Let $A$ be a commutative ring with identity. Then we have an isomorphism $$A \rightarrow \textrm{Hom}_{A-mod}(A,A), a \mapsto (x \mapsto ax)$$, whose inverse is $$\textrm{Hom}_{A-mod}(A,A) \rightarrow A, \varphi \mapsto \varphi(1).$$Then applying this isomorphism, we get $$\textrm{Hom}_{A-mod}( \textrm{Hom}_{A-mod}(A,A) ,A) \cong \textrm{Hom}_{A-mod}(A,A) \cong A.$$

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 a field. Let $\mathcal U$ be a nonprincipal ultrafilter on $\Bbb N$. Define$$(a_1,a_2,\cdots)\sim(b_1,b_2,\cdots)$$ when $$\{n\in\Bbb N\,\vert\, a_n=b_n\}\in\mathcal U.$$The quotient $F = P/\sim$ is a field strictly bigger (in the sense of cardinality) than $\mathbb{C}$, which is called ultraproduct. And the inclusion $\Bbb C\longrightarrow F$ is obvious.Method 2: Given any field, we can always construct bigger fields. If our field isn't algebraically closed, we can adjoin new roots of polynomials, otherwise we can adjoin transcendental elements (which is equivalent to forming a field of rational functions). In fact, every field extension is an algebraic extension of a purely transcendental extension.(As $\Bbb C$ is algebraically closed, it has no algebraic extensions, so no finite ones.)In particular, $\Bbb C(T)$ (the field of rational functions in the variable T𝑇 with complex coefficients) is bigger than $\Bbb C$ in the sense of set-theoretic inclusion. However, the algebraic closure has the same cardinality as $\Bbb C$ itself, and therefore is abstractly isomorphic to $\Bbb C$, which means there exists a way to embed $\Bbb C(T)$ inside $\Bbb C$. If we want to go bigger in the sense of cardinality, we can form the field of complex-coefficient rational functions in $\kappa$-many variables, where $\kappa$ is a cardinal number bigger than the continuum $\mathfrak{c}=|\mathbb{C}|$. This is certainly bigger!Note that there is no subring of the polynomial ring $\Bbb C(T)$ that is a field and which strictly contained $\Bbb C$ itself, for if there were then it would contain some nonconstant $f(T)$, hence contain the nonpolynomial element $f(T)^{−1}$, which is impossible inside $\mathbb{C}[T]$.Note also that fields of different characteristic are incompatible: fields of different characteristic can never be contained inside a common field. So not only is there not a field to end all fields, but the "class of all fields" juts out in different, mutually exclusive directions (one for each prime, and zero).

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 with one object $*$ and for each $g\in G$ a morphism $\varphi_g\colon *\to *$, where composition of morphisms is defined via the group operation and $\operatorname{id}_* = \varphi_{e}$ for $e\in G$ the identity element.Given a poset $(P,\le)$ we can form a category with set of objects $P$ and exactly one morphism $x\to y$ for each $x,y\in P$ with $x\le y$.The homotopy category of topological spaces, where the objects are topological spaces and a morphism $X\to Y$ is a homotopy class $[f]$ of a continuous map $f\colon X\to Y$.Answer 2: I think the issue begins here:It is well known that morphisms between the objects of the category are structure-preservingThis is not the case. The notion of a category generalises the notion of 'sets-with-structure and structure-preserving functions', such as groups and homomorphisms, or topological spaces and continuous maps.But the extent of the generality is extreme: the objects of a category need not even be sets, and the morphisms of a category need not be functions.For example, every monoid can be considered as a one-object category, where the elements of the monoid are the morphisms from the single object to itself. In this case, the 'object' is just a placeholder—it has no notion of 'structure'—and the morphisms are certainly not functions (in general).Categories that 'look like' sets-with-structure and structure-preserving morphisms are called concrete categories. What this means is that the category $\mathcal{C}$ comes equipped with a faithful functor $U : \mathcal{C} \to \mathbf{Set}$. An object $A$ of $\mathcal{C}$ can be thought of has having 'underlying set' $U(A)$, and a morphism $f : A \to B$ can be thought of as having 'underlying function' $U(f) : U(A) \to U(B)$. However, concrete categories are still more general than sets-with-structure and structure-preserving morphisms. There may not actually be any structure to speak of.This question was asked on MathStackExchange on February 19, 2019. At that time, I was in high school and had to balance the college entrance examination and interest. So I didn't have much time to study mathematics, and my math level at that time was not satisfactory, but I couldn't do anything about it.

