Mathematics - 2023-12-19 (page 1 of 2)

leslie townes

00:16

Dec 1 at 21:33, by Ted Shifrin

OK, so what order do you take the tangent vectors so that their cross-product gives that orientation?

indeed.

Ted Shifrin

The search meister rears his electronic head.

leslie townes

00:42

beep boop

Thorgott

01:24

@Ted I think I found an index error in Griffiths & Harris!

Ted Shifrin

01:37

@Thor I have pages and pages of scribbled notes on errors in G&H. There are plenty.

Although I did type up errata for Guillemin & Pollack, I never did so with G&H, because I knew I had far from a complete list.

Thor: Sanity check. What are $H^1(S^3,\Bbb Z_2)$ and $H^4(S^3,\Bbb Z_2)$?

Thorgott

$0$, no matter the coefficients

Ted Shifrin

Of course. What is FShrike doing?

Thorgott

the cohomology ring of a wedge sum behaves very simply in general (if the spaces aren't completely bad), so I agree with Kevin's comment. I don't think there's any subtlety no matter if the sphere's dimensions are even or odd

Ted Shifrin

But thanks :) Yeah, I don't either. .... I don't get the point of the question, honestly.

What error are you referring to?

Thorgott

@TedShifrin probably just slightly confused himself

Ted Shifrin

01:47

That's the trouble with being precocious and doing graduate level mathematics in high school :)

Thorgott

@TedShifrin in the inductive proof of Giambelli's formula, the indices in the middle rows are all off by $2$

(p. 206)

Jakobian

@TedShifrin "Oxford undergraduate"

Ted Shifrin

@Jakobian Only recently, yes.

@Thor I've never been through the general proof of Giambelli. I can't help.

As I said, my list of errata comes only from the stuff I've actually taught or read carefully.

Thorgott

it's a very easy to fix error, to be fair

Giambelli is more or less easy once you have Pieri, I'll present it in a seminar talk tomorrow (talk about working your results out in time...)

Ted Shifrin

02:09

You’re hours ahead.

robjohn

02:47

Hey @Ted!

Ted Shifrin

Hi again, @robjohn.

robjohn

The rain I mentioned earlier has dribbled in. Not really much to speak of yet.

Ted Shifrin

None here … despite predictions. But 80-94% chances the next three days. We shall see.

robjohn

Wednesday, we're supposed to get heavy rain and a bit of thunder.

Ted Shifrin

How enlightning.

Thomas Finley

05:26

Does Krull's Theorem: "Let $R$ be a commutative ring, and let $I$ be a ideal of $R$ that is distinct from $R.$ Then there exists a maximal ideal $M$ of $R$ containing $I$" assume that R has an identity $1\neq 0$ ?

I know nowadays a ring is defined to have be commutative having an identity, but I follow mostly Herstein's book and there a ring may not necessarily be commutative and even might not have an identity.

So, this is why my confusion arose.

oscarmetal break

05:59

It normally assumes additive identity

I remembered the ring without 1 is called rng, there are some theorems that are able to extend to this but I never concern about this.

But I just read the proof, it seems like you don't need the identity to prove the theorem with Zorn's lema

In this case, having commutative is convinent so that you don't need to worry about left or right ideal. However, I would say it is more common to encounter ring that is not commutative than has no identity in Herstein's book. In addition, I don't the definition of commutative is necessarily connected to that the ring has identity though you can always join one.

leslie townes

"the" proof of that result? it might help focus the discussion if you linked to "a" proof of that result, so the hypotheses and argument and conclusion are clearer.

e.g. the version at proofwiki.org/wiki/Krull%27s_Theorem assumes the existence of a 1 in its rings, and uses it in the proof, in arguing that the union of a chain of proper ideals is proper.

oscarmetal break

06:14

Yes, that where I read. Oh I see. it does assume 1.

leslie townes

who knows what other proofs are out there. i expect that some kind of hypothesis is necessary, whether on there being a 1 in the ring, or on I not being just an ideal but some special kind of ideal, if only because if not, i'd expect a proof on a place like proofwiki to avoid the use of the hypothesis. but who knows.

it might be more instructive to focus on particular examples of "rings" that do not have 1 but whose ideal structure can be determined, if not entirely, enough to answer the question for that "ring."

a natural example from analysis would be R the ring of real valued functions on the reals having limit 0 at both +/- infinity, and I the set of all functions in R that have compact support. I is clearly a proper ideal in R. is I contained in a maximal ideal? [what are the maximal ideals in R?]

oscarmetal break

emm, can I do it with basic algebra and analysis knowledge that I have known?

leslie townes

possibly? i'm not specifically asking you, or anybody in particular. i'm just thinking out loud. i think this example would be tractable. it is probably even on MSE somewhere.

oscarmetal break

Let me google it somehow. Looks innocent. We never talk a lot of rng on class.

leslie townes

06:35

this example at least shows how the idea used in that proofwiki proof might fail in general. if you think of I as a non-unital "ring" in its own right, and let I_n denote the subset of I consisting of those functions with support in [-n, n], n a positive integer, then {I_n} seems to be a chain of proper ideals in I whose union is all of I.

so that particular use of zorn's lemma is definitely not going to work in general.

oscarmetal break

07:02

Yes, indeed

Koro

@Thorgott it didn't work.

