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12:00 AM
Well, I'm glad you told me ... and I'm glad you've taken some pressure off yourself to keep quiet.
 
We spent about two lectures computing integrals to show the Eichler-Selberg trace formula that relates the traces of Hecke operators to equivalence classes of positive-definite quadratic forms of certain discriminants.
Now we have an exercise to use this formula to compute some numbers of equivalence classes for certain discriminants.
But you can show that this numbers are class numbers of imaginary quadratic fields
 
Oh, surely the professor wants you to use the techniques that he'd developed, @Mathein.
I don't like giving full credit when students avoid what I want them to be practicing.
 
yeah, I'm just handing this in as an alternative solution, I have done the proof using the trace formula, too. I know the guy who makes and grades the exercises, he's a cool guy and I'm sure he finds this interesting
 
Oh, fine, then ... good to show off :P
 
sure thing :P
it's amazing who connected this stuff is
quadratic forms were already studied by Legendre and Gauss and now we're throwing modular forms and algebraic number theory at them
Hi @MikeMiller
 
12:07 AM
@Ted I don't know how many walls get knocked down. But the support is important.
Hi @Mathei
 
@MikeM: It's incremental and cumulative.
I look at it from a very long-range perspective, being ancient.
 
I would have this conversation elsewhere, but not here.
 
Fair enough.
 
I'm too grumpy for public.
 
I don't want to stomp on Fargle's exhilaration. :)
 
12:10 AM
what does "knock down walls" mean?
 
Try to lessen prejudices and discriminatory laws.
 
Ah, I see
 
@TedShifrin LOL
 
@MikeMiller do you remember the thing about Galois cohomology and real division algebras? Apparently, you can classify some non-associative algebras with non-abelian Galois cohomology (Probably just generalized octonion algebras or something like that) and the coefficient group is an exceptional algebraic group which is pretty crazy. I don't know the details, this is just what a prof said when I asked
 
I meant that sincerely, Fargle.
 
12:14 AM
Exhilaration is quickly giving way to exhaustion.
It is hot.
 
I guess...
 
As one of my best friends likes to say, Fargle, "Wait 'til you're my age!" :D
 
@MatheinBoulomenos huh. What is the group? Do you know why?
@Ted You're saying you only got hotter as you aged?
2
 
ROFL ... I'd best keep quiet.
 
@MikeMiller no, the prof didn't seem to have much time when I asked. But surely this can be found somewhere. I'll see what I can find when I have time
 
12:19 AM
I miss Balarka
 
I do too. Did he get lost on the plane going home?
 
He's not headed home yet. He should be at TIFR now
 
oh, I didn't know when the month started.
Remember when he used to annoy you so much, Mike? :D
(and often me too)
 
Yes. I remark that to him sometimes.
 
lol, now I annoy him. The cycle continues.
(and both of you :D)
 
12:27 AM
Different annoy.
 
LOL ...
I am very proud of how Balarka and others here have matured.
Yes, Fargle, you've annoyed me for years, but I tolerate you :D
 
@MikeMiller I suppose that's fair. I don't remember much in this chat beyond the past year. In fairness, I've also been at least 18 for my whole time here.
@TedShifrin I'll take it. :^)
 
I really wonder why amazon offered this new €50 springer book for €6.50
 
It's like a whole other side to my professor existence here, watching people grow up and learn huge amounts of math. Alessandro, Balarka, Meow, you ... It makes me proud.
Even Mike counts ... I knew him long before graduate school started. I don't take much credit for anything with him, though.
 
@MatheinBoulomenos Really?
 
12:31 AM
 
It's not used or remaindered or something?
 
A few minutes after I bought it, it was back at €51,99
 
A computer glitch? You'd better check your credit card record on line.
 
I already got the order confirmation with that price
 
Well, good for you.
 
12:33 AM
but yeah, better safe than sorry
 
There was definitely a glitch.
 
I am surprised Benson co-authored that
 
The real play would be publishing another version of the book with author line
"R. Keith Dennis
Benson Farb"
 
Didn't Keith Dennis used to be a topologist too?
I've met him, I think.
 
oh no, did I buy a topology book?
 
12:38 AM
ROFL
no wonder they sold it to you cheap
 
@MikeMiller I looked at his page, and he has some papers like "FI-modules over Noetherian rings"
 
Could you please help me a little q about mutually independent? math.stackexchange.com/q/2830527/390226
 
Bye @Ted @MikeMiller
 
see ya later, @Mathein.
@Niing: You already got a good answer. The computation of $E[X_i]$ has nothing to do with any of the other random variables — independent or not. It's just additivity of expectation.
 
But for computing each $E[X_i]$, why the probability of $t_i$ is always $p$? Wouldn't it change by those $t_{i'}, i' < i$ already been performed?
 
12:51 AM
no, @Niing. The random variable knows nothing about anything that happens previously or subsequently. There's no conditioning here.
 
@MatheinBoulomenos Yeah, so he can study homology of mapping class groups and symmetric groups
 
@TedShifrin So when I perform $n$ trials, which are not mutually independent, then the probability of success of each trial are all the same, i.e. $p$?
 
I still don't understand how/why Springer gave Mathein such a humongous discount. :)
@Niing, yes, you just don't know the probability of joint success.
 
@TedShifrin Wow... thank you, that is completely cool... I need some time to digest it...
 
Expectation being additive is so powerful, @Niing.
 
1:01 AM
@TedShifrin I was thinking about that for the probability of joint success is not the same as the product of each single event, then the probabilities of some of them of success might be changed...
 
That's right, @Niing. But that's not what expectation is doing.
You don't want to do a giant inclusion/exclusion computation when you don't even know what the joint probabilities are.
 
@TedShifrin I'm thinking about that since it's a series of $n$ Bernoulli trials, in timeline there may be some events have happened before the right trial is to perform, so why its probability is the same as nothing happened before it?
 
But with Bernoulli trials they are independent. How the coin flips on the $n$th time has nothing to do with what's happened before.
 
@TedShifrin Wait a minute... so Bernoulli trials are independent then why the question says it's not (mutually) independent?
 
Oh sorry, @Niing. I forgot my definitions.
I am not an expert.
 
1:10 AM
No worries, I really appreciate your help, I just want to solve this...
 
Well, as I said, the answer that's there is correct.
 
I trust the answer in my book, but I just don't know what's wrong in my reasoning...
 
I guess I was thinking of binomial distributions, as the case of independent Bernoullis.
I haven't seen your reasoning get to anything quantitative.
You can't do a calculation with worrying about dependence, as you don't know joint probabilities.
 
I just learned about conditional probability, so in my current understanding if a series of trials are not mutually independent, then for some $i$-th trial, $i\ge2$, in the series, the probability of $i$-th trial would not be $p$. So what's wrong in this statement?
 
No, that's not right.
Dependence/independence has to do only with joint probabilities.
 
1:20 AM
So why conditional probability is out of consideration here? why the probability of success of $i$-th trial is not $p(i\text{-th trial success}\ |\ \text{the outcome(s) of trials performed before})$ as a conditional probability?
 
That's not the right formula. The formula is $P(X) = P(X|Y)P(Y)$.
And with dependence, you don't know the conditional probability unless you have further information.
 
1:33 AM
 
Don't know why you posted that. That's what I've been saying.
 
If Bernoulli trials are mutually independent then the success of any given trial is $p$. So without the condition of mutually independent, the probability is still $p$?
 
I don't think independence should be mentioned. See Wiki. It's just a yes/no $p/1-p$ thing.
 
hi @TedShifrin :)
 
I'm just leaving, whoever you are, Jalapeno.
 
1:42 AM
Oo :(
ok no worries - have a great night!
 
Dinnertime here.
 
ah, gotcha - enjoy your dinner!
 
@vzn I read Gowers article. Its quite interesting in several side aspects as well. In a sense it describes that in areas like "combinatorics" one needs to use "the way of chemistry" to digest large bodies of knowledges. Since that way ("I have been trying to counter the suggestion that the subject of combinatorics has very little structure and consists of nothing but a large number of problems.
@vzn While the structure is less obvious than it is in many other subjects, it is there in the form of somewhat vague general statements that allow proofs to be condensed in the mind, and therefore more easily memorized and more easily transmitted to others.)" is a very good way to describe what is going on inside chemistry as well.
 
2:00 AM
@Daminark hi
 
Hello!
 
you should see this
yesterday, by Leaky Nun
@MatheinBoulomenos @loch what the actual
 
With tensor products... What the
 
what
 
@Daminark just look at the theorem that comes afterward
 
2:10 AM
This is some level 7 stuff we got here
 
That really is a slick proof holy crap
 
lol
 
@MikeMiller not that kind of hot. Ted Shiffrin's body goes up 1000 degrees in temperature every day. By the end of his life, he will literally become as hot as a star.
@Daminark I believe they just used a nuclear bomb to pound in a nail.
 
