorphism $Hom[R,A] \cong A$ for every $R$-module $A$. I don't know if we have such isomorphisms for rings without identity. You still have them for $R=2\mathbb{Z}$ by taking $f\mapsto f(2)$ because $f(2n)=n.f(2)$ but maybe it fails in general. Maybe you could embed your ring without identity in a ring with identity to recover this natural isomorphism, I don't know.

David Michael Roberts (Mar 29 2023 at 23:32):

You can always freely adjoin an identity element to a ring, but I didn't check if this results in the natural iso.

Jean-Baptiste Vienney (Mar 29 2023 at 23:34):

You need to decide what f(1) is then!