I have to show that in $A[X]$, the nil and the Jacobson radical coincide. We always have $N\subseteq J$.

Now, if something is not nilpotent in $A[X]$, I know that either the indep. coefficient is not a unit, or some of its nonindep. coeff. are not nilpotent. I proved that earlier. I am not sure if I should use that.

Sorry.

I meant nilpotent.

Each one nilpotent.

I think I can do it.

@Karl I think Pedro can do all these on his own.

Let's make fun of him when he asks questions.

@Mike Booo Mike.

Boooo.

@Mike Sounds good.

Hey guys, what Latex editor do you use?

I currently use Lyx, but I'd like to move onto something better.

That is, if you think I could benefit from another editor.

notepad :P

fantastic

:p

Seriously, what do you want in a latex editor?

@eXtremiity I use TeXWorks.

sup

So it depends on what I want from one? Well, recently I wanted to compare pictures in columns and I had to resort to all this "floating" and "box" nonsense. In addition, I wanted to draw some nice arrows but could not do it. I read there is a package called lyx etc.

HAI,.

I'm doing some AM exercises @FernandoMartin

Cool

I'll try to finish half of the first section.

Which one?

I've done everything up to ex. 9

@FernandoMartin At the moment doing 5.

I have a measure theory problem I came up with

ORLY

I don't know where it stands on the trivial-impossible spectrum

Suppose I have a $\sigma$-algebra $\Sigma$ over a set $X$

Is there some topology $\tau$ over $X$ such that $\Sigma$ is generated by the open sets of that topology?

OK.

i.e. is every $\sigma$-algebra Borelian?

I think not, but I haven't come up with a proof yet.

Did you think about a finite ($\sigma$-)algebra?

Finite $\sigma$-algebras are Borelian

Right, yes.

(they are topologies)

Yes.

I'm tempted to say this is not true in general, since you don't necessarily have arbitrary unions in a $\sigma$-algebra

@FernandoMartin Right.

@FernandoMartin Well, it suffices you find a $\sigma$-algebra that is not closed under arbitrary unions right...?

I don't think it suffices to show that

@FernandoMartin Oh.

Generated as a $\sigma$-algebra.

Yes

No idea at the moment.

Yup, I've tinkered with it for a while with no results. I just wanted to know if it was a known result or something like that

Hey, I have a question. So I am trying to find the convergence and existence of the function, $f(x)=\sum_{k=1}^{\infty}\frac{x^{k}}{k^{2}}$. At first, I would say the domain can be anything. What such $x$ would make $f(x)$ explode? But after doing some research I see that the function I have defined is actually the Dilogarithm Function. Furthermore, this function is said to only be defined where $|x|\leq 1$, such that $x \in \mathbb{C}$ (sorry for not using $z$). Is this true in GENERAL?

Or just the fact that - we say this Dilogarithm function is defined only for |z| $\leq$ 1.

students should be able to see the radius of convergence before they here about the dilogarithm function

@eXtremiity

your function is a power series

x^k is exponential in k whereas k^2 is only polynomial in k

Where do power series converge?

ratio test that shit, bro

that's three markedly different approaches to helping with the same problem

Hey @Mike

does the problem I wrote above ring any bell to you?

Ahhh, maybe the $\sigma$-algebra generated by singletons in an uncountable set works.

Let's see

@FernandoMartin How did you do 5.(ii)?

I was thinking induction on the index of nilpotence.

the one about nilpotent formal series?

Aha.

Alright, I'll think about it. But how about the domain of $f$? Is it also dependent on the radius of convergence?

@FernandoMartin Tried that, just let the singletons be closed.

Is it clear that $a_0$ is nilpotent? @Pedro

@FernandoMartin Sure.

Ok, and you know the sum of nilpotents is nilpotent

Ah.

Yes.

hence $f-a_0$ is, too

@Karl disregard me

@FernandoMartin Mike didn't look at this.

I think it is a nice solution. =)

I don't remember what Baire's theorem said

I are dumb

@FernandoMartin Anyway, it's been asked on MO. mathoverflow.net/questions/87838/…

@FernandoMartin Suppose $M$ is a complete metric space. If $O_1,\ldots$ are open and dense, then $\bigcap O_n$ is dense.

So it was actually pretty difficult I guess... Thanks @Karl!

Woah, the asker was Joel David Hamkins

@Pedro right

@Karl If that guy can't answer a set theory question then I definitely can't answer it

:)

@FernandoMartin My measure theory background is just strong enough to do other stuff.

I can integrate.

But I don't think too hard about sigma-algebras.

@Mike: That's ok, the problem turned out to be harder than I expected

@FernandoMartin I've been trying to work on a problem that's harder than I expected

Can I understand it?

Given about an hour you can have the same background I do.

It's terrible, though.

Don't think about it.

