Mathematics

Saving time, yeah :P Let me draw things.

You're not?! O_o

do the units of a ring form a group under ring multiplication?

I share Daminark's surprise

@Ted Oh you're doing the cohomology SS? I guess it doesn't matter; the differentials just go opposite way

5:02 PM

what am I asking, of course they do.

Yeah, they should

@Astyx I thought that's what units mean

I thought I knew how to read

Welcome to the club @Astyx! @Balarka and I are here too!

@Balarka: I only understand Euler in terms of cohomology :)

5:03 PM

yep

The word "read" is hard to read, ironically.

It can be pronounced either read or read.

MickLH

@LeakyNun I've heard that some people define ring with the multiplicative identity being optional

So maybe be sure that $1$ exists :P

@Akiva I know what you meant but that wording was rather unfortunate

@Balarka: I actually think (to add to my stooopidity that Jack Lee caught) that I always thought a closed submanifold represented homology. (The question of when homology is represented by submanifolds is a much subtler question, as we know.)

@MickLH I thought that's what unit means

5:04 PM

Or perhaps entirely deliberate

@mickLH Well those people are wrong

angery reacts only

("Angery"?)

@TedShifrin Yeah from our previous discussions only oriented closed submanifolds are.

I like to call ideals subrings, ok?

shivers @AlessandroCodenotti

That's very insightful to me. I am glad we talked about this.

MickLH

5:05 PM

> Most or all books on algebra[17][18] up to around 1960 followed Noether's convention of not requiring a 1. Starting in the 1960s, it became increasingly common to see books including the existence of 1 in the definition of ring, especially in advanced books by notable authors such as Artin,[19] Atiyah and MacDonald,[20] Bourbaki,[21] Eisenbud,[22] and Lang.[23] But even today, there remain many books that do not require a 1.

I'm not saying it's a good thing or that I'm proud of it

I'm just saying "don't die unexpectedly"

And do you call subrings ideals then ? @Alessandro

Pah @Alessandro

@AkivaWeinberger I also like $A\times B$ having a subring isomorphic to $A$ and a subring isomorphic to $B$

@MickLH Wut

@MickLH Only die if you're expecting it

5:06 PM

@Astyx nah, not every subring is an ideal

@AlessandroCodenotti …fair…

Lol

Hm, wait, that brings a point:

And @Akiva it's Facebook memetics "anger" + "angry"

I figure that if Mike Artin says all rings have identity, that's good enough for me. :P

5:06 PM

Fact: If $N$ is a normal subgroup of $A_n$ ($n \ge 3$) and $N$ contains a $3$-cycle, then $N = A_n$. Problem: Show that $A_4$ has no subgroup of order $6$. Proof: Suppose $C$ is a subgroup of order $6$. Then it has an index of $2$ in $A_4$, and therefore it must be normal. Futhermore, for every $x \in A_4$, we have that $(xC)^2 = C$ which implies $x^2 \in C$ for every $x \in A_4$. Since $(132)$ is the square of $(123)$, $C$ must contain this $3$-cycle, a contradiction.

Does this seem right?

A ring that doesn't have identity is just a rng ;)

MickLH

lol

If we define rings to include identity, then does "subring" not just mean "subset that's a ring"? Or is it "subset that's a ring and has the same identity as the original set"?

The latter I think

Strange.

5:07 PM

And if you remove + inverses (negative elements), you get a rg

And if you get rid of gravity you get rin

I mean you generally have to have that the subsets of the ring stay a ring under that operation

@Daminark It could be the same operation but different identity.

@Akiva there can only be one identity any way right ?

Consider $\Bbb R\times\{0\}\subseteq\Bbb R^2$.

$(1,0)$ "acts" like an identity on the left set, but not on the right.

5:09 PM

Huh

@TedShifrin So all the higher differentials $d_3, d_4$ etc vanish because they come from the 3rd and higher rows which are all zero, which means the $E^3$ page is the same as the $E^\infty$ page. Now what's that page... you get a $H^1(S^1)$ in the $(0, 0)$ coordinate, and a $H^1(\Sigma)$ in the $(1, 0)$ coordinate. So I guess I want to understand $d_2 : (0, 1) \to (2, 0)$, just like you said.

The unique identity of $\Bbb R^2$ is $(1,1)$.

nods @Balarka

Would someone mind reading the proof I gave above? It's rather short; I just want to be sure that it is correct.

Not complex multiplication

5:10 PM

@Daminark I'm doing $\Bbb R^2$, not $\Bbb C$.

@Dami I think you use ring product

Define $(a,b)(c,d)=(ac,bd)$.

@AkivaWeinberger as a ring?

OK, @user193319. Give me a while to read all that.

@LeakyNun Yeah. $\Bbb R^n$ is a ring for any $n$. Fields are harder.

5:10 PM

Reretract then

@TedShifrin Okay. Thanks!

Wikipedia says a subring is a submonoid

@AkivaWeinberger it isn't a field because there are zero divisors.

So same identity I guess

If you want an example that isn't a product of rings: consider $M_2(\mathbb{R})$, and the subring generated by elements where all entries are necessarily identical

5:12 PM

@LeakyNun Yep. Don't care for rings, though.

So there's this theorem which says you can only get a division algebra on $\mathbb{R}^{1,2,4,8}$, and only $1,2$ could get you a field

The identity of $M_2(\mathbb{R})$ is, of course, the $2$-dimensional identity matrix

quaternions and octonions

But that subring has the matrix with every element $1/2$ as identity

The latter you can also find in food

5:13 PM

Yeah, @user193319, you don't need normality, do you? Any group of order $6$ must have an element of order $3$, so your fact applies immediately. ... Of course, I like thinking of $A_4$ as the group of symmetries of a tetrahedron, and then it's geometrically pretty clear.

leaves dramatically

@Astyx: stage left or stage right?

dramatic bye

See here @Akiva

(OK I'm seeing what my friend that one time said that many wishlist propositions in ring theory are sneaky because people only ever think of $\mathbb{Z}_n$ and polynomial rings...)

5:13 PM

As long as the audience cries ... @Ted

rolls 5 eyes @Astyx

@TedShifrin Oh, yes. You are right. Thanks for the help!

(That answer is from my abstract algebra prof, who never assumed that rings have identities)

OK I think there's only one good way to settle this

Gonna check Aluffi and Mac Lane. If they agree it's settled on that. If they don't, Lang is the tiebreaker

@TedShifrin Stage front.

5:15 PM

For a simple example of a subring with a different identity take the $2\times 2$ matrices and the subring of matrices with entries only in the upper left corner

resigns from the room until this ring/rng nonsense is over

Heh

lol

@Ted I am forgetting how to extract $H^*$ of the total space from the $E^\infty$ page.

I'll have an analysis question for you later @Ted, when I get to my laptop