My question: Let $(\Gamma,+,\leq)$ be an ordered abelian group. We know that archimedean property can be stated as: for all $a,b\in\Gamma$ with $a>0,b\geq0$, there exists $n\geq0$ such that $b\leq na$. However, if we consider the multiplicative case, namely $(\Gamma,\cdot,\leq)$ is the ordered abelian group. Is there exists Archimedean property written multiplicatively? I think there is. And I state that as follows: for all $a,b\in\Gamma$ with $b<1,a\leq1$, there exists $n\geq0$ such that $b^{n}\leq a$. Is it correct?In fact, I failed to show that it is equivalent to $\Gamma$ having convex rank 1.Answer: You have correctly stated the multiplicative version of the Archimedean property.Let $\Gamma$ be an ordered multiplicative group with the Archimedean property.Suppose $H$ is a convex subgroup of $\Gamma$ with $H\ne \{1\}$. Let $1\ne x\in H$. Then $\{x,x^{-1}\}\subset H$ and one member of $\{x,x^{-1}\}$ is $>1$. So WLOG let $1<x\in H$.(i). If $1\le y\in\Gamma$ there exists $n\in \Bbb N_0$ with $y\le x^n\in H $. But $H$ is convex with $\{1,x^n\}\subset H$ and $1\le y\le x^n$ so $y\in H$.(ii). If $1>z\in \Gamma$ then $1<z^{-1}$ so $z^{-1}\in H$ by (i), so $z\in H$.So $H=\Gamma$.So the only convex subgroups of $\Gamma$ are $\Gamma$ and $\{1\}$.Appendix. It was unnecessary to assume $\Gamma$ is Abelian. Non-Abelian ordered groups do exist. But by elementary (but not brief) methods we can show that if $\Gamma$ is an ordered group with the Archimedean property then there is an ordered-group-isomorphism from $\Gamma$ to a subgroup of the additive Reals, implying $\Gamma$ is Abelian.

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:$$\begin{align} &S = 1 + 1 + 1 + \dots \\ &\Rightarrow 2S = 2 + 2 + 2 + \dots \\ &\Rightarrow 2S = (1+1) + (1+1) + (1+1) + \dots \\ &\Rightarrow 2S = 1 + 1 + 1 + \dots \\ &\Rightarrow 2S=S \\ &\Rightarrow S = 0 \end{align}$$Typically the extra conditions involve requiring all but a finite number of the terms to be $0$ ("almost all" in mathematical shorthand) or convergence conditions to make sure that the sum has a limiting value.This question was asked on January 22, 2020, when I was in my senior year of high school. My questioning was very poor 😅. This is incomparable to someone like Peter Scholze who already known spectral sequences in high school 🙃。

