No accptances though. I am jealous @Brian.

Almost perfect timing.

oops, I missed one, make that 3 votes

@robjohn I think that my all-time highest point-getters were non-acceptances.

@BrianMScott yeah, that might be, but without acceptances, you are limited to 200.

So a capped day with acceptances beats a capped day without. :-)

True, and I got only a fraction of the nominal value, especially on one that went mildly viral.

I’m happy to reach 200 whether it’s capped or not!

I know that Epic is judged on points alone, whether acceptances or not. Is there anything that counts only non-acceptance/bounty points?

I don’t know: I’ve paid very little attention to the badges, though I was pleased (and a little surprised) when I got Epic. The only ones that I actually thought about before I got them were Fanatic and Copy Editor.

I was going for the 100-day straight (Fanatic) badge, and did not know that simply working on an answer was not enough to count. I worked on an answer for over a day and lost the continuity.

I wrote a program btw that will output data automatically for the coin problem

it won't scale for large n and c but it is good for low values

@WhatsInAName which coin problem?

@BrianMScott I'm half Epic

a problem Brian was advising on

@robjohn Ep, or ic?

@BrianMScott probably the ic

perhaps the pi

take my half out of the middle.

Middle halves cantor set, base 4 rep with 0s and 3s

That’s cherry-picking, so it must be cherry pi.

If pi is half epic, does that make tau epic?

You’ve lost me with that one, I’m afraid.

tau = 2pi <- Wikipedia

Really?

That’s one I’d not encountered before.

@BrianMScott yeah, there is a movement to replace $\pi$ with $\tau=2\pi$

@robjohn Silliness.

The video that really kicked off the so-called "movement"

@BrianMScott of course.

But speaking of food, I’ve a guest coming in a bit, so I should go make a few preparations. I’ll see you folks later.

@BrianMScott have a good time.

Thanks!

@t.b I gave your argument on sections of TS^2 to my students today —I seem to have convinced them :)

but she is talking about pie=$\pi e=8.53973422267357$

And that is confusing.

Her point that $e^{i\tau}=1$ misses the entire point of Euler's identity.

@WhatsInAName Now come on! It's Dick Palais's son Bob (first link gives the original intelligencer article). Maybe Vi (whom I like a lot) resurrected the movement...

Sorry -- right you are

@MarianoSuárezAlvarez Great! Glad to hear that :)

@tb Yes, that is not the seminal work in the $\tau$ movement.

the video was the thing that made its rounds all throughout FB and brought forth "awareness" among students

@WhatsInAName maybe recently...

I see.

$e^{i\tau}=1$ so what, $e^0=1$ -_- big woop

I am completely OK with Pi, personally XD

@robjohn Is she related to him?

Hmm... Hart = Hartl?

It's close, but not close enough

I dunno, youtube handles are often shortened names.

unless she changed her name out of embarrassment ;-)

She wrote her name as "Vi Hart"

OK, then. Whichever is objectively better, pi or tau, introducing a new constant can either fail or create a mess.

And then she wants a whole pie to be 1 tau? that doesn't even make sense :-)

@ymar She is his daughter. He's a retired professor from Stony Brook and currently associated with the momath.

@ymar I think most people view $\tau$ with mild amusement and put all the $\tau$ supporters in the same room with the alien abductees.

Perhaps only alien abductees support $\tau$. Could this be the first part of an alien invasion?

Ah, I have to go and take my dog to the park. I hope I am not abducted and start worshipping $\tau$.

@robjohn she has way better videos on her blog. Check out her math doodling

Goodbye I'm going too.

Bye, see you later.

@tb I will do that when I return.

No hurry. See you later

Guys, can you help me with a notation?

1.0649618067087347e-22 means 1.0649618067087347^-22, right?

no

It means $\cdots \times 10^{-22}$.

What syntax is this one?

LaTeX.

It's a variant of Scientific notation See the part on E-notation

Yep, i've found it. In the Ada, C++, FORTRAN, MATLAB, Perl, Java[3] and Python programming languages, 6.0221418E23 or 6.0221418e23 is equivalent to 6.0221418×10^23.

Thanks, guys.

@ZhenLin: What exactly do you want to avoid in your question on derived functors?

