@blue Ah, I was going with $7$.

=P

?

$(2+\sqrt{-10},7)$

ah. same difference.

I know.

I guess it is easier to work with yours.

mm, no

the exact same argument both ways

OK.

the rank 1 R-submodules of R are preisely the [-----]s?

If the ideal is principal, it is free, since it is torsionfree and finitely generated. Sorry, I am a bit off today. I had three midterms in one week, and I am having another tomorrow. I'm snoozing off.

Let me think about it.

@blue Today in fact I had combinatorics. I did 4 out of 5 problems, I didn't have the energy to finish the 5th. =P

I had 2 more hours.

you want to reason the other way around: if J is rank-1 free inside R, then it is a principal ideal. conversely if J is not a principal ideal in a domain then if it's free it must have rank >1. suppose R^n -> J is an R-module isomorphism. argue it has nontrivial kernel. (any two nonzero elements in an ideal satisfy a nontrivial R-relation)

@PedroTamaroff ah. how many hours did you have total?

@blue 5.

our exams are never that long

Well, if you are well prepared it shouldn't take that long.

I "finished" this in 3hs, but it was very computational.

Algebra I did in full in 2hs. Same with Calculus, (maybe a tad longer) because I have to prove something slightly tedious.

@blue How long do your exams last?

not more than 2.5 in my experience.

my diff equ exams I remember I knew how to solve every problem immediately, and was writing the whole time, and still always only barely managed to finish in the alloted time. must have been even worse for the other students.

I'm a little heartbroken fellas...

I'm still ragey

@blue what about?

@blue Yes, I remember that, I don't know why you posted it.

he gets 350+ rep I was aiming for

ah

using the same solution I had got

where?

he crossed the finish line before the blue shell could get him

@blue =)

dis

DAYUM

that looks nice

I had the exact same formula written down, $$|{\rm GL}_n(k)\sum_{e\vdash n}\frac{|{\rm nilEnd}_{k[T]}(N(e))|}{|{\rm Aut}_{k[T]}(N(e))|} $$

@blue =O What is $N(-)$?

the module $k[T]/(T^{e_1})\times\cdots k[T]/(T^{e_r})$

Oh. OK.

I don't know if this is over my head, but it seems pretty detailed work.

Also, this is the first problem I had in the midterm today. Let $a_n$ denote the number of words of length $n$ in $A,B,C$ such that no $B$ is adjacent to a $C$. One had to find a recurrence for $a_n$ (with a hint), then find the GF for $a_n$, then find a closed formula. The recurrence was $$a_n=2\sum_{i=0}^{n-1}a_i-a_{n-1}+2$$ for $n>0$. The GF was rational, and then some partial fractions did the work.

@blue Are you there?

yeah

I don't think I understood your idea on showing $(7,2+\sqrt{-10})$ is nonprincipal.

if it were principal, then 7 and 2+sqrt(-10) would share a nonunit divisor. since they are irreducible, this would imply they are associate. so (2+sqrt(-10))/7 would be a unit in Z[sqrt(-10)]. but it isn't even in that ring.

Yes, that is my idea too, but I didn't get this: "you want to reason the other way around: if J is rank-1 free inside R, then it is a principal ideal. conversely if J is not a principal ideal in a domain then if it's free it must have rank >1. suppose R^n -> J is an R-module isomorphism. argue it has nontrivial kernel. (any two nonzero elements in an ideal satisfy a nontrivial R-relation)"

could you be more specific about which part you don't get?

I don't get where that idea is coming from...

You said "the rank 1 R-submodules of R are preisely the [-----]s?"

the principal ideals

Then I said "If the ideal is principal, it is free, since it is torsionfree and finitely generated. "

Then you said that.

right. you're showing principal implies free. but we're trying to determine whether a nonprincipal ideal is free or not, so your implication is irrelevant, because it goes in the wrong direction.

"principal implies free" is of no help in determining if a nonprincipal ideal is free or not

@blue But that is not what I wanted to do? I wanted to determine if some ideal was principal.

