so if $f:A \to B$ is an epimorphism, then $B \to B \otimes_A B$ is an isomorphism (choose either component), but this map is just $f:A \to B$ tensored with $B$ over $A$

so if $f$ is faithfully flat, $f$ is an iso

thanks for the insight. bye for now

bye!

@Semiclassical agreed

Here's a result that I can't actually verify:

$$\int_0^\pi \int_0^\pi \sqrt{\sin ^2(\alpha ) \sin ^2(\alpha -\beta )+\sin ^2(\beta ) \sin ^2(\alpha -\beta )+\sin ^2(\alpha ) \sin ^2(\beta )}\,d\alpha\,d\beta \stackrel{=}? \frac{5\pi}{2}$$

Numerically, that's right on the dot

should've been $\stackrel?{=}$ I guess

But analytically, I've no clue whether it's actually true

@MatheinBoulomenos is the following reasoning also acceptable? That $f$ is epic is equivalent to $A\to A\otimes_\Bbbk A$ being a $\Bbbk$-algebra isomorphism. In particular this is a $\Bbbk$-linear isomorphism, so $\dim _\Bbbk A = \dim_\Bbbk A\otimes _\Bbbk A=(\dim _\Bbbk A)^2$ whence $\dim _\Bbbk A=1$. Now since $f$ is a $\Bbbk$-linear injection between $\Bbbk$-linear space of the same dimension, it's bijective.

it doesn't work if $A$ is infinite-dimensional

Right. ah well :)

@MatheinBoulomenos I read the following nice proposition today: a finite type ring morphism is epic iff its (scheme-theoretic) fibers are empty or isomorphisms. The reference was EGA, but I thought you might see a self-contained approach :)

(The previous fact feels related)

$R \to S$ is epic iff $R_{\mathfrak{p}} \to S_{\mathfrak{p}}$ is epic for all primes.

scheme-theoretic fibers are just $\kappa(\mathfrak{p}) \to \kappa(\mathfrak{p}) \otimes_R S_{\mathfrak{p}}$

Yes, I'm just surprised because it seems you do not use the finite type hypothesis, which is cool

well- for one direction we don't

but for the other direction there's some Nakayama application going on I think

Yeah. Guess the other direction is sneakier.

Do you get that feeling from geometric intuition or just because of going local-to-global?

just commutative algebra experience

Nakayama implies that if $\kappa(\mathfrak{p}) \to \kappa(\mathfrak{p}) \otimes_{R_{\mathfrak{p}}}S_{\mathfrak{p}}$ is surjective, then so is $R_{\mathfrak{p}} \to S_{\mathfrak{p}}$

so from the condition that scheme-theoretic fibers are empty or isomorphisms, we even get that $R \to S$ is surjective, not just epic

How did you conclude $R\to S$ is surjective? Is surjectivity also stalk-local? That seems strange to me

yes, surjectivity is stalk-local

for modules actually

it follows from the exactness of localization (=> commutes with cokernels) and the fact that being zero is stalk-local

Sorry if this is silly confusion, but I seem to get a strange conclusion. If $R\to S$ is epic then its fibers are either isomorphisms or empty. In particular its fibers are surjections. Conversely if $R\to S$ is of finite type and has surjective fibers then you've shown $R\to S$ is itself surjective. Doesn't this imply that any finite type epimorphism is surjective?

it does and this is correct

very cool! Here's where I read about that proposition, if you're interested to see what's done in EGA.

@Semiclassical your surface area integral is also 1/2 times the integral over $[0, 2\pi]^2$, and you can straightforwardly replace that with two contour integrals over $|z| = 1$ and $|z'| = 1$, which don't look too bad except for the square root and the branch points ...

@MatheinBoulomenos do you happen to have a geometric intuition for the fact surjectivity is stalk-local for ring (even module) morphisms? For sheaves over topological spaces this isn't true, and I guess it holds in the affine case because some cohomology vanishes, but I don't really understand the picture.

@student hmm, yeah, that's an interesting idea

it's from a question i put on the main site, by the way: math.stackexchange.com/questions/3068735/…

@Arrow not sure

I'm more of an algebraist than a geometer

Okie :) i'll live to bug you another day then

Back under 100 tabs

Some day I'll fix this

Is it true that $\mathbb{R}\mathbb{P}^1$ is homeomorphic to $S^1$?

Why can't wolfram alpha find the inverse of this $x^3+y^3+(1-x)^3+y^3=1$

isn't it just $2x^3+y^3+(1-y)^3=1$

@CaptainAmerica16

[Random]

Hmm... I just realise that sets with restricted number of subsets is actually pretty common in social context

For example, consider a group of 8 people, and compute its powerset. Then there are clearly some subsets of people that are impossible because those people are incompatible

and thus need to be excluded

Of course, this can be easily done within the usual framework of set theory as follows:

Let the set by $A$. Then the required set of subset is:

$f(\mathcal{P}(A))$

where $f$ maps from powerset of $A$ to itself, getting some of its subsets

Therefore it should not be far fetch to have an axiom where there exists some set $S$ such that:

$|S|=|T|,\mathcal{P}(S) < \mathcal{P}(T)$

Suppose a person wanted to start learning how to type up files in Latex. What would be an advisable utility to use?

overleaf.com/learn/latex/Learn_LaTeX_in_30_minutes

Overleaf is a good place especially it is online so you can access it anywhere with an internet connection

Alright, danke

@Daminark hey dami

@Daminark what do you know about algebraic groups?

got a Q that I could not make sense of -.-'