# 数学代写|交换代数代写commutative algebra代考|A Fundamental Notion

#### Doug I. Jones

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## 数学代写|交换代数代写commutative algebra代考|A Fundamental Notion

A ring $\mathbf{A}$ is called coherent if every linear equation
$$L X=0 \text { with } L \in \mathbf{A}^{1 \times n} \text { and } X \in \mathbf{A}^{n \times 1}$$
has for solutions the elements of a finitely generated $\mathbf{A}$-submodule of $\mathbf{A}^{n \times 1}$. In other words,
$$\left{\begin{array}{c} \forall n \in \mathbb{N}, \forall L \in \mathbf{A}^{1 \times n}, \exists m \in \mathbb{N}, \exists G \in \mathbf{A}^{n \times m}, \forall X \in \mathbf{A}^{n \times 1}, \ L X=0 \Longleftrightarrow \exists Y \in \mathbf{A}^{m \times 1}, X=G Y . \end{array}\right.$$
This means that we have some control over the solution space of the homogeneous system of linear equations $L X=0$.

Clearly, a finite product of rings is coherent if and only if each factor is coherent.
More generally, given $V=\left(v_1, \ldots, v_n\right) \in M^n$ where $M$ is an A-module, the A-submodule of $\mathbf{A}^n$ defined as the kernel of the linear map
$$\breve{V}: \mathbf{A}^n \longrightarrow M, \quad\left(x_1, \ldots, x_n\right) \longmapsto \sum_i x_i v_i$$
is called the syzygy module between the $v_i$ ‘s. More specifically, we say that it the syzygy module of (the vector) $V$. An element $\left(x_1, \ldots, x_n\right)$ of this kernel is called a linear dependence relation or a syzygy between the $v_i$ ‘s. When $V$ is a generator set of $M$ the syzygy module between the $v_i$ ‘s is often called the (first) syzygy module of $M$.

By slight abuse of terminology, we indifferently refer to the term syzygy to mean the equality $\sum_i x_i v_i=0$ or the element $\left(x_1, \ldots, x_n\right) \in \mathbf{A}^n$. The $\mathbf{A}$-module $M$ is said to be coherent if for every $V \in M^n$ the syzygy module is finitely generated, in other words if we have:
$$\left{\begin{array}{c} \forall n \in \mathbb{N}, \forall V \in M^{n \times 1}, \exists m \in \mathbb{N}, \exists G \in \mathbf{A}^{m \times n}, \forall X \in \mathbf{A}^{1 \times n}, \ X V=0 \Longleftrightarrow \exists Y \in \mathbf{A}^{1 \times m}, X=Y G \end{array}\right.$$

## 数学代写|交换代数代写commutative algebra代考|Let M be a coherent A-module

Any homogeneous system of linear equations $B X=0$, where $B \in M^{k \times n}$ and $X \in \mathbf{A}^{n \times 1}$, has the elements of a finitely generated $\mathbf{A}$-submodule of $\mathbf{A}^{n \times 1}$ as its solution set.

D The general proof is by induction on the number of linear equations $k$, where the procedure is as follows: solve the first equation, then substitute the obtained general solution into the second equation, and so on. So let us for example do the proof for $k=2$ and take a closer look at this process. The matrix $B$ is composed of the rows $L$ and $L^{\prime}$. We then have a matrix $G$ such that
$$L X=0 \Longleftrightarrow \exists Y \in \mathbf{A}^{m \times 1}, X=G Y .$$
We now need to solve $L^{\prime} G Y=0$ which is equivalent to the existence of a column vector $Z$ such that $Y=G^{\prime} Z$ for a suitable matrix $G^{\prime}$. