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# 数学代写|交换代数代写Commutative Algebra代考|Modules over Principal Ideal Domains

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## 数学代写|交换代数代写Commutative Algebra代考|Modules over Principal Ideal Domains

In this section we shall study the structure of finitely generated modules over principal ideal domains. It will be proved that they can be decomposed in a unique way into a direct sum of cyclic modules with special properties. Examples are given for the case of a univariate polynomial ring over a field. We show how this decomposition can be computed by using standard bases (actually, we need only interreduction).

Theorem 2.6.1. Let $R$ be a principal ideal domain and $M$ a finitely generated $R$-module, then $M$ is a direct sum of cyclic modules.

Proof. Let $R^m \rightarrow R^n \rightarrow M \rightarrow 0$ be a presentation of $M$ given by the ma$\operatorname{trix} A=\left(a_{i j}\right)$ with respect to the bases $B=\left{e_1, \ldots, e_n\right}, B^{\prime}=\left{f_1, \ldots, f_m\right}$ of $R^n, R^m$, respectively. If $A$ is the zero-matrix, then $M \cong R^n$, and we are done. Otherwise, we may assume that $a_{11} \neq 0$. We shall show that, for a suitable choice of the bases, the presentation matrix has diagonal form, that is, $a_{i j}=0$ if $i \neq j$. For some $k>1$ with $a_{k 1} \neq 0$, let $h$ be a generator of the ideal $\left\langle a_{11}, a_{k 1}\right\rangle$, and let $a, b, c, d \in R$ be such that $h=a a_{11}+b a_{k 1}, a_{11}=c h$, $a_{k 1}=d h$ (we choose $a:=1, b:=0, c:=1$ if $\left\langle a_{11}\right\rangle=\left\langle a_{11}, a_{k 1}\right\rangle$ ). Now we change the basis $B$ to $\bar{B}=\left{c e_1+d e_k, e_2, \ldots, e_{k-1},-b e_1+a e_k, e_{k+1}, \ldots, e_n\right}$. $\bar{B}$ is a basis because $\operatorname{det}\left(\begin{array}{cc}c & -b \ d & a\end{array}\right)=1$. Let $\bar{A}=\left(\bar{a}{i j}\right)$ be the presentation matrix with respect to this basis, then $\bar{a}{11}=h$ and $\bar{a}{k 1}=0$, while $\bar{a}{i 1}=a_{i 1}$ for $i \neq 1, k$. Note that the first row of $A$ and $\bar{A}$ are equal if and only if $\left\langle a_{11}\right\rangle=\left\langle a_{11}, a_{k 1}\right\rangle$. Doing this with every $k>1$, we may assume that $a_{k 1}=0$ for $k=2, \ldots, n$.

## 数学代写|交换代数代写Commutative Algebra代考|Tensor Product

Let $A$ be a ring, and let $M, N$, and $P$ be $A$-modules. Let $B(M, N ; P)$ be the $A$-module of bilinear maps $M \times N \rightarrow P$. In this section we want to construct a module $M \otimes_A N$, the tensor product of $M$ and $N$, together with a bilinear map $M \times N \rightarrow M \otimes_A N,(m, n) \mapsto m \otimes n$, such that this map induces a canonical isomorphism
$$B(M, N ; P) \cong \operatorname{Hom}_A\left(M \otimes \otimes_A N, P\right)$$
of $A$-modules, and study its properties. The tensor product reduces the theory of bilinear maps to linear maps, for the price that the modules become more complicated.

Let $\sigma: M \times N \rightarrow P$ be a bilinear map, that is, for all $a \in A, m, m^{\prime} \in M$, $n, n^{\prime} \in N$
(B1) $\sigma(a m, n)=\sigma(m, a n)=a \sigma(m, n)$,
(B2) $\sigma\left(m+m^{\prime}, n\right)=\sigma(m, n)+\sigma\left(m^{\prime}, n\right)$,
(B3) $\sigma\left(m, n+n^{\prime}\right)=\sigma(m, n)+\sigma\left(m, n^{\prime}\right)$.
To obtain the isomorphism above, the elements of type $m \otimes n$ of the module to construct have to satisfy the following properties:
(T1) $\quad(a m) \otimes n=m \otimes(a n)=a(m \otimes n)$,
(T2) $\left(m+m^{\prime}\right) \otimes n=m \otimes n+m^{\prime} \otimes n$,
(T3) $m \otimes\left(n+n^{\prime}\right)=m \otimes n+m \otimes n^{\prime}$,
for all $a \in A, m, m^{\prime} \in M, n, n^{\prime} \in N$. The properties (T1)-(T3) imply the bilinearity of the $\operatorname{map}(m, n) \mapsto m \otimes n$.

## 数学代写|交换代数代写Commutative Algebra代考|Tensor Product

$$B(M, N ; P) \cong \operatorname{Hom}_A\left(M \otimes \otimes_A N, P\right)$$

(B1) $\sigma(a m, n)=\sigma(m, a n)=a \sigma(m, n)$;
(B2) $\sigma\left(m+m^{\prime}, n\right)=\sigma(m, n)+\sigma\left(m^{\prime}, n\right)$;
(B3) $\sigma\left(m, n+n^{\prime}\right)=\sigma(m, n)+\sigma\left(m, n^{\prime}\right)$。

(T1) $\quad(a m) \otimes n=m \otimes(a n)=a(m \otimes n)$，
(T2) $\left(m+m^{\prime}\right) \otimes n=m \otimes n+m^{\prime} \otimes n$;
(T3) $m \otimes\left(n+n^{\prime}\right)=m \otimes n+m \otimes n^{\prime}$，

## MATLAB代写

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