在西班牙巴塞罗那过中秋节

【📍🇪🇸 巴塞罗那-哥特区/中秋节】

月是故乡明，而且还是超级月亮

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Grothendick经典同调代数文章：Some aspects of homological algebra

这是Grothendick著名的关于同调代数的文章Tôhoku paper的英文翻译版，原文是法语版，标题为Sur quelques points d'algèbre homologique。英文翻译为：Some aspects of homological algebra。该文章概述了很多同调代数的重要概念，其中基本都跟代数几何有联系，并且里面不少概念其实是Grothendick本人提出来的，如abelian categories。可以说这篇文章是同调代数的经典文章，在数学圈内也时常有人推荐看这篇文章，毕竟这可是祖师爷亲自从同调代数的基础概念一步步讲起，这对学同调代数或者代数几何的人都有很大裨益。我收藏这篇文章的时候都2021年了，现在拿出来推荐给大家！之后我还会把法语原版也发出来。

A question about smooth invariance of domain

My question: Theorem 22.3 (Smooth invariance of domain). Let $U \subset\mathbb{R}^n$ be an open subset, $S \subset\mathbb{R}^n$ an arbitrary subset, and $f : U \rightarrow S$ a diffeomorphism. Then $S$ is open in $\mathbb{R}^n$.I can't understand why the set $S$ is not automatically open in $\mathbb{R}^n$. The mapping is a diffemorphism, which means it is continuous in both directions, so $S$ is open.Answer: All you know a priori, is that open sets $V$ of $U$ satisfy: $f(V)$ is open in $S$, not ...

Where could I find a math pen pal?

My question: I'm a high school student. I love mathematics and I read a lot of maths books. I always like to discuss some mathematical problems with others.But unfortunately, there aren't anyone in my school tending to truly love mathematics, they just care about the exams and they often ignore me when I'm talking mathematics with them.This maybe because they can't understand what I'm talking about. So I often feel bloody lonely, it's painful. Could anyone help me? I wanna find a pen pal to comm ...

Is it okay if I am slow in reading math?

I am in final year of my undergraduate in mathematics from a prestigious institute for mathematics. However a thing that I have noticed is that I seem to be slower than my classmates in reading mathematics. As in, how muchever I try, I seem to finish my works at the last moment and I rarely find any time for extra reading. Is there any suggestions or tips that I could try that you know of? Or is it advisable to skip details in favour of saving time?Answer 1: The question and information given is ...

Why infinite sum need to be made sense ?

My question: For example, the sum in the partition of unity, and the polynomial expression in abstract algebra.Answer: Sums with an infinite number of terms (or "series" in more formal terms) need some extra conditions to make sure they are "well behaved". Otherwise you can get paradoxes like the following:$$\begin{align} &S = 1 + 1 + 1 + \dots \\ &\Rightarrow 2S = 2 + 2 + 2 + \dots \\ &\Rightarrow 2S = (1+1) + (1+1) + (1+1) + \dots \\ &\Rightarrow 2S = 1 + 1 + 1 + \dots \\ & ...

Must the morphisms of the category be structure-preserving？I found something different in a textbook.

My question: It is well known that morphisms between the objects of the category are structure-preserving, but I found that in a textbook it said that morphisms are often structure-preserving. Does this mean that there can be a morphism that is not structure-preserving?Answer 1: A category doesn't have to consist of sets with some additional structure and maps between those preserving the structure.Examples for categories that are not of this kind are:Given any group $G$, we can form a category ...

How to carry out induction in abstract algebra?

My question: I fail to understand how the induction step is carried out in this proof. Can anyone help? Thanks!Answer: Fix $n = deg B$. They are proving the statement by inducion on $m = deg A$. The base case is $m < n$. If $m \geq n$, then they find another polynomial $A'$, in this case $A' = A - B a_m X^{m - n}$, which has degree smaller than $m$, so we can deal with it by induction hypothesis. The quotient-and-remainder representation of $A'$ is used to find that of $A$.I suppose there are ...

How to understand $\mathbb{Q}_{p}(p^{1/p^{\infty}})$ ?

My question: It is known that $\mathbb{Q}_{p}(p^{1/p^{\infty}})$ is defined to be $\bigcup_{n>0} \mathbb{Q}_{p}(p^{1/p^{n}})$, which means adjoining all $p$-power roots of $p$ to the mixed characteristic field $\mathbb{Q}_{p}$. However, I have problem understanding the symbol $\mathbb{Q}_{p}(p^{1/p^{n}})$. How can this relate to the $p$-power roots of $p$? Why in the symbol, the power of $p$ is $1/p^{n}$? I think that $\mathbb{Q}_{p}(p^{1/p^{n}})$ is a cyclotomic extension of $\mathbb Q_p$, w ...

Surjectivity of representable sheaves

My question: Let $S$ be a base scheme and let $(Sch/S)_{fppf}$ be a big fppf site. Let $U$ be a scheme over $S$. Suppose that there is a surjective morphism $\Phi_{U}:U\rightarrow U$. Then can we show that the induced morphism of sheaves $h_{U}\rightarrow h_{U}$ is locally surjective? It seems that this is false.Note that $h_{U}={\rm{Hom}}(-,U)$ is a representable sheaf. A map of sheaves $F\rightarrow G$ on $(Sch/S)_{fppf}$ is locally surjective if for every scheme $U\in{\rm{Ob}}((Sch/S)_{fppf}) ...

Is any essentially surjective morphism of categories an epimorphism?

My question: Let $\cal{C},\cal{D}$ be categories (resp. stacks). Let $F:\cal{C}\rightarrow\cal{D}$ be an essentially surjective functor, i.e. surjective on isomorphism classes of objects. Then is $F$ an epimorphism in the category of small categories (resp. stacks)?Answer: No. For example, any functor between categories with one object is essentially surjective, but e.g. if $M_1, M_2$ are two nonzero monoids then the inclusion $M_1 \to M_1 \oplus M_2$ of a direct summand, thought of as a functor ...

How to construct a field that is larger than the complex numbers $\mathbb{C}$?

In this article, we discuss the following questions:Can we extend the complex numbers in any way such that $\mathbb{C} \subset\mathbb{C}[a]$ ? Or is $\mathbb{C}$ the extension to end all extensions?Surrounding these questions, we will provide two methods that extend the complex field $\mathbb{C}$.Method 1: The cartesian product of fields $$P = {\Bbb C}\times{\Bbb C}\times\cdots$$ isn't a field because has zero divisors: $$(0,1,0,1,\cdots)(1,0,1,0\cdots)=(0,0,0,0,\cdots).$$But a quotient will be ...

If the colimits of two objects are isomorphic, then are these two objects isomorphic?

Let $A,B$ be commutative rings in characteristic $p$. Let $\phi_{A}:A\rightarrow A,\phi_{B}:B\rightarrow B$ be the Frobenius morphisms, i.e. the $p$-power maps. If we have ${\rm{colim}}_{n\in\mathbb{N}}A\cong {\rm{colim}}_{n\in\mathbb{N}}B$, where the transition maps are Frobenius morphisms, can we show that $A\cong B$ ?Answer: No. Recall that an $\mathbb{F}_p$-algebra $R$ is perfect if the Frobenius map $\varphi : R \ni r \mapsto r^p \in R$ is an isomorphism. The colimit of powers of the Froben ...

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