« first day (2519 days earlier)   

12:00 AM
If I localize a ring at just the neutral element, I get the same ring again, right?
 
12:11 AM
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
 
12:16 AM
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
 
12:23 AM
man I have a confrontation today with my supervisor
 
12:42 AM
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
 
12:54 AM
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
 
12:59 AM
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 $\sigma$?
 
1:14 AM
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
 
1:27 AM
1
2
.
(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
 
1:48 AM
@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
 
2:34 AM
If I had to guess, they're just suppressing the $z$-dependence for the purposes of their argument.
 

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