@user48929 that's not a subgroup

"A sheaf theoretic approach to measure theory" pure gold

Also hello everyone!

@MatheinBoulomenos you dont know that for sure

Hi Demonark

@TedShifrin the "international version" of your book is less than $50, is that bootleg?

multivariable

@GFauxPas: Probably.

NEW NEW, SOFTCOVER INTERNATIONAL EDITION. DIFFERENT ISBN AND COVER IMAGE BUT CONTENTS ARE SAME AS US EDITION. EXCELLENT CUSTOMER SERVICE.

sketchy

ALL CAPS MEANS LEGIT

@user48929 but $(1,0)$ is not containedin the cyclic subgroup generated by a member of $\{(0,0),(0,1),(1,1)\}$

@GFauxPas there is an international verysion made by the same company that not avalible in the us thats hardcover and reasonably prices

thats actually a legit copy

Let $X \subset G$ be a subset. Define $\langle X \rangle = \cap_{X \subset H \subset G} H$ where we take the intersection of all subgrups of $H$ that contain $X$

YEah, @Faust, but the author gets no money from this. Just the bootleggers.

So not legit.

no not that one

there an international version of you book it has the same publisher as the us edition

it says for sale in canada uk etc only

@Daminark hi

"Take a subset and smash all the elements together until you get a group"

its still like 100$

I'll have to look more carefully, @Faust.

but it less than the us edition

Oh I've heard of this kind of thing happening

not the rip off indian international scam editions

McGraw-Hill I think has some "international editions" of stuff?

@user48929 $H$ ranges over all subgroups of $G$ that contain $X$

@Daminark imagine you just give a course and you just call it "measure theory". Then on the first lecture you start doing topos theory

mcgraw hill is evil

Yeah, now that you mention it, I think there's something on my statements like that, Demonark and @Faust. I'll look later.

ahlfors costs like 180$ lmao

i found a 15$ copy in a local bookstore lol

All publishers are evil at this point.

I got Ahlfors for like 50 on Amazon :)

Except Dover with its older things.

I got one book new for 18€ where the official price is 100€ (same edition)

tinyurl.com/ycmuxobx I'm looking at this but it seems sketchy and feels bootleggy

Ted

Well, I support Dover in spirit much as I usually don't buy them anyway because I'm cheap

Seems like Springer's prices are much better and they give some free PDFs

(Much better than the $150 nonsense)

if you have access to the pdfs through Springer, you can buy a MyCopy edition for 25€

They may also be evil but we take what we can get

@GFauxPas: The list a publisher for one, but not the others. I dunno. Suspicious.

Perhaps but in the presence of a pdf, I'll probably stick with the pdf. Also if I only need a book short run, I'll get it from the library. Otherwise... that money could be going toward tuition

Springer is also evil, but they have sales and it seems that the universities can buy e-book access or reasonable conditions (not sure, but at least there has been no protests on that afaik) and if you have electronic access, you can buy a copy for 25€ which seems fair

@Mathein re measure theory, that'd be a fun prank to pull but... keeping that up might be controversial :P

I once took a part time job at Pearson and the job description made it sound like I'd be helping them grade tests for machine learning or something exotic but it was just data entry so I quit

@Daminark in a similar spirit: offer a course "introduction to analysis.", then on day one you start with "Let $K$ be a field and $\Gamma$ a totally ordered abelian group. A valuaion on $K$ with values in $\Gamma$ is ..." and if someone complains you just say "nobody said it was about Archimedean analysis"

lel

@TedShifrin for some reason i can't find it online but i have it it look identical to the online picture but in the bottom left hand cover it says in black writing international edition and inside it says not its illegal to buy or sell it in the US

Lol I didn't even know anything about non-Archimedean stuff. Does that include p-adics?

@user48929 yeah

@Daminark yeah

tfw there was no good horror movies this year that you haven't watched so you have to watch M Night Shyamalan instead

sad violin

@TedShifrin seems its no longer a thing but here it is

amazon.ca/Multivariable-Mathematics-Theodore-Shifrin-2004-09-03/…

was it the movie about aliens who get killed by touching a drop of water and who decide to invade a certain blue planet

complete with prophecy book

Ahah, so it MIGHT not be a chinese bootleg

guess I'll buy it and cross my fingers

Hm, so let's say $D\ne 1$ is a square-free integer and $a\in\mathbb{Q}\setminus\{0\}$. Is it obvious that $x^4 - a^2D$ is irreducible?