Ok

I'm great at not thinking about stuff

@anon why are you an anime eye?

@Mike there is no why, only feels

@anon holy monkey

you appeared out of thin air

you just mixed up ghostbusters and a 4chan meme

i'm so confused

Ah. Figured it out. Thanks @anon

@Mike.

And the other guy :p

what don't I get about ehsan's answer?

Fernandooo. Nice name. My uncle has that name ^_^

@anon the original movie blew, by the way

@Mike ghostbusters or ghost in the shell (the eye)?

the latter

if anyone says ghostbusters blew i will end them

heh

ghostbusters blew

So since f(x) converges to whatever value it does, it can not exist for values to the "side" where convergence starts and finished. Yah

I WILL END YOU

@anon You can argue by dimensions, since the map is surjective, the kernel must be prime but not maximal.

what the hell kind of person doesn't like ghostbusters?

someone being ironic to annoy me, i assume.

the truth is actually worse than that

I haven't watched ghostbusters

@KarlKronenfeld so the kernel must contain the irreducible (x-y)^2-x (which generates a prime ideal), the kernel itself has to be prime but can't be maximal, but how does that imply that -is- the kernel? (I haven't worked with dimensions unfortunately.) There are prime ideals containing ((x-y)^2-x) that aren't maximal (and whose quotients are iso to C[t])

@FernandoMartin i won't end you.

you should just end yourself.

:(

@anon It's just that there can be no chain of prime ideals $(0)\subset P_1\subset P_2\subset P_3$ in $k[x,y]$. In the general form, it's a tough theorem though.

i have deemed you not even worthy of ending

lol

@KarlKronenfeld but we're trying to rule out $(0)\subset ((x-y)^2-x)\subset P$ which doesn't have that many terms, no?

honestly the best thing about math and science isn't that it's given us such technology or any of its inherent beauty or anything like that

it's how hilarious cranks are

Let $P$ be any maximal ideal containing $(f)$, @anon. We know it can't be $P$, so it's some ideal between (possibly containing) $(f)$ and $P$.

cranks are funny only when there are none in your courses @Mike

it gets old pretty fast

@FernandoMartin there are cranks in your courses?!

@KarlKronenfeld oh, so P_3 is the maximal (hence prime) ideal we can assume contains a putative P_2 containing P_1, I see

public uni is public

none of that theory is referenced in the answer though...

omg i love it

@FernandoMartin WHAT.

Who?

Facebook maybe.

I want in on this convo

back in high school in my bible class we had a guy who believed in santa claus and vehemently argued against us all that he existed

kind of a sheltered kid

@Mike I'll tell you on FB.

anon, do you know vixra?

@anon No, he was referencing the direct computation in that other answer, which imo is unnecessarily complicated. I said "you can" on purpose.

here's a great recent vixra article. make sure to check out the footnote.

wat

Oh wait no. :'( Ratio test doesn't work.

It ends up giving me |x|.

@eXtremiity um

Ohhhhh, which converges is |x| < 1

yes, that's the point

lol

But, thats IF |x|<1 it will converge.

conjecture: Giuseppe Rauti = Karl Kronenfeld

See, what it comes to finding the domain of f(x)=x^k/k^2, I'm a little stumped.

I'm sure I can put ANY x into there.

But f(x) will only converge if |x|<1

uh... you just found its natural domain.

i don't understand what you don't like about it.

f(50) for example should have a value.

@Mike yes I showed my phycisist buddy it the other day and he was elated

Should it not?

no it shouldn't

@eXtremiity Nope.

Ok, that's the part I do not understand.

it diverges to hell and back

@Mike Shit, don't blow my cover.

To hell and back hahaha ^_^.

@eXtremiity how don't you understand it? you just proved it.

Oh bloody hell.

Wait..

f(50) = \infty , then we say it does not exist?

we say that the series diverges

@Mike Oh, wait. Mine was published on April 1. Phew.

Ok.

and we want a function $U \rightarrow \Bbb C$, where $U$ is some open subset of $\Bbb C$

Uh-huh.

Ok. That was the thing I was missing. If f(x_1) diverges we say x_1 is not in the domain of f.

@Mike: well, Pedro doesn't agree on the crankiness of the guy

@FernandoMartin I WANNA HEAR

There's an older guy in our comm. algebra course

@KarlKronenfeld all the things posted in april 1 are too much effort to be you

@anon

Will you answer a question for me?

@Mike lol you actually searched?

@PedroTamaroff ?

@karl i was already in the area.

Hi @Mike @Pedro @Karl @Fernando

@TedShifrin hi

And @anon

hi ted

Hi @Ted

@anon I have to show that the contraction of a maximal ideal in $A[[X]]$ is maximal, and that it is generated by the contraction and $X$. I have an idea.

