Alex Youcis

Berkeley, CA

ayoucis.wordpress.com

Age: 23

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

2d
comment Is "being an integral domain" a local property?
@FredRohrer I just now saw this comment. I can't fathom why you would be so rude, and about as something as insignificant as the degenerate case of the empty scheme. Also, the whole "Noetherianess obfuscates the true nature of things" argument usually leaves me wanting. When one is sufficiently interested in scheme theory that they care about Noetherianess conditions, they probably don't need to be told that theorem ____ only requires qc or something. And, if they are not that advanced, then an extra Noetherian condition, if for simplification or just comfort, is not poison in baby's milk.
Apr
12
comment Is is true that $\zeta$ has finite order?
No, take almost any algebraic number an normalize it.
Apr
12
comment Appropriateness of using an interval in $\Bbb{R}$ as the parameter to a continuous path in space $X$.
As for your third question, what do you mean "doesn't really work"? Are you asking are there spaces for which, say, another choice of $H$ (in my above notation) gives a 'better' notion of a group structure than $\pi_1(X,x)$?
Apr
12
comment Appropriateness of using an interval in $\Bbb{R}$ as the parameter to a continuous path in space $X$.
It depends on what you're looking for. One possible generalization would be to consider $[(H,h),(X,x)]$ for various pointed-spaces $(H,h)$, and ask which of these have group structures. A natural class of examples is when $H$ is a so-called "co-$H$-space" (an $H$-space is roughly a topological group 'up to homotopy'). Then, for example, $S^1$ has a co-$H$-space structure, and $[(S_1,0),(X,x)]$ is just $\pi_1(X,x)$. As for your other questions, they are a bit too vague. The resulting object won't, in general, be the fundamental group, and so will obviously change it.
Apr
12
comment Does $I(J\cap K)=IJ\cap IK$ hold in a Dedekind ring?
@KatrinaXiao If you're happy with this answer, you can upvote it :)
Apr
10
revised Let $K$ be a Sylow subgroup of a finite group $G$. Prove that if $x \in N(K)$ and the order of $x$ is a power of $p$, then $ x \in K$.
added 9 characters in body
Apr
10
answered Let $K$ be a Sylow subgroup of a finite group $G$. Prove that if $x \in N(K)$ and the order of $x$ is a power of $p$, then $ x \in K$.
Apr
10
comment How can I compute $Tor\left(Z_{p},Z_{q}\right)$?
@CYC Because one of the properties of Tor (or more generally of derived functors) is that it is independent of which resolution you take.
Apr
10
answered How can I compute $Tor\left(Z_{p},Z_{q}\right)$?
Apr
8
awarded Popular Question
Apr
8
comment Does $I(J\cap K)=IJ\cap IK$ hold in a Dedekind ring?
@KatrinaXiao Right, exactly.
Apr
8
comment Does $I(J\cap K)=IJ\cap IK$ hold in a Dedekind ring?
@KatrinaXiao I don't quite follow. Containment and divisibility are reversed. So, the ideal minimal with respect to being divisible by both $K$ and $J$ should be the ideal maximal to containment in both $I$ and $J$.
Apr
8
comment Does $I(J\cap K)=IJ\cap IK$ hold in a Dedekind ring?
@KatrinaXiao This is just because dividing and containment are the same thing for Dedekind domains. Thus, the intersection of two ideals is the ideal maximal among containment. So, $I\subseteq K$ and $I\subseteq J$ if and only if $I\subseteq K\cap J$. So, this says that $K\mid I$ and $J\mid I$ if and only if $K\cap J\mid I$.
Apr
8
comment What's the intuition for the fact that $\mathscr{O}(-k)$ and $\mathscr{O}(k)$ are so different?
They are a wily bunch. Thank heavens we have the non-algebraic geometers like you to safeguard us from their devissage and their pro-etale sites. :)
Apr
8
answered Does $I(J\cap K)=IJ\cap IK$ hold in a Dedekind ring?
Apr
8
comment What's the intuition for the fact that $\mathscr{O}(-k)$ and $\mathscr{O}(k)$ are so different?
intuitive, despite using ideas (algebraic/complex geometry) which are, technically, unneeded. Regardless, this is a good answer, clarifying some of the ambiguous comments (including mine!) about the differences between the real and complex cases. +1!
Apr
8
comment What's the intuition for the fact that $\mathscr{O}(-k)$ and $\mathscr{O}(k)$ are so different?
Georges, while your answer is certainly 100% correct content wise, I disagree that the original question has absolute nothing to do with algebraic geometry. While it is certainly true that the Chern classes tell the bundles apart, it is hard to understand intuitively what the difference in Chern classes (beyond "they have different orientation twisting") tells us. It's a kind of technical fact which is, as hard, if not harder, to reconcile than the global sections argument. I do think that understanding why the bundles are specifying different pole data, and so shouldn't be the same, is more
Apr
8
comment When does a field extension canonically determine a morphism of schemes?
Of course, the answer to your titular question is "yes": $\text{Spec}(L)\to\text{Spec}(K)$. Now, it is possible to phrase your question like this: does every map of fields occur as the map between generic points of a map of positive dimensional schemes. Of course, this is generically possible (except for finite fields!), as is implicit in Georges's answer. But, as Georges also points out, there is not canonical choice.
Apr
8
comment What's the intuition for the fact that $\mathscr{O}(-k)$ and $\mathscr{O}(k)$ are so different?
@user40276 You're welcome :)
Apr
8
comment What's the intuition for the fact that $\mathscr{O}(-k)$ and $\mathscr{O}(k)$ are so different?
as stringent of conditions on what global functions must satisfy. So, we should expect that bundles should behave different there.
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