well I guess it's kind of dumb, but you can't do it if $f = 1$ right?

Oh sorry

I meant an element $f\in p$

oh ok

ok well, broadly, the answer should be "yes" since otherwise $p = (f)$ so it would be principal

exactly

cool cool

Ok, now we also would like $f\notin (g)$.

What do you think about that?

isn't it the same thing? if you always landed $f \in (g)$ no matter what $f$ you initially chose, then $p = (g)$?

Well not quite

ok lemme think

(thanks for all this effort, btw)

hmm do you... just continue finding another $h$ that would work that $g$ doesn't divide and you do something like infinite descent?

I remembered the slogan "to contain is to divide" and I think it might apply here

so you couldn't keep dividing forever

that's why I suggested the idea

hmm that might need this to be a euclidean domain or something

@ShaVuklia I think this is a constant sheaf, so it's flabby (and this works because irreducible schemes have a generic point)

In particular a global section is the same as a section on any open

That's a good idea BigSocks

@Astyx whaaat hehe that's cool

if g|f then deg g < deg f (if the degrees are equal, then (f) = (g) contradicting how we picked g)

right so $f \in (g) \Rightarrow g \vert f \Rightarrow deg(g) < deg(f)$

but we picked $g$ such that $g \notin (f)$ so $f \nmid g$

Not quite

now you want to replace f by g and start over

yeah I kept staring at that last bit

yeah exactly, we pick $h \notin (g)$

but we also want $g \notin (h)$

if $g \in (h)$, then $h \vert (g)$, so $deg(h) < deg(g)$

Right

you can only keep building $... < deg(h) < deg(g) < deg(f)$ so long

So you look at the degree of successive elements and you get a strictly decreasing sequence of positive integers.

Exactly

neat, very neat

So eventually this process will not work any more, ie we'll find f and g such that $g\notin (f)$ and $f\notin (g)$

Now we want stronger than that, we want no common irreducible factors

right, so we can think about $(f,g)$

this is where we're going to use the fact that p is prime

write f= hu and g = hv such that u and v have no common irreducible factors

since $p$ prime, either $h$ or $u$ is in $p$, and either $h$ or $v$ is in $p$

Right, let's consider the case where h is not in p first

what can you say then

$u \in p$

and?

oh, also $v \in p$

$uv, u + v$

So now we have u and v in p, and by hypothesis they have no common irreducible factors

So we win, right?

@Astyx I don't think it's a sheaf in the first place, actually

but I solved it I believe

what why

using affine opens and the universal property of $\Omega^1_{\mathcal O_X(U)}$

@Astyx aha

It depends what your definitions are @Sha. You can always consider it a sheaf on a space

@BigSocks Cool cool. So now we need to deal with the case where h is in p

then $f,g \in (h)$ contradicting our descent thing?

Once again we're going to argue with the degree

Kinda

@Astyx Well, I quoted my definition

but sure

You have deg h< deg f and deg h < deg g (why?)

So you can start over by taking f = h and use the degree descent argument again

@Astyx yeah this is what I was thinking

wondering what happens if $u, v$ are constants...

they technically share no irreducible factors right?

@ShaVuklia My point is that $\Omega^1_K(X)$ can be seen as an $\mathcal O_X$-module, and you can define it as a sheaf on $(X, \mathcal O_X)$. Since K(X) is a field, localization doesn't do anything so it's constant (ie the same on every stalk). Since X is irreducible it has a generic point so all open set intersect, so in fact every section on a (nonempty) open U $\Omega^1_X(U)$ is iso to global sections (this is the same as asking the restrictions $\Omega^1_X(U)\to \Omega^1_X(V)$ to be ...

... surjective. such a sheaf is called flabby or flasque)

damn, nice cliffhanger

@BigSocks if u is constant and nonzero, then $g\in(h) = (f)$

right

@Thorgott The sequel next wednesday

Don't miss it!

