### 数学代写|交换代数代写commutative algebra代考|MTH2121

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• Statistical Inference 统计推断
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• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|交换代数代写commutative algebra代考|Faithfully Flat Descent of Modules and of their Mor-phisms

Let $R$ and $R^{\prime}$ be two rings and $p^{}: R \longrightarrow R^{\prime}$ a ring homomorphism. The somewhat exotic notation $p^{}$ for a ring homomorphism has been chosen with care; it is motivated by the algebraic-geometric background of affine schemes. As we have already pointed out in Section $4.5$, the category of affine schemes can be interpreted as the dual of the category of rings and vice versa. This way, $p^{}$ corresponds to a well-defined morphism $p: \operatorname{Spec} R^{\prime} \longrightarrow \operatorname{Spec} R$ between associated affine schemes, where as a map of sets on the underlying prime spectra $p$ is given by $\operatorname{Spec} R^{\prime} \ni \mathfrak{p}^{\prime} \longmapsto\left(p^{}\right)^{-1}\left(\mathfrak{p}^{\prime}\right) \in \operatorname{Spec} R$
In fact, our notation suggests to take the morphism $p$ as point of departure and to interpret $p^{}$ as the associated dual. As we will see in Chapter $5.4$, the ring homomorphism $p^{}$ admits a natural interpretation as a so-called pull back of functions on $\operatorname{Spec} R$ to functions on $\operatorname{Spec} R^{\prime}$, just by composition with $p$. In the same spirit we define for $R$-modules $M$, hence objects living on the level of Spec $R$, their pull-backs under $p$ by
$$p^{} M=M \otimes_{R} R^{\prime}$$ and, likewise, for morphisms of $R$-modules $\varphi: M \longrightarrow N$ their pull-back under $p$ by $$p^{} \varphi=\varphi \otimes \mathrm{id}{R^{\prime}}: M \otimes{R} R^{\prime} \longrightarrow N \otimes_{R} R^{\prime} .$$
The notations $p^{} M$ and $p^{} \varphi$ have the advantage that they spell out in an explicit way, how to view $R^{\prime}$ as an $R$-module in the occurring tensor products. For questions of descent, this is of essential importance, as we will see below. Also note that if $p^{}: R \longrightarrow R^{\prime}$ is decomposable into the product $r^{} \circ s^{}$ of two ring homomorphisms $r^{}$ and $s^{}$, there is a canonical isomorphism $p^{} M \simeq r^{}\left(s^{} M\right)$ for $R$-modules $M$ by $4.3 / 2$. In the following such isomorphisms will occur as identifications, just writing $p^{} M=r^{}\left(s^{*} M\right)$. In the same way we will proceed with pull-backs of morphisms.

Now let us start descent theory by discussing the descent of module morphisms.

## 数学代写|交换代数代写commutative algebra代考|Homological Methods: Ext and Tor

Consider a short exact sequence
() $0 \longrightarrow M^{\prime} \stackrel{\varphi}{\longrightarrow} M \stackrel{\psi}{\longrightarrow} M^{\prime \prime} \longrightarrow 0$ of modules over some ring $R$ and fix an additional $R$-module $E$. Then it follows from the right exactness of tensor products $4.2 / 1$ that the corresponding sequence $$0 \longrightarrow M^{\prime} \otimes_{R} E \stackrel{\varphi \otimes \operatorname{id} E}{\longrightarrow} M \otimes_{R} E \stackrel{\psi \otimes \operatorname{id} E}{\longrightarrow} M^{\prime \prime} \otimes_{R} E \longrightarrow 0$$ obtained by tensoring with $E$ over $R$ is right exact, but maybe not exact in terms of a short exact sequence. Indeed, unless $E$ is flat over $R$ (see $4.2 / 3$ ), the map $\varphi \otimes \mathrm{id}{E}$ might have a non-trivial kernel. For example, let us look at an exact sequence of type $$\text { (*) } 0 \longrightarrow R \stackrel{-a}{\longrightarrow} R \longrightarrow R /(a) \longrightarrow 0$$
where ” $\cdot a$ ” indicates multiplication by some element $a \in R$ that is not a zero divisor in $R$. Tensoring with $E$ over $R$ and identifying $R \otimes{R} E$ with $E$ produces the right exact sequence
$$0 \longrightarrow E \stackrel{-a}{\longrightarrow} E \longrightarrow E / a E \longrightarrow 0,$$
where similarly as before the map ” $a “$ is multiplication by $a$ on $E$. In particular, we get
$$\operatorname{ker}(E \stackrel{\cdot a}{\longrightarrow} E)={x \in E ; a x=0},$$
the latter being called the submodule of a-torsion in $E$. Clearly, this torsion module can be non-trivial, as the example $E=R /(a)$ shows.