What is contraction? @Pedro

Intersecting with the subring $A$ @TedShifrin

@TedShifrin If you have a morphism of rings $f:A\to B$, the contraction of a ideal in $B$ is it's preimage in $A$.

Oh, sigh.

@KarlKronenfeld I misread.

@PedroTamaroff 7?

@PedroTamaroff stop that

The one you told me earlier? Yeah, I know.

@KarlKronenfeld I also mistyped.

@Mike Ah?

stop misreading

Unlike @Pedro to be so amiss

Duplicate.

@PedroTamaroff I'm not sure why you want me to duplicate it.

I wonder if those guys are in the same class.

@Pedro @Mike, a guy downvoted me for not giving him a whole grad course on Chern classes in the answer to his question.

@TedShifrin I would too.

Who better to get a grad class on Chern class from than you?

@Mike True dat.

@TedShifrin I just read it. Sounds so fancy.-

You found it?

"Finally, let me preemptively answer the hardcore algebraists which I expect to respond along the lines of "read this book which contains all the rigorous definitions": I've done that, but I still find it difficult to produce examples."

He's pre-emptively being spiteful towards people for some reason...

I'm far from a hardcore algebraist. Farrrrrerr. But he knows no topology, alg geo, or diff geo, so what should we do?

@TedShifrin Ignore.

I didn't read the question. How do you know he doesn't? What are his difficulties?

Hell if I know. But hard to jump into third year stuff not knowing first or second ...

Well, I just didn't know how you knew he didn't know first or second. But I didn't read the question, so I'll do taht ebfore asking :P

I don't know much. Just annoyed.

Closer to retiring :)

Wanna see something more annoying?

I think annoying has no sup.

How about people asking questions in bad faith?

@Mike You want Ted to retire so you can have him all to yourself.

I don't think I'll get him

I might get his books, though.

Could someone please verify my answer: i.imgur.com/vYa4PQu.png

I'm not sure it's bad faith ... Argumentative maybe. I wouldn't spend all that effort answering.

@TedShifrin It definitely is, if you look at all the comments. I wouldn't suggest you do look at it, though unless you're sick like me and enjoy it.

Why do you ask, @Mikrop?

You sick @Mike?

@TedShifrin WebWork will not accept my answer.

@TedShifrin I enjoy reading argumentative answers and stuff by the cranks. :P

Because you typed cost, not cos(t)

You're a crank-in-training, @Mike?

@TedShifrin Changed it, still not being accepted. Are my answers correct?

So mean!

maybe also (cos(t))^2 rather than cos^2(t)

@МикроПингвин You also wrote sint.

Change everything, @mikrop.

I have changed everything -.-

:confus:

@TedShifrin I had to spend an hour and a half doing homological algebra in front of a board today. It was horrifying.

@МикроПингвин screenshot or it didn't happen

Then it should be ok ...

Here is what WebWork sees: i.imgur.com/ovQCdIq.png

In front of a board?

@KarlKronenfeld Karl.

My algebraic topology class is me presenting, and the professor correcting.

@МикроПингвин Maybe switch and have $x=\sin(t)$, etc

@PedroTamaroff yes

The math is right ... Do you put in commas or are there separate blanks? Do you need $\langle blah \rangle$ for it to recognize a vector?

@TedShifrin No, that is included already. And there are separate blanks. So, I guess I will just email my instructor and see if he inputted the answers wrong.

@KarlKronenfeld Suppose $\mathfrak m$ is maximal in $A[[X]]$.

He didn't do it. The publisher hired someone, but there are known mistakes. I've written hundreds of WebWork problems for my students, so I've got it about right after 4 years ...

It might be the case $\mathfrak m\cap A$ is zero, is it not...?

For example, if $A$ is a field.

sure

But then it is not maximal.

it is

Oh. LOL.

Sorry about that.

Nevermind.

Ugh...my mom came into my room today while I was at school and she threw away all my math textbooks...good thing most of the books I need are on my laptop which I carry with me

WHAT

WHAT.

What I can do is first show $\mathfrak m$ is generated by $X$ and the contraction @KarlKronenfeld. Then showing the contraction is maximal should be easier.

Hi @Mike

How are you doing today

@PedroTamaroff right, yeah show that $\mathfrak m$ contains $X$.

@user127001 You're going to get a new mom.

lol

@user127001 That's not nice

I have to deal with it until I graduate

Why would she do that...?

She's a very devout Christian and dislikes anything outside of the Bible. She doesn't like me majoring in mathematics

i find this hard to believe

Not if you grew up with her

@user127001 "Anything outside the Bible." is so broad.

@user127001 What did we do to her?

I'm so so sorry, @user127001 ... Good thing she doesn't know there's pornography in all the math texts I've written!

@PedroTamaroff I challenge you to a duel

Damn stars ...

