@DanielF Why is every open set in $\Bbb R^n$ a union of countably many open balls? I know this should be trivial. I assume it's from second-countability.

@MikeMiller In particular, that you have a countable basis consisting of open balls.

Oh, right, all open sets can be written as a union of basis elements. Whoops.

(Thanks!)

17:41 avreage of 100, 16:55 avreage of 12, 15:87 avreage of 5, happy rubiks cube day

What are the scores

Is this answer truly worth 2 downvotes and no upvotes? In fairness, I did receive the downvotes prior to some edits for clarification. But wow.

What's a ring with multiple additive identities / inverses?

not possible @EnjoysMath

that's good news

don't shoot me down bra

:D

I read the claim that by the substitution $t=a\tan\theta$, $$\int_0^{\pi/2}\frac{\mathrm d\theta}{\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}}=\int_0^{\infty}\frac{\mathrm dt}{\sqrt{(a^2+t^2)(b^2+t^2)}}$$ I can get close, but not exactly that. Does anybody see a way to manipulate the denominator?

@Karl Why are you interested?

@MikeMiller Why do you ask?

@KarlKronenfeld hi .. its direct change of variables .. (no extra manipulations required) :)

@Karl I want to know if I'm interested.

@r9m Oh, I accidentally wrote it as $(a^2+a^2t^2)$ etc on my paper leading to sillyness. I will see if I agree now.

@MikeMiller Something about fast computations of elementary functions.

@r9m Yeah, you're right. Thanks

facepalm

@KarlKronenfeld :D

Thanks, @Karl

Not interested, I imagine.

I'm not uninterested. But yes, you're right.

@DanielFischer

@AlexanderGruber

@Karl I should mention if I haven't that the proof of Serre duality is eminently skippable. It's such a different flavor.

Hey guys

I need a reality check

suppose $M$ is an $A$-module

$A$ commutative with unit

I define the following sets of prime ideals of $A$

@MikeMiller Ah, ok. :)

the set of associated prime ideals, $AP(M) = \{P : P = Ann(x)$ for some $x\in M \}$ (thanks Karl)

and the support of $M$, $S(M) = \{P : M_P\neq 0\}$ where $M_P$ is the localization at $P$

I'm being asked to prove that the minimal elements in those two sets are the same

I think I have a counterexample to that

You need some finiteness condition

yeah, the internet says noetherian

@robjohn aha .. there is a version for locally compact .. smooth :D .. thanku ! :)

What's your counterexample?

@Karl: I'm not sure about my counterexample, since it is noetherian. I think I'm ignoring something stupid

let $A=M=\Bbb Z$

then $AP(M)$ is just $0$, since $\Bbb Z$ is an integral domain

ahh I'm stupid

:P

well, even then

the exercise didn't say anything about finiteness, so I guess the result is either false or hard to prove

It should be false.

I'm checking an example of mine which has no associated primes.

that should work

since a module with no support has to be $0$

Yeah.

$A=k[x_1,x_2,\dots,x_n,\dots]$ where $k$ is a field of characteristic $2$. Set $I=\langle x_1^2,\dots, x_n^2,\dots\rangle$. Then $A/I$ is an $A$-module with no associated primes.

Thanks @Karl

I have a proof of that in my typed notes on this topic. Want them?

Is it very hard? I was about to try to prove it myself

No, not hard.

Ok, I'll give it a try and ask you if I get stuck

thanks!

I would like to see your notes anyway after I finish this if you feel like sharing

sure

I think the condition $M$ is finitely generated may be necessary as well, though I don't have any counterexamples.

Ok, I think I proved it

annihilators are of the form $\langle x_{a_1},\dots, x_{a_n}\rangle$

and none of those are prime

(I proved that for $k=\Bbb Z/2\Bbb Z$, I don't know if what I said holds in general)

That's not true.

annihilators are ideals generated by finitely many products of finitely many different variables

those still aren't prime

Yeah, I followed basically the same process when I found/proved it.

:)

Thanks @Karl

Can I take a look at your notes now?

yep. megafileupload.com/en/file/531774/associatedprimes-pdf.html

thanks once again

>May 18, 2014

you compiled the .tex again or were you actually typing this up today?

I recompiled it after finding a typo

I typed it up on the 11th of April

I usually work the exercises and throw them in the notes where they fit naturally, don't think I did that here though

Which book did you follow/did exercises from?

Matsumura's section 6

haha, your proof is much slicker

than what I said

@KarlKronenfeld Hello.

