@DonLarynx If you want to do anything analytic, especially something in geometry, then knowing PDEs is pivotal. A lot of modern mathematics in the realm of, say, differential geometry can be aptly described as reducing everything to PDEs. So yes, PDEs as a general subject is a useful thin to know. Whether or not it is the best use of your time, or if you would be prepared for it, I am unable to say.

@AlexYoucis hey

@Grothendieck Hey man, long time no talk :)

yea

@AlexYoucis I lost your google profile

Like my gmail?

yea

it's just my name with a . between first and last

@AlexYoucis Just sent you a message on hangout

Cool :)

I got to bolt dude, but now we cna more easily get in touch!

yea sure

@AlexYoucis c ya :D

See ya!

but anway

for sylow p subgroups

if $P,P' \in Syl_p(G)$

it it necesarry for $P \cap P' = \{e\}$?

is it*

or can the intersection have more than just the identity

intersections of conjugate subgroups, in particular of sylows, are generally nontrivial

ok

evening all

Anybody know where I can find a good quality PDF of Euclid's Elements? Preferably one that's not a photocopy.

I may have a copy in my dropbox. 1 sec

@Marie here?

Can anyone add to Marie's response here on topological rings: math.stackexchange.com/questions/542067/…

dropbox.com/s/hqjpn0leuyxoemf/The_Elements_of_Euclid.pdf It's the google scanned version

@agent154 bookos.org/s/?q=euclid%27s+elements&t=0

Thanks

@EnjoysMath that sends me to citysex.com

weird. Just go to bookos.org and search

@EnjoysMath what about it?

@anon Maries response? Where does a proof that $R$ forms a topological ring with unions of ideals fail?

she already said where, did you not read what she said?

if {0} is open then all other singletons {x} are open, but they are not unions of ideals

Yeah, I did, but that's a contradiction of a property proved after you know something is a topo group under addition.

um, what?

it contradicts the claim that the unions of ideals as open sets makes R a topological ring

so that claim is false

I need to stop answering questions by newbies.... last 5 answers not a single upvote/accept.

Let $U$ be open in topo group $(G, +)$. Then $+^{-1} ( g + U) = \{(x,y) : x + y \in g + U \}$ How is that open?

@DanielRust :-(

Then again, some of them are just hints, so I guess I can let those slide :P

@anon see above

@anon I'm interested in what properties that $(R, \tau)$ fails to have that are used to prove that translation maps open sets to open sets

simple: $\tau$ fails to be closed under translation

note that $\{(x,y):x+y\in g+U\}$ is the union of $(a+U,b+U)$ over all pairs $a,b$ such that $a+b=g$

what I mean, what basic properties (more basic than translation of open sets), are required to proved translation of open sets that $R$ fails to have?

lol, this is pointless. I'm not making sense

translation invariance of open sets is an irreducibly basic concept

@anon irreducibly basic. I like.

copyrights

no, it had to be proved

what do you do with the pieces of a broken heaaART?

@anon Darn.

@EnjoysMath let me try to crack what it is you want. your definition of topological ring includes the axiom that $R\times R\to R:(x,y)\mapsto x+y$ is continuous, where $R\times R$ is given the product topology, and from this you're trying to show that for all $x\in R$, the map $a\mapsto x+a$ is continuous. am I right?

No, trying to show that $a + U$ is open for each open set $U$.

In a topological ring?

so that the map you speak of is an open map

that's what Marie said didn't she?

Showing $a+U$ is open for each open set $U$ and each $a$ is equivalent to showing $x\mapsto -a+x$ is continuous for each $a$.

(Since $a+U$ is the preimage of $U$ under the subtraction-by-$a$ map.) So what you said and what I said are the same thing.

What is your definition of a topological ring?