@mercio i do not know of such a movie

you are very lucky my good sir

it seems like a kawaii anime

i could dig it

I guess $a = \frac{p}{q}$, then this has the same roots as $q^2x^4 - p^2D$

so basically, $a^2D$ is a non square rational ?

$f(S)$ should be called "the image of $S$ under $f$" rather than "the image of $f$ under $S$" who's with me

@Daminark multiply with denominator of $a^2$ then Eisenstein I think

the image of $S$ over $f$

the way I think of it is $f$ is like a beam of light and as it passes through a set you get the image set

the forward image of $S$ by $f$

so when $f$ shines through $S$, you're getting the image of $S$ under $f$.

whom do I send a letter to

I would like it if words made it more clear if we are taking images of elements or of subsets of the domain

The image of $S$ athwart $f$

mercio there actually is a seperate notation for that but it's not universal

It bugs me that "the image of $3$ by $f$" and "the image of $\{3;4;5\}$ by $f$" are both used

$f[S]$ for $\{ f(s): s \in S\}$

I have literally never seen that before

when $f$ rumplestiltskins at $S$ we get the triumphant penguins with a mirror dancing

i've seen it in uhh whats it called

some linear algebra book

@Mathein well, the issue is that a priori, $D$ could divide $p$ or $q$, so Eisenstein doesn't seem like it's automatic

and proofwiki uses it but it's not common

but it makes sense

there was a discussion about which notation would be best for $\{f(s): s \in S\}$ and all of them were obscure but we decided on $f[S]$

there's also $f^\to(S)$ iircc

Daminark maybe we should find a prime number $p$ congruent to $1$ mod $4$ for whichyour nonsquare rational $a^2D$ is not a square

with the notation suggesting "forward image"

then your polynomial would be irreducible mod $p$ I believe

and with that notation $\{x: f(x) \in S\}$ becomes $f^{\leftarrow}(S)$

For reference (since this may not be the best way of going about my grand goal), this is in the context of showing that if $\mathbb{Q}(\sqrt{a\sqrt{D}})$ is Galois, then $D=-1$. Well, technically $1$ is square free and that breaks it but we put that off to the side. Now I'm good as long as I know that the extension is degree 4, but I need to worry about the possibility of degree 2

a lattice always comes with a scalar product

Actually hmm

if not, maybe it's a subset of $\Bbb R^n$ and then you can just pick the euclidean scalar product

probably is implied when the lattice is defined?

If $\Bbb Q(\sqrt{a \sqrt D})$ is Galois, then $i$ should be in it, right ?

tbh I've never seen that phrase used in this context before

latice

oh, this kind of lattice

it's a poset of a vector space $V^2$?

I know that to be true as long as $\mathbb{Q}(\sqrt{a\sqrt{D}}) \ne \mathbb{Q}(\sqrt{D})$

en.wikipedia.org/wiki/Lattice_(order) <- that kind

en.wikipedia.org/wiki/Lattice_(group) <- not that kind

and your definition of dual is for the other one

isn't it just reversing the order relation ?

idk really I'm just trying to read the wikipedia article

oh this is a different kind of dual

I was thinking of dual of a vector space

if someone asks you something that you don't know the definition of, maybe you should ask him

what's sub

and what's the order ?

well if I'm reading WP correctly $(\mathbb N_0, \text{divisible by})^{\text{dual}}= (\mathbb N, \text{divides})$?

the inclusion ?

not sure where the inner product notation fits in

there is no inner product for order lattices

@user48929 used the notation $\langle x,y \rangle$

I bet this is just associating $n \in \Bbb N_0$ with $n\Bbb Z$ and saying that $n$ divides $m$ if and only if $m\Bbb Z$ is a subset of $n\Bbb Z$

or turn your paper upside down

wait,no, that doesnt work

o..o

lol

I guess we could try solving $a\sqrt{D} = (x+y\sqrt{D})^2 = x^2 + Dy^2 + 2xy\sqrt{D}$

Oh okay

That would mean $D = -(\frac{x}{y})^2$ which is shit

for free?

lol u got a troll

I downvoted it fo ru

Faust thanks for the pep talk

I feel more motivated now :)

Cars really was a bext movie