### 数学代写|交换代数代写commutative algebra代考|MTH2141

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## 数学代写|交换代数代写commutative algebra代考|The Tor Modules

In the following we want to apply the constructions of Section $5.1$ in order to study the functor
$$\cdot \otimes_{R} E: R-\operatorname{Mod} \longrightarrow R-\operatorname{Mod}, \quad M \longmapsto M \otimes_{R} E$$
for a fixed $R$-module $E$. Note that this is a covariant additive functor which is right exact due to $4.2 / 1$.

Definition 1. The nth left derived functor of $\otimes_{R} E$ is denoted by $\operatorname{Tor}{n}^{R}(\cdot, E)$. Thus, if $M{} \longrightarrow M$ is a projective homological resolution of an $R$-module $M$, we have $$\operatorname{Tor}{n}^{R}(M, E)=H{n}\left(M_{} \otimes_{R} E\right), \quad n \in \mathbb{N},$$
and the latter is called the $n$th Tor module, associated to $M$ and E. In particular, $\operatorname{Tor}{0}^{R}(M, E)=M \otimes{R} E$ by $5.1 / 11 .$
Rewriting $5.1 / 12$ in our special setting, we arrive at the following fact:
Proposition 2 (First long exact Tor sequence). Let $E$ be an $R$-module. Then every short exact sequence of $R$-modules
$$0 \longrightarrow M^{\prime} \longrightarrow M \longrightarrow M^{\prime \prime} \longrightarrow 0$$
induces a long exact sequence
$$\begin{gathered} \cdots, \operatorname{Tor}{2}^{R}\left(M^{\prime \prime}, E\right) \ \longrightarrow \operatorname{Tor}{1}^{R}\left(M^{\prime}, E\right) \longrightarrow \operatorname{Tor}{1}^{R}(M, E) \longrightarrow \operatorname{Tor}{1}^{R}\left(M^{\prime \prime}, E\right) \ \longrightarrow M^{\prime} \otimes_{R} E \longrightarrow \otimes_{R} E \longrightarrow M^{\prime \prime} \otimes_{R} E \quad \longrightarrow 0 . \end{gathered}$$

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

Let $M$ and $N$ be $R$-modules. Then we can look at the covariant additive functor
\begin{aligned} \operatorname{Hom}{R}(M, \cdot): R-\operatorname{Mod} & \longrightarrow R-\operatorname{Mod}, \ E & \longmapsto \operatorname{Iom}{R}(M, E), \ \left(f: E \longrightarrow E^{\prime \prime}\right) & \longmapsto\left(\varphi \longmapsto f \circ \varphi, \varphi \in \operatorname{Hom}{R}(M, E)\right), \end{aligned} which is easily seen to be left exact in the sense that it transforms exact sequences of type $$0 \longrightarrow E^{\prime} \longrightarrow E \longrightarrow E^{\prime \prime}$$ into exact sequences $$0 \longrightarrow \operatorname{Hom}{R}\left(M, E^{\prime}\right) \longrightarrow \operatorname{Hom}{R}(M, E) \longrightarrow \operatorname{Hom}{R}\left(M, E^{\prime \prime}\right) .$$
On the other hand, Hom can be viewed as a functor in the first variable as well,
\begin{aligned} \operatorname{Hom}{R}(\cdot, N): R-\operatorname{Mod} & \longrightarrow R-\operatorname{Mod}, \ E & \longmapsto \operatorname{Hom}{R}(E, N), \ \left(f: E^{\prime} \longrightarrow E\right) \longmapsto &\left(\varphi \longmapsto \varphi \circ f, \varphi \in \operatorname{Hom}{R}(E, N)\right), \end{aligned} and we see that $\operatorname{Hom}{R}(\cdot, N)$ is an additive, in this case contravariant functor, which is left exact in the sense that it transforms exact sequences of type
$$E^{\prime} \longrightarrow E \longrightarrow E^{\prime \prime} \longrightarrow 0$$
into exact sequences
$$0 \longrightarrow \operatorname{Hom}{R}\left(E^{\prime \prime}, N\right) \longrightarrow \operatorname{Hom}{R}(E, N) \longrightarrow \operatorname{Hom}_{R}\left(E^{\prime}, N\right)$$
just use the Fundamental Theorem on Homomorphisms 1.4/6.
We already know from Section $5.1$ that an $R$-module $P$ is called projective if for each surjective morphism of $R$-modules $E \longrightarrow E^{\prime \prime}$ the associated map

$\operatorname{Hom}{R}(P, E) \longrightarrow \operatorname{Hom}{R}\left(P, E^{\prime \prime}\right)$ is surjective, a property which is characterized by the following diagram:
Passing to the “dual” diagram
we obtain the notion of an injective $R$-module.

