If I localize a ring at just the neutral element, I get the same ring again, right?

Wait, by neutral element, do you mean the additive or multiplicative identity? Just to be clear.

Sorry, I mean the multiplicative identity

for the additive identity we get the zero ring

Right, right.

I should think so, actually.

okay, thanks!

At least it's something isomorphic to it I think

I sometimes don't really know if I should say "it's the same" or "it's isomorphic"

Checking the definition, you want $(m,1)$ and $(n,1)$ to be equivalent if $1(1n - 1m) = 0$, or $m = n$, so you're creating a ring isomorphic to $R$ whose elements could be written as $\frac{r}{1}$.

Unless you're still working with the exact same objects, saying "it's isomorphic" is more precise.

Okay

In the case of localization, you're constructing a new thing--namely, a ring structure on $R \times S$, so it's different, but isomorphic.

Yeah

I think my Professor is not so strict with the notation etc.

In that case, you'd be fine either way. After all, two isomorphic objects are basically the same thing, up to relabeling, so conceptually you wouldn't be wrong.

Just that there's a pedantic way to look at it.

Yeah

and I guess if we kept track of the correct notation all the time, then algebraic geometry wouldn't be much fun

That I can't speak to as much--I don't have the right background for algebraic geometry yet. Though I really want it!

Haha me too

it's a tough way though :D

man I have a confrontation today with my supervisor

why?

He was always rude to me so I changed to someone else

you mean you had a confrontation?

yeah

He was being super rude to me and not being friendly

so I wanted to see if I could change him and I did end up working with someone else

lol, the tense conveys meanings other than time

if you have a confrontation, it sounds like you planned it

yeah i guess

Damn, sorry

Hope your new adviser is better

me 2 I know one guy who I want to work with, so I hope he accepts me.

Oh, wait so you don't yet know if you'll get that new adviser? What happens then?

@Daminark my old advisor was very rude to me and if I made a small mistake he would like throw a fit and wouldn't explain to me what is wrong or anything like that.

@Daminark I will see what will happen. I just don't think about it until this potential advisor emails back.

Alright, good luck

thanks

Which year are you in grad school now?

Or I guess, which did you just finish?

@Daminark yeah I am in grad school atm

No I mean, are you a first year student, second year, what?

first year @Daminark

wtf is $\sigma$?

I see

@Twink a function

anyway I am happy I changed now I wouldn't be able to take it to work with him @Daminark

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(Remind me to send the first link to my English teacher when I wake up.)

Yeah for sure

Send the first link to your English teacher when you wake up

@Twink If you like, $\sigma(w)\equiv \dfrac{f(z+w)-f(z)}{w}-f'(z)$

which goes to $f'(z)-f'(z)=0$ as $z\to 0$.

why they only seem to think of it as a function of $w$...hrm.

I can't quite make sense of what they're doing, alas.

blah, should have been "as $w\to 0$" above

@Twink It's the terms (to be multiplied by $w$ later) in the Taylor expansion of $f(z+w)$, excluding the first 2 terms (one of which cancels $f(z)$ and the other is just $f'(z)w$

wait

I'd need to see this

why are they even doing this. I've not seen such a derivation of Cauchy Riemann equations before

If I had to guess, they're just suppressing the $z$-dependence for the purposes of their argument.