@TedShifrin I see. So the determinant of $[E_i. E_j]$ can be written as some $BB^t$

@TedShifrin yeah, in that case, I was trying to understand curvatures.

I don't know how to preserve orientations in the case of cylinder.

8 hours ago, by Koro

The definition that I use for a map f:S--> \tilde S to be orientation preserving is the following: If for positively oriented bases v_1, ...,v_n\in S_p, the vectors dv_1,..., dv_n are positively oriented.

I have one confusion in the definition also: If for some positively oriented basis in S_p, the vectors df(v_1),...,df(v_n) are +ly oriented, can we have the same conclusion for all positively oriented basis in S_p?

Ted Shifrin

No, we discussed the normal to get orientation, too. Remember these are surfaces, not $n$-dimensional.

Koro

07:19

I understand that for the cylinder. But this current situation seems different to me as it is about "orientation preserving of a map".

which has the definition as above.

but our cylinder discussion in past was about orientation on the cylinder, no?

Suppose that $f:S\to \tilde S$ is a smooth map between two oriented n- surfaces. We say f is orientation preserving (o.p.) if for all $p\in S$ and for all positively oriented bases $v_1,...,v_n$ in S_p (the tangent space to S at p), the vectors $df(v_1),..., df(v_n)$ are positively oriented in $\tilde S_{f(p)}$.
Suppose that for a smooth $f: S\to \tilde S$, we are only given that for some positive oriented bases in $S_p$, the image vectors under df are positively oriented. Can we conclude by this information that f is positively oriented?

Thomas Finley

07:34

0

Q: Let $R=\{\begin{pmatrix} 0 && a\\0 && b\end{pmatrix}:a,b\in\Bbb Q\}$, be a ring and $I=\{A\in R:A^2=0\}$ an ideal of $R.$ Show that $R/I\cong \Bbb Q.$

Show that $R=\{\begin{pmatrix} 0 && a\\0 && b\end{pmatrix}:a,b\in\Bbb Q\}$, is a subring of $M_2(\Bbb Q)$ and $I=\{A\in R:A^2=0\}$ is an ideal of $R.$ Finally, show that $R/I\cong \Bbb Q.$ I was able to show that $R=\{\begin{pmatrix} 0 && a\\0 && b\end{pmatrix}:a,b\in\Bbb Q\}$, is a subring of $M...

I need some help with this, please.

Koro

07:51

@ThomasFinley Did you try considering the map $f: R\to Q$ defined as $f(A)=f([a_{ij}])= a_{22}$?

i.e. A getting mapped to its (2,2)th entry.

leslie townes

the question is answered now on main.

Koro

Ooo

Thomas Finley

@Koro That was the only thing left. So frustrating

Let $R$ be a commutative ring with identity. Prove that there exists an epimorphism from $R$ to a field $F$.

Any idea how to solve this question?

leslie townes

in retrospect, what may have been the missing link between the attempt and the solution was an explicit identification of I as a set of matrices in more concrete terms than its definition.

thomas: this is basically that krull thing from earlier. consider a quotient map R -> R/m, m a maximal ideal.

Koro

is it true for any field F?

Thomas Finley

07:58

@leslietownes Then, the quotient map is an epimorphism. But where does the field comes to play?

Koro

can we have epimorphism $Z\to \mathbb R$?

Thomas Finley

@Koro Yes, I think the question is valid so...

leslie townes

koro: no, there's is an implicit quantifier there. prove there exists a field F such that there is an epiomorphism R to F.

Koro

@ThomasFinley I don't know the answer to your question off the top of my head..

Thomas Finley

@Koro maybe, it's not R it may be some other field

Koro

08:00

@leslietownes I see, thanks. It seems that Thomas misstated the question.

leslie townes

thomas: fill in the blank: R/m is a field if and only if the ideal m is [blank].

Thomas Finley

@leslietownes [blank] =maximal

leslie townes

koro: i don't know about 'misstated,' certainly it could have been phrased better. this is the kind of thing textbooks often do suboptimally.

Koro

@leslietownes Ahh I see. I misunderstood the question.

it is to be interpreted as 'Show that there is field F such that there is an epimorphism $R\to F$'.

Thomas Finley

@Koro yes

Koro

08:04

Okay. Then your question is answered right?

F= R/m as Leslie advised.

and the epimorphism is the quotient map.

Thomas Finley

@Koro ,@leslietownes and such a maximal ideal exists from Krull's Theorem, right?

Krull's Theorem:Let $R$ be a commutative ring with a identity, and let $I$ be a ideal of $R$ that is distinct from $R.$ Then there exists a maximal ideal $M$ of $R$ containing $I$.

leslie townes

there's maybe something kinda interesting to be said there, about how sometimes an omitted quantifier whose meaning can be inferred from context is an existential quantifier, although sometimes, perhaps even more often, it is a universal quantifier.

i had to teach an 'intro to proof' class once where the textbook went to great pains to belabor stuff like that, but also wasn't written very carefully in its own exercises. i wanted to kick its author. we did not have much choice over the textbook.

Koro

@ThomasFinley yeah, Zorn's lemma basically.

I didn't know it's called Krull's theorem.

