Alex Youcis

Berkeley, CA

ayoucis.wordpress.com

Age: 24

I am a second year graduate student at the University of California, Berkeley.

My interests are generally in the field of arithmetic geometry. More specifically, I am interested in various aspects of the Langlands program. My advisor is Sug Woo Shin.

2d
awarded Good Answer
Dec
17
answered Describing $Spec(\mathcal{O}_K[X])$
Dec
14
awarded Enlightened
Dec
14
awarded Nice Answer
Dec
12
comment Module of differentials in the functorial approach to schemes and quasi-coherent modules
Serious question (please don't take this facetiously)--what is the use of this point of view? Thanks! :)
Dec
11
comment What do we lose in Projective Spaces?
@John Smith I guess the operative thing is that there is no algebraic statute on the projective line which plays nicely with the underlying structure of the variety. In other words, the projective line cannot be made into a group variety, like the affine line can.
Dec
8
awarded Caucus
Dec
6
comment What is the etale fundamental group of Z((x))?
@oxeimon Of course! How embarrassing :) I'll leave this up for posterity though! This seems like a strange thing to care about then. Looking it up, it seems as though you're choice needn't be Noetherian even if $R$ is, in which case $\pi_1$ gets wonky. Why these rings?
Dec
6
comment cohomology homomorphism induced by classifying map
Dear RSQ, I'm glad you have a question not about H-spaces. :) That said, you can stop tagging these algebraic geometry? :) That is, unless i am missing some algebro-geometric connection?
Dec
6
revised cohomology homomorphism induced by classifying map
edited tags
Dec
6
answered Counterexample in Dedekind domains
Dec
6
answered What is the etale fundamental group of Z((x))?
Dec
2
answered Why does taking completions make number fields simpler?
Dec
1
answered Special case of Kronecker–Weber theorem.
Dec
1
comment Why does taking completions make number fields simpler?
Just to point out, I don't think you answered the question of 'why', even though you provided some other instances where it's easier. :)
Dec
1
comment Why does taking completions make number fields simpler?
because we have infinitely many primes whose square roots we can adjoin! I could explain more/differently in an answer if this approach is appealing at all to you.
Dec
1
comment Why does taking completions make number fields simpler?
correspond to maps $\text{Spec}(\widehat{\mathcal{O}_{\text{Spec}(L),\mathfrak{q}}})\to\text{Spec}(\­widehat{\mathcal{O}_{\text{Spec}(K),\mathfrak{p}}})$ which is something like a supped up version of the derivative, and so, once again, should not care about primes of $\mathcal{O}_L$ and $\mathcal{O}_K$ except the point where the derivative is taking place. Your second bullet can easily be interpreted from the case of $\mathbb{Q}$ when, once again, you cut away possible extensions coming from 'other primes'. Think, for example, of quadratic extensions. There are infinitely many, but that is
Dec
1
comment Why does taking completions make number fields simpler?
Dear Mathmo123, I don't know what your background in algebraic geometry is, but if you have interest, you could look at this blog post of mine: ayoucis.wordpress.com/2014/07/22/… Since $\mathbb{Z}_p$ is just $\widehat{\mathcal{O}_{\text{Spec}(\mathbb{Z}),p}}$, the intuition of just keeping differential data at $p$ goes through. In particular, it should give you intuition about why other primes are no longer relevant. For the third bullet, the same intuition applies since extensions of, say, local fields
Dec
1
awarded algebraic-number-theory
Dec
1
revised Is the ring of p-adic integers of finite type over the ring of integers?
deleted 20 characters in body
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