这是一份umass麻省大学 MATH 611作业代写的成功案例
Proof. Let $M$ be the submodule of $I(G)$ generated by all $x_{i}-1$. We show by induction on the length of the reduced word $x \in G$ that $x-1 \in M$ for all $x \in G$ : indeed, if $x \neq 1$, then either $x=x_{i} y$ or $x=x_{i}^{-1} y$, where $y$ is shorter than $x$, and the induction hypothesis yields either $x_{i} y-1=x_{i}(y-1)+\left(x_{i}-1\right) \in M$ or $x_{i}^{-1} y-1=x_{i}^{-1}(y-1)-x_{i}^{-1}\left(x_{i}-1\right) \in M$. Then $k=\sum_{x \in G} k_{x} x \in I(G)$ implies $k=\sum_{x \in G} k_{x}(x-1) \in M$, since $\sum_{x \in G} k_{x}=0$. Thus $I(G)=M$ is generated by all $x_{i}-1$.
To prove that $\left(x_{i}-1\right){i \in I}$ is linearly independent in $I(G)$ we show that, for every $G$-module $A$ and $a{i} \in A$, there is a module homomorphism $\varphi: I(G) \longrightarrow A$ such that $\varphi\left(x_{i}-1\right)=a_{i}$ for all $i$. In particular, there is a homomorphism $\varphi_{j}: I(G) \longrightarrow \mathbb{Z}[G]$ such that $\varphi_{j}\left(x_{j}-1\right)=1$ and $\varphi_{j}\left(x_{i}-1\right)=0$ for all $i \neq j$; hence $\sum_{i \in I} k_{i}\left(x_{i}-1\right)=0$ implies $k_{j}=\varphi_{j}\left(\sum_{i \in I} k_{i}\left(x_{i}-1\right)\right)=0$ for all $j$. (Alternately, $\left(x_{i}-1\right)_{i \in I}$ has the universal property that characterizes bases.)
MATH611 COURSE NOTES ：
Proof. Since all $P_{m}$ are projective, Theorem and the exact sequences $0 \longrightarrow K_{0} \longrightarrow P_{0} \longrightarrow A \longrightarrow 0,0 \longrightarrow K_{m} \longrightarrow P_{m} \longrightarrow K_{m-1} \longrightarrow 0$ yield exact sequences that are natural in $B$, for every $k, m \geqq 1$ :
$$
\begin{gathered}
0 \longrightarrow \text { Ext }^{n+1}(A, B) \longrightarrow \text { Ext }^{n}\left(K_{0}, B\right) \longrightarrow 0 \
0 \longrightarrow \text { Ext }^{k+1}\left(K_{m}, B\right) \longrightarrow \text { Ext }^{k}\left(K_{m-1}, B\right) \longrightarrow 0 .
\end{gathered}
$$
There is also a uniqueness result for syzygies.