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# 数学代写|同调代数代写Homological Algebra代考|MAST90068 Tensor Products

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## 数学代写|同调代数代写Homological Algebra代考|Tensor Products

We are going to introduce a fundamental construction, the tensor product, which is absolutely necessary for higher algebra. In particular this is the abstract tool used for induced modules and Mackey’s formula, which in turn are the most important tools in the representation theory of finite groups over an arbitrary field.

Furthermore, tensor products are in some respect counterparts to homomorphisms, in a sense which can be made very precise. The precise formulation is given in Lemma 1.7.9 below, which will play an extremely important role in the subsequent material.

1.7.1 The Definition and Elementary Properties
The first concept to introduce is a free abelian group generated by a set.
Definition 1.7.1 An abelian group $A$ is free on a subset $S$ of $A$ if for every abelian group $B$ and every mapping $\varphi_S: S \longrightarrow B$ (as a set !) there is a unique homomorphism $\varphi: A \longrightarrow B$ such that the restriction $\left.\varphi\right|_S$ of $\varphi$ to $S$ equals $\varphi_S$.

Consider the abelian groups $\left(\mathbb{Z}^n,+\right)$ for any integer $n$. The group $\left(\mathbb{Z}^n,+\right)$ is a free abelian group on the set
$${(1,0,0, \ldots, 0),(0,1,0, \ldots, 0), \ldots,(0,0, \ldots, 0,1)}$$
in the definition below. Indeed, the above set is a basis of $\mathbb{Z}^n$ in the sense that for any abelian group $B$ and any $n$ elements $b_1, b_2, \cdots, b_n$ of $B$ there is a unique homomorphism of abelian groups $\varphi: \mathbb{Z}^n \longrightarrow B$ such that
$$\begin{gathered} \varphi(1,0,0, \ldots, 0)=b_1 \ \varphi(0,1,0, \ldots, 0)=b_2 \ \ldots \ldots \ldots \ldots \ldots \ldots . \ \varphi(0,0, \ldots, 0,1)=b_n \end{gathered}$$
$\varphi\left(a_1, a_2, \ldots, a_n\right)=\sum_{i=1}^n a_i b_i$ is a group homomorphism and is the unique one with the above properties.

Suppose $A$ is a free abelian group on a set $S_A$, and suppose $B$ is another free abelian group on a set $S_B$. If $S_A$ and $S_B$ are of the same cardinality (i.e. there is a

bijection $\beta: S_A \longrightarrow S_B$ ) then $A$ and $B$ are isomorphic. Indeed, $\beta$ defines a unique homomorphism of groups $\widehat{\beta}: A \longrightarrow B$ restricting to $\beta$ on $S_A$, and $\beta^{-1}$ defines a unique homomorphism of groups $\widehat{\beta^{-1}}: A \longrightarrow B$ restricting to $\beta$ on $S_B$. Now, the identity on $S_A$ is the unique group homomorphism $A \longrightarrow A$ restricting to the identity on $S_A$. But, $\widehat{\beta^{-1}} \circ \widehat{\beta}$ is a group homomorphism restricting to $\beta^{-1} \circ \beta=i d_{S_A}$ on $S_A$. Therefore, by the unicity, $\widehat{\beta^{-1}} \circ \widehat{\beta}=i d_A$. Analogously, $\widehat{\beta} \circ \widehat{\beta^{-1}}=i d_A$. Hence, $\widehat{\beta}$ is an isomorphism.

## 数学代写|同调代数代写Homological Algebra代考|Immediate Applications for Group Rings

If $E$ is an extension field of $K, A$ is a $K$-algebra and $M$ is an $A$-module, then by Lemma 1.7.12 $E \otimes_K A$ is again an algebra. Moreover, by Lemma 1.7.8 the tensor product $E \otimes_K A$ is an $E$-vector space. The multiplication inside $E \otimes_K A$ implies that the elements $e \otimes 1$ for $e \in E$ are all central, and hence $E \otimes_K A$ is an $E$-algebra. The module $E \otimes_K M$ is an $E \otimes_K A$-module by Lemma 1.7.14. This procedure is called a “change of rings”, meaning change of base rings. We observe that the discussion in Sect. 1.4.3 is precisely a down-to-earth description of this abstract “change of rings”.
The second occasion where tensor products are used is a generalisations of this: induction of modules. We start with the observation that given a commutative ring $R$, an $R$-algebra $A$ and an $R$-subalgebra $B$, the ring $B$ acts on $A$ by right multiplication. In other words, restricting the action of $A$ on the regular right $A$-module $A$ to $B$ gives $A$ the structure of a right $B$-module. Also, $A$ can be regarded as a regular left $A$-module. Both actions commute. That is, $A$ is an $A-B$-bimodule if we define $a \cdot x \cdot b:=a x b$ for all $a, x \in A$ and $b \in B$.