Thomas Finley

@Koro , @leslietownes Thanks a ton!

leslie townes

thomas: yes. applying that result to the proper ideal I = {0} would return a maximal ideal in any commutative ring with identity, although as koro notes it is zorn magic and not something you generally have a lot of control over.

Thomas Finley

08:10

Got it.

leslie townes

for specific examples R you might not need the full force of zorns lemma of course. it's that "for all commutative rings with identity" that needs zorns lemma.

Koro

08:26

probably a silly question: Why doesn't probability tally with the truth here?
Say there's a town A and town B with population 100k, 200 k respectively. A crazy virus enters the picture and 'it makes people to throw stones randomly at one another'. In which town, would you be safer? In town A, right? Intuition suggests that.
But if we calculate the probability of getting hit by a stone in the towns, we get that the probability of getting stone hit in town A is about 1/(100 k) and in B it is 1/(200 k)!! So town B is safer??

please ignore this.

Thomas Finley

@leslietownes Zorn's lemma is precisely this:" If R is a commutative ring with identity then it has a maximal ideal", isn't it?

Joe

09:17

@ThomasFinley: No, Zorn's lemma is an assertion in order theory. It states that if $X$ is a partially ordered set, and every chain in $X$ is bounded above, then $X$ has a maximal element. However, one uses Zorn's lemma to prove a number of results, including the ideal theorem

Thomas Finley

@Joe "If R is a ring (not necessarily commutative) with identity then it has a maximal ideal" is true in general, right?

Does this theorem have a name?

Joe

@ThomasFinley: I have sometimes heard it being called the "Maximal Ideal Theorem". As for whether it is true in general, the answer is "yes, if you are working in $\mathsf{ZFC}$" (i.e. you are happy to freely use the axiom of choice). However, the Maximal Ideal Theorem is independent of $\mathsf{ZF}$ (assuming that $\mathsf{ZF}$ is consistent).

Thomas Finley

@Joe "if you are working in $\mathsf{ZFC}$"- What is $\mathsf{ZFC}$?

I don't know what $\mathsf{ZFC}$ stands for.

Jakobian

@ThomasFinley it fails for the zero ring

@ThomasFinley Zermelo-Frankel with choice, the set theory system everyone assumes for granted

Joe

@Jakobian: Heh, very true.

@ThomasFinley: Are you familiar with set theory? $\mathsf{ZFC}$ is the standard axiom system used in set theory, and is the most common foundational theory in mathematics

Thomas Finley

09:31

@Joe oh! I never had any idea about this axiom

Joe

@ThomasFinley: In that case, I would shorter my answer to "yes, every (nonzero) ring has a maximal ideal".

Jakobian

@Koro $\mathbb{Z}\to \mathbb{Q}$ is an epimorphism

Joe

It's great fun learning set theory. Every mathematical object, even numbers, can be realised as certain sets, and virtually every theorem in classical mathematics can, in principle, be proven in formal proofs in $\mathsf{ZFC}$.

About your specific query: $\mathsf{ZFC}$ includes an axiom called "the axiom of choice", and some people, for whatever reason, have felt uncomfortable using this axiom. However, the axiom of choice is widely accepted in mainstream mathematics these days. And, assuming the axiom of choice, the maximal ideal theorem is provable in set theory. But we can't prove the maximal ideal theorem in $\mathsf{ZF}$, which is $\mathsf{ZFC}$ with the axiom of choice omitted.

The $\mathsf{C}$ in $\mathsf{ZFC}$ stands for "choice"

Thomas Finley

@Joe So to sum up things: 'Maximal Ideal Theorem' states that: If R is a ring (not necessarily commutative) with identity $1\neq 0$ then it has a maximal ideal . We must note that $1\neq 0$ automatically implies $R$ is not a zero ring. The theorem fails if $R$ is a zero ring i.e if $R=\0\}$.

Am I correct?

Joe

Yes, you are correct

I would advise you not to worry about these thorny set-theoretic issues, until you feel the desire to learn about these thorny set-theoretic issues

Thomas Finley

09:40

@Joe Thanks a ton!

(Krull's Theorem:) Let $R$ be a commutative ring with a identity $1\neq 0,$ and let $I$ be a ideal of $R$ that is distinct from $R.$ Then there exists a maximal ideal $M$ of $R$ containing $I$.

This theorem is phrased correctly, right? Krull's theorem( as I wrote above fails again, if $R$ is a zero ring i.e if $R=\0\}$), right? @Joe , @Jakobian

Joe

@ThomasFinley: What you wrote is correct, although I think that the theorem is also true for noncommutative rings

I'll try to find a reference

Koro

2 hours ago, by Koro

Suppose that $f:S\to \tilde S$ is a smooth map between two oriented n- surfaces. We say f is orientation preserving (o.p.) if for all $p\in S$ and for all positively oriented bases $v_1,...,v_n$ in S_p (the tangent space to S at p), the vectors $df(v_1),..., df(v_n)$ are positively oriented in $\tilde S_{f(p)}$.
Suppose that for a smooth $f: S\to \tilde S$, we are only given that for some positive oriented bases in $S_p$, the image vectors under df are positively oriented. Can we conclude by this information that f is positively oriented?

the answer is - yes, we can.