@FernandoMartin Let me see what I did so far...

@PedroTamaroff hey yo

@KarlKronenfeld I have this problem: prove that if $G$ is an abelian $p$-group and $G/pG\sim \Bbb Z_p$; then $G$ is cyclic.

I have a solution which is uses Frattini.

Was wondering if a more elementary solution could be produced.

Basically I am using Burnside Basis's theorem.

@KarlKronenfeld: why is it that $m^2 = 0$?

where $m=\langle x_1, \dots\rangle$

since $f^2=a^2$ for all $f$

yes but $x_1x_2 \in m^2$

righ' looks like I messed up

$\mathfrak m^2\subset J$ is definitely unnecessarily strong.

@FernandoMartin suppose $J$ were a prime ideal properly contained in $\mathfrak m$. Note that some $x_i$ does not belong to $J$, yet $x_i^2=0$.

yup, I was thinking in the same thing

great.

@PedroTamaroff Shit I forgot you asked me about this.

@KarlKronenfeld Hehe, I'm in no hurries don't worry. I'm not interested in an answer now.

@FernandoMartin

From PIII I've done E1,E2,E3,E4.

sup @Pedro

E1 from the book?

or from Alicia's questions?

Nah, the problem sheet.

The first four exercises.

well E1 from the problem sheet is wrong :P

Which part?

E1.2 needs noetherianness to hold

There are three bulletpoints.

how did you prove it?

@FernandoMartin What's wrong with it?

Karl gave me a counterexample

What does the problem say again?

The problem says that minimal associated primes are the same as minimal primes in the support

Right.

Karl gave me an example of a non-zero module with no associated primes

since being 0 is a local property, any non-zero module can't have no support

@FernandoMartin I'm thinking.

the example is not obvious at all

No, not the example.

off topic. Can someone define a "free ring"? I can't seem to find a hits by googling this or in Artin's text.

a ring is a $\Bbb Z$-algebra

so a free ring is just a free $\Bbb Z$-algebra

@FernandoMartin Let's suppose there are associated primes.

Does the claim hold?

I don't know

Probably not @PedroTamaroff

@KarlKronenfeld Maybe there are no minimal primes at all?

that can't happen

Recall the prime ideals of a ring do satisfy the DCC :)

@KarlKronenfeld I didn't know that.

derp, no you didn't

that's exercise 1.8

@KarlKronenfeld I mean ideals minimal respect to inclusion, that happen to be prime.

Is that what Alicia is saying? @FernandoMartin

Maybe we're reading the question off wrongly?

Yes, that's what Alicia is saying

I mean, suppose ${\rm Ass}\; M$ is not empty.

ok

how does concentration of measure work for ellipsoids? say you have a very thin ellipsoid with major axis = 1, and the rest of the axes equal to epsilon, say. where does the mass concentrate in high dimensions?

@EricGregor Not our thing, I think. =P

whose thing?

who is this "our"?

We proved a prime $\mathfrak p$ is in ${\rm Sup}\; M$ iff $(0:x)\subseteq \mathfrak p$, @FernandoMartin.

yup

there are just no primes of that form $(0:x)$

OK, now suppose $\mathfrak p$ is minimal w.r.t. to inclusion.

wait

the condition is

being minimal with respect to inclusion amongst associated primes

not minimal in general

@FernandoMartin But if we take a minimal prime in ${\rm Supp}\;M$ shouldn't it be of that form?

Maybe she mixed the things?

What's "that form"?

An annihilator.

it can't be

unless we have extra hypotheses

if not, Karl's example works

My point is this. ${\rm Supp}\; M$ is always nonempty.

yes

(unless $M=0$)

Yes, yes.

An element $p$ is in supp if it contains some $(0:x)$. This is an ideal.

Right

If $p$ was minimal, then $(0:x)\subseteq p$ forces $(0:x)=0$ or $=p$.

Ah

So it must be the case that $(0:x)=0$ for any $x$. Is that the case in Karl's example?

(I don't know what his example is)

Nah, I was interpreting the problem as "minimal primes in Supp M"

as in

there's no smaller prime in Supp M

Oh. I don't think she means that.

not that the ideals were minimal in the whole ring

Still it seems wrong?

but the statement is dead wrong for any other interpretation

no crafty counterexamples needed

@KarlKronenfeld Hehe, sorry, I'm a complete dewb at this.

How did you prove the other inclusion @Pedro?

@FernandoMartin I didn't.

@KarlKronenfeld Could you enlighten me a bit? =D

In what sense?