(or rather universal delta-functors)

Well, I'm trying to fill in a gap in my argument that the universal delta functor "must" be constructed via satellites.

I need some help identifying a pattern

@WhatsInAName Fire away.

@ZhenLin I'm sorry, if I'm thick, but I guess it wouldn't satisfy you to say: choose for each $A$ a short exact sequence $0 \to A \to I \to A'' \to 0$ and to put $T^1(A) = T(A'')/\operatorname{Im}T(I)$, show that this is independent of the choice of the short exact sequence and verify that you get an effaceable (hence universal) $\delta$-functor this way?

(and iterating this)

Well, if we assume that $T^1 (I) = 0$, then it follows quickly from the definition how $T^1$ and the connecting maps must be defined in general.

The difficulty is in justifying the assumption...

Let me try to be more clear: I define the functor $T^1$ as I said above and for this $T^1(I) = 0$ follows from its well-definedness, and then you can prove by induction that this functor is universal.

Yes, that works. But that's like knowing the answer beforehand and just proving that it's correct!

That's why I guessed that you wouldn't be satisfied by that :)

But sometimes math is this way. Once you know the answer it's clear that it is the answer but how on earth should one come up with it?

Perhaps the difficulty in answering the question has to do with the case where there aren't enough injectives. Because as soon as there are all this is forced...

You need considerably less than enough injectives... I think even Grothendieck says somewhere that he has yet to encounter the functor that you can't derive.

tau is not a bad idea, it's just a few centuries too late

$\frac{\tau}{2}r^2$

the primitive of $\tau \,r$

oh lordie. they look like practically the same symbol.

zzzz'

Can someone answer a quick question about Galois Extensions?

it depends on the question

Unsurprisingly

Can anyone help me identify a pattern? I think it involves combinatorics

Suppose I have $K=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$, and that $|\text{Aut}(K/\mathbb{Q})|=8, and that the three generators of $ \text{Aut}(K/\mathbb{Q})$ are all of order 2 in $\text{Aut}(K/\mathbb{Q}$. Then is it the case that $$\text{Aut}(K/\mathbb{Q})=\{1,\sigma_{1}\}\times\{1,\sigma_{2}\}\times\{1,\sigm‌​a_{ 3 }\}$$ where $\sigma_{2}(\sqrt{2})=-\sqrt{2}$ etc..

you get the idea.

weird bug

@MarianoSuárezAlvarez - Did you notice that the Math jax in the preview of asking question or answering the question doesn't work?

@WhatsInAName - What is your question?

@Victor: For answering the preview works for me.

@anon - it seems both doesn't work for me.

don't

@WhatsInAName - Maybe you should put in the main site so that all people could see it.

(just running the program for two different configurations here of n and c)

alright nvm

What I'm really trying to ask is: if I know that $\sigma_{1}$, $\sigma_{2}$ and $\sigma_{3}$ are all of order 2, and that the Galois group is of order 8, can I just write $$\text{Aut}(K/\mathbb{Q})=S_{2}\times S_{2}\times S_{2}$$ where $S_{2}$ is the symmetric group of order 2?

@DavidK - If the question is too advanced, please put it on the main site, i don't think anyone could spend time to explain it to you for over an hour on this.

@Victor Sorry. I didn't realize it could take so long to answer.

actually the question looks like it'd be easy for someone more familiar with group theory

@anon I feel like this is correct. It's just weird because in D&F they don't discuss this kind of representation until much later in the text from where this problem is.

@DavidK: Do the $\sigma$'s commute? If so, there is a canonical isomorphism from the RHS to the LHS $(a,b,c)\mapsto abc$ (the domain/codomain of the homomorphism have the same cardinality)

@anon Yes they do commute. I see. So yes, this is correct?

Yes. Do you see my reasoning?

@anon Sweet!

@anon Yes. I just hadn't thought of it quite like that.

@anon Thanks.

There's probably a more powerful way to look at the question that doesn't require the generators commute..

@anon Well, the way I defined the $\sigma$'s, commutativity is trivial. I just had to think about it.

General problem: Suppose $X$ is a generating set for a finite group $G$ (not necessarily abelian), and $$|G|= \prod_{a\in X}\operatorname{ord}_G(a).$$ Then is it true that $$G\cong \prod_{a\in X}\langle x \rangle?$$

@anon Ok, that's better. $\sum\neq\prod$ !

I think $S_3$ is already a counterexample. Might be wrong.

No, I think that works. Because any $2$-cycle together with any $3$-cycle should generate. And the order equation you give is satisfied. But $S_3$ is not a product.

Cool.

Word.

@DavidK - Curious about what grade of college you are in?

@Victor I'm a senior. Sort of. I'm in my 4th year of my undergrad, but I've delayed graduation to work with one of my professors and take some graduate courses. I'll be finishing my undergrad next spring and then (hopefully) heading off to grad school.

well S3 IS the "set-product" <(1 2)><(1 2 3)> (it's of the form HK, for two subgroups H and K), it's just not a direct product. i think the notation $\prod_{a \in X} \langle x \rangle$ is ambiguous

Hi

I think if you want to get picky then every group is a product in some interpretation of that word. But I think my meaning was clear. Relatively, at least :)

what i meant is it's not clear what anon meant. i understood you.

It is somewhat ambiguous. Some authors (I think Isaacs does this) have tried to push the big \times for the categorical product but I don't see that so often.

you mean the universal cartesian construction?

I just meant the direct product of groups as opposed to this "multiply the elements and see what you get" product.

I wonder about that interpretation. The order is going to matter, in that case.

So maybe that symbol isn't good after all.

and what i meant is "whatever you get where you have two canonically defined epis to the factors"

Same thing, yeah.

as you pointed out, S3 doesn't work because there's no epimorphism to the 3-cycle subgroup

Right. If it were a direct product of two groups, then it would have to be abelian.

i've often wondered: why is it in this world we live in, that it matters what you do first?

i feel as if it ought not to matter whether i put on my shoes first, or my socks first, as long as i get both on eventually, but...it doesn't work like that

and...given that it DOES matter, i find it utterly bizarre that addition in the natural numbers should be commutative.

It's hard to imagine things being otherwise. But there's no shortage of noncommutative objects, if that's what you're looking for.

(The notion of noncommutative natural numbers does remind me of some viXra papers...)

well the closest reason i have for "why" is: all "1's" are the same....although common sense leads one to believe otherwise: in our language, "this" is different from "that"

or: the free monoid/semigroup on one letter is necessarily commutative, but that's just because there's only one letter...add another and the nice-ness goes bye-bye

Hi @David

@RajeshD Hi!

there was a confusion...I got confused with David Wheeler and David Wallace, the former is on math.SE while the later is on SO, thats why i once asked David Wheeler about his profile on SO

which is why David Wallace asked that i change my name to Zebediah

lol

Hey, would this be the appropriate forum to ask about math reus?

definitely not

what is reus BTW..........If he want proper answers, he should ask proper questions....you can't be lazy and say 'reus' without giving its full form...am I right ?

@Rob

@RajeshD In some ways you are right, but as far as politeness you don't answer "definitely not" before you hear more about the question.

@RajeshD reus are Undergraduate Research programs

ok then i am not the appropriate person

to say "definitely not"

in this forum

nice removal

Sigh, Algebra. Anyone able to answer a quick automorphism question?

hahhahhahhahhha

NOT funny.

rudeness never is...

Rob, are you Skullpatrol? I am asking because I don't recognise the gravatar.

@Srivatsan Yup.

@IsaacSolomon I don't think it's a great place for it, only because it seems much more helpful to speak with someone who knows more about your situation and what they are talking about.

Hm, I am surprised how much we (i.e., I) depend on gravatars...

Me to...

If you're at UCLA, then you are surrounded by people who are qualified to talk about this sort of thing.

@Srivatsan This gravatar is my way of trying to "make peace with Asaf."

That said, I don't see any problem with trying here. But I wouldn't expect much. And I would definitely try other avenues.

@Rob Oh. I don't remember that you had any "conflicts" with him recently...

@DylanMoreland Yes. And chances are that the question would be closed here as too localised. I guess it depends on how the post is framed (and also on the moods of the readers ;))

Oh, I thought he was asking about chat.

My mistake.

The main site would be even worse, I think.