Joe

09:57

@ThomasFinley: The theorem appears to also be true for noncommutative rings. If $R$ is a not necessarily commutative ring with $1\neq0$, and $I$ is a proper (two-sided) ideal in $R$, then there is a maximal (two-sided) ideal including $I$. In particular, letting $I=\{0\}$, we see that $R$ has a maximal (two-sided) ideal.

But I don't like having to work with noncommutative rings anyway, since I still have trouble telling my left from right.

Thomas Finley

10:44

@Joe Thanks for the clarification!

Koro

@Koro probably no.

Jakobian

This is a more accurate statement that Dieudonne is proving

With interpretation that what comes after fixing $U_0$ is consequence of fixing an appropiately small $U_0$

So it shouldn't be moreover, more like continuation of properties that $U_0$ satisfies

Perfect

Thorgott

11:09

@Koro then you must've made an error

Thorgott

11:23

@Koro there is only one ring homomorphism $\mathbb{Z}\rightarrow\mathbb{R}$ and it is not an epimorphism

Jakobian

I forgot to write $u(x_0) = y_0$

Thorgott

that said, I feel like if an introductory algebra lecture uses the term epimorphism, it's just a mistake and they intended to say surjective. nobody talks about epimorphisms of rings without calling specific attention to their subtle nature.

Jakobian

Omitting a lot of details this is more or less the proof

Except for continuous differentiability of $u$

It mainly runs on mean value theorem and Banach fixed point theorem

eigenvalue

12:02

Hi there!
I have a connected Riemannian manifold M (not necessarily compact) with boundary. What conditions/ constraints do I need to impose on M in order to ensure that a non-vanishing vector field exists that is transverse at the boundary?

Thorgott

did you mean nowhere-vanishing?

eigenvalue

Yes, nowhere vanishing. However, it seems you can push the zeros to infinity (in the non-Compact case). But even then it’s not clear if that’s feasible

Thorgott

12:19

a nowhere-vanishing vector field on $M$ exists iff
a) $M$ is compact, without boundary and $\chi(M)=0$
or
b) $M$ is non-compact or has non-empty boundary

so if $\partial M\neq\emptyset$, there exists a nowhere-vanishing vector field and I think you should always be able to make it transverse to the boundary

by perturbing it with a small vector field that is normal the boundary and supported in an arbitrarily small nbhd of the boundary

eigenvalue

Is there any reference I could use? Seems like if you push the zeros to infinity in the non-compact case, then you still don’t know if there are infinitely many zeros at infinity - or in general how infinity looks topologicaly once you move All zero there. Or am I overthinking this?

Also, in the non-compact case, do we need to take the Euler characteristic into account? Or is it irrelevant?

Thorgott

"infinity" is not a place that exists, that's precisely why we can push zeros there and why the idea does not work in the compact case (in the compact case, there is no "infinity", so nowhere you can push the zeros to)

a nice explanation of these ideas is here: math.stackexchange.com/questions/3961266/…

the Euler characteristic in the non-compact case might not even be defined

what you actually want to take into account is the so-called Euler class, which just so happens to be equivalent data to the Euler characteristic in the compact case

however, the Euler class vanishes in the non-compact case

Sine of the Time

I was thinking about uniform convergence, and there's the well known theorem that states that given $\{f_n\}$ where $f_n$ are continuous, and $f=\lim f_n$ then $\int_I f=\lim \int_I f_n$ where $I$ is a compact interval. I was wondering if this is true if $I$ is unbounded. I found a counterexample where $f_n$ are not continuous but I didn't find anything if $f_n$ are continuous.

eigenvalue

12:34

@Thorgott thanks! Your answer is very helpful. Just to summarize: if my manifold has a non-empty boundary and I want such a vector field as indicated, then i just need to restrict myself to “b) $M$ is non-compact or has non-empty boundary”?

Thorgott

sure

the part that needs verification is that the perturbation argument I sketched actually works

Jakobian

@SineoftheTime on $[1, \infty)$ consider $f_n(x) = 1_{[1, n]}/x$

You can make $f_n$ smooth if you want

Thomas Finley

12:50

0

Q: Let $p$ be a prime. Show that there exists two non isomorphic rings with $p$ elements. (Is my solution correct?)

Let $p$ be a prime. Show that there exists two non isomorphic rings with $p$ elements. I feel that there are two rings of order $p.$ They are $(\Bbb Z_p,+,.)$ And $(\Bbb Z_p,+,×),$ where $.$ is multiplication modulo $p$ and $×$ is such an operation so that $a.b=0,\forall a,b\in \Bbb Z_p.$ I took ...

abstract-algebra ring-theory ring-isomorphism

I don't know how to show that $R\cong (\Bbb Z_p,+,×)$ if $R$ does not have a unit element, i.e $1\neq 0,$ where $×$ is such an operation so that $a.b=0,\forall a,b\in \Bbb Z_p.$

Thorgott

13:05

stop calling them just "rings" when you don't assume the existence of units

2

it invites an absurd amount of misunderstandings

also, what's the issue? you have written down two not necessarily unital rings. one has a unit, the other does not, so they are not isomorphic. you're done.

Keane

If we know the action of a Lie algebra $\mathfrak{g}$ on a vector space $V$, is there a way to canonically construct a Lie algebra action $\mathfrak{g}\oplus \mathfrak{g}$ on $V$?

Jakobian

@Thorgott I don't think thats the problem. The problem is not clarifying definition of ring

After that, I think they're free to call them rings

one potato two potato

All rings are assumed to be commutative

Jakobian

I like to assume that rings are commutative, but thats only because I don't want to think of "sides"

Thomas Finley

@Thorgott No, I wanted to prove that there exists two rings upto isomorphisms that are of order $p$.

Xander Henderson

13:18

@Jakobian Anyone is free to use any word to mean anything they like. That doesn't mean that they are communicating effectively.

Jakobian

left-sided ideal, right-sided, I don't care. Just give me ideals

Thorgott

@ThomasFinley you're missing a word like "exactly" in that sentence

or "only" in the title to your question

Thomas Finley

@Thorgott Oh, I see how the misunderstanding occurs.

Jakobian

@XanderHenderson Yes. And that's solved by clarifying definition of a ring

one potato two potato

choose your side Jakobian

Thomas Finley

13:21

@Thorgott Oh, I see how the misunderstanding occured

I have fixed the title now.

But then again I don't know how to show that $R\cong (\Bbb Z_p,+,×)$ if $R$ does not have a unit element, i.e $1\neq 0,$ where $×$ is such an operation so that $a.b=0,\forall a,b\in \Bbb Z_p.$

@XanderHenderson Hello, I hope you are doing well.

@Thorgott Can you please help me solve this problem?

I have mentioned my definition of rings.

Also, I have included the word, "only" in my post.

Jakobian

Since this is a ring of prime size, I think it'd be natural to consider some generator $a$ of its additive group

Take $b = n\cdot a$ for $1\leq n\leq p$, and suppose that $a\cdot b = 0$

I think the right approach is to then try to deduce that $xy = 0$ for all $x, y$ in your ring

Thomas Finley

@Jakobian But I am considering $1\neq 0$ is not in $R$.

Jakobian

You wanted to consider non-unital rings?

Thomas Finley

I already proved that if $R$ has a unit, then it is isomorphic to $\Bbb Z_p$ with the usual operations.

@Jakobian yes

Jakobian

@Jakobian Then taking integers $z, t$ such that $zn+tp = 1$. Then $a^2 = zn a^2 + tp a^2 = 0$

This shows that for this ring, $xy = 0$ for all $x, y$

Thomas Finley

13:33

@Jakobian What is $p$ ?

I get that $R=<a>$

Jakobian

@ThomasFinley A prime?

you gave some prime $p$

Thomas Finley

And $1\leq n\leq p$. Ok, I overlooked the notations from your previous comment.

Jakobian

Anyway, if $a\cdot b\neq 0$ for all $1\leq n\leq p$, then if $x, y$ are non-zero, then $xy = (ma)(na) = mn \cdot a = ka$ for some $1\leq k\leq p$ so that $xy$ is non-zero

Thomas Finley

@Jakobian I did not get how you concluded this. Till now, we have only shown, $a^2=0,\forall a\in R.$

Jakobian

All elements in $R$ are of the form $n\cdot a$ for some $n$

so their product is a multiple of $a^2 = 0$ by an integer

Thomas Finley

13:37

@Jakobian oh! Right.

Jakobian

actually I made a mistake

Those should be $1\leq n\leq p-1$ and $1\leq k\leq p-1$

Thomas Finley

@Jakobian where?

Jakobian

Its a minor mistake, I just wrote $p$ instead of $p-1$

Thomas Finley

@Jakobian Why do you even consider $a.b\neq 0$ ? You already proved $ab=0$

Jakobian

I've proved that if $ab = 0$ for some $b$ of chosen form, then the ring must be the ring for which the multiplication is trivial i.e. $xy = 0$

Now we have to consider case when $ab\neq 0$ for all $b$ of the chosen form, and prove its a field

Thomas Finley

13:41

@Jakobian ok

Jakobian

In particular this leads to the property akin of that of integral domains, $x\neq 0, y\neq 0$ implies $xy\neq 0$, but we don't know that this ring has unit $1$ yet, so we can't conclude its a field

Alright, the map $f:x\mapsto ax$ is injective, so a bijection, then $f, f^2, ...$ is a sequence of maps $f^n(x) = a^nx$

Since there is only finite amount of maps from this ring to itself, we must have $f^n = f^m$ for some $m < n$, thus $a^n x = a^m x$, so that $a^{n-m}x = x$ for all $x$

Now the element $e = a^{n-m}$, which is non-zero, is a unit of this ring

Thus the ring is actually an integral domain, so since finite integral domains are fields, its a field

Thomas Finley

Wait, I am getting confused. What I could make out was( I am quoting your statements):

Since this is a ring of prime size, I think it'd be natural to consider some generator $a$ of its additive group

Take $b = n\cdot a$ for $1\leq n\leq p$, and suppose that $a\cdot b = 0$

Then taking integers $z, t$ such that $zn+tp = 1$. Then $a^2 = zn a^2 + tp a^2 = 0$

This shows that for this ring, $xy = 0$ for all $x, y$

Anyway, if $a\cdot b\neq 0$ for all $1\leq n\leq p$, then if $x, y$ are non-zero, then $xy = (ma)(na) = mn \cdot a = ka$ for some $1\leq k\leq p$ so that $xy$ is non-zero

Jakobian

For $1\leq n\leq p-1$ and for some $1\leq k\leq p-1$

Thomas Finley

Ok, but how $a^2=0$ ?