2:42 AM
So it seems
 
 
1 hour later…
4:02 AM
@AlessandroCodenotti do you know any good differential algebra texts?
and hi
 
I don't even know what differential algebra is, sorry
 
yeah it's a niche thing
apparently ive been unknowingly tinkering with it
 
hi @AlessandroCodenotti @loch
 
@LeakyNun you ever do any differential algebra stuff?
 
no
 
4:15 AM
darn
 
I'm ca-tching u-p with mah-SEHLF (and the heat goes on!)
What did I miss?
 
@BalarkaSen I learned that I need to learn differential algebra. Know any good sources?
 
4:31 AM
Yo @BalarkaSen! How've you been? Also hey everyone!
 
I'm good my m8
How is it on your end
 
Everything's alright, thanks!
 
vzn
@Rudi_Birnbaum glad you enjoyed it, think its a classic, re combinatorics/ gowers/ thm proving/ "empirical math" etc another classic of his cited/ more thoughts here, he has remarkable ideas on AI + math vzn1.wordpress.com/2013/05/03/…
 
4:52 AM
mind=blown
"What is an exact sequence of sets?!"
oh man vakil is terrific
 
Mayer Vietoris my dude
Cech yo shit
 
@BalarkaSen btw I don't know if you would understand, but
yesterday, by Leaky Nun
@MatheinBoulomenos @loch what the actual
 
I saw but I don't want to read
 
ok
 
@BalarkaSen I'm not a drowning man! I'm not a burning building! I'm a tumbler!
 
4:57 AM
FIRE CANNOT HURT A MAN
(not the government man)
 
David "wacky boi" Byrne strikes again
 
Byrne is great. You should listen to American Utopia if you haven't already (did I say this exact sentence to you already? In any case I'll say it again (have I said that I have suggested that album to you already? ... ))
Honestly you suggesting me "Remain In Light" was a life-changer :P
I listen to that album on a weekly basis
 
Too groovy for this planet
 
also yeah I heard AU, it was pretty alright and I'll have to again
 
5:05 AM
Ah cool
 
@Fargle apparently that's supposed to be pure gold
 
:|
 
I facepalmed some twenty seconds later reading that message
 
Lmaooooooooo
 
The 119-th element of the periodic table is "Fu" and is dedicated to Daminark for trolling us
It stands for Fuck you
 
5:09 AM
I've always hoped for one that's just "Oh"
The abbreviated form of "Oh"
 
@BalarkaSen FYI that'd be Fy, not Fu
 
But that's not l33t enough
 
Tru
 
It's a Balarka!
 
Hi @Ted!
 
5:15 AM
Mike and I were mentioning earlier than we missed you :P
rehi Fargle (who I thought would be asleep). And hi, Demonark.
 
I'm hono(u)red
 
Rehi @Ted. I should go to bed soon, but I'm waiting on the proverbial jury.
 
You were, you mean
 
Jury?
 
hi @Loch
 
5:17 AM
Hey Ted!
 
(Unless you're stiiiil waiting)
 
Was supposed to be a play on the phrase "the jury's still out on that one"
 
I'm acquainted with the phrase ...
or even the sentence.
 
What are sentences if not complete phrases?
 
phrases are typically incomplete clauses ... fragments of sentences ... like "a turn of phrase"
 
5:21 AM
Are they also totally bounded phrases?
 
unless they're open-ended, Demonark
 
Someone get Daminark into a dungeon
 
I've been saying that for two years, a Balarka
 
He's a danger to the public
 
Ah just a year and a half
 
5:22 AM
"But who's counting?"
 
Probably the combinatorialists
combinatorists?
combinsgfgkhhbhts?
 
You were right the first time.
 
Combinatoricationalizers
2
 
throws everyone in the dungeon and jettisons the key
 
Lol
combobobodalgharaghtakamminarronnkonnbronntonnerronntuonnthunntrovarrhounawnskaw‌​ntoohoohoordenenthurnuk
as Joyce would say
 
5:26 AM
OK, that thing I said about missing a Balarka. Erase it.
 
:(
@TedShifrin Silliness aside, I have been learning about stratified stuff a lot
 
Oh yeah?
 
I have been trying to find the right way to define Nash blowup intrinsically after our discussion a week ago.
By which I mean, take a topological space $X$ such that $\Sigma \subset X$ is a closed subset with a nice neighborhood around it such that $X -\Sigma =: M$ is a (dense in $X$, say) open manifold. $E$ be a vector bundle on $M$. I think there's a way to "blowup $X$ along $\Sigma$" to get a larger topological space $Y$ that contains an embedded copy of $M$ with a vector bundle $\tilde{E}$ over $Y$ which restricts to $E$ over $M$.
 