Definition 1.7.20 Let $G$ be a group, let $R$ be a commutative ring and let $H$ be a subgroup of $G$. Then for every $R H$-module $M$ one defines the induced module $M \uparrow \uparrow_H^G$ to be $R G \otimes_{R H} M$ where for all $g \in G$ and all $x \in R G, m \in M$ we define $g \cdot(x \otimes m):=(g x) \otimes m .$
The induced module is of extreme importance in representation theory.
Remark 1.7.21 Tensor products also appear in more sophisticated structures, socalled Hopf-algebras. For a very brief introduction see Sect.6.2.1 or e.g. Montgomery’s book [10]. Group rings are Hopf algebras. For a commutative ring $K$ and a group $G$ as well as two $K G$-modules $M_1$ and $M_2$ we get that $M_1 \otimes_K M_2$ is a also $K G$-module by putting $g \cdot\left(m_1 \otimes m_2\right):=\left(g m_1 \otimes g m_2\right)$ for all $g \in G$ and $m_1 \in M_1$, $m_2 \in M_2$. Indeed, for all $g \in G$ the mapping
\begin{aligned} g \cdot: M_1 \times M_2 & \longrightarrow M_1 \otimes_K M_2 \ \left(m_1, m_2\right) & \mapsto g m_1 \otimes g m_2 \end{aligned}
is $K$-balanced and so it induces a mapping
$$\text { g. }: M_1 \otimes_K M_2 \longrightarrow M_1 \otimes_K M_2$$

$m_1 \otimes m_2 \mapsto g m_1 \otimes g m_2$
which in turn induces a $K$-linear action of $G$ on $M_1 \otimes_K M_2$, since $1_G \cdot=i d_{M_1 \otimes_K M_2}$ and $\left(g_1 g_2\right) \cdot=\left(g_{1^*}\right) \circ\left(g_2 \cdot\right)$ for all $g_1, g_2 \in G$ as is immediately checked.

However, a warning should be given here. Elements $g_1+g_2$ in $K G$ do not act diagonally on $M_1 \otimes_K M_2$. In fact, for $m_1 \in M_1$ and $m_2 \in M_2$ we have
\begin{aligned} \left(g_1+g_2\right) \cdot\left(m_1 \otimes m_2\right) &=g_1 \cdot\left(m_1 \otimes m_2\right)+g_2 \cdot\left(m_1 \otimes m_2\right) \ &=\left(g_1 m_1 \otimes g_1 m_2\right)+\left(g_2 m_1 \otimes g_2 m_2\right) \end{aligned}
whereas
\begin{aligned} \left(g_1+g_2\right) m_1 \otimes\left(g_1+g_2\right) m_2=&\left(g_1 m_1 \otimes g_1 m_2\right)+\left(g_1 m_1 \otimes g_2 m_2\right) \ &+\left(g_2 m_1 \otimes g_1 m_2\right)+\left(g_2 m_1 \otimes g_2 m_2\right) . \end{aligned}
We now continue to examine induced modules.

## 数学代写|同调代数代写同质代数代考|张量积

$${(1,0,0, \ldots, 0),(0,1,0, \ldots, 0), \ldots,(0,0, \ldots, 0,1)}$$

$$\begin{gathered} \varphi(1,0,0, \ldots, 0)=b_1 \ \varphi(0,1,0, \ldots, 0)=b_2 \ \ldots \ldots \ldots \ldots \ldots \ldots . \ \varphi(0,0, \ldots, 0,1)=b_n \end{gathered}$$
$\varphi\left(a_1, a_2, \ldots, a_n\right)=\sum_{i=1}^n a_i b_i$是一个群同态，并且是具有上述性质的唯一的一个

## 数学代写|同调代数代写Homological Algebra代考|群环的直接应用

\begin{aligned} g \cdot: M_1 \times M_2 & \longrightarrow M_1 \otimes_K M_2 \ \left(m_1, m_2\right) & \mapsto g m_1 \otimes g m_2 \end{aligned}

$$\text { g. }: M_1 \otimes_K M_2 \longrightarrow M_1 \otimes_K M_2$$

$m_1 \otimes m_2 \mapsto g m_1 \otimes g m_2$
，这反过来诱导了$G$对$M_1 \otimes_K M_2$的$K$ -线性动作，因为$1_G \cdot=i d_{M_1 \otimes_K M_2}$和$\left(g_1 g_2\right) \cdot=\left(g_{1^*}\right) \circ\left(g_2 \cdot\right)$对于所有$g_1, g_2 \in G$立即被检查

\begin{aligned} \left(g_1+g_2\right) \cdot\left(m_1 \otimes m_2\right) &=g_1 \cdot\left(m_1 \otimes m_2\right)+g_2 \cdot\left(m_1 \otimes m_2\right) \ &=\left(g_1 m_1 \otimes g_1 m_2\right)+\left(g_2 m_1 \otimes g_2 m_2\right) \end{aligned}

\begin{aligned} \left(g_1+g_2\right) m_1 \otimes\left(g_1+g_2\right) m_2=&\left(g_1 m_1 \otimes g_1 m_2\right)+\left(g_1 m_1 \otimes g_2 m_2\right) \ &+\left(g_2 m_1 \otimes g_1 m_2\right)+\left(g_2 m_1 \otimes g_2 m_2\right) . \end{aligned}

## MATLAB代写

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