So hold on @Astyx

Return of the flabby Sheaf.

we now know about all the prime ideals of $k[x,y]$

why?

but I'm still wondering about the prime ideals of $k[x,y]/(f)$ for irreducible $f$

wait

then the only prime ideals are $(0)$ and the maximal ones huh...

we don't know everything about prime ideals of k[x,y]

w h a t

We've shown they are either principal or contain (u,v) for some u,v with no irreducible factors

We haven't shown those are of the form (x-a, y-b)

well we know they look like $(0), (f), (h,g)$ for $f$ irreducible and for $h,g$ with no common irreducible factors

@Astyx r i g h t

Let's do that

ok

To do that we're finally going to use that k is algebraically closed

It involves a small trick

We want to use the euclidean algorithm in k[x,y], but we can't

it's not a euclidean domain?

Right

Instead, we're going to notice that $k[x,y]\subset k(x)[y]$

And in k(x)[y] we have a euclidean algorithm

Since k(x) is a field

bc k[y] is a euclidiean domain

Yes, for any field k

(still not using alg. closed)

Ok, so in this ring, we can look at the ideal generated by f and g

We've assumed f and g did not have common irreducible factors, using the euclidean algorithm, we can find some element of k(x) in the ideal generated by f and g in k(x)[y]

Say I have a space $X$ and $A,B\subset X$. I think there is a LES $...\rightarrow H^k(X,A)\rightarrow H^k(X,B)\rightarrow H^k(A,A\cap B)\rightarrow...$. 1. Am I going crazy? 2. If not, what's the conceptual explanation of this?

What are the maps in that LES?

@BigSocks do you agree ?

Question: How are eigenvalues calculated "in the wild"?.................I could imagine calculating determinants to be very computationally intense. Do you attempt to row reduce until the matrix is upper triangular and then take the diagonal?

@Astyx I don't see this so easily. what if instead it might be in $k(y)$ or something?

@Astyx I don't really know

I mean, I can define them by a construction, but they're convoluted

@Astyx Right, I think I'm still lacking flexibility in thinking about (certain) notions in terms of sheaves automatically. Thanks for the explanation, that's a nice pov

glad to help

@Thorgott I'm afraid I can't help then

@dc3rd that's possible, but I think usually it's done numerically

@BigSocks k(x)[y] is a euclidean domain, in particular it's principal

So (f) + (g) is a prime ideal (h)

@Thorgott as in just take the determinant of the characteristic polynomial as is? so $det(A - \lambda I)$?

Let's assume h non constant (this means that $h\notin k(x)$ in this context - constant elements are elements of k(x))

no, that's something you almost surely never do

taking determinants is very ineffective computationally

Then, up to multiplying by a polynomial in x, we have that h is an element of k[x,y]

with degree in y>=1

and even then, you'd have to solve for the Eigenvalues as roots of a polynomial, for degrees >4, this is in general not feasible to do algorithmically

@Astyx Wait, but when we're considering stalks, we're localising $\mathcal O_X(X)$ right, not $K(X)$?

@Thorgott, can definitely see that....I'm here messing up a damn $3 \times 3$ so I could only imagine what could happen in larger realms

@ShaVuklia We're looking at $\Omega^1_{K(X)}\otimes \mathcal O_{X,x}$

I think you usually also wouldn't row reduce

QR decomposition is preferred, I think

@Astyx Oh, hm, nvm I'm afraid

Gaussian elimination/LU decomposition is twice as fast, but not well-conditioned iirc

I guess at some point what you said will pop up/make sense

I think I'm lacking some small construction, but it's fine

You shouldn't be, all I'm saying is that $K(X)\otimes \mathcal O_{X,x} = K(X)$

Because everything is invertible in K(X)

Ye, that's fair

but I don't know why we're considering that tensor product in the first place

That's how localization of a module works

I meant

You localize the ring of scalars

I don't know why we're localising that way

Thing is

in my reader they wrote that

if you have a (pre)sheaf $F$ on $(X,\mathcal O_X)$

Then $F_x=\mathcal O_X(X)_x$

So I don't see yet where $K(X)$ comes in

Wait, you define is as the constant sheaf?

So $U\mapsto K(X)$?

Yep, unless $U=\emptyset$

Ah ok, now your tensor product makes sense to me

rather $U\mapsto \Omega_{K(X)}^1 = \Omega^1\otimes K(X)$

(note that this works only because there is a generic point, otherwise you don't get a sheaf but a presheaf)

You mean $\Omega^1_{\mathcal O_X(U)}\otimes K(X)$?