Okay, so it's that $+$ is continuous $R \times R \to R$ is the basic property you're saying?

yes

Well I think I've proved that it is indeed continuous. Let me sketch up the proof

wait wait

o_O

O_o

do you mean "I've proved in any topological ring, $+:R\times R\to R$ is continuous," or do you mean "I've proved in my ring, with topology given by the unions of ideals, addition $+:R\times R\to R$ is continuous"?

the first doesn't make sense, because it is part of the definition of topological ring

the latter

if $+:R\times R\to R$ is continuous then it necessarily follows that translations of open sets are open

however in your topology, {0} is open and its translates are not, hence your topology does not make $R$ a topological ring

Okay, an open set in $R$ is a union of ideals $I = \bigcup I_i$. And $+^{-1}(I) = \bigcup +^{-1}(I_i)$

Yeah I know so I want to see where my proof failed

your question doesn't have a single + sign in it

does it talk about addition?

I never wrote down that part of the proof

$+^{-1}(I_i) = \{ (x, y) \in R^2 s.t. x + y \in I_i \}$

and how do you write that as a union of ideals in RxR?

I conjecture this equals $\bigcup_{K + J \subset I_i} (K \times J)$

are K and J ideals?

where the union is over all ideals $K,J$ such that $K + J \subset I_i$

i.e. $(1,i-1)$ for any $i\in I$. hence your conjecture is false.

wait a min

in fact your union is just $I\times I$, that's not the inverse image of $I$ under $+$

$\bigcup_{K + J \subset I_i} (K\times J) \subset +^{-1}(I_i)$ clearly right?

yes

like I said the union is $I\times I$

since $K+J\subset I$ implies $K,J\subset I$ implies $K\times J\subset I\times I$

@anon Question.

assuming by $\subset$ you mean what I would write as $\subseteq$

go on pedro

pre-emptive hello @Charlie

@anon @EnjoysMath "hence your conjecture is false" ooooooooooooooo buuuuuurn!!!!!!!!!

Wait.

Okay, well $x + y \in I \implies Rx + Ry \subset RI = I \implies (x,y) \in Rx \times Ry \subset \bigcup (K \times J)$

@anon is that a miracle? You pinging me? :O

@anon, see my reverse inclusion holds, therefore $+$ is continuous

BOOYAH!!! BABY

@anon Nevermind, I was completely ignoring a definition.

@EnjoysMath $x+y\in I$ does not imply $Rx+Ry\subseteq RI$

takes away his booyah

what?

@Charlie heh

R(x + y) \subset I

@EnjoysMath did you read my example? let $i\in I$ and $I$ be a proper ideal. then $(1,i-1)$ is in the preimage of $I$ under $+$. but $R1+R(i-1)=R$ is not contained in $I$.

okay, Rx + Ry \subsetneq I, got it

There's the culprit then

I thought it looked right originally

@PedroTamaroff gcd and lcm can be defined in commutative rings with identity. And they are unique modulus being associates

@anon what fun things have you been doing?

Thx for helping me suss it out, kthxBai1

@anon Thanks

@Charlie got steak with a friend for his bday, saw capt philips, been looking at alg num thry, watching old cartoons, yada yada

@anon sounds exciting

y tu?

@anon Aw, le anon uses spanish.

But Charlie is brazilian! =D

@anon reading about pointe shoes manufacturer, watching Paquita, nothing very interesting

de colores?

@PedroTamaroff no one knows "e você", no one cares for my idiom :(

I do

:D

@Charlie I do.

I mean, I know "e você"

@PedroTamaroff good, if you didn't that would be sad

@leo do you speak portuguese? :)

@Charlie no

ainda

It's a beautiful language

:-)

I think so

:D

@anon

yes

The fact that a polynomial in $p\in R[x]$, $R$ a UFD is primitive i.e. gcd(coefs)=1 implies that we cannot write $p=r\cdot q$ where $r$ is an element in $R[x]$

This means that if we obtain a factorization into irreds in $R[x]$, we obtain one in $F[x]$.

For the irreds must be nonconstant.

So this will be a "genuine" factorization in $F[x]$

Here $F$ is ${\rm frac}\; R$.