Anyway, my point is that Isaac is surrounded by good sources of information on this topic. If you have that in hand, then I don't yet see why you'd ask a bunch of yahoos on the internet.

Yahoooooo...

@Srivatsan I agree, whether it stays alive depends on who's around.

Hey guys

Yo

Hi Benjamin.

@DylanMoreland Have you seen this? math.stackexchange.com/questions/125795/…

It's an exercise from Dummit and Foote

I don't understand 20

I saw it. I started typing up an answer but the whole thing was so boring and I knew that Arturo would soldier on.

If you look at Arturo's answer, is the method described in question 20 saying that we can compute a polynomial satisfied by $1 - \sqrt{2}$ of degree 4 over $\Bbb{Q}$?

For example the characteristic polynomial of Arturo's matrix is $[(x-1)^2 - 2]^2$

Yes, this is what pops out.

But then how is it "efficient" when one can very easily compute a polynomial with $1 -\sqrt{2}$ as a root

It isn't, in this particular case.

sigh...

Or for example if we want to compute a polynomial with coefficients in $\Bbb{Q}$ with $\sqrt{3}$ as a root

I mean, the degree is obviously not optimal.

suppose we were stupid and did not know the minimal polynomial of this over $\Bbb{Q}$

With this method, if we wanted to do it with $1 + \sqrt[3]{2} + \sqrt[3]{4}$

But for the example given at the end of that question, with $\alpha = 1 + \sqrt[3]{2} + \sqrt[3]{4}$, it isn't so obvious how to get a polynomial. But this is a mindless way of doing it. It's probably good for a computer, for example.

What extension to do I need to go above $\Bbb{Q}$?

Adjoin $\sqrt[3]{2}$. This has degree $3$ over the rationals.

Ok yeah because then $\sqrt[3]{4}$ is in there

And it contains everything you need for that element to make sense. You've got a basis $\{1, \sqrt[3]{2}, \sqrt[3]{4}\}$ over $\mathbf Q$ and it should be easy to see what multiplication by $\alpha$ does to this.

I am computing that now

I believe the matrix you get is

$\left[ \begin{array}{ccc} 1 & 2 & 2\\ 1 & 1 & 3 \\ 1 & 1 & 1 \end{array}\right]$

Why not (2, 2, 1) for the last column? Maybe I'm groggier than I thought.

It's 2 not 3 sorry

I am the one that is groggy...

This is a duplicate of a question I asked before here

Just sayin

@DavidK We are talking about question 20 not 19b

I have a field theory exam coming up next week

Been computin' like a bunch of minimal polynomials

@DylanMoreland Someone told me that a proper field extension of $\Bbb{C}$ is $\Bbb{C}[x]$

@BenjaminLim Oh sorry. I came in in the middle.

But then $\Bbb{C}[x]$ is not a field I'm guessing he meant $\Bbb{C}(x)$

That is probably what was meant.

@DylanMoreland Our lecturer likes to ask us to name crazy examples in assignments

Like one time he asked to name a field extension of $\Bbb{Q}{(\sqrt{2})}$ of degree 1234567891

What does "name a field extension" mean?

Is it synonymous with find or give?

Sorry I should have said find

@Srivatsan Interesting exercise: Prove that if an integral domain $R$ contains a field $F$ and $[R:F]$ is finite, then $R$ is actually a field. Furthermore give an example to show that the result is false if the hypothesis that $R$ is an integral domain is dropped.

@IsaacSolomon Hi, if you check back on the transcript you'll find some helpful advise from Dylan.

Hey Rob. Yes, I'm just reading the transcript now.

Aha. Perhaps my question wasn't well-phrased. I wasn't planning on posting a question on the site about REU programs. I was just wondering if this chat room would be an appropriate place to see if anyone might have some experience with a particular REU program.

You could try robjohn

@DylanMoreland Hey can one prove that if $\operatorname{Hom}(M',-) \cong \operatorname{Hom}(M,-)$ then $M' \cong M$ (talking about modules now) without the Yoneda Lemma?

Anyhow, Dylon's advice is good. I'll see if I can contact people connected with the program directly.

Good evening, gents.

@IsaacSolomon hello

where are you now in the world?

Now is anyone able to answer a quick galois/automorphism question?

