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# 数学代写|交换代数代写Commutative Algebra代考|MATH612

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## 数学代写|交换代数代写Commutative Algebra代考|Basic definitions

Suppose $(M,+)$ is an abelian group. For any $m \in M$ and any integer $n$, one can make sense of $n \bullet m$. If $n$ is a positive integer, this means $m+\cdots+m$ ( $n$ times); if $n=0$ it means 0 , and if $n$ is negative, then $n \bullet m=-(-n) \bullet m$. Thus we have defined a function $\bullet: \mathbb{Z} \times M \rightarrow M$ which enjoys the following properties: for all $n, n_1, n_2 \in \mathbb{Z}, m, m_1, m_2 \in M$, we have
(ZMOD1) $1 \bullet m=m$.
(ZMOD2) $n \bullet\left(m_1+m_2\right)=n \bullet m_1+n \bullet m_2$.
(ZMOD3) $\left(n_1+n_2\right) \bullet m=n_1 \bullet m+n_2 \bullet m$.
(ZMOD4) $\left(n_1 n_2\right) \bullet m=n_1 \bullet\left(n_2 \bullet m\right)$
It should be clear that this is some kind of ring-theoretic analogue of a group action on a set. In fact, consider the slightly more general construction of a monoid $(M, \cdot)$ acting on a set $S$ : that is, for all $n_1, n_2 \in M$ and $s \in S$, we require $1 \bullet s=s$ and $\left(n_1 n_2\right) \bullet s=n_1 \bullet\left(n_2 \bullet s\right)$.

For a group action $G$ on $S$, each function $g \bullet: S \rightarrow S$ is a bijection. For monoidal actions, this need not hold for all elements: e.g. taking the natural multiplication action of $M=(\mathbb{Z}, \cdot)$ on $S=\mathbb{Z}$, we find that $0 \bullet: \mathbb{Z} \rightarrow{0}$ is neither injective nor surjective, $\pm 1 \bullet: \mathbb{Z} \rightarrow \mathbb{Z}$ is bijective, and for $|n|>1, n \bullet: \mathbb{Z} \rightarrow \mathbb{Z}$ is injective but not surjective.

## 数学代写|交换代数代写Commutative Algebra代考|Finitely presented modules

One of the major differences between abelian groups and nonabelian groups is that a subgroup $N$ of a finitely generated abelian group $M$ remains finitely generated, and indeed, the minimal number of generators of the subgroup $N$ cannot exceed the minimal number of generators of $M$, whereas this is not true for nonabelian groups: e.g. the free group of rank 2 has as subgroups free groups of every rank $0 \leq r \leq \aleph_0$. (For instance, the commutator subgroup is not finitely generated.)
Since an abelian group is a $\mathbb{Z}$-module and every $R$-module has an underlying abelian group structure, one might well expect the situation for $R$-modules to be similar to that of abelian groups. We will see later that this is true in many but not all cases: an $R$-module is called Noetherian if all of its submodules are finitely generated. Certainly a Noetherian module is itself finitely generated. The basic fact here which we will prove in $\S 8.7$ – is a partial converse: if the ring $R$ is Noetherian, any finitely generated $R$-module is Noetherian. Note that we can already see that the Noetherianity of $R$ is necessary: if $R$ is not Noetherian, then by definition there exists an ideal $I$ of $R$ which is not finitely generated, and this is nothing else than a non-finitely generated $R$-submodule of $R$ (which is itself generated by the single element 1.) Thus the aforementioned fact about subgroups of finitely generated abelian groups being finitely generated holds because $\mathbb{Z}$ is a Noetherian ring.
When $R$ is not Noetherian, it becomes necessary to impose stronger conditions than finite generation on modules. One such condition indeed comes from group theory: recall that a group $G$ is finitely presented if it is isomorphic to the quotient of a finitely generated free group $F$ by the least normal subgroup $N$ generated by a finite subset $x_1, \ldots, x_m$ of $F$.

## 数学代写|交换代数代写Commutative Algebra代考|Basic definitions

(zmod1) $1 \bullet m=m$．
(zmod2) $n \bullet\left(m_1+m_2\right)=n \bullet m_1+n \bullet m_2$．
(zmod3) $\left(n_1+n_2\right) \bullet m=n_1 \bullet m+n_2 \bullet m$．
(zmod4) $\left(n_1 n_2\right) \bullet m=n_1 \bullet\left(n_2 \bullet m\right)$

## MATLAB代写

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