@agent154 Well, note that you get $xy=1$.

This means $x,y$ are invertible in $\Bbb Z$.

What are the invertible elements of $\Bbb Z$ (i.e. wrt to multiplication?)

The only one should be $1$, since $\mathbb{Z}$ isn't a cyclic group...

What about $(-1)(-1)=1$?

@agent154 (I don't know why you're bringing up cyclic groups, though.)

Ah, yes...

Z is cyclic

I'm only bringing it up because I just recently had an assignment in that chapter of my book

so it's kinda fresh in my mind

@agent154 And as anon said, $(\Bbb Z,+,0)$ is cyclic.

OK, so then... if $x=y=1$, then $a=b$... and if $x=y=(-1)$, then $a=-b$ and $b=-a$...

@agent154 Yiss.

Ahh, ok. Thanks

@Charlie To aqui endo essa zuêra.

Hey guys, I am working on a problem: If $A$ is a compact operator (linear and bounded), is it possible that $\overline{\text{Range } A} = H$ where $H = l^2(Z_+)$

I found an example in a textbook

basically, if you let $A$ be the diagonal operator, $A (x_n) = \frac{(x_n)}{n}$

but I am not sure how this implies that the image is dense.

@TedShifrin Ha!

Who ha?

@TedShifrin Ah?

Silly boy!

trix are for kids

Happy almost Halloween @anon

@TedShifrin hi Ted, nice to see you again. Any insight/explanation on my problem?

Hello. Can anyone help me with the following problem:

Hi @masfenix ... What is $x_n$?

I'm trying to figure this out for some time now and I have showed what I have done.

I don't know how to proceed.

$(x_n) \in l^2(Z^+)$

sorry I have a type, the image under A, ie $A((x_n)) = x_n/n$

But what sequence is it?

specifically? could be any arbritrary sequence.

well it has to be squar esummable

Oh, I see. For any element you send the sequence $(x_n)$ to the sequence $(x_n/n)$.

yes. precisely, my question is how does this imply that $\overline{\text{Range }} A = l^2$, ie its dense

well I mean, first of all I am assuming this operator is compact.

Are you thinking that has dense image?

My functional analysis is rusty. Compact operators are in some sense the closure of finite-dimensional ones ...

Intuitively, we get $(1/n)$ as the limit of the image of $(1,1,\dots,1,0,\dots)$.

Where did @Pedro go?

@TedShifrin I'm here! Reading some algebra, to complement for Ireland and Rosen.

Ah, good, then. You haven't missed me :D

@TedShifrin hmm okay. I am not getting it but i'll think about it a bit more.

@masfenix: I suspect you get any element of $\ell^2$ as a limit of images of finite sequences in such a manner.

@anon This is a typo, right? "We may further assume that $p$ is not a unit in $R[x]$, i.e. ${\rm deg}\; p>0$."

Shouldn't it read "constant"?

@DanielRust I don't like proofs by contradiction when unnecessary.

For example, $2\in \Bbb Z[x]$ is not a unit, but has zero degree.

@Pedro too bad :P It's how I think.

@DanielRust Just giving my opinion.

I deleted. I read $F$ :)

I'm really confused about Mhenni's answer though. He just gave the problem statement as an answer...

I have seen too many people invent contradictions out of the blue, just because they have been shown proofs by contradiction one too many times, but the contradiction is either non existent, it is not a proof by contradiction, or is simply nonsense.

@TedShifrin ;)

I tell students to approach proofs by reasoning by contradiction when they have no other way to approach ... But that that isn't always what they should write as their final proof.

And yeah @PedroTamaroff, I know some people prefer constructive proofs. I just tend to start writing down definitions and see where it gets me.

@DanielRust Yeah, Mhenni is one strange lad.

does it really take abstract algebra to solve some of the exercises in ireleand and rosen? @PedroTamaroff

@user60887 Exercises, I don't know. They use it in their work.

