Problem: If $u$ is a unit and $r$ nilpotent, then $u+r$ is a unit.

I have been working on this for quite some time. I could use a hint.

Can you prove it for u=1?

@AlessandroCodenotti I have to learn chemistry ...

hrlp

@Daminark I'll give it a try.

Hello, if i consider $$\Omega_r=\{(x,y)\in\mathbb{R}^2\mid x^2+y^2-2x+2y+2\geq r^2\}, \quad r\geq0$$ what is $\bigcup_{r>0}\Omega_r$ ?

@AkivaWeinberger happy new year

Happy New Year

Oh, lord, that equation looks horrible

Actually, wait

We can make that look a lot simpler

$(x^2-2x+1)+(y^2+2y+1)\ge r^2$

$$\Omega_r=\{(x,y)\in\mathbb{R}^2\mid (x-1)^2+(y+1)^2\geq r^2\}, \quad r\geq0$$

Right

So geometrically it looks like the complement of a circle around the point $(1,-1)$

yes

$\Omega_0=\mathbb{R}^2$

Yup

But you want $\bigcup_{r>0}\Omega_r$, not $\bigcup_{r\ge0}\Omega_r$.

yes

Do you think you could guess at the answer, from here?

@Daminark Well...I would be able to do it if I were able to factor $r^n+1$...I want to factor it into $(r+1)$ and something else, but for some reason I am blanking on what that other factor must be....Embarrassing

@user193319 Try polynomial division

@user193319 Have you ever thought about what $(1 + x)^{-1}$ looks like?

$\frac{r^n+1}{r+1}$ is actually a famous formula

@AkivaWeinberger $\bigcup_{r>0}\Omega_r=\bigcup_{r\geq0}\setminus\Omega_0=\emptyset$ ?

No

@AkivaWeinberger Already tried that.

@Akiva isn't that related to geometric series?

No, $\bigcup_{r\ge0}\Omega_r$ and $\Omega_0$ are very much not disjoint, so you can't do that @Vrouvrou

@CookieToast Yes

$r$ is a nilpotent element.

$(1-x)^{-1}=1+x+x^2+x^3+\dotsb$

@AkivaWeinberger but $\mathbb{R}^2$ is one of $\Omega_r,r\geq0$ then the union is $\mathbb{R}^2$

@AkivaWeinberger If it's an infinite series, then how can I prove that $r+1$ in some commutative unital ring is a unit?

Multiply $(1 - r + \cdots - r^{n-1})(1 + r)$ by hand.

$\bigcup_{r>0}\Omega_r$ and $\Omega_0$ are very much not disjoint, I meant

Akiva is giving a heuristic argument

@BalarkaSen Erm. Sign error.

Thanks

@user193319 $1+x+x^2+x^3+\dotsb$ isn't an infinite series if $x$ is nilpotent, is it?

It's just $1+x+x^2+\dotsb+x^{n-1}+0+0+0+\dotsb$

@AkivaWeinberger i don't understand how i can find it

@Vrouvrou Consider the point $(1,-1)$. Is it in any of the $\Omega_r$ (for $r$ nonzero, positive)?

What about (say) $(0,0)$?

no

$(0,0)$ is in any $\Omega_r$

No. It's not in $\Omega_{10}$.

oh yes

Because $(0-1)^2+(0+1)^2\not\ge10^2$.

But it is in $\Omega_{\sqrt2}$, and in any $\Omega_r$ with $r$ less than $\sqrt2$.

Since $(0-1)^2+(0+1)^2\ge\sqrt2^2$

ok, but why we do this ?

So $(0,0)$'s in $\bigcup_{r>0}\Omega_r$ and $(1,-1)$ is not.

(Because something's in $\bigcup_{r>0}\Omega_r$ if and only if it's in at least one of the $\Omega_r$ sets.)

you say that $\bigcup_{r>0}\Omega_r=\mathbb{R}^2\setminus\{(1,-1)\}$ ?

Yes.

how i can prove this ?

Show that, if something is in $\bigcup_{r>0}\Omega_r$, then it's in $\Bbb R^2\setminus\{(1,-1)\}$. And, show that if something's in $\Bbb R^2\setminus\{(1,-1)\}$, then it's in $\bigcup_{r>0}\Omega_r$.

That first direction should be easy; the second direction is the more interesting one.

yes

i do the first

(Perhaps it's best in the contrapositive direction) @Vrouvrou

(i.e., if something's not in $\Bbb R^2\setminus\{(1,-1)\}$, then it's not in $\bigcup_{r>0}\Omega_r$)

if i take $(x,y)\in \mathbb{R}^2\setminus\{(-1,1)\}$ the question is is there exists $r>0$ such that $(x,y)\in \Omega_r$ that $r>0, (x-1)^2+(y+1)^2\geqr$ @AkivaWeinberger

@AkivaWeinberger ah ok

Yeah

@Vrouvrou I meant the contrapositive direction for the first direction

I don't see the executed version of your messages when you code them using mathjax

@AkivaWeinberger i thinked that if $(-1,1)\notin \Omega_r,\forall r>0$ then directly $\cup\Omega_r\subset \mathbb{R}^2\setminus \{(-1,1)\}$

Do I need to change browser settings or something?