Not if some nonprincipal ideal was free.

you seem to have a memory blackout

after we finished that, I asked you to prove that it's not free

Maybe. I just drank some lemoncello.

Oh. OK.

Let me think.

Hello everyone. I'm trying to patch this argument up to make it work: math.stackexchange.com/questions/863790/…

I feel like I'm very close, but there's currently a problem with it that I've mentioned as a comment.

@blue Isn't it trivial to show it is not free?

@PedroTamaroff how so?

(I would consider the argument I gave pretty trivial, but...)

I have this basic problem: How many ways can 11 boys on a soccer team be grouped into 4 forwards, 3 midfielders, 3 defenders, and 1 goalie? If anyone knows a Q&A I would be grateful for a link, I could not find one but I'm sure it's there somewhere.

@blue Sorry, I was being stupid.

@Pickett Where are you stuck?

@blue OK. Suppose $\eta:R^n\to J$ is an isomorphism, $n>1$.

@KajHansen you can narrow it down to 10,20,30,40 since 5 and 15 are cyclicity-forcing. go the rest of the way using sylow theory to force normal subgroups, for which the quotients will be small enough to examine by hand

I was thinking along those lines @blue, but damn. Lots of cases.

Say $2+\sqrt{-10}=\sum a_i\eta x_i$, $2=\sum b_i\eta x_i$.

@KajHansen eh.

What do you mean by "cyclicity-forcing"

?

@KajHansen Those groups are cyclic.

A group of order $5,15$ is cyclic.

Oh I see. Yes, I agree.

@KajHansen n is cyclicity-forcing if the only group of order n is cyclic

Then $\sum ((-5+\sqrt{-10})b_i-\sqrt{-10}a_i)\eta x_i=0$.

If one hasn't seen the proof for the $15$ case, then that's not necessarily trivial in itself. That takes a little bit of thought.

There must be a way to eliminate at least a couple more.

or just notice $\eta(e_2)e_1-\eta(e_1)e_2\in\ker\eta$, @Pedro

@blue My point was that any two thins in $R$ are always $R$ dependent.

But I thought I was being goofy.

@PedroTamaroff I don't know where to start.

I can use the binomial coefficient to figure out how many ways there are to group the boys into four groups. But I don't know how to add the constraint about how many boys should be in each group.

@Pickett arrange the 11 boys in sequence. make the first four forwards, the next three midfielders, the next three defenders, and the last one a goalie. by a factor of how much are we overcounting?

Aha! Burnside's theorem would eliminate $20$.

@Pickett this is where the multinomial coefficient enters

@Pickett Well, you have 11 guys. You need to assign them to 1,4,3,3. You can pick the first guy in 11 ways. The second, in $\binom{10}4$ ways, the third in $\binom 64$ ways, &c.

@KajHansen seems high-falutin

And $40$.

Note I am doing $\binom {11}1\binom {10}6\binom 6 4$. Some cancelling will occur, and multinomials will appear as @blue says.

It does. But at the same time it eliminates 2 of these :P

Oh look, @Ted is here.

@TedShifrin Hello!

heya messieurs @pedro @Kaj

@Kaj, @blue is calling you high-falutin'?

So it seems!

Thanks @PedroTamaroff and @blue I will try this.

I won't even ask, @Kaj

@Pickett Equivalently, @Pickett. You have 11 letters. You want to shuffle them as follows A|BCD|EFG|HIJK.

Since {B,C,D}, {C,B,D}, &c are the same, you get $11!/3!3!4!$.

this symmetry argument is basically orbit-stabilizer :)

Sorry, the last should be $\binom 63$.

Orbit-stabilizer is the best

It's gotten me out of many jams.

but you're well-preserved, @Kaj

Well-preserved?

do I really have to explain?

@PedroTamaroff In your last example, I don't get what the logic of the example is. $11!/(\binom{6}{3})$ - what does 6 and what does 3 represent?

How....what...? Someone is asking about the solvability of certain quintics, but hasn't seen the Sylow theorems before?

