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00:00
That is, the assumption I started with doesn't work.
@AkivaWeinberger Right, that's why you work characteristic 0 for fields. At least on integral domains for rings.
This always works for characteristic 0?
(And integral domain means the product of two nonzero constants is nonzero, right?)
I guess the real question is, is it possible to have a polynomial have zeroes on the entire ring and not algebraically be the zero polynomial
(for $\Bbb Z_2$, such a polynomial is $x^2+x$)
And I guess the answer is no.
Yes to the second. Hmm, are polynomial ring on char 0 fields integral domain? I think so
I think it is. Induct on the number of variables.
00:03
Isn't the polynomial ring over an integral domain an integral domain?
$R[t_1,t_2,t_3,\dots]=R[t_1][t_2][t_3]\dots$, right?
So, we just need to show that if $R$ is an integral domain, then $R[t_1]$ is. And you can just look at the term of highest degree.
You're right. I am sleep deprived.
That's not the question anyway.
As for nonzero polynomials not having zeroes everywhere, I guess the problem (for integral domains) is that it would have to have a factor of $(x-k)$ for all integers $k$
(This breaks if you break either the integral domain part or the characteristic zero part)
@AkivaWeinberger Not on algebraically closed fields, anyway.
@BalarkaSen Why do you need algebraic closedness?
Oh, so you could add 1.
Yeah, I think that's not necessary
'cause of the infinitely many $x-k$ factors I mentioned
00:06
It's not necessary, for sure.
So, consequently, $\forall x,f(x)g(x)=0$ would then be the same as $fg=0$; $(\forall x,f(x)=0)\lor(\forall x,g(x)=0)$ would be the same as $f=0\lor g=0$. And the equivalence of $fg=0$ and $f=0\lor g=0$ comes from the integral domain-iness of the polynomial ring.
Right, that you can't have $f(x) = 0$ on $R$ but $f \neq 0$ is just from being an integral domain. Just factor out roots one after another (which you can do, 'cuz, integral domain)
On an infinite integral domain.
Ah, right, so it's not a characteristic 0 thing, it's an infinite ring thing
Ya
Whatever man, stick to $\Bbb C$
Only ring ever :D
00:15
@BalarkaSen might need some commutativity, you can have more roots than degree otherwise I think
@Alessandro Every ring is commutative, because every ring is C or R (thanks, Mike)
Ah, wait, integral domains are assumed to be commutative usually?
And C and R are both, as you know, commutative.
@Alessandro I don't think so, but who knows?
My dear friend, the matrix ring, would like a word with you
Strange friends you have. Stick with a good company.
00:19
lol
Soon we'll be arguing that all spaces are metrizable and separable
(Just like the analysis people)
Every space is a manifold
And yes, every manifold is Hausdorff and second countable and paracompact - like they should be
Nah, I have some good friends which are nonmanifolds. That was a joke.
Cayley–Hamilton only makes sense with commutative rings, anyway
Ok, I'm off to bed, bye everyone
G'night
I have a couple of days before exams and my sleep schedule has fubar
Yay
Hi, Ted
00:26
Hi DogAteMy
Do you know if it's possible for a nonzero polynomial in an infinite noncommutative ring to equal zero everywhere?
Is the pooch ok?
Yeah, she's fine now :)
@Alessandro My book defines Hilbert spaces as being separable
Mentions in a note that some people drop the condition to have non-separable Hilbert spaces, but nah
@BalarkaSen You do realize you were just teaching me about the Zariski topology
00:27
And hey @Ted
I have no idea !
Hilbert spaces are called separable if they're separable!
Hi Demonark.
Yeah, normally that's what I'd have thought, but I guess this book didn't care about non-separable Hilbert spaces so it just said everything is. ('-')\
Sorta like Artin assuming all rings have 1.
I think Keerthi does that too
If Eric was around now we could've asked him to confirm, but yeah
Of course. Rings are commutative and have 1
00:35
Krijn is my mathematical brother
aka "homeboy"
Brochacho.
I'm afraid matrices are too important to dismiss.