## 数学代写|交换代数代写commutative algebra代考|The Ext Modules

In the following we want to introduce Ext functors as right derived functors of the Hom functor. To do this, let $M, N$ be two $R$-modules. Choosing a projective homological resolution $M_{} \longrightarrow M$ of $M$, we can apply the functor $\operatorname{Hom}{R}(\cdot, N)$ to it. Since the functor is contravariant and additive, it transforms $M{}$ into a cochain complex
$\operatorname{Hom}{R}\left(M{}, N\right): 0 \quad-\operatorname{Hom}{R}\left(M{0}, N\right) \quad-\operatorname{Hom}{R}\left(M{1}, N\right) \quad-\ldots$
and we can define
$$\operatorname{Ext}{R}^{n}(M, N)=H^{n}\left(\operatorname{Hom}{R}\left(M_{}, N\right)\right), \quad n \in \mathbb{N},$$
as the $n$th Ext module associated to $M$ and $N$. Of course, we have to check that $\operatorname{Ext}{R}^{n}(M, N)$ is well-defined. Proceeding as in the proof of $5.1 / 10$, consider a second projective homological resolution $M{}^{\prime} \longrightarrow M$ of $M$. Then we conclude from $5.1 / 9$ that the chain complexes $M_{}$ and $M_{}^{\prime}$ are homotopy equivalent. As an additive functor, $\operatorname{Hom}{R}(\cdot, N)$ transfers this equivalence into a homotopy equivalence between $\operatorname{Hom}{R}\left(M_{}, N\right)$ and $\operatorname{Hom}{R}\left(M{*}^{\prime}, N\right)$. Indeed, as the latter complexes are cochain complexes, we adapt the definition of a homotopy, known from $5.1 / 3$ for chain complexes, to our situation as follows:

Definition 1. Let $f, g: C^{} \longrightarrow D^{}$ be homomorphisms of cochain complexes. $A$ homotopy between $f$ and $g$ is a module homomorphism $C^{} \longrightarrow D^{}$ of degree $-1$, in other words, a system of $R$-module homomorphisms $h^{n}: C^{n} \longrightarrow D^{n-1}$, such that the maps of the diagram
satisfy the relation $f-g=h \circ d+d \circ h$. Just as in the setting of chain complexes, $f$ and $g$ will be called homotopic in this case.

Notice that the above diagram coincides with the one given in $5.1 / 3$, except for the fact that in chain complexes passing through arrows of boundary maps

decreases module indices, whereas the contrary is the case in cochain complexes. In any case, we thereby see that the Ext modules are well-defined. In particular, we have
$$\operatorname{Ext}{R}^{0}(M, N)=\operatorname{Hom}{R}(M, N)$$
since $\operatorname{Hom}{R}(\cdot, N)$ is left exact. Alternatively, we can take an injective cohomological resolution $N \longrightarrow N^{}$ of $N$ and consider the cohomology modules $$\operatorname{Ext}{R}^{\prime n}(M, N)=H^{n}\left(\operatorname{Hom}_{R}\left(M, N^{}\right)\right), \quad n \in \mathbb{N} .$$

## 数学代写|交换代数代写commutative algebra代考|The Tor Modules

⋅⊗R和:R−反对⟶R−反对,米⟼米⊗R和

0⟶米′⟶米⟶米′′⟶0

⋯,托尔⁡2R(米′′,和) ⟶托尔⁡1R(米′,和)⟶托尔⁡1R(米,和)⟶托尔⁡1R(米′′,和) ⟶米′⊗R和⟶⊗R和⟶米′′⊗R和⟶0.

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

0⟶和′⟶和⟶和′′成精确的序列

0⟶他⁡R(米,和′)⟶他⁡R(米,和)⟶他⁡R(米,和′′).

0⟶他⁡R(和′′,ñ)⟶他⁡R(和,ñ)⟶他R⁡(和′,ñ)

，我们得到单射的概念R-模块。

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