In that generality, I highly doubt it.
 
Right, I should have been specific. I think there are conditions on $\Sigma \subset X$ and $E$ that allows this.
if $\Sigma$ has a standard cone neighborhood like in Whitney stratified spaces, then this should be possible
 
5:37 AM
Well, surely it's gonna depend on what the bundle $E$ is.
Whitney is all about tangent bundles.
 
One replaces the cone points with the link, and the fibers are defined by "pulling along the strands of the bundle".
@TedShifrin There might be some regularity condition required. Since $E$ is a v.b. on an open manifold, one needs to define a notion of properness of the bundle (like, it's nice enough at the endspace)
Trivialization outside of a compact set does it but that's too restrictive
 
But if $E$ is totally unrelated to the geometry of $\Sigma\subset X$?
 
Nash blowup doesn't take in the information about the geometry of the higher stratums and how it contains the lower stratums in it's frontier. Having a Whitney stratification imposes a condition on the Nash blowup.
 
Maybe then it's easy, actually. Maybe I need $E$ crazier than what's going on with Whitney strata.
BTW, plural is strata.
 
Namely, that "tangent bundle of the lower stratums are contained in the frontier bundle you get by extending via Nash blowup"
Hah good point
I need to get used to using that plural
 
5:42 AM
Basic second declension Latin.
 
The stratified category is very technical
 
Yup.
 
I'm getting lost trying to read like two pages of any references I have on them
 
Although at a certain level, it all makes basic sense. I've forgotten most of it.
I'm going to go to bed, Balarka, but you can tell me more later ...
 
It has a classy feel of the Polish point-set topology school, tbh
Good night! I will!
 
5:46 AM
G'night.
 
The last conversation really helped, so looking forward to more
Sleep well
 
See you Ted!
 
hi
 
good night
 
5:49 AM
Yo Meow
 
6:34 AM
@AlessandroCodenotti Most of modern topology is rooted in that principle.
I think it's a style that's worth getting used to.
I'm reading Goresky-MacPherson's (one of the) magnum opus on stratified spaces, and basically you're going to die if you want to read the formal stuff, then the motivation OR the motivation, then the formal stuff
It's supposed to be an inseparable mix of the two
 
Those guys collaborated with Deligne to lay a completely algebraic groundwork for their theory as well
Perverse sheaves, etc etc
 
By the way I started reading about homology (simplicial and singular) so I'll have a lot of questions
 
Very cool
I'll be happy to talk to you about stuff
 
Oh, how was your interview?
 
6:40 AM
Bizarre
 
That's unexpected
 
They asked me about my visit to TIFR and reading courses/workshops I took part in, then asked me a very simple calculus question which I was able to answer in a few minutes, and told me the interview was over
I don't know what that means
It's supposed to be usually, on average, half an hour long. Mine was barely ten.
 
Ah, I see, that is indeed weird
Did you have any chance during the interview or the application process to show that your level is "a bit" higher than the average high school student?
 
Well, I wrote in my application form about my visit to TIFR and the reading courses and the workshops I attended
That's why they asked
So we talked about that for a bit
The short length of the interview either means (1) I did good, they immediately thought I was worth being a student of this university, and hence asking further questions were pointless (2) I did horrendous, they immediately thought I had fat chance, and hence asking further questions were pointless
 
I'd be very surprised if it were the latter. When will you know for sure?
 
6:47 AM
In a week probably
 
I see, I don't think you should be too worried about it
And hi @Dami
 
What do vector fields in $\Bbb R^3$, that are gradients of a scalar function in $\Bbb R^3$ look like? Is there a way to develop an intuition for that?
 
7:28 AM
For any positive integer n  prove by induction that :
$$ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{n+1\sqrt{n}}<2$$
the author  said   it  is  sufficient to prove  that
$$ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{(n+1)\sqrt{n}}<2-\frac{2}{\sqrt{n+1}}$$
why ?
where this $\frac{2}{\sqrt{n+1}}$  term come from?
 