Illustration: Trivial linear algebra exercise the first week. Prove that if $x\cdot y=0$ for all $x\in\Bbb R^n$, then $y=0$.

@TedShifrin Pick $x=e_1,\ldots,e_n$.

are you reading their elementary number theory book? some of their exercises were really tough.

You get $y_1=y_2=\cdots=y_n=0$. No contradiction! =D

Better: Pick $x=y$, but students get that after thinking about contradiction.

@user60887 It is called "A classical introduction to modern number theory." Dunno if they have a book called "Elementary Number Theory."

@TedShifrin Sum of squares is nice too.

I prefer coordinate-independent proofs.

oh yeah thats an advanced book. they do have a book for elementary number theory.

@TedShifrin Oh, you had another idea?

@DanielRust Le sigh

Votes to delete

@Pedro I'm confused with his train of thought. I've decided to ignore it.

@DanielRust Wise.

What just got deleted?

@TedShifrin Some spamish answer.

"[----] is a homework tutoring marketplace. If you're a student seeking smart solutions...."

That's the second time I've seen that today

you have enough rep to see the revision history @Ted

@anon You agree on the typo, yes?

probably

@anon Kay.

@anon what's the rep cap? I thought that was just a default option.

Hmm ... I gave you +1 for telling him he's an idiot, @Pedro :)

@TedShifrin Hehehe. He has a PhD!

@DanielRust "rep cap" refers to the max rep you can gain in a day (which is 200, not counting accepts or bounties you obtain after the cap is reach). the rep threshold for seeing deleted posts is 10k I believe.

@anon sorry should have said threshhold.

@TedShifrin What proof did you have in mind of $x\cdot y=0$ for every $x \implies y=0$?

That doesn't impress me, @Pedro. He just wrote more sloppy crap.

@TedShifrin I know. I was just saying that in an "holy crap, what?" voice.

Lots of mediocre Ph.D.

I think @Ted mentioned $y\cdot y=0\Rightarrow y=0$ by definition of inner product.

though not by definition of dot product (which one should establish is an inner product at some point)

true

@anon Was about to say that.

@TedShifrin Did you make them prove that $\lambda v=0$ and $\lambda\neq 0\implies v=0$?

And similarly $\lambda v=0$ and $v\neq 0\implies \lambda =0$.

That is in the list of things we prove immediately (using coordinates). But thereafter I prefer to eschew coordinates.

@TedShifrin Can it be proven without coordinates? (in that course)

it is by definition in a vector space, but one should establish $\Bbb R^n$ is a vector space at some point

which would of course use coordinates, because you're proving things about coordinate vectors

Well, we can use $1/\lambda$ for the former, dot product for the latter :)

Yeah, I tell them the properties in an exercise but do not stop to verify them. Toooo tedious.

Night night, all.

@anon Delete?

Night

This too.

Just because an answer is wrong doesn't mean it should be deleted imo. Only if it doesn't attempt to answer the question in the OP.

@DanielRust Usually, the answer are wrong or don't answer. In the first case, Mhenni makes no effort to correct it, or to carefully read what the corrections should be. Moreover, he claims they are correct.

> Why the upvote? For the plot, maybe?

snort

@anon just read that too haha

@DanielRust This, for example, Daniel.

@pedro I still think incorrect answers have their place, and it's made clear they are incorrect by downvotes/comments.

I am perplexed at this guys rep though...

bad behavior, unteachable poster, misleading erroneous answer, unnecessary drama - if there is no positive aspect at all I think an answer can be deleted. even incorrect answers can have positive qualities ("have their place").

Oh, my. I might as well cry for a while now.

btw @anon thanks for the favourite list you linked on your profile page... I've just devoured two brilliant questions.

mmhmm

math.stackexchange.com/questions/362446/… some great examples. I might try and add my own.

it's about time to prune my list

I have a bad habit of favouriting questions which I just want to know if they've had any recent activity, and not necessarily because I like the question.