You need the plug-in known as ChatJax. See this link: math.ucla.edu/~robjohn/math/mathjax.html

@AkivaWeinberger Hm, Eliashberg-Mishachev's version of sphere eversion is about an annular neighborhood of the $S^2$ instead of $S^2$. Namely, let $V$ be a $\delta$-nbhd of the unit sphere in $\Bbb R^3$. Then the inclusion map $i : V \to \Bbb R^3$ and $r \circ \text{inv} \circ i : V \to \Bbb R^3$ are regularly homotopic, where $\text{inv}(x) = x/\|x\|^2$ just switches the exterior and interior of the unit sphere in $V$ and $r(x, y, z) = (x, y, -z)$ is reflection along the $xy$ plane.

This makes more sense, kinda

@AkivaWeinberger to prove that $\mathbb{R}^2\setminus\{(-1,1)\}\subset \bigcup\Omega_r$ it is sufficient to choose $r=(x-1)^2+(y+1)^2$ right?

I guess the "usual" version about embeddings of $S^2$ in $\Bbb R^3$ follows from restricting to the equator of $V$ and building the antipodal map as a composition of the three reflection maps.

Kind of a trivia

When I have three sets $A,B,C$ how is the operation $A \cap B \setminus C$ done? First the $\cap$ or first the $\setminus$?

@Daminark here?

@AkivaWeinberger are you here ?

@philmcole isn't $(A\cap B)\setminus C=A\cap (B\setminus C)$?

here is a question. I know that if k goes to infinity $f_k(x)$ goes to 1

this is how they show it

you haven't asked a question, and it's unclear how that proof relates to that claim since you didn't say what $f_k(x)$ is

I understand the equation made here, but how does it imply the convergence is 1? Is it because RHS is becoming 0 as k goes to infinity?

@anon can you help me on topology ?

@anon This is the same $f_k(x)$ as the one screenshotted above your A \cap B minus C message I suppose

oh

am blind

so am i

nobody saw anything

what's going on :D

@Adeek yeah but now classes have started so I won't be consistently on here at this point

@LeylaAlkan then, yes, |f_k(x)-1| tending to 0 implies f_k(x) tends to 1

Also hello nerds

At least, for a fixed x, so you know f_k pointwise converges to 1

For uniform convergence, if it's relevant to you, you might have to work harder

@Daminark eyy it's the one-thonk man

hi @Daminark

can we discuss this last point by Ravi Vakil ?

i.e the furthermore comment ?

let me make some tea first okay ?

@anon Ok, thanks so much

@BalarkaSen and thank you too

@Daminark

actually let us discuss it tomorrow.

@Vrouvrou I think $\sqrt{(x-1)^2+(y+1)^2}$, 'cause it says $r^2$ in the problem

But yes, that should do it

@anon Oh you're totally right.

Thanks

I think you trick it to get a vector field on $\Sigma$ with total index the same as degree of $G$ somehow

Ah, no, it should be half the Euler characteristic.

@Adeek today I just had the first lecture of algebra 2 where we defined a ring, I'm not sure if I'll quite be able to pull that off

@AkivaWeinberger if i Want to define a topology on R2 using Omega_r how I do

The emptyset with all union of Omega_r

@Daminark that's the first lecture, what's on the syllabus?

Or emptyset and all Omega_r with R2 - (1,-1) @AkivaWeinberger

We don't quite have a syllabus yet, our professor isn't around so another lectured in his place. It's generally ring/module theory

K, I guess I just take a normal vector $n \in S^2$ which is a regular value of $G$ and look at projection of $n$ upon $T\Sigma$. That gives a transverse-to-the-zero vector field which vanishes at twice as many points in $G^{-1}(n)$ (both on that and the preimage $G^{-1}(-n)$).

Poincare-Hopf noscopes it off to heaven

@Daminark so in the general direction of commutative algebra?

@Daminark I am sure the definition ringed a bell

/end my life/

Hmm, I seem to remember that the fact I stated gives a proof of Gauss-Bonnet theorem somehow.

I guess I get to write it as $\displaystyle \int G^*\omega = \frac12 \chi(\Sigma) \int \omega$ where $\omega$ is the volume form on $S^2$.

Ah... $\det(dG)$ is the Gaussian curvature

@AlessandroCodenotti we're gonna focus more on commutative rings I think, though I'm not sure if we'll reach the fancy bits of commutative algebra

So $\displaystyle \int K dA = \frac12\chi(\Sigma)4\pi = 2\pi\chi(\Sigma)$

Cool I like this proof, which I once used to know but forgot

Hello :) I have problem with $ \lim_{n \to \infty } \int_{\RR^+}(1+ \frac{x}{n})sin^n(x)d\mu$ where $\mu$ is Lebesgue measure

@AkivaWeinberger are you here?

@Daminark commutative ring is very interesting

@Daminark I always prefer self-study than lectures.

I think lectures are useless

unless Ted is teaching it

@anon maybe you can help me with this simple fact?

I find lectures to be pretty good for the most part. If anything I prefer to relegate more of the interesting parts of self-study away from things that are gonna happen in class already

I always learned stuff by myself, so maybe that is the reason.

But I don't like lectures personally I think you get more if you go through a book and prove things very diligently by yourself and solve the exercise than lectures.

@Daminark math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf

Vakil notes work through it

it should take 1 year to finish

@adeek you dislike lectures because you prefer to work at your own pace?

@AkivaWeinberger sorry are you there

@AkivaWeinberger if I consider tau as the emptyset and all Omega_r is it a topology?

@Vrouvrou Yes

If you include $\Omega_0=\Bbb R^2$, I guess, since you need $\Bbb R^2$ in your topology

Note that $\Omega_r\subseteq\Omega_s$ if $r\ge s$.