@Mike How do you like having a popular name? mathoverflow.net/questions/162762/…

Challenge him to a dual, @Mike

@KarlKronenfeld I already showed the Jacobson radical of $A[[X]]$ is generated by $X$ and the Jacobson radical of $A$. =)

@Mike OK.

@TedShifrin That would be unfair. I'm a dual master at this point.

@PedroTamaroff That makes it easier. :)

@KarlKronenfeld Indeedz.

@Mike So?

@KarlKronenfeld I commented just so I could ping him with @mike

What's this duel about?

I dunno, seems fun.

@KarlKronenfeld I think I should delete my comment, before somebody at MO yells at me and tells me the question is perfectly research-level.

@TedShifrin When people paste their questions on MSE, and their question is an imperative, I like to just comment "No."

It's fun.

Is it?

yes.

They never say "write a proof". They say "prove". So, when I'm in that mood, I just say, "ok, done".

What does it mean when a question is "an imperative"

Sorta like I write GRR and GRR$^2$ on homeworks.

@user127001 "Do this." is imperative.

"Find a 99% confidence interval"

Oh

@KarlKronenfeld If $\mathfrak p$ is prime in $A$, it is the contraction of some $\mathfrak P$ prime in $A[[X]]$.

yuppers

Let me think a second.

Grisha Perelman documentary in russian with english subtitles

meta.math.stackexchange.com/questions/12883/…

I have to go @PedroTamaroff.

@Mike I know that if $\mathfrak p$ is prime in $A$, $\mathfrak p[X]$ is prime in $A[X]$.

@karl

@eXtremiity any good?

It's alright. There was not anything new to take from it in terms of his achievements and life.

good to know

@Mike If $\mathfrak p$ is prime in $A$, $\mathfrak p[[X]]$ is prime in $A[[X]]$, right?

Because the quotient is a domain.

I don't wanna think about it.

Probably.

The quotient is $(A/\mathfrak p)[[X]]$.-

Err.

erm

if that's true then I assume you already proved that $R[[X]]$ is a domain for $R$ a domain

@Mike Well, consider the map $A[[X]]\to (A/\mathfrak p)[[X]]$ that sends a series to the obvious series of coefficients $a_n+\mathfrak p$.

Clearly surjective.

Kernel is $\mathfrak p[[X]]$.

Yeah, OK. It works. But proving that thing is a domain must be done with a little care.

@Pedro: That ex. was tricky

I didn't prove that the quotient was a domain though

@FernandoMartin Tricky in what sense?

Help on math.stackexchange.com/questions/744561/… please

@FernandoMartin No? Suppose $fg=0$, $g\neq 0$. Let $m$ be the least natural for which $g_m\neq 0$.

long argument

Which one?

AERHGAHEGRHGERHGFHGSHGD thunder

Wait, are you talking about 5.v?

No, you're not

lel

Which one is it?

@FernandoMartin Yes.

Was it 5v?

I am saying that if $D$ is a domain, so is $D[[X]]$.

Anybody could help?

Is this your homework?

That follows by looking at the smallest coefficients, right?

@FernandoMartin Yes.

@mike - That was a random problem from a foreign forum

But all people give some wrong answer in that forum

MSE forum is best forum

SIGHHHHHHHHHHHHHHHHHHHHH

I can not seem to find anyway to show $\sum \frac{sin(k)}{k^2}$ converges. I have used the ratio test, root test and W-Test.

mse is not a forum

So do I conclude that it diverges or am I so fried that I am missing something :'(

@eXtremiity Have you used the W-test...?

Because $|\sin k|\leqslant 1$, dear.

I have treid that, and IFFFFFFFFFFFFFFFF

I need to have a break. I am unbelievable. How could I miss that

lolo

Sorry

@FernandoMartin this is tragic.

@Mike MAIK

wanna do some hangouts

wtf is a hangout

google hangouts brah

sounds like some real nerd shit

indeed @Mike

So I have finally shown that my function, $f(x)=\frac{x^{k}+sin(k)}{k^{2}}$ converges for $|x| < 1$ via Ratio and W Tests. Now, I have to determine, "In what sense does the series converge". In class I have studied 4 types of convergences - Uniform, Pointwise, $L^{2}$ and $L^{1}$. My question is -> what I have done with the ratio and W test above has guaranteed what? I've just done some working out an I think it is the same as Uniform convergence.

Hello!

Hello Andy.

How are you Pedro?

Not good.

I hope everything is going all right. By the way, would you happen to know anything about this: math.stackexchange.com/questions/744677/…?

Hi Andy

How do we get our brackets and parenthesis etc to span for the whole fraction?

\| .. something something

What do you mean

$|\frac{\frac{a}{c}}{hi}|$

See how my absolute value does not go from top to bottom.

I know there is a way to make it that way. I just do not know the command.

\left\lvert\frac x y\right\rvert

Thank you.

:)