Maybe there ought to be a "follow question" option then... So it doesnt' get marked as a favorite, but people can watch its activity?

@agent154 There is the "star question."

Isn't that favoriting though?

yeah

It is a bit of a strange system. Someone's probably made a meta post about it.

There could be a text link under the post saying "Follow" or "Watch" instead, which doesn't make the question look like a favorite when it's not.

Hey guys, how do I go about showing $Ax = y$ has a solution $x$ for every $y \in H$ if and only if $$ \sum \frac{1}{\lambda_n^2} |<y, \phi_n>|^2 < \infty$$ . What I know: $H$ is infinite-dimensional Hilbert space, $A$ is a bounded compact operator that is diagonizable. Ie, $\{ \phi_n \} $ is a orthonormal basis of $H$ consisting of eigenvectors of $A$ with corresponding eigenvalues $\{ \lambda_n \} $

For statistics purposes at the very lesast

least

Hey hey hey

so la da da di we like to party

dancing with Miley

@agent154 I suppose you then run the problem of having two different, similar systems with non-obvious differences.

doing whatever we want

I'm stuck on this -- If $a\mid b$ and $b>0$, then $a\leq b$... So I assume $b=ax$. Since $b>0$, $a>0$ and $x>0$ or $a<0$ and $x<0$. If $a,x<0$, there's nothing to prove.. But if $a,x>0$, then how do I prove that $a\leq b$?

@FernandoMartin WOP WOP WOP

@agent154 Note that $x\geqslant 1$.

Multiply by $a$.

I was considering that, but it looked like I'd need to break it up into like 6 cases... I'm clearly over thinking it

@FernandoMartin

@agent154 $a(x-1)\geq0$ because $a>0$ and $x\geq 1$ so $ax-a\geq 0\Rightarrow ax\geq a\Rightarrow b\geq a$.

@PedroTamaroff's proof is quicker

OK, it's super simple now that you mention that. $x\geq 1\Rightarrow x\cdot a\geq 1\cdot a\geq a$

Why would I need to multiply both? $\mathbb{Z}$ is commutative

oh sorry didn't see your edit

Assuming I dont' have the wrong term

Man.... I am really liking Number Theory, but I hate that I'm slow to pick up on small details... God help me if the prof asks for a proof tomorrow on my test that I can't come up with quickly.

It's all about definitions :)

I wish I had more time to study it and just do practice problems... but I'm doing Analysis and Linear Algebra 2 as well, which are both also proof heavy, and Analysis is kicking my ass

The questions in my book are out to lunch though. The chapter covers the material you need, but you need to have such a deep understanding of the concepts to put together the pieces to do the questions.

Like... "Prove that the Diophantine equation $ax+by+cz=e$ has a solution if and only if $(a,b,c)\mid e$". I know how to do this now after getting help, but it wasn't clear in the chapter. There didn't appear to be any examples to go by.

Question.. what is the convention for gcd$(x,0)$? Is it $0$?

where $x\neq0$

I think the convention is $\gcd(x,0)=x$

@KarlKronenfeld

@PedroTamaroff yo

@KarlKronenfeld I need to clear something in a proof.

ok

Precisely, about R a UFD implies R[x] a UFD.

Let F denote the field of fractions of R.

The authors proved the following first.

If p is red in F[x] then it is red in R[x]

More precisely, if p = AB in F[x] then p = ab in R[x] where a= r A and b = s B with r,s in F.

Then they prove that if p is primitive, i.e. gcd(coeffs)=1 then it is red in F[x] iff is it red in R[x]

In particular, this is true for monic polynomials.

@KarlKronenfeld So far so good?

@PedroTamaroff mmhm

@KarlKronenfeld OK. Now, onto the proof of R UFD then R[x] UFD.

Pick p in R[x]. Then we can write p = d p' where d is the gcd of the coeffs of p and p' is thus primitive.

Since R is a UFD, d factors uniquely into irreducibles, so we may assume that p' is primitive.