Yeah, something like that

hm ok, it sounds fair, but I won't say that I fully grasp it yet

but thanks anyways

Neither do I

x'D

@TedShifrin isn't $f$ simply a sum of quadratic polynomials in each coordinate?

\o @robjohn

with positive ($k$) lead coefficient?

@user85795 hello!

o/

How are you doing?

sometimes we all forget how utterly insane everything we do is, to other people. A girl I know from high school asked me about statistics/probability for some very basic statistics course for, idk, psychologists. Explaining the calculation for binomial coefficients alone is a rough task lol

Not counting pascals triangle here :D

before some smartass here comes with that example

@Astyx hey, mb, class started- currently in class

@BigSocks no worries, I'm stumbling on my proof

virtual, but yeah, they give us little things to prove and I was focusing on that

class is more important, I'll ping you when I figure stuff out

you mean with $k[x,y]$?

oh ok, no worries. thanks again for all the instructive questions, man

yeah, math education has its own faculty @user2103480

@TedShifrin: sorry, I thought you were still in the room. Reading too far back in the log.

@ShaVuklia where are you reading this stuff?

@user85795 let's not talk about that hahaha not a big fan of the way math is taught in schools

And many educators are poorly educated themselves

@Bigsocks Ok I got it

nice nice

k[x,y] is a UFD, and we get f = u/v h ie vf = uh, so h divides vf. Take i an irreducible factor of h with degree in Y >0. Then i divides vf. obviously i doesn't divide v (because it's in k[x]) so it divides f. with the same argument, it divides g, contradicting that f and g have no common irreducible factors

This means that h is and element of k(x), which we can take in k[x]. We have fu + gv = h for some u,v in k[x,y].

So $h\in p$

@Thorgott I want to prove that for every field $\Bbb F$, and any finite CW complex $X$, the euler characteristic for homology with coefficients in $\Bbb F$ is equal to the usual one. Any smarter way than making a case distinctions between characteristics and, in the char $0$ case, working with the prime field, i.e. $\Bbb Q$, to handle tensor products in the universal coefficient theorem

Since p is prime, one irreducible factor of h is in p. Since k is alg closed this is an (X-a) for some a

A similar argument gives $(Y-b)\in p$ for some b

In particular $(X-a, Y-b)\subset p$, but since (X-a, Y-b) is maximal (which we haven't proved yet), $(X-a, Y-b) = p$

fields are torsion-free, so universal coefficents directly give you $H_n(X;\mathbb{F})=H_n(X;\mathbb{Z})\otimes\mathbb{F}$

I don't see what you need a case distinction for?

Ok I expected the first part

How's that immediately follow though

$H_n(X;\mathbb{Z})$ is a f.g. abelian group, so you can write as $\mathbb{Z}^r\oplus T$ for some natural $r$ and $T$ a torsion group. Tensoring this with $\mathbb{F}$ yields $\mathbb{F}^r$.

ah well I should've just looked up "torsion-free". man I'm so inept at algebra

@robjohn I said there was an elementary proof, but it's a good practice problem for compactness/max value thm.

clearly fields are torsion-free

@Thorgott Tensor product with which coefficients?

over Z

@TedShifrin okay, just checking ;-)

no wait, I'm actually full of shit

I would've also used that the homology group is finitely generated but what happens when T contains a $\Bbb Z / p\Bbb Z$ and our field is the same

yeah exactly

you were right

this needs a case distinction, let me think for a sec

at least there clearly exists a field that satisfies this, by the proof

ok, I was being overly silly

just look at the cellular chain complex

the cellular chain complex is a finite complex of finite rank free abelian groups, so the alternating sum of their ranks (which is the alternating sum of the number of cells) is the Euler characteristic

Ah yeah and I can just tensor this with the field

but the cellular chain complex for coefficients in $\mathbb{F}$ is just obtained by tensoring the cellular chain complex for $\mathbb{Z}$ coefficients with $\mathbb{F}$

right, and then the dimensions are just the ranks, no torsion to worry about

@Thorgott They have stuff for that.

over the counter even

thanks for the effort, that helped

@Thor is being very multifarious these days!

he masters all the elements

👏🏼👏🏼👏🏼

am I?

feels like all I'm doing is topology

and poorly so

@Thorgott not as poorly as I am, evidently

But it's also my own fault. Logic and probability also wrecked me for some time, but I put in the grit until it started being natural. For topology I haven't done that enough so yeh exam will be fun

Thor:

@user2103480 pure algebra

Let $(X,\mathcal O_X)$ be an algebraic variety. If we sheafify a presheaf of $\mathcal O_X$-modules, does that become a sheaf of $\mathcal O_X$-modules?