I get that $zna^2=0$

But what about $tpa^2$ ?

Jakobian

$px = 0$ for all $x$

This is because the additive group is of size $p$

Thomas Finley

13:50

Ah, ok.

'Anyway, if $a\cdot b\neq 0$ for all $1\leq n\leq p$, then if $x, y$ are non-zero, then $xy = (ma)(na) = mn \cdot a = ka$ for some $1\leq k\leq p$ so that $xy$ is non-zero"--- Why is (ma)(na)=mn(a)?

Jakobian

@ThomasFinley I see I have a typo here too. It should be, if $x, y$ are non-zero then $x = ma, y = na$ for some $n, m$ co-prime to $p$, so $xy = mn a^2$ where $mn$ is also co-prime to $p$< so $xy\neq 0$

Thomas Finley

@Jakobian ok

Jakobian

If $x$ is non-zero in this ring, then $kx\neq 0$ for any integer $k$ co-prime with $x$

this uses only the additive group structure

Thomas Finley

Next portion: "I've proved that if $ab = 0$ for some $b$ of chosen form, then the ring must be the ring for which the multiplication is trivial i.e. $xy = 0$

Now we have to consider case when $ab\neq 0$ for all $b$ of the chosen form, and prove its a field"

Ok, till now I grasped everything.

Next comes: "In particular this leads to the property akin of that of integral domains, $x\neq 0, y\neq 0$ implies $xy\neq 0$, but we don't know that this ring has unit $1$ yet, so we can't conclude its a field
Alright, the map $f:x\mapsto ax$ is injective, so a bijection, then $f, f^2, ...$ is a sequence of maps $f^n(x) = a^nx$
Since there is only finite amount of maps from this ring to itself, we must have $f^n = f^m$ for some $m < n$, thus $a^n x = a^m x$, so that $a^{n-m}x = x$ for all $x$
Now the element $e = a^{n-m}$, which is non-zero, is a unit of this ring

"Alright, the map $f:x\mapsto ax$ is injective, so a bijection"---- So you are considering a mapping $f:R\to R$ such that $f(x)=ax,\forall x\in R.$

Jakobian

Yes

Thomas Finley

13:57

If $x,y\in R$ such that $f(x)=f(y)\implies ax=ay\implies a(x-y)=0\implies a=0$ or $x-y=0$. But $a\neq 0$ so, $x=y.$ This proves $f$ to be 1-1.

Jakobian

Sure

Thomas Finley

Let $y\in R$ then, $\exists n\in \Bbb Z$ such that $na=y.$

But then how $f$ is onto?

Jakobian

since $f$ is from $R$ to $R$ and $R$ has finite cardinality

Thomas Finley

@Jakobian Oh! Ok.

Jakobian

This is just a property of finite cardinals

Thomas Finley

14:01

Next, you consider a sequence of composite mappings "$f, f^2, ...$ is a sequence of maps $f^n(x) = a^nx$"

Jakobian

@Jakobian Or, er, what is the same, finite ordinals. This should basically follow from the fact that no finite ordinal is a limit ordinal

Thomas Finley

Ok, as $R$ has order $p$ so, $a^p=a\implies f^p=f$ so, the sequence is finite and repeating.

Jakobian

@ThomasFinley No, why

$R$ with multiplication is only a semigroup, you can't claim anything about $a^p$

Thomas Finley

Well $f^p(x)=a^px=ax=f(x)$ so?

@Jakobian oh gosh

Yes

Sine of the Time

@Jakobian 👍

Thomas Finley

14:03

So why is "we must have $f^n = f^m$ for some $m < n$, thus $a^n x = a^m x$, so that $a^{n-m}x = x$ for all $x$"?

Jakobian

$f, f^2, f^3, ...$ is an infinite sequence of maps

Thomas Finley

Because $R$ is finite, right?

Jakobian

but there is only a finite amount of maps $g:R\to R$ because $R$ is finite

Thomas Finley

@Jakobian yes and as $R$ is finite so in the sequence $a,a^2,a^3,...$ there are only finite number of elements. This means $\exists m,n\in\Bbb Z$ such that $a^m=a^n$ due to which $f^n=f^m.$

Right?

Jakobian

Thats another way of seeing it, yes

its important for $m, n$ to be distinct

Thomas Finley

14:07

@Jakobian yes.

"$a^n x = a^m x$, so that $a^{n-m}x = x$ for all $x$"

Now, why is this?

Inverse of $a^m$ might not exist?

Jakobian

Because $a^m\cdot (a^{n-m}x) = a^m\cdot x$ and $a^m$ is non-zero

from the property akin to that of integral domains we obtain $a^{n-m}x = x$

its important here that $n > m$

Thomas Finley

@Jakobian Ok

@Jakobian yes , WLOG, I presume.

"Now the element $e = a^{n-m}$, which is non-zero, is a unit of this ring
Thus the ring is actually an integral domain, so since finite integral domains are fields, its a field"

Ok, so $R$ is a field.

