Alex Youcis

Berkeley, CA

ayoucis.wordpress.com

Age: 23

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

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 $R=\mathbb{Z}[2\alpha_1,\ldots,2\alpha_n]$.
Aug
23
comment If field has a prime field isomorphic to $\mathbb{Q}$, sufficient condition for every subring being integrally closed domain
Do you have any motivation for this? I can't think of a natural field which satisfies this property, and most likely, there is none. For finite extensions $K$, and non-maximal order in $\mathcal{O}_K$ works, and for infinite extensions, taking a trasncendence basis, and then appending this basis to something like a non-maximal order should work--although details must be checked there.
Aug
23
comment insight into the definition of intersection multiplicity for two plane curves
For your first two questions, this is purely intuition. Is this (math.stackexchange.com/questions/882382/…) at all helpful? If not, I can try and explain more. For the third question, would you mind defininig $\mu_p$?
Aug
22
comment Example of a module such that every proper submodule is finitely generated but the module is not.
Quang Hoang's notation is not standard, usually $\mathbb{Z}_{(p)}$ denotes the localization of $\mathbb{Z}$ at the prime $(p)$, which what he wrote certainly is not. But, another way to phrase $\varinjlim \mathbb{Z}/p^n\mathbb{Z}$ which may be more amenable to your palette is as $\left\{e^{\frac{\pi i}{p^k}}:k\in\mathbb{N}\right\}$--i.e. all $p^\text{th}$-power roots of unity in $\mathbb{C}$.
Aug
22
comment sections of birational proper morphism over an etale cover
@Cantlog Oh, of course. How silly of me. Thanks :)
Aug
22
comment Regular function on a variety which is not globally rational
What are the $X_i$? By first principles, giving a morphism $f:V\to \mathbb{A}^1_k$ is the same as giving an element of $\mathcal{O}_V(V)$.
Aug
22
answered For a discrete valuation ring to be a PID, must it have an element of valuation 1?
Aug
21
comment sections of birational proper morphism over an etale cover
No, what about $Y=\mathbb{P}^1\sqcup\mathbb{P}^1$ and $X=\mathbb{P}^1$? Am I misunderstanding your question?
Aug
20
comment Weil restriction for schemes
What usage of torsor are you using here?
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