Well, it seems you have counterexamples galore.

Eightfold path?

What exact statement do you want a counterexample for? @PedroTamaroff

@KarlKronenfeld That a minimal prime with respect to inclusion that is also an element of the support needn't be an associated prime.

I gave that above.

minimal primes need not be annihilators

@PedroTamaroff It's tricky, though.

@PedroTamaroff See this message chat.stackexchange.com/transcript/message/15595746#15595746

@Karl: pedro is asking about minimal primes (minimal in the whole ring) that are in the support, but aren't asscoiated

pick any minimal prime that isn't an annhilator

then pick a non-torsion $x$

then $Ann(x)=0\subseteq P$

so $P$ is in the support, but not associated

I don't think an explicit counterexample would be hard to come up with

I may not be agreeing with that last sentence in the following minutes

Is it possible to measure subsets of $\Bbb R^2$ using the 1D lebesgue measure function?

Yeah, not trivial, since you need to have a non-Notherian example

But just take $A/I$ as an $A/I$-module.

why aren't there noetherian examples?

(it's not the same question as before!)

since a minimal prime that belongs to the support is necessarily a minimal element of the support

ah I'm stupid

right

Damn, can't fix my misspelling of Noetherian

one last question @Karl

and I'll leave you in peace :P

ok

I'm not sure I understand your proof that $Ann(f)$ can't be prime

from the notes

It cannot be $\mathfrak m$ since it does not contain one of the generators of $\mathfrak m$.

Ah right, you're using the characterization you proved before

great-.

Thanks!

No prob

@FernandoMartin So we add Noetherian and everyone is happy?

No

We add Noetherian and we think hard about why it holds

Hm

(but it does hold)

@FernandoMartin Karl has a proof. Hehe.

It should hold.

Yes, but Karl having a proof != me having a proof :)

@FernandoMartin Yes, but = you having a proof from Karl.

=D

Yes, but after some time in which I bang my head against my desk

I will probably study from those notes, @KarlKronenfeld.

@FernandoMartin I'm still not convinced about sitting for the final in CA.

Why not?

You can take it some time later anyway

@FernandoMartin I don't think I'm up for it.

@Pedro please

you know more than half of the class does

stop whining

@FernandoMartin Well, I didn't say I wouldn't pass.

I just think I don't know this stuff well enough.

and?

I like to learn stuff properly. =P

You keep on learning after you leave the course

@PedroTamaroff Ah :) If you find errors (there's one in the proof of prop 2 that Fernando found that I already fixed) or typos please let me know. :)

The final test is just to give you a grade

I'm sure you'll get a good grade, and good grades are good - they help you get travel grants and make you a better candidate for TA-ing

If you don't feel you have learnt enough, keep reading after the course is finished

but that's no reason for not getting an easy 10 :P

@FernandoMartin Feels dirty.

There's nothing dirty about it.

You're taking the same test as the other students are

@FernandoMartin Yeah, I know.

I think you underestimate what you already know

@Pedro Is the abelianization of $GL(k)$ just $k$?

@MikeMiller No.

Darn.

Oh, stupid.

Are we talking ${\rm GL}(n,k)$ here?

It couldn't possibly have been.

Yes, I was lazy.

It was a stupid question, defer you ever heard it @Pedro

forget*

@MikeMiller Don't worry.

There are cools results about the commutator group of ${\rm GL}(n,k)$.

For $k$ a finite field.

It is ${\rm SL}(n,k)$, unless $(2,2)$.

And repeating, i.e. taking $G''$, we get ${\rm SL}(n,k)$ again, unless $(2,2)$ or $(2,3)$.

Where $(n,q)$ means size and size of field.

@MikeMiller So, for the finite cases the abelianization is mostly ${\rm GL}/{\rm SL}$.

@FernandoMartin

@Pedro

@FernandoMartin Did you know that if $S\subseteq {\rm Spec}\,A$ is any set and we take $\Delta_S=\bigcap S$, then ${\rm cl}\; S=V(\Delta_S)$? That's pretty useful.

why the strange notation for the intersection?

Ask Jacobson.

Classic Enjoy Maths =D

Hi @Fernando @Pedro @Karl @Mike

@TedShifrin Hello.

Hi @Ted

Hi Ted

we tried to respond in order, but Pedro ruined it

He usually does :D

welp, just emailed Alicia about the exercise

fun

@KarlKronenfeld Karl.

