Alex Youcis

Berkeley, CA

ayoucis.wordpress.com

Age: 23

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

2h
comment trivialising cover for etale morphisms
$\text{Spec}(\mathbb{C})$ is connected, as a scheme over $\text{Spec}(\mathbb{R})$, but when you base change to the alg. closure of $\mathbb{R}$ it becomes two points. In fact, $\text{Spec}(L)$ will have $[L:K]$ many discrete points in base change to the alg. closure of $K$. Am I misunderstanding you?
3h
comment trivialising cover for etale morphisms
What about when $X$ is the spectrum of a field $k$? The only etale covers are of the form $\text{Spec}(L)$ for $L/K$ finite separable. But the only such geometrically connected such scheme is $\text{Spec}(K)$ itself.
Sep
11
comment Interpretation of $\Omega_{A/k} \simeq A \otimes_k I/I^2$ for affine group schemes
I mean it comes down to both sides being the invariant differentials. Have you seen this definition?
Sep
10
answered $\dim (D-P)=\dim (D)-1$
Sep
8
comment Is a twist of a connected torsor connected?
I apologize, what is your definition of torsor here? Do you mean principal homogenous space?
Sep
7
answered Exercise 3.20 in Hartshorne on dimension of integral schemes of finite type over a field.
Sep
4
comment Shtukas?$\mbox{}$
There is a currently a course being taught at Berkeley by Scholze, accompanied by supplementary lectures by Jared Weinstein. The course is essentially how he is generalizing the notion of shtukas, and how it relates to the Langlands program. They are both being recorded, I believe (certainly Scholze's is). You should try and see if you can get a hold of the lectures.
Sep
2
comment Tensor product of injective ring homomorphisms
First, it seems all you've shown that $R\to A\otimes_R B$ has kernel in the nilradical of $R$, since this is what is equivalence do dominance.
Aug
24
awarded Nice Answer
Aug
24
comment Compute principal divisor for a rational function on a curve
@GeorgesElencwajg How is that the correct homogenization? Am I missing something?
Aug
24
comment Cardinality of variety
Oops! I missed that conversation. Apologies. :)
Aug
24
comment Cardinality of variety
+1 You could also use Noether normalization :)
Aug
24
comment Embedding an affine curve in a proper curve.
For a) what about taking the normalization of the projective closure?
Aug
24
comment Does every group have a 'cyclization'?
@KCd Oops, stupid initial object in $\mathbf{Ring}$. :)
Aug
24
comment Does every group have a 'cyclization'?
@KCd Ah, I see. That's a fair point. It is nice to note that hidden in your argument, and as in mine, there are no non-zero maps to $\mathbb{Z}$ from $\mathbb{Z}_p$. Regardless, nice answer. +1
Aug
24
revised When is a polynomial map proper?
deleted 11 characters in body
Aug
24
comment Does every group have a 'cyclization'?
@KCd Concerning your edit, why is what I wrote not sufficient? Every element of $\mathbb{Z}_p$ is $q$-divisibile for a prime $q\ne p$-- namely for any $x$, we have that $x=qy$ for $y=q^{-1}x$. Thus, if $f:\mathbb{Z}_p\to\mathbb{Z}$ is a group map, then for any $x\in\mathbb{Z}_p$ we'd have $f(x)=f(qy)=qf(y)$ for all primes $q$ except $p$. But then this implies that $f(x)=0$.
Aug
24
comment Does every group have a 'cyclization'?
@KCd He almost certainly thought you meant $\mathbb{Z}/p\mathbb{Z}$. That said, the argument I said above works.
Aug
24
comment Does every group have a 'cyclization'?
@Bryan That's not right, $\mathbb{Z}_p$ is torsion free. That said, every element of $\mathbb{Z}_p$ is $q$-divisible for any prime $q\ne p$, and so its image in $\mathbb{Z}$ must be zero. Or, in other words, $\text{Hom}(\mathbb{Z}_p,\mathbb{Z})=0$.
Aug
23
comment If field has a prime field isomorphic to $\mathbb{Q}$, sufficient condition for every subring being integrally closed domain
As zcn showed there, we can assume that $k$ is algebraic over $\mathbb{Q}$. Suppose that $k\ne \mathbb{Q}$. Let $\mathcal{O}_k=\mathbb{Z}[\alpha_1,\ldots,\alpha_n]$ be a minimal set of generators. Consider then $R=\mathbb{Z}[2\alpha_1,\ldots,2\alpha_n]$.
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