@TedShifrin It's a good one.

you don't need Sylow theory for that

@Pickett now look what you did pedro, you made him combine the two totally different arguments into one nonsense one

heh, did I misunderstand the correction?

@Pickett You can pick the goalie in 11 ways. Now you have 10 people. You pick a subset of size 4. Now you have 6 elements. You pick a subset of size 3. Now you're done. Party.

@TedShifrin, how would this argument be made without the Sylow theorems? math.stackexchange.com/questions/863790/…

@Pickett you misunderstood that there are two different arguments on the table

That's the $\binom {11}1\binom {10}4 \binom 63$ one.

Canceling shows that is the same as $\binom{11}{1,3,3,4}$ which is the second argument.

solvability has nothing to do with Sylow, @Kaj

That's true, but it would be a big, big help in this case.

how so?

You split your group as 1|234|567|891011 and shuffle. Since inside the brackets, order is irrelevant, and brackets have size 3,3,4,1, you get $$\binom{11}{1,3,3,4}$$

@TedShifrin sylow can help you divine normal subgroups, then we can focus on the quotient, a smaller group

...Showing that a group of order $40$ is solvable?

not needed to prove $S_5$ isn't solvable, @blue !!

oh, where did a group of order 40 come from?

Oh! I think you're misinterpreting what we're talking about.

@KajHansen ERMAGHERD Kaj when you said Burnside you meant $p,q$-Burnside!

That explains it.

That is high falutin.

@TedShifrin we're not trying to prove S5 is not solvable

LOL, yes @PedroTamaroff.

sorry, @blue ... I have no idea what I walked into.

@PedroTamaroff I had to read your answers a couple of time because I'm slow but I got this now, thank you.

@TedShifrin, we're talking about this question: math.stackexchange.com/questions/863790/…

@Pickett No problem.

@KajHansen hey, you still need to check that the sylow theory stuff actually works, I never tried

It does @blue. I looked.

oh, I see, if the Galois group is smaller than $S_5$ we want the group to be solvable.

@KajHansen okay good

@TedShifrin, I see no way of doing that, at least easily, without Sylow stuff.

But don't the groups have to embed in $S_5$ if the polynomial is irreducible?

classifying groups of small / arithmetically fortuitous order makes up like all of the exercises after a dose of sylow theory in a text

@TedShifrin yes

So no group of order 40 !

That is one approach. I don't think $S_5$ contains a subgroup of order $40$.

But how to show?

And you would need to do that each time. I.e. for groups of order $10$, $20$, and $30$ as well.

Unless it's in $A_5$ (which we know it isn't), what is its intersection with $A_5$?

itself

well

We know the subgroups of $A_5$ ... at least, @Kaj does.

normal subgroups, or subgroups in general?

Um, does 40 divide 60?

@blue I have to show that if $M$ is a f.g. module over a PID, it is torsion iff ${\rm Hom}_A(M,A)=0$.

Of course not, but how would we now that $G \cap A_5 = G$?

@KajHansen What is $G$?

@PedroTamaroff group of order 40

This purported subgroup of $S_5$ of order $40$.

And note also that this only takes care of one case. How would you even attack it when $|G| = 10, 20, 30$?

I think $S_5$ is so concrete, you can attack directly.

I mean, even Groupprops is telling me that $S_5$ contains a subgroup of order $30$ that I've never heard of before.

Or geometrically in terms of symmetries.

@KajHansen 10 is easy, the 5-syow must be cyclic, it's index 2 so normal, the quotient has order 2 so is cyclic

@blue, we're avoiding Sylow though.

Yeah, 30 sounds right, I'll have to think.

@KajHansen and Cauchy?

UNLESS we take Cauchy for granted.

Let's see...how is that proven.

I have to think.

You can prove Cauchy by group actions by quite elementary means. It's an exercise you should have done, @Kaj.

I'm pretty sure I nailed it on our final @Ted. It's been a few months is all :P

Act on $G^p$ in an obvious way.

@KajHansen Dude, if the OP is reluctant to use Sylow, its his/her problem.