Returning to the general case of a short exact sequence $(*)$ as above, one might ask if there is a natural way to characterize the kernel of the morphism
$$\varphi \otimes \mathrm{id}{E}: M^{\prime} \otimes{R} E \longrightarrow M \otimes_{R} E .$$
Of course, keeping $E$ fixed for a moment, $\operatorname{ker}\left(\varphi \otimes \mathrm{id}_{E}\right)$ will be determined by $\varphi$ and thereby depends on both, $M^{\prime}$ and $M$. This dependence is most naturally clarified by the so-called long exact Tor sequence 5.2/2. Indeed, the striking fact we will discuss in $5.1 / 10$ and $5.2 / 1$ is that there exist so-called left derived functors $\operatorname{Tor}{i}^{R}(\cdot, E)$ of the tensor product $\cdot \otimes{R} E$. These are functors on the category of $R$-modules such that any short exact sequence (*) will canonically lead to an infinite exact sequence of $R$-modules

## 数学代写|交换代数代写commutative algebra代考|Complexes, Homology, and Cohomology

Let $R$ be a ring. In the following we will consider $R$-modules and homomorphisms between these, indicating the latter by arrows, as usual. We have already mentioned in Section $1.5$ that a complex, or as we like to say, chain complex of $R$-modules is a sequence of $R$-module homomorphisms
$$\ldots \longrightarrow M_{n+1} \stackrel{d_{n+1}}{\longrightarrow} M_{n} \stackrel{d_{n}}{\longrightarrow} M_{n-1} \longrightarrow \ldots$$
satisfying $d_{n} \circ d_{n+1}=0$ when $n$ varies over $\mathbb{Z}$. One calls
$$Z_{n}=\operatorname{ker}\left(M_{n} \stackrel{d_{n}}{\longrightarrow} M_{n-1}\right) \subset M_{n}$$
the submodule of $n$-cycles,

$$B_{n}=\operatorname{im}\left(M_{n+1} \stackrel{d_{n+1}}{\longrightarrow} M_{n}\right) \subset M_{n}$$
the submodule of n-boundaries, and
$$H_{n}=Z_{n} / B_{n}$$
the $n$th homology or the $n$th homology module of the complex. We will use the notation $M_{}$ for such a chain complex and write more specifically $H_{n}\left(M_{}\right)$ for its homology. In many cases, complexes will satisfy $M_{n}=0$ for all $n<0$.

Viewing $M_{}$ as a direct sum $\bigoplus_{n \in \mathbb{Z}} M_{n}$, we can combine the homomorphisms $d_{n}: M_{n} \longrightarrow M_{n-1}$, also known as boundary maps, to yield an $R$-module homomorphism $d: M_{} \longrightarrow M_{*}$. The latter is said to be of degree $-1$ since its application lowers thee index of eaach dirēct summand by 1. In most cãsess, we will just write $d$ instead of $d_{n}$.

Passing to the dual notion of a chain complex, we arrive at the notion of a cochain complex $M^{}$. It is of type $$\ldots \longrightarrow M^{n-1} \stackrel{d^{n-1}}{\longrightarrow} M^{n} \stackrel{d^{n}}{\longrightarrow} M^{n+1} \longrightarrow \ldots,$$ where its $n$-cocycles are given by $$Z^{n}=\operatorname{ker}\left(M^{n} \stackrel{d^{n}}{\longrightarrow} M^{n+1}\right) \subset M^{n}$$ its $n$-coboundaries by $$B^{n}=\operatorname{im}\left(M^{n-1} \stackrel{d^{n-1}}{\longrightarrow} M^{n}\right) \subset M^{n}$$ and its $n$th cohomology module by $$H^{n}\left(M^{}\right)=Z^{n} / B^{n}$$

## 数学代写|交换代数代写commutative algebra代考|Faithfully Flat Descent of Modules and of their Mor-phisms

p米=米⊗RR′同样，对于态射R-模块披:米⟶ñ他们的回调p经过

p披=披⊗一世dR′:米⊗RR′⟶ñ⊗RR′.

## 数学代写|交换代数代写commutative algebra代考|Homological Methods: Ext and Tor

()0⟶米′⟶披米⟶ψ米′′⟶0一些环上的模块R并修复一个额外的R-模块和. 然后从张量积的正确精确性得出4.2/1对应的序列

0⟶米′⊗R和⟶披⊗ID⁡和米⊗R和⟶ψ⊗ID⁡和米′′⊗R和⟶0通过张量获得和超过R是正确的，但就短的精确序列而言可能不精确。确实，除非和是平的R（看4.2/3）， 地图披⊗一世d和可能有一个不平凡的内核。例如，让我们看一下类型的精确序列

(*) 0⟶R⟶−一个R⟶R/(一个)⟶0

0⟶和⟶−一个和⟶和/一个和⟶0,

## 数学代写|交换代数代写commutative algebra代考|Complexes, Homology, and Cohomology

…⟶米n+1⟶dn+1米n⟶dn米n−1⟶…

n-边界的子模块，和

Hn=从n/乙n

…⟶米n−1⟶dn−1米n⟶dn米n+1⟶…,它在哪里n-cocycles 由下式给出

Hn(米)=从n/乙n

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