原本我以为微分几何跟代数几何仅仅是数学两个有关联的分支，结果是我之前肤浅了。随着对代数几何深入的学习了解，我发现代数几何跟微分几何也有很深的联系。因此我完全可以说微分流形的理论是深入学习代数几何的necessity，或许你懂抽象代数、交换代数，甚至同调代数，但是若你不懂一些manifold的理论，你完全没有机会去学习étale cohomology、Hodge theory等代数几何更深层次的理论。当然想要学习代数几何最高深、最先进的部分，仅仅懂abstract algebra、homological algebra、manifold是完全不够的。以我自己为例，我的方向是算术几何，这意味着你还需要懂elliptic curve、modular forms、$\ell$-adic representation、algebraic topology等更深层次的知识。还没完，你觉得你所学的东西就真的在研究的过程中用得上吗？在看书的过程中，你还得不停地看文献，就像定制一台机器一样定制自己所需要学习的知识，这样才能保证自己学到有用的东西，否则就是浪费时间，人的脑容量是有限的，用不上的东西时间长了就会忘记。很多人想做数学研究，结果却把大量的时间浪费在无谓的学习上，其实我更加提倡边做边学的做法，先找到个问题，然后尝试去做它，在做的过程中不断学习自己所需要的知识，这样效率是不是高很多呢。但说这么多都没用，很多人本身没有这么强烈的motivation，动机是前提，连最基本的动机都没有，谈再多的方法都没用。————————————————————本文原发布于2020年10月14日

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 two things that you may find bothering, and why you didn't recognize the induction. First, that the base case is not just one case but a bunch of them. Notice that here this was fundamental: the induction step in the proof only works for $m\geq n$. Also, notice that in this case proving it for $m=1$ is not less work that proving it for $m<n$: it was a one-line proof for all of these cases.The second thing you may find bothering is that we are not only using the induction hypothesis for $m−1$, but for any polynomial with degree strictly smaller than $m$. This is called complete or strong induction: where in the induction step, you use the hypothesis that the statement is true up to $m−1$, and not just for $m−1$. On the wikipedia page for induction it's explained well.

我的提问：一个整环$R$中的元素$p$是素的，如果$p$不是零或者一个单元，并且$p|ab$意味着$p|a$或者$p|b$（等价的$ab\in Rp$意味着$a \in Rp$或者$b\in Rp$）。一个整环$R$的元素$q$是不可约的，当$q$不是零或者一个单元，并且$q = ab$意味着$a$或$b$是一个单元。那么素数在整数整环中是素的吗？然后素数都是不可约的吗？回答1：这两个问题的都是对的。根据基础数论的事实，$\pm 1$是唯一可逆的整数，除$\pm 1$以外的整数可以唯一地表示为不同素数的乘积加上$\pm$，每个素数的幂都是正整数，这两个结果都很容易得到。回答2：素数在整数整环中既是素的，也是不可约的。根据定义，它们就是不可约的。为了证明它们是素的，请回顾一下欧几里得算法，该算法用于找到两个整数的GCD（并同时证明任意两个整数都有一个在等价意义下唯一的GCD，其中并不涉及素数的分解）。根据欧几里德算法可以得出，如果$d=\gcd(a,b)$对两个整数$a,b\in\mathbb Z$，则存在整数$u,v\in\mathbb Z$使得$d=ua+vb$。（贝祖特性。）现在，想象$p$是普通意义上的一个质数，意味着它是不可约的。假设$p|ab$，然后令$d=\gcd(p,a)$，$d=up+va$。因为$d|p$，（从不可约性）可以得出，要么$d$是单元，要么$d$等价于$p$，因此我们能将$d,u,v$与一个单元相乘，将其缩减到到$d=1$或者$d=p$的情况。如果$d=1$，那么$1=up+va$，即$b=upb+vab$并且右侧可被$p$整除，所以$p|b$。如果$d=p$，其中$d=\gcd(p,a)|a$，于是我们有$p|a$。在一个一般的整环$R$中，素就意味着不可约。相反的蕴含关系是有效的每当$R$是一个欧几里得整环（即它在某种意义上允许欧几里德算法），或者如果它是一个主理想整环（即每个理想都是主的——由单个元素生成），或者它是一个贝祖整环（即每两个主理想之和是一个主理想）——证明与$\mathbb{Z}$的情形是相同的。原文地址：Are the prime numbers prime in the domain of integers?