user616128

@Rithaniel You already can't star your own message right? Unless you're a moderator

@loch Apparently I didn't read this properly at all... Thanks (for some reason I took it to be the max ideal of O_Y,y and it contraction, rather than max ideal of O_X,x and its extension - oh well I better sleep before I make more mistakes)

Oh, I just realised there are 4 of them lol

Since they self upvoted their question, and they got an answer

@LeakyNun Well they got what they wanted I guess

Thanks for your help @loch

No wonder i'm super confused

For some dumb reason, I've just been checking closed points

Crap! Good point

What more basic definition is there than the exterior measure here??

Just noticing the instant starring, and same typing style

/ all working on the same homework assignment?

Bhowmick, Hansie, RockDock, are you all the same person

But your statement about dilation holds in R^d with cubic covers, and the same argument described above

By cubic cover, I mean you are computing the exterior measure, where in one dimension, a cubic cover is just a cover by closed intervals

And then you can see that the cubic covers have the same volume

and showing that multiplying by k^{-1} sends cubic cover to cubic cover

@RockDock You can show this by taking a standard cubic cover

Why not, with the definition on wiki for unramified en.wikipedia.org/wiki/Glossary_of_algebraic_geometry#U

Then that seems satisfied too

and giving K\to O_{A^1,x}/m_x as a separable field extension

But if I take unramified as locally of finite type, which seems to be satisfied

Why is A^1_K not etale over Spec(K)?

The unramified contains relative dim 0 somehow

How do I see that to be equivalent to smooth and unramified?

How the hell have I not seen that definition :'(

@loch So $Y$ has to have dimension $0$???

I guess I expect there to be lots of other etale morphisms, like many smooth schemes such as $\Bbb A^n_K$, but perhaps someone knows one extra condition that makes the separable field extensions the entire class

If $L$ is a finite separable field extension of the field $K$, then $Spec(L)\to Spec(K)$ is etale. Does this actually classify all etale morphisms $Y\to Spec(K)$ to a field? Or are there high dimensional etale covers etc?

@user193319 does $\Bbb{Q}(\sqrt{2},\sqrt{3})$ contain $\sqrt{6}$. Does $\Bbb{Q}[\sqrt{2},\sqrt{3}]$?

Since cohomology of spaces with Z coefficients is just homology of spaces with torsion shifted up a degree, is it true that rational homology and rational cohomology agree in all degrees

Is the topology on the manifold that the regular value gives you the subspace topology?

I wonder why this room is becoming less active over time. I guess there are many other places to chat now about mathematics?

@Rithaniel Things will be much worse in due time. Your language will undergo highly nontrivial deformations with probability measure 1.

What are F and G?

Do you mean in reality, or in logic, or what?

@Secret What are these different shades of nothingness?

?

Like you are forbidding saying stuff like $a\in A\cap (B\cup C)$ so $a\in A$ and $a\in (B\cup C)$

You don't want to prove it by studying where elements live? @nurunnesha

^

btw who is -> who'is -> who's

eyes here

and look at that graphic

please consider my responses before writing any more proofs

It's not right for the three reasons I list above

Also, there may be (and are) points that aren't critical between a and b, such that f''>0 and f''<0 (i.e. you have to be careful with using your criterion when f'\ne 0)

Also, the function is decreasing for some time, and then it is increasing for some time, does your description of f' agree with this intuition?

@user330477 did you get f''>0 with the sign the wrong way, and this carried through everything else?