Daniel McLaury

@jamesqf: You vastly underestimate how well-known this incident is. It was probably the second-most notable event of the decade in American popular culture; #1 was of course Janet Jackson's Super Bowl performance.

@Schmuddi I'm a little skeptical about the existence of "I'mna," but "I'm going to," "I'm gonna," "I'munna," and "I'mma" are all very commonly heard.

It seems like a university that would admit such students should have its accreditation called into question. It would be a betrayal of the public trust to award a bachelor's degree to someone lacking an 8th-grade education.

As far as I can tell nobody here ever made the claim that dark pools violate Reg NMS.

Getting filled at or better than the NBBO doesn't mean you got good execution. If it did, you could just send market orders for everything.

whereas Spec k[e1, e2] doesn't have any

and the difference is that Spec k[x,y]/(x,y) has a couple of canonical directions at (0,0)

or I guess a direction is any vector of T_x X up to scaling

up to scaling, maybe

nonzero vector, maybe

Let's say given a scheme X and a point x of X, a "direction" is any canonical choice of a vector in T_x X

At least*

This is weird to me, which I guess means I don't really grok what a tangent space is, and least outside the smooth case.

Then you have no way to distinguish any two vectors in the tangent space to the closed point

However if you take two Spec k[e]s and glue them at the closed point, forgetting the inclusion maps from the Spec k[e]s

For example if you take the tangent cone at the singular point it is not the entire tangent space

So if you take two A^1s, glue them at a point, then forget the inclusion maps of the original A^1s, then the resulting space still has privileged directions

I believe I'm here, though I'm on mobile

@AlexYoucis I agree with that but I think the situation at hand is different. $\operatorname{Spec} k[x,y]/(xy)$ inherently has two privileged directions; $\operatorname{Spec} k[\varepsilon_1, \varepsilon_2]$ does not. But both are formed by gluing "lines" at a point; it's just that one is an actual line and the other is an "infinitesimal line."

How are you defining $T_{(0,0)} \mathbb{R}^2$ in such a way that it doesn't come preequipped from the outset with a map taking curves through $(0,0)$ in $\mathbb{R}^2$ to vectors in $T_{(0,0)} \mathbb{R}^2$?

"Really, unless you make an explicit identification of $T_{(0,0)} \mathbb{R}^2$ with $\mathbb{R}^2$ there is no real labeling of the axes" Well, curves through $(0,0)$ in $\mathbb{R}^2$ determine tangent vectors in $T_{(0,0)} \mathbb{R}^2$. That doesn't require making any extra identifications.

@lulu the question is different from the one you linked in that it includes arbitrary divisors rather than prime ones.

If Joe changes history, is historynecessarily changed in such a way that everything Joe said in that alternate history was true?