Don't suffer because of his/her ignorance.

@PedroTamaroff, not reluctant. The problem was assigned before they ever learned Sylow. And I like the challenge regardless.

mr @Pedro: Did you note that I sent you an email earlier?

@TedShifrin Yes. This week has been a bit messy, but next week I will have some quiet time and probably book for August the 3rd, is that fine?

Mike told me not to book on weekends.

I suggested you might want to look at prices ... it may be cheaper later in the week altogether.

You may be shocked at this point and decide to punt ...

airport-ewr.com/index.cfm

I'm trying to buy there. I don't think it's the correct place.

no, you should use Kayak to get pricings

I don't know what that is. =P

google it

It is in the middle of the night here 4:34 in the morning.

How did you get your BA->Newark tickets, @Pedro?

@TedShifrin I am doing BA --> JFK actually.

makes more sense

but you did that through an airline, not an airport

Aha.

@TedShifrin (!)

U$S 300. That's not bad, is it?

Well, on AirTran it might be cheaper than that ...

@TedShifrin The airport is Hartsfield Jacobson?

Hartsfield jackson, yeah

Hehe, that.

I read too quickly.

@TedShifrin US Airlines is 299, Southwest airlines is 269.

I don't care about some extra dollars if it is a better service.

They're all sort of the same, although a few others are worse. What about AirTran?

Yes, AirTran spitted out Southwest airlines for 269.

ok ...

Agh, this is not nice.

2 --> 5 is 299.

But 3 --> 6 is 370.

:confus:

what about 5->8 or something else?

That's what Mike and I were warning you about.

I am checking.

4-->7 is 411

>:(

ah, apparently it's cheaper staying over Saturday night. It used to be that way.

I'll still be totally jetlagged, but I suppose we can do it.

Let me see if I can buy it right now.

let me know times

WAIT NO.

That's just part of the flight.

I have to change planes. =(

oh, hell, how many stops? :(

It isn't such a long trip...?

Just one.

no, I always do non-stop

At Charlotte.

US Airways?

Yes.

yeah, they'll do that to you ...

At any rate, might be the reason it is cheaper.

In fact nonstops are about 800.

because you're so late doing this ...

and flights are all booked ...

WOW.

@TedShifrin Sorry.

Reminds me of my procrastination, this does :)

indeed, @Kaj ... At this rate, you might never find out if you get to meet @Pedro.

@TedShifrin Is $299 expensive? I don't care about stops. At any rate, I'll suffer. =P

haha, indeed

I just don't want to get ripped off.

Air travel has gotten expensive, with fuel prices up, and then summer is the worst

^Fascinating question

Do there exist? :D

I would bet money the answer is no.

But I'm no expert.

I would too. It looks like Qiaochu produced a proof.

Ah, Chebotarev density is what proves that the generic polynomial that is irreducible/$\Bbb Q$ is reducible mod every $p$.

@TedShifrin As per communication; I think we need something like whatsappñ

Huh? @Pedro

Do you know what "Whatsapp" is?

Nope.

Oh. But do you happen to have a "smartphone"?

@TedShifrin, I saw that one problem in your Algebra text. I remember having to put it down and admit defeat a while back.

Nope.

Oh. I will have to activate roaming or something like that.

Yeah, @Kaj, but generically it's the case. Totally stunning.

Woah, I didn't even see "generically"

That's fascinating

"Generically"?

Also, hi.

LOL @Andrew "also"

:P

How's your projective geometry coming, @Andrew?

Slowly. I took a bit off to spend with friends.

@TedShifrin What's the difference between a coach seat and a preference seat?

Working on it again tonight though.

Preference? What's that? @Pedro

I am choosing the seats.

US AIrways.

Some charge like 29 extra bucks.

They charge more for whatever that is. I do coach. You pay more for preferred or whatever ...

Maybe its becuse they are front.

You get a bit more legroom.

Airplanes have got very tight.

Email me the dates/times, @Pedro.

@TedShifrin 9:38 PM Aug 2nd.