We may assume also that p is nonconstant, since R is a UFD.

Thus, start over with p primitive and deg p > 0.

Since F[x] is a UFD, we can factor p there uniquely into irreds.

And we know this factorization gives rise to one in R[x] with F-multiplies of the irreds in F[x]-

But p is primitive, so all this F-factors must be 1.

@KarlKronenfeld Yes?

@PedroTamaroff Right

@KarlKronenfeld OK, now I have written p = p_1 p_2 ... p_r with the p_i irred in F[x]

I have to prove they are irred in R[x]

we know this is true if they are primitive.

Just do a proof by contrapositive to show that they are all primitive when p is primitive.

@KarlKronenfeld I know that if p,q are primitive, then pq is primitive.

I have proven that.

This proof is easier. :)

DERP.

=)

Ah, I need the cone of shame.

Ice cream?

@KarlKronenfeld That would be too much.

@KarlKronenfeld Karl.

@PedroTamaroff uh huh

@KarlKronenfeld Today, this came up.

Note Robert's valid point, before I added "with the usual order."

Yes

Alex said "For me, a field is Archimedean with respect to some norm if and only if the integers are unbounded."

What is a field norm in this context?

an absolute value

@anon OK, and that gives rise to an ordering in $F$?

|x|=0 iff x=0, |xy|=|x||y|, |x+y|<=|x|+|y|

@anon Aha.

no, it does not give an ordering on F

(for instance, the complex numbers are not ordered)

@anon OK. How do we talk about Archimedianity?

I.e. what does it mean for a normed field to be Archimedean?

it is non-archimedean if |x+y|<=max{|x|,|y|} for all x,y and archimedean otherwise

for motivation, look up the "archimedean property"

@anon Right, and Alex said "integers are unbounded".

This means $\{|n|:n\in \Bbb Z\}$ is unbounded in $\Bbb R$?

if |-| is archimedean then |-| restricted to the prime subring is unbounded, yes

@anon It is not an iff?

and if it isn't archimedean, then |-| restricted to the prime subring is bounded

it is an iff

Ah, guessed so.

And what does bounded mean?

@Pedro: It means that the image of $\mathbb{Z}$ in your ring has valuation less than or equal to 1

Well, actually that it is bounded

but it is the case that they fall into the unit ball

for this equation (x^2 - 4)/(x^2 - 9) where the horizontal asymptote is the highest powers in numerator and denominator, so x^2/x^2 = 1 which makes horizontal asymptote 1. Why does that work? How come dividing the highest powers forces the lines to not cross that point on y axis

For instance, all finite fields are non-archimedean, since the only valuation you can define over them is the trivial one

The problem with algebra is they tell you to do something and it gives you the result but it doesn't explain why it works

@JohnMerlino that's the same as (1-4/x^2)/(1-9/x^2), which tends to (1-0)/(1-0) in the limit

the reason is that x^2-4 and x^2-9 get bigger and bigger and bigger, but their difference remains constant, so the ratio of their difference to one of them tends to zero, so their ratio tends to 1

thanks for explaining that, I understand now

and the moral of the story is ... ?

in classes like prealgebra or intermediate algebra or college algebra, there is a reasonable worry on the part of the teachers that if you try to explain why and how everything works you'll just lose and confuse and cognitively overload students and they end up learning even less than if they just memorized rote formulas and methods. sometimes this worry is wrong though, in that explanations, if understood, can make facts easier to remember.

@anon If I give you a ring and write $R^{\times}$; how do you read that?

group of units

@anon OK. =)

why?

I was going with "multiplicative" group of units, but multiplicative is rather redundant.

Also, I see some authors use U(R) instead of R^x

anyone know where I can learn to prove properties of the floor and ceiling functions?

But in any case, the former carries naturally a mult. group structure from the monoid (R, . ,1 )

@user60887 For example?

like floor(x+n)=floor(x)+n where n is a natural number. But I just want to learn how to get better at proving these.