Or maybe more specifically, we know that $\Omega^1_X$ is a sheaf (as it's a sheafification of the presheaf $U\mapsto\Omega^1_{\mathcal O_X(U)}$, where $\Omega^1_{\mathcal O_X(U)}$ is the module of Kähler differentials of $\mathcal O_X(U)$

And each $\Omega^1_{\mathcal O_X(U)}$ is a $\mathcal O_X(U)$-module

oh wait

I know what to do

I think I'm just going to use the explicit construction of sheafification

the sheafification of a sheaf of $\mathcal{O}_X$-modules should by definition be a sheaf of $\mathcal{O}_X$-modules. why would one call it a sheafification otherwise. so I guess the question is whether, if we regard a sheaf of $\mathcal{O}_X$-modules as a sheaf of abelian groups and sheafify it as such, the sheafification as sheaf of abelian groups possesses a natural $\mathcal{O}_X$-module structure turning it into the sheafification as $\mathcal{O}_X$-module.

@Thorgott you cannot imagine the satisfaction when my lecturer proved the classification of finite-dimensional real division algebras with topology

I'm exaggerating a bit

isn't the sheafification of a sheaf just the sheaf?

Or should I say isomorphic to the original sheaf

oh, you did K-theory?

I forgot a couple of pre- in my message above

ah ok

I haven't checked the details, but the answer should be clearly be a "yes"

and the proof should just be glorified bookkeeping

I mean, it's trivial by the universal property right?

@Astyx There are weird things with the sheaf of constant functions versus locally constant functions on a disconnected domain. Or something.

yeah, I'm not sure whether the sheafification of a sheaf is just the sheaf itself

usually composing two adjoints does not give you the identity

Well, sheaf of constant functions isn't a sheaf on a disconnected domain

So the presheaf of the sheafification needn't be the presheaf.

I should keep quiet.

yes, of course, if a presheaf isn't a sheaf it's sheafification isn't the presheaf, since it has to be a sheaf

(that's a lot of sheaf)

perfect sentence

algebra

Good evening to everybody into chat...

Just an humble comment on my question

and I hope that someone give me an answer. But the downvotes are of the haters of why I not put the proof how to get the formulas?

@Astyx of course you were right

sheafs is a full subcategory of presheafs, so if $i$ denotes the inclusion and $s$ sheafification, for sheafs $F,G$, we have $Hom(F,G)=Hom(iF,iG)$ since it's a full subcategory, and $Hom(iF,iG)=Hom(siF,G)$ by universal property, both equalities being natural isomorphisms, so Yoneda gives $F=siF$

Uh what am I missing. Why is the sequence of chain complexes exact? That would imply trivial homology

yeah it isn't

has to be a typo

Also, in the last exact sequence, shouldn't that be either $C_{n+1}$ and $d_{n+1}$ or the inclusion of $B_n$, and why do I obtain that the rank of $Z_{n}$ is equal to that sum. I can see that the second sequence splits by freeness of $B_{n-1}$ but in the third I dont see it

The point is to make the two short exact sequences out of the complex.

Yeah, the professor's notes aren't too carefully crafted.

But this is standard and is in every text.

yeah, should be inclusion of B_n in Z_n

I also am not fond of using $d$ for $\partial$, for obvious reasons.

the point is that in a SES 0->A->B->C->0, we have rkA+rkC=rkB

Okay that is good to know

which is essentially just rank-nullity or the first isomorphism theorem, depending on how you wanna go about it

Yes. If you have a long exact sequence of vector spaces (say), the way you prove that the alternating sum of dimensions is $0$ is precisely to break it up into lots of little short exact sequences, just like this.

The second and third listed sequences are exact.

By definition of the things involved

ye ye this I get

@Thorgott this fact is what I needed to know

The typo with $d_{n+1}$ is a bit upsetting, but ... shrug.

I don't have the energy to get upset about typos.

me neither I'm satisfied with catching them