Is that fine?4

Sure, I'll be dead :)

And you'll be starved.

And departure?

Travel insurance?

Do send me flight #, etc., so I can check up on progress.

OK.

@Ted does "generic polynomial" mean anything like points in general position?

I never do travel insurance. Unless you think you might cancel, it's a rip-off.

I won't cancel, no.

@Andrew, it means except for a Zariski-closed set of coefficients it's true

Zariski-closed means cut out by polynomials, yeah? So, excepting when there are polynomial relationships between the coefficients or something?

@TedShifrin I am about to buy for August the 2nd.

Confirm? @TedShifrin =)

And departing when?

August 5th.

Is that fine?

When?

Oh.

8:50 AM.

I haven't bought it yet.

If it is too early, I can change it.

Yikes. Remember that it's 1 1/2 hours from my house to the airport ...

And you have to be there at least an hour early.

You said the airport was far away.

Yes, that's why.

UGH.

DANG IT US AIRWAYS.

Even if we use the shuttle, you'd have to get up 5 AM.

I don't care much, I think. I just don't want to pester your days.

Is there nothing later?

I changed to 6th, says take off is at 5:45a...

LOL, what?

YAS.

Pilots must be crazy.

That's way worse.

It's all for business travel.

With the 8:50 time, we're going to have to get you to the airport shuttle at 5:30 AM.

Found an AirTran for 320 at 4pm.

That's nice.

2 --> 6.

That's a little less crazy.

UGH but it has a 3hr stay at Chicago.

Oh geez.

Chicago is nice though, right? =D

I'll be travelling light.

I'm ok with the early one on the 5th if you want to get outta here at that absurd hour.

F**k it, I'm OK with this one.

What time does it get you back to Newark? Might not be so good for your sister.

I'll take the train on my own.

That's fine.

That late? Be careful.

She's not coming back till August the 8th.

Trains don't run that late at night ...

I'd be arriving at 9.25 pm.

Really? =/

I didn't recall trains didn't work at night.

I'm not sure which train you're talking about. But it's worth checking before you do craziness.

Amtrak, IIRC.

No, sorry.

They don't run all that often ...

NJTransit.

Ah, that should be ok at that hour.

Cool. =)

Montclair is not a shady town. Dunno about Newark's airpot.

Newark is not all peaches and cream.

It might be good to run this by her and make sure you're not missing something crazy.

She's in Asia at the moment.

LOL, oh

Damn procrastinator.

Well, I'm not sure how much energy I'll have, but we'll get you to do math with @Kaj and others ...

@TedShifrin I need to make sure trains work at the time I'll be leaving and departing.

Or buses.

Right, that was my point.

The problem is most of the options either depart too early or arrive too early.

hi @Karl

Hey @Ted

@TedShifrin Ted, I'm super exahausted now.

LOL, I bet

Yeah, I had the combinatorics make up exam today.

I have a doctor's appointment at 8 am, so I'm going to bed anyhow.

I'm pretty sure I passed.

@TedShifrin Good luck.

Excellent.

But still, one more midterm tomorrow.

Then I can hibernate.

Oh, I heard back ... it's not anything super serious.

Ah, I'm glad.

Well, I'm sorry to add stress ... but I need to know what's going on before I leave on my trip :P

When are you leaving?

like 5 am Tuesday

My plan is to arrive on Aug 2nd midday and leave Aug 6th noon.

I will try to stick to that.

OK?

well, keep me posted. I can handle those if you work it out. Can't make it to the airport Aug 2nd before noon, though.

Oh. When are you coming back from your trip?

Was it the 1st?

about 1 am August 1st ...

Yeah, I'm gonna be pretty dead.

OK.

Maybe I can do 3rd. I will email you ASAP.

Bye.

Oy ... What an ordeal.

How've you been, @Karl?

@TedShifrin Fine, I just played some tennis with my uncle.

Nice ... Another tennis player :)

Indeed :)

I am pretty bad though

It's all relative. I expect Pedro will beat me 6-0 6-0.

heh