@Karl: Did you come up with that counterexample yourself? It's quite tricky

@PedroTamaroff yo

@FernandoMartin yeah

@KarlKronenfeld Nevermind. Hehe.

Nice.

@FernandoMartin Someone told me about this one some time ago. .

We were incidentally talking psychology some days ago, too.

I have that

Depression seminar?

@TedShifrin WAT.

except that I actually am an impostor

@FernandoMartin "My real name is..."

=O

an impostor in the academic sense

or in that article's sense

@FernandoMartin What do you mean? Fake ID?

Did you read the article?

@FernandoMartin Yes, I thought you were being funny.

>those with the syndrome remain convinced that they are frauds and do not deserve the success they have achieved.

I am being tongue in cheek, but I'm not being 100% funny

I don't think I deserve half the grades I've got

Well, I think everyone feels that way sometimes.

it's been a combination of unbelievable good luck and generous teachers

=)

@PedroTamaroff Btw, a group $G$ such that $G/pG=C_p$ can be shown "elementarily" to be cyclic.

@KarlKronenfeld Hit me.

If you had me as a prof, you wouldn't say that @Fernando ;)

I don't know if I should smile or not @Ted :)

You'd have earned a good grade, not been given it @Fernando ...

@TedShifrin Awww...

Yes, but that could have also been interpreted as "you wouldn't have got a good grade at all" :P

Most of my students are very proud :) The ones with Ds and Fs, probably not,

Yes, @Fernando, it could. But you don't impress me as that sort.

@Pedro: Alicia answered

haha

What did she say?

I'm surprised she's awake and checking her email this late at night...

classic Alicia

Apurada = ?

@TedShifrin In a hurry.

Ah ...

@KarlKronenfeld

I don't understand the first paragraph of proposition 2.

I used the correspondence between the prime ideals containing $\mathfrak a$ and the prime ideals of $A/\mathfrak a$.

(For any $A$, $\mathfrak a$)

@KarlKronenfeld Yes.

@KarlKronenfeld But why do you say "this is because..."?

So that I can quotient by $\operatorname{ann}_A(B)$.

:( No one noticed my question

Is it possible to measure subsets of $\Bbb R^2$ using the 1D lebesgue measure function?

@MarioCarneiro Yes, google "product measure."

I know there is a 2d lebesgue measure function

That is the product measure of the 1d Lebesgue measure.

My question is if you can emulate that given only the 1D function

@KarlKronenfeld Sorry, I am lost.

That would then be 2D lebesgue measure @Pedro

I have $A$. I take $B=A/I$.

This is an $A$-module in the usual way.

Why are you taking that quotient?

Only by integrating, @Mario.

@MarioCarneiro Why not take the product measure?

That was my first thought

@PedroTamaroff So that I can work with $B$ as a module over $B$.

Rather than $A$.

It's unnecessary, in hindsight, since I could have just written the statement of the proposition differently.

@KarlKronenfeld I thought you meant $B$ as a $B$-module in the usual way.

If you take $\int \mu(A_x)d\mu$, do you get the same measure as the joint 2D lebegue measure

$B$ is a ring too after all.

@PedroTamaroff They're the same.

that's Fubini's theorem, @Mario.

@Pedro Sorry, still reading the wiki on product measure

@KarlKronenfeld Right... sorry.

=)

Ahh, pulp fiction. Good morning all.

Good night, @Studentmath :)

Quote from Fubini page on wikipedia: "The product of two complete measure spaces is not usually complete. For example, the product of the Lebesgue measure on the unit interval I with itself is not the Lebesgue measure on the square I×I. There is a variation of Fubini's theorem for complete measures, which uses the completion of the product of measures rather than the uncompleted product."

How odd is that, two parts of the world with opposite times

@Studentmath It's almost morning here. :D

Yes, @Mario ... A technical subtlety because of additivity ...

@PedroTamaroff No need to apologize.

What is sigma-finite?

A countable union of sets of finite measure

@Karl Sun is up and shining here :P

@Studentmath Hey! You stole my Sun!

Give it back.

In... 12 hours or so.

oh... alright...

It's like taking shifts over it.

We need it in 7 hours ...

Well, isn't $\Bbb R$ sigma finite then?

Yes, @Mario

why wouldn't fubini hold then

It does ...

I'm puzzled by the above quote that the joint lebesgue measure is not the product measure

It seems like it should be the same since the product measure is unique, since R is sigma-finite

@Karl: Suppose I have $k[x_1,x_2,\dots]/\langle x_1^2, x_2^2,\dots\rangle$