8:05 AM
If you use Mathematica sometimes, please fill this survey: community.wolfram.com/groups/-/m/t/1362353
 
 
2 hours later…
10:10 AM
Does anyone know any good resources on the continuum of exotic 4-spaces?
I've been trying to find out if there exists a parametrization of it
 
10:30 AM
Yes, the original proof by Taubes exhibits an exotic R^4, let's call it X, and a number R so that every $B(0,r) \subset X$ for $r > R$ are pairwise non-doffeomorphic.
The uncountable parameter being the radius, then.
This does not cover all exotic R^4s. I would be surprised to see any way of making the whole family of them 'tractable' in any sense.
There is at least a 'universal exotic R^4' in which all other exotic fellas are open subsets, but your won't get them by something so easy as the family of balls of radius r.
 
10:44 AM
Wow, that's crazy awesome, thanks!
 
@Rodolvertice You might like section 3 of this
 
11:45 AM
Thank you, I will read that
 
12:03 PM
wb pal @BalarkaSen
 
 
1 hour later…
1:11 PM
This diagram represents $x^2=mx+c$. For $L_1$, discriminant $m^2+4c$ is positive while for $L_2$ it is negative. I guess discriminant is $0$ iff line '$L_3$' is tangent to graph of $x^2$. Am I right?
 
@Silent Seems right. Zero discriminant for a quadratic implies a double root, which in this case can only occur with a tangent line
 
@Fargle you mean single root?
 
By "double root" I mean something like $(x-1)^2$--there's only one root, but it's of multiplicity two, hence double.
I've never stopped to consider whether that's standard notation though
 
oh! thank you very much
 
you can also take it as 'how many terms in the taylor series vanish at that point'
 
1:21 PM
ok, i am going to try that
 
e.g. $f(x)$ has a double root $\implies f(x)=\frac12 f''(x_0)(x-x_0)^2+\cdots$
(equivalently, it means that $f(x)$ and $f'(x)$ both have a zero at $x=x_0$)
 
That was really helpful, thank a ton
 
Can I please get some help with number 9?
8 I mean...
 
1:44 PM
Oh I get it we can just some sum from $k=1$ to $n$ over both sides
 
@FuzzyPixelz 0, 4, 10, 18, 28, 40, 54, 70, 88, 108, 130,
maybe
 
Then after some simplifications the answer will be $(n-1)n$
Which is wrong apparently
 
@FuzzyPixelz Don't you need the number 2 as a multiplying factor somewhere?
No I am wrong.
a(n) = n*(n+3)
according to the OEIS
 
Huh
 
Am I wrong?
 
1:48 PM
But the first term clearly doesn't match
 
shift the index one step
and it matches
 
Perfect, how could we deduce is the 'traditional' way?
 
I don't know.
 
@MatsGranvik In this example they just made a conjecture and proved it by induction...
 
@FuzzyPixelz What if you add $+1$ to each $n$ in each index and each $n$ in the last equation I wrote, repeatedly, and see if the resulting sequence has a pattern?
$a_{n+1+1}=2(n+1)+a_{n+1}$
But I don't know. This is the first time I am trying to solve a recurrence. You might want to ask someone with more experience.
 
2:09 PM
Why is separability so important in functional analysis
 
Substitute $a_{n+1}=2n+a_n$ so that it becomes:
$a_{n+1+1}=2(n+1)+2n+a_n$
and so on...
 
2:22 PM
@quallenjäger Separable spaces are generally more well behaved
 
In which sense?
Because of the cardinality of the space and thus I can better find a metric or norm structure on the space?
 
For example the weak topology on an infinite dimensional Banach space $B$ is never metrizable, but if $B^*$ is separable then there is norm on $B$ inducing the weak topology on all bounded sets. There are also some pathological spaces where a sequence converges weakly iff it converges strongly, this can't happen in spaces with a separable dual (note that having a separable dual is stronger than just being separable)
 
Good point, thanks..
Is separability also an useful concept in ODE theory?
 
I know nothing about ODEs (or PDEs), sorry
 
Hi @AlessandroCodenotti
 
2:30 PM
Ok, thanks.
 
Hi @Mathei
 
Hi @Mathei
 
Hi @quallenjäger
 
hi @MatheinBoulomenos
 
Hi @LeakyNun
 
2:33 PM
@MatheinBoulomenos is there a connection between the picture of functions sending each prime ideal $\mathfrak p$ to an element of $A/\mathfrak p$, and the picture of functions sending each prime ideal $\mathfrak p$ to an element of $A_{\mathfrak p}$?
well, I guess the former isn't even a presheaf
but it's an important example of thinking about a ring as a ring of functions
 
I don't know what you mean. You send a prime ideal to an element of $A/\mathfrak{p}$? which element in $A/\mathfrak{p}$ do you have in mind?
 
of course when you say function you mean, pick an element $a \in A$ and map $\mathfrak{p}$ to $a$ mod $\mathfrak{p}$
 
right
 
and not any arbitrary set theoretic function
 
oh okay
 
2:37 PM
@loch is finally here!
is that picture worth anything?
 
the second picture? idk lol
 
of sending $\mathfrak p$ to $a + \mathfrak p \in A/\mathfrak p$
 
Elements of $S^{-1}A$ are partically defined functions on $\mathrm{Spec}(A)$, they are only defined on the "subset" $\mathrm{Spec}(S^{-1}A)$, but in general that subset is neither closed nor open. For $A_{\mathfrak{p}}$, there's also the germs of functions interpretation
where $S$ is any multiplicatively closed subset
 
I'm not talking about that
I'm talking about treating $a \in A$ as a function sending each prime ideal $\mathfrak p$ to $a + \mathfrak p \in A/\mathfrak p$
 
oh - then sure

in the classical world, so let's say $A=k[x]$ (but you can replace this with any fin gen. $k$-algebra - reduced if you want)

then picking an element $a\in k[x]$ (a polynomial), the map you're talking about is the map $\mathbb{A}^1_k \rightarrow \mathbb{A}^1_k$ that this polynomial is supposed to be in the classical case
i.e. the element $x+1$ maps the prime ideal $(x)$ to $1$
(i.e. sends 0 to 1)
etc.
 
2:42 PM
can I not recover the classical picture with the Spec way instead?
 
sure - im just saying what's happening when we think like normal people

Let $A$ be a $k-$algebra. Then picking an element $a\in A$ is the same as giving a $k-$algebra hom from $k[x] \rightarrow A$, where you map $x\mapsto a$.

This induces a map $\operatorname{Spec}(A) \rightarrow \mathbb{A}^1_k$
which if you translate back to classical terms is what we want
 
is there nothing else we can do treating $a \in A$ as a function sending each prime ideal $\mathfrak p$ to $a + \mathfrak p \in A/\mathfrak p$?
 
actually now that i wrote that im not sure if that answers your question
 
I mean, in "my view", i.e. treating $a \in A$ as a function sending each prime ideal $\mathfrak p$ to $a + \mathfrak p \in A/\mathfrak p$, we can recover the classical picture of a polynomial in n variables being a function $k^n \to k$
my question is, is there more we can do with my view?
 
you can now think of $2$ as some kind of a function on the primes of $\mathbb{Z}$ taking values in different fields ($\mathbb{F}_3, \mathbb{F}_5, \ldots $) :)

this is in vakil somewhere early on - but right now off the top of my head im trying to think what's a nicer way to say this
 
2:52 PM
that's just an example of my view
but I'm asking whether my view has any significance
 
but you're right that that's how you're supposed to view the elements of ring as functions
 
instead of the Spec view?
 
not "instead"
what do you have in mind when you say the "Spec" view - if i ask you 'how do you view your ring as functions?'
 
So you're taking the family of maps $A \to \prod_{\mathfrak{p}} A/\mathfrak{p}$ and then interpret that product as a function of sets.
You can do the same thing with other families of maps, i.e. $A \to \prod_{\mathfrak{p}} A_{\mathfrak{p}}$ and $\prod_{\mathfrak{p}} A_{\mathfrak{p}} \to \prod_{\mathfrak{p}} \kappa(\mathfrak{p})$ and $\prod_{\mathfrak{p}} A/{\mathfrak{p}} \to \prod_{\mathfrak{p}} \kappa(\mathfrak{p})$
In general you have commutative diagram with $A$, $A/\mathfrak{p}$, $A_\mathfrak{p}$ and $\kappa(\mathfrak{p})$ and I suppose you can give interpretations to all those objects
 
rip latex
@loch the Spec view is sending each prime ideal $\mathfrak p$ to an element in $A_{\mathfrak p}$
 
2:56 PM
??
 
the structure sheaf
every element $a \in A$ corresponds to the function sending $\mathfrak p$ to $\frac a1 \in A_{\mathfrak p}$
is the Spec view
 
you should think of $A_{\mathfrak{p}}$ is some small neighbourhood/germ around $\mathfrak{p}$

so - i mean- of course you can think of your element $a$ as a function which maps a prime ideal (point) to some germ about your point

but that's not really how people view them
 
I thought that's how the structure sheaf is defined
the sheaf of locally regular functions
 

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