The Terrible Puddle

Why don't we write $a : A$ instead of $a \in A$ yet?

@topologicalorientablesurface Thanks, appreciate it. Do I just type your name in the chat or?

@topologicalorientablesurface yeah I see that $a\notin aR$ if $R$ does not contain the mult. identity

Okay, I don't really get what I'm starting with or what I should end with

nothing really

How should the proof be?

thanks

@topologicalorientablesurface Okay, I think I can wrap my head around it

@topologicalorientablesurface Okay, that seems easy. But if it is not principal though...

Ok, I'll do that

Atleast not on these courses

I barely have any

Yeah, about that

Maybe there is something I just don't know...

Oh for every teacher who have had office hours it has been 1 a week, max.

No, but I don't think most of them would mind if you crash by their office at some time

If they have them, yes

20-50

Depends on the popularity of the course. Changes from year to year.

Hmm I don't think they know

And what advice could they give me?

Don't know, everybody else are doing what I'm doing

Well my next course will be an introduction to logic

Just math, third year

I realize I can't do it with anything

Well, the whole degree is too advanced for me then

Every time I think I might understand something it happens that I don't

@TedShifrin That's the most difficult thing, I don't know whether I understand something or not

@EnjoysMath Ok

Anything else I can do?

Yes...

I was just given that so I'm not sure. I'm not sure how to find this basis thing...

yeah so for $n=1$ it's $\{1\}$ right so dimension is one

count?

@TedShifrin I wrote out $(x,y),(x,y)^2,(x,y)^3$

So the thing I started with was to compute $\dim_\mathbb{C}(\mathbb{C}[x,y]/(x,y)^n)$ as a function of $n$, where $n$ is a positive integer.

Hmm.. I'll give it a try

Don't think I can contain it in my head

Is there not an easy way to think about generated ideals?

So $xy^3$ is in the ideal and as $x^3$ is in the ideal then $x^3+xy^3$ is in the ideal.

$y^2$ and $xy$ are in the ideal...

I don't see it

Wait, it was wrong?

Oh because $x\in \mathbb{C}[x,y]$...

It's a subgroup, for every element in the ring and every element in the ideal the product is in the ideal

And where do I find it?

No

Apply the operations