Jakobian

yep

Thomas Finley

Now what?

Jakobian

We are done

Thomas Finley

14:11

How is $R\cong (\Bbb Z,+,×)$ ?

Jakobian

There is a unique field of size $p^n$ for prime $p$ and $n \geq 1$ a natural number

Thomas Finley

So, precisely, we have proved the result, "If R is commutative having p elements then R is a field", correct?

Jakobian

incorrect

Thomas Finley

Why?

We were proving this all the time!

Or did I miss something?

Jakobian

we've proven that "If $R$ is a ring with $p$ elements then either the multiplication is trivial or $R$ is a field"

nowhere did we assume commutativity

Thomas Finley

14:17

@Jakobian oh, even better.

So, if $R$ is a field then $R\cong Z_p$

This is due to the fact that:There is a unique field of size $p^n$ for prime $p$ and $n \geq 1$ a natural number

I think the above fact is some kind of a theorem, right?

If the multiplication in $R$ is trivial, then how can we show that $R\cong (Z_p,+,×),$ where $×$ is such an operation so that $a.b=0,\forall a,b\in \Bbb Z_p.$

For this, I think we still have to exhibit an isomorphism between $R$ and $Z_p$ explicitly. Isn't it?

What would that isomorphism be?

Jakobian

For fields of size $p$ - a prime, you can just write the isomorphism as $f(n\cdot 1_R) = n\cdot 1$

since any element will be an integer multiple of $1_R$

for other powers of $p$ it might be harder, I don't remember the argument

Thomas Finley

@Jakobian Ok, but what if the multiplication is trivial?

Jakobian

for a field?

Thomas Finley

@Jakobian no, we had two cases. 1. R is a field , 2. Multiplication in R is trivial.

If case 1 occurs then R is isomorphic to $\Bbb Z_p$

Jakobian

Oh, I misunderstood you, I thought you're still going over the case of field

Thomas Finley

14:28

Now, what happens for case 2?

Jakobian

@ThomasFinley Take any non-zero $a$ and then $f(n\cdot a) = n$ is an isomorphism

Thomas Finley

@Jakobian Wait, by $a$ you mean a generator for $R$ right?

Jakobian

Any non-zero element of $R$ is a generator for its additive group

It doesn't matter

Thomas Finley

To be precise, if $R=\langle a\rangle$ we take, $b\in R.$ Then, there exists $n\in \Bbb Z$ such that $b=na$ . $f$ maps $b$ to $n.$ So, $f(na)=n_p,\forall n\in\Bbb Z.$

Correct?

Xander Henderson

Use \langle and \rangle.

Jakobian

14:34

Its not pleasant to go over things so elementary

you can confirm your own doubts

Thomas Finley

@Jakobian confirmed it and yes what I wrote is correct

Thanks a lot! I got it completely!

Thomas Finley

15:39

@Jakobian Till now, I have seen many solutions for this problem. But I can confidently say this seems to be the best to me. Thanks again!

Ted Shifrin

18:22

Howdy @Thor. How did your seminar lecture go?

Thorgott

18:37

hi, everything went smooth!

(although most Schubert varieties aren't smooth, of course)

slightly over the time (I always am), but nothing I'm significantly unhappy with

had a very enlightening discussion on whether the proof of Pieri's formula in Eisenbud & Harris is wrong or not afterwards

Ted Shifrin

What was the conclusion? I don't know that book, sadly. I'm sure it has tons of gorgeous stuff in it.

Koro

Say $\phi: U\to R^3$ is smooth, $U\subset R^2$ is open. Define $v_1=d\phi/dx, v_2= d\phi/dy$. Look at det$(v_1, v_2, v_1\times v_2)$. Here think of , as a new row. How in the world is that determinant equal to $(v_1\times v_2).(v_1\times v_2)$?

=$\|v_1\|^2\|v_2\|^2- (v_1.v_2)^2$

Ted Shifrin

I think I've told you this before, @Koro. It's actually the same computation you were doing yesterday, for which I gave you the linear algebra fix. But, independently, $\det(v_1,v_2,b) = b\cdot (v_1\times v_2)$.

This is in essence the definition of cross product.

leslie townes

18:53

Triple product

In geometry and algebra, the triple product is a product of three 3-dimensional vectors, usually Euclidean vectors. The name "triple product" is used for two different products, the scalar-valued scalar triple product and, less often, the vector-valued vector triple product. == Scalar triple product == The scalar triple product (also called the mixed product, box product, or triple scalar product) is defined as the dot product of one of the vectors with the cross product of the other two. === Geometric interpretation === Geometrically, the scalar triple product...

Thorgott

there is a very weird step in the proof where they quote the technical and very annoying lemma 4.5. the conclusions are
a): it is unclear as it is written how that step in their proof invokes lemma 4.5. however, i have been convinced that there is a way of applying the lemma and making choices in the right sequence that makes the proof work.
b): the entirety of lemma 4.5 is, as far as Pieri's formula is concerned, entirely unnecessary and can be replaced by elementary linear algebra

other than that, though, I really like their proof. it's very geometric, unlike the less enlightening more algebraic proof in Griffiths & Harris that I only skimmed

Ted Shifrin

Interesting. I have known Eisenbud and Harris for 4 decades or more. Eisenbud always struck me as more careful with details.

