Nicole
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 Jul21 comment What are the sections of $\mathbb{C}\rightarrow \mathbb{C}[[x,y]]^{alg}[\frac{1}{xy}]$? Qiauchu, I wonder what you realized that made you remove your last comment. I am now quite confused myself by this... Jul21 comment What are the sections of $\mathbb{C}\rightarrow \mathbb{C}[[x,y]]^{alg}[\frac{1}{xy}]$? What makes this question interesting is that $\mathbb{C}[[x,y]]^{alg}$ (as well as without $alg$) is a $2$-dimensional ring, and when you invert $xy$ it becomes $1$-dimensional (because there is only one maximal ideal, which we remove). So there may be many maximal ideals in $\mathbb{C}[[x,y]]^{alg}[1/xy]$ but I don't know how to tell what they are, and whether or not they induce a geometric point (a section $\mathbb{C}[[x,y]]^{alg}[1/xy]\rightarrow \mathbb{C}$). Jul21 comment What are the sections of $\mathbb{C}\rightarrow \mathbb{C}[[x,y]]^{alg}[\frac{1}{xy}]$? The $\mathbb{C}[[x]]$ case is much easier: this is a $1$ dimensional ring with only two prime ideals. The geometric point $Spec(\mathbb{C})\rightarrow Spec(\mathbb{C}[[x]])$ must go to (possibly a base extension of) the generic point of some subscheme, and so it must either go to the maximal point (what you called the trivial section) or the generic one. But the latter thing cannot happen since that would mean that the section $Spec(\mathbb{C})\rightarrow Spec(\mathbb{C}[[x]])$ would have to factor through $Spec(\overline{\mathbb{C}((x))})$ which it does not. Jul21 revised What are the sections of $\mathbb{C}\rightarrow \mathbb{C}[[x,y]]^{alg}[\frac{1}{xy}]$? added 23 characters in body Jul21 revised What are the sections of $\mathbb{C}\rightarrow \mathbb{C}[[x,y]]^{alg}[\frac{1}{xy}]$? added 14 characters in body Jul21 asked What are the sections of $\mathbb{C}\rightarrow \mathbb{C}[[x,y]]^{alg}[\frac{1}{xy}]$? Jul11 awarded Teacher Jul11 awarded Commentator Jul6 comment Comparing Category Theory and Model Theory (with examples from Group Theory). My description of model theory is also outdated (Shelah's work is far beyond my comprehension, for example). But those were the original applications of those theories. It was important to note that they are not competitors. Jul6 comment Comparing Category Theory and Model Theory (with examples from Group Theory). isomorphism in that category will not imply a homeomorphism. But as you see from this example, the whole point is that "usual technique" is not a well defined term. What you're asking is: is a categorical isomorphism always what you expect? But you can expect many things. As I said in the answer, if you are in a concrete category (as groups) and you want to define an isomorphism as a morphism which is 1-1 and onto (as do want to define for groups) then the categorical isomorphism is indeed what you expect. Jul6 comment Comparing Category Theory and Model Theory (with examples from Group Theory). Regarding Q1 you are right, if you care about categories up equivalence rather than up to isomorphism of categories (I have never met anyone who cared about iso.'s of categories) then you can pick however many of any copy. Strictly speaking, when people say "all finite groups" you take all of them. About Q2: surely if you take a non-concrete category then there is no "usual technique". But for concrete examples (maybe someone can come up with an easier one) if you look at the category of topological spaces, with morphisms continuous maps where you identify any two homotopic ones, then an Jul5 answered Comparing Category Theory and Model Theory (with examples from Group Theory). Jul5 comment What is the precise relationship between connections in differential geometry and Kähler differentials? I wanted to put this as a comment, but I don't have any reputation points yet. I'm a little confused by John's answer. It is true (in differential geometry) that you can think of $\nabla$ as a map from $M$ to $M \otimes T*(M)$ (the cotangent bundle). How does the cotangent bundle relate to the module of Kahler differentials? I'm finding it hard to connect the dots.. May25 awarded Nice Question Dec25 accepted Does the Langlands program preserve CFT's distinction between local and global theories? Dec25 asked Does the Langlands program preserve CFT's distinction between local and global theories? Dec22 comment Is an unramified cover of the p-adics determined by its degree? Right you are. I guess the right definition of degree here should be profinite size. Dec21 comment Is an unramified cover of the p-adics determined by its degree? Well, all infinite extensions are just unions of finite ones. This would imply the infinite case, unless I'm mistaken. Dec21 accepted Is an unramified cover of the p-adics determined by its degree? Dec21 comment Is an unramified cover of the p-adics determined by its degree? Good enough for me. Thanks!