We're not dismissing them as unimportant. Just dismissing them as a ring :)
Next you will say all groups are abelian.
00:39
All groups are locally compact abelian groups!
(Disclaimer: I only know what that word means because the last chapter in my analysis book from first quarter was about Fourier analysis on such groups)
And all Riemannian manifolds are flat, all spaces simply connected.
All closed forms are exact
And we assume all fields to be of char 0
Well, we've just trivialized PhD qualifying exams.
No, just joking about the rings.
I always assume they are Dedekind domains
00:42
At least we're gonna do well on them when we get around
I can't work without some kind of primes around
Dedekind domains have too much structure.
Composite primes?
On the other hand I prefer locally Dedekind
so you have any 1-dimensional affine algebra in there I guess
aka curves
00:46
Actually better thing to do on phd exams if you're stuck
Just leave this link
@Krijn Are local rings which are also Dedekind the same thing as DVR?
@BalarkaSen Yeah
Hmm, wait
If it's not a field
Oh wait I am confused. Dedekind means Noetherian and localization at each max. ideal is a DVR, right?
That's already what I want.
00:49
@BalarkaSen There are like, three definitions that people use
I vaguely remember the number-theoretic one, that every ideal factors into prime ideals.
If it's equivalent to what I said above that means Dedekind domains are the right notion of curves in the algebra world
It's equialent
Got it.
Or another definition would be integrally closed, noetherian and krull dimension one
I.e prime ideal is maximal ideal
Except 0 ofc
Oh, great, so these are Krull dimension 1. Curves.
How do you prove that? Hm
00:53
@BalarkaSen But really this one is the one I intuitively use
If I take the first thing I said as defn actually that's obvious. DVR's are dimension 1, and you can pass to localization.
@Krijn I think the interaction between the geometry and the number theory is very interesting here. I don't claim to understand it though.
@BalarkaSen It is, which is why I'm trying to find out more about it
There's this weird world of number theory and function fields and varieties and stuff that I don't understand anything of
Mysterious objects
And one of those mysterious objects is that counterexample that I've been trying to find all day
How does the proof of equivalence between the number theoretic definition and the other ones (any one you prefer - both of the other are geometric) go? Do you remember it off the top of your head?
00:59
@BalarkaSen No not out of the top of my head
I'll look it up, to see if its long
@BalarkaSen math.ru.nl/~keune/CFT/cft_02.pdf First three pages
@Daminark Lol I think this thing just proved P != NP. "The proof is trivial! Just view the problem as a sesquilinear ultrafilter whose elements are simplicial complexity classes." Too bad it lacks rigor (and correctness)
hi @Semiclassical
What do you mean it lacks correctness??!
01:01
I swear I meant to have a "probably" in there somewhere
Daminark's hurt and heartbroken by that insult you made to thisproofistrivial
I think even a trivial proof for P vs NP wouldn't really be trivial at this point
:'( How could you @MickLH?
@BalarkaSen Well! It just Fermat's last theorem'd us how are you so calm about this!
@Krijn Ahh simple enough
Fun because that's just a little algebraic manipulation you're doing, nothing geometric.
01:05
Afterwards comes the fun stuff
Which also explains where the term ideal comes from
You can see that these ideals are a substitution for numbers, they are ideal numbers in our domain, and prime ideals are the ideal primes in this domain.
Although ideal numbers was a different term invented byKummer
It works better in Dutch :(
Sure, I understand that you use the prime factorization of ideals instead of prime factorization for these chaps
I think Kummer proved a weaker version of FLT (for regular primes) using these
@BalarkaSen And some class field theory with that
Interesting.
There's loads of interesting stuff in there :)
Of course, if you are working with a PID, all these prime ideals are just (p) where p is a prime
01:10
But for other rings of integers of $K$ over $\mathbb Q$ there's a thing that measures how much of a PID you aren't
Ideally you spend most of your time working in PIDs
Don't take "measure" too seriously here
Class number something something
@BalarkaSen Which is the number theoretical equivalent to the Picard something something
Is that the order of the ideal class group?
01:11
@BalarkaSen Yes!