@DavidK I am less pro than you. Can I ask you a question?

@BenjaminLim Sure.

I don't understand something: $\Bbb{Q}$ is a perfect field. So according to the primitive element theorem, every finite extension $F$ of $\Bbb{Q}$ is $\Bbb{Q}(a)$ where $a \in F$

Does this mean then that $\Bbb{Q}(\sqrt{p_1}, \ldots \sqrt{p_n})$ is $\Bbb{Q}$ adjoined with one element?

the $p_i$ are distinct primes

@DavidK

@BenjaminLim What is the primitive element theorem? I have D&F in front of me if you want to refer to a pg #.

It is: Let $K$ be a separable finite extension of an infinite field $F$. Then $K = F(a)$ for some $a \in K$

@BenjaminLim I'm not really sure. That doesn't seem right though.

What does not seem right?

The statement of the theorem or what I said about $\Bbb{Q}$ and the roots of primes?

$\mathbb{Q}$ adjoined one element.

yes

But $\Bbb{Q}$ is perfect

so that the extension $\Bbb{Q}(\sqrt{p_1}, \ldots \sqrt{p_n})$ is separable

furthermore it is a finite extension of $\Bbb{Q}$ (degree is $2^n$)

So shouldn't the primitive element theorem apply?

Oh. Right... perfect $=\text{char}(\mathbb{Q})=0$. So yes. The above finite extension is simple.

That's amazing!

Now, since I am clearly not "more pro" than you, I have a question. @BenjaminLim

@DavidK I don't know $\textit{any}$ Galois Theory.

What about just automorphisms?

urggg if you talking about $F$ - algebra isomorphisms then ok

Actually, nevermind. I think I got it.

good :D

Now can you help me understand something?

@BenjaminLim Maybe.

It's about the primitive element theorem

We want to show that if $a,b \in K$ an extension of $F$ that is finite and separable

Then $F[a,b] = F[c]$ for some $c \in K$

Let $f,g$ be the minimal polynomials of $a$ and $b$ respectively over $F$.

Let $E$ be the splitting field of the polynomial $fg$ ($f$ times $g$)

@BenjaminLim Surely you mean $F$ adjoin $a$ and $b$ right?

Is that not what I said?

Well in my experience, $F[a,b]$ is the polynomial ring in the variables $a$ and $b$ with coefficients in $F$.

Now the claim is that if $a = a_1$, $a_2, \ldots a_n$ and $b= b_1$, $b_2\ldots b_n$ are the roots of $f$ and $g$ respectively

@DavidK $F[a,b]$ is the same thing as $F$ adjoined with $a$ and $b$.

Now back to the problem

Got it.

The claim is that $a + \alpha b \neq a_i + \alpha b_j$ for all $i$, for all $j \neq 1$ and all but a finite number of $\alpha \in F$

I don't understand this claim

@BenjaminLim Me neither...

The part on "for all but a finite number" is confusing me

@Rob The one that is deleted?

@robjohn May I post another answer?

In particular this one.

@Rob I will consider the title change, but without any means of real enforcement, most rules suggested in a post are viewed as guidelines anyway. Thus, adding emphasis by calling them rules vs guidelines is reasonable.

@Rob the only real enforcement a room owner has is calling in a moderator, and they will go with whatever they see as proper.

@robjohn Asaf did try to "enforce" his rules by creating "The Bin" for comments he disapproved of.

@Rob He could have enforced guidelines as well. With rules/guidelines, at least you have an idea what the room owner sees as inappropriate.

@Rob if the room owner behaves inappropriately, they can be flagged just as any user can be (as we have seen).

Indeed. Thank you for your attention.

@Rob: when you click on a user's gravatar, do you see the option to "hide posts"?

Yes.

@Rob Okay. Thanks, I wasn't sure if that was something that was an owner ability (hiding someone's posts to the room). It is obviously not.

I don't know how Asaf moved a comment to "The Bin". I don't see an option to do so.

Ah, I found it. :-)

@robjohn Please do not pursue this option further, it's not worth it in my opinion. As you said calling in a moderator, and letting them go with whatever they see as proper is a much more reasonable approach.

@robjohn BTW have you considered putting a "crown" on your mean square gravatar? ;-)

@Rob :-)