Koro

@leslietownes ohh yes, I remember it now :). Thank you so much.

leslie townes

having pasted that (which covers the topic but is also a good example of wikipedia spew) my only other observation is, there is a school of thought under which examples in R^3 and analogies with stuff like the cross product improve and solidify the intuition for general dimensions, and another school of thought where the number of different notions collapsing into the same notion in R^3 is seen as a peculiar distraction from general ideas. i am probably more in the latter camp.

Ted Shifrin

@Koro Isn't this what I just wrote you above?

leslie townes

18:55

and also, as always, i wonder what dieudonne would dieu

Koro

I studied this thing many many years ago. We called it 'box product'.

Ted Shifrin

@leslie I never included cross product in my linear algebra courses. I agree with you. But it's too important for multivariable calculus and physics in $\Bbb R^3$ not to pay attention to it. :)

Koro

@TedShifrin yes. I saw the triple product and I remembered it and could relate to your comment too.

Ted Shifrin

But, Koro, this actually ties exactly into that linear algebra we were discussing with the determinants and $AA^\top$.

Take your two vectors and then put the normal in the third row/column.

So that normal, is of course, up to scalars the cross-product. Proceed.

Anyhow, to see it explicitly, @Koro, look at p. 40-41 of my diff geo text.

Thorgott

It's not necessarily a detail issue. I oftentimes have this experience that things get iffy when algebraic geometers start using the word "generic" a bit too freely. When I brought up my issue with the proof after the seminar, though, others immediately saw the fix, so it's not like it's grave or anything.
Explicitly, the point is that we have Schubert cycles $A,B$ and $C$ in complementary codimension. Picking general flags with respect to which we define them, we can assume $A$ and $B$ meet generically transversely so that $A\cap B$ represents the intersection product. Then, they make a cl…

@leslietownes I am in the exterior algebra camp, we're really just observing that $v_1\wedge v_2\wedge\ast(v_1\wedge v_2)=\lVert v_1\wedge v_2\rVert^2\mathrm{vol}$ by definition of the Hodge star!

Ted Shifrin

19:10

Ah, right, that is a typical situation with flags. ... McCrory and I wrote our first paper together being totally rigorous about what "generic" means in a situation with surfaces in $\Bbb P^3$ and doing some Thom-Boardman loci. Arnold had published the same formula for an intersection number with no rigor whatsoever about what "generic" means (nor justification).

Thorgott

oh, right

Ted Shifrin

Yeah, all sorts of horrendous linear algebra stuff amounts to elegant stuff about the inner product on $\bigwedge^k\Bbb k^n$.

Thorgott

especially Laplace expansions

however, I don't have an elegant exterior algebra explanation for the $\det(1+vv^t)$ we had yesterday

Koro

@TedShifrin noted, thanks.

The determinant of AA^\top is the determinant of matrix with i,j th entry E_i. E_j.

Thorgott

@TedShifrin yeah, I think issues with what "genericity" means are as old as time. professionals will have no issue parsing the argument, but as someone who is not quite an algebraic geometer, it stumped me

Ted Shifrin

19:14

Ah, but people throw it around without careful proofs. McCrory wrote very detailed proofs and defined exactly what generic meant. It took many pages.

The generic we have for transversality theory (complement of measure zero) is not nearly precise enough for most algebraic geometry.

Koro

@TedShifrin noted. I see how it defines the triple product (scalar). I put $b= v_1\times v_2$ to see that.

Ted Shifrin

Cool, @Koro.

@Thorgott I don't remember that discussion, but this is a classic thing that shows up lots of places.

Thorgott

"general" for me usually means "holds on an open+dense", but I wager there's more sophisticated variants for sophisticated applications

Ted Shifrin

Not to mention the difference between the Zariski topology and the usual :)

This discussion reminds me. We haven't seen @Balarka around in many months.

Thorgott

@TedShifrin yeah, but I've never actually found a proof I found fully satisfying

Ted Shifrin

19:23

Where was this in yesterday's discussion? I recall figuring out a simple proof years ago.

Sine of the Time

19:34

Hi, I have a quick question. Here, in the section "Specifying a transformation by three points", the uniqueness of the Mobius transformation is up to a multiplicative constant right?

Ted Shifrin

No. Unique, period.

If you're talking about $a,b,c,d$, those are unique up to multiplying by a non-zero constant.

Sine of the Time

ok makes sense

thank you

Ted Shifrin

This is why the group of Möbius transformations is $PSL(2,\Bbb C)$.

Sine of the Time

I'm not familiar with this notation. For curiosity, what's $PSL$?

Ted Shifrin

The P means projective, so you mod out by (nonzero) scalars.

$SL$ means matrices of determinant $1$.

Sine of the Time

19:39

👍

Jakobian

19:52

People don't study analytic functions over infinite-dimensional Banach spaces?

Ted Shifrin

I don't know how to write convergent power series in infinitely many variables.

Thorgott

around here chat.stackexchange.com/transcript/message/64878597#64878597

Ted Shifrin

Oh, that was not $\det (I+vv^\top)$, though. :)

In particular, we had $[I|v]$ and I put $[-v^\top|1]$ on the bottom row.

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