So the Picard group of $Spec(O_K)$ is the ideal class group of $O_K$
Fractional ideals are projective submodules of $\mathcal{O}_K$, am I right?
Ehh, don't know
Just submodules of $\mathcal{O}_K$ such that there is an integer with which you can multiply s. t. you end up in $\mathcal{O}_K$
Ah, wikipedia has the answer
A (nonzero) fractional ideal is invertible if, and only if, it is projective as an R-module.
What does invertible mean? That you can tensor with another fractional ideal and get the trivial ideal?
So yeah, for dedekind omains this is the same
Trivial meaning (1)
01:18
Oh apparently we have found another defenition for Dedekind domains
@Krijn Ok, great. By Serre-Swan these are exactly line bundles on Spec(O_K)
an integral domain is a Dedekind domain if, and only if, every non-zero fractional ideal is invertible
So it makes sense that it'd be the Picard group (which is the group of line bundles upto tensor product I think)
fun. so all this algebra means studying line bundles on curves. :P pretty cool
If the domain is PID the ideal class group becomes trivial; so I guess under this dictionary PID's are curves of trivial 1st homology.
Oh, I wonder if the ideal class group of O_K has anything to do with the 1st Galois cohomology of K
@BalarkaSen I have never looked into Galois cohomology
I should, one day
Me neither, really.
01:24
Wikipedia is interesting: The current theory of Galois cohomology came together around 1950, when it was realised that the Galois cohomology of ideal class groups in algebraic number theory was one way to formulate class field theory
Actually, I take that back. Galois cohomology is not the right thing to look at.
If fundamental group correspond to absolute Galois group, what will 1st homology correspond to? Hm
I guess the abelianization
Ok, google-fu gives this. That's what you want. Ideal class group of K is canonically isom to Gal(E/K)
E is the maximal abelian unramified (whatever that last word means) extension of K
Yeah, HCF is nice
Unramified in the sense of spec
Ah, so the covering map Spec E --> Spec K is unramified?
Although using prime ideals its much more clear, I think
Look at the pdf I send you for the proof, just a small section ahead
I'm sorry, proof of what?
01:30
The different definitions of Dedekind domain being equivalent
Ah, alright.
$E$ unram. over $K$ means that no prime of $K$ ramifies in $E$
Makes sense. I have vaguely heard of that story
Now, this is impossible for $\mathbb Q$
But then again, $\mathbb Q$ is PID
Cool, so that theorem about the isomorphism agrees.
01:33
A better example is probably $\mathbb Q(\sqrt{-5})$
I don't know it's class number or whatever unramifies over it
Class number is 2
And I believe HCF is $\mathbb{Q}(i, \sqrt{5})$
Yeah probably, any other extension would rammify I guess
Interesting.
There's a really interesting theory for imaginary quadratic fields to find the HCF
I can prove Q(sqrt(-5)) is not principle (look at 2 x 3 = (1 + sqrt(5)) x (1 - sqrt(5)), so (2, 1 + sqrt(-5)) should work) but not that it's class no. 2
@Krijn Mhm?
01:37
@BalarkaSen Yeah, you use elliptic curves associated to such a field , and adjoin the $j$-invariant
Ohh. Right.
I have heard of it but I know nothing about this. How interesting
@BalarkaSen This is a quite a lot of calculation to do right now
Don't bother, I believe you
There are formulas (and a lot of research goes into these formulas)
Also, unramified of a field also means over the infinite primes
I.e. real fields stay real
Well, it's almost morning now so I should get some sleep. That was a nice conversation. There's just too many cool things to learn.
01:41
Yeah I'm off to bed as well
Sleep tight
Night!
Bye @Balarka
01:55
Quick question about quarternions and complex numbers. I can represent a quarternion as 4 numbers. Some computer languages have a built in complex number type. It might be faster to represent a quarternion as 2 complex numbers. Can I do that?
@StevenStewart-Gallus Severely doubt this will be fruitful. Complex numbers are not hardware accelerated.
So even if it were a good fit (It doesn't obviously seem like one to me either.), it wouldn't have any fundamental benefit
Why didn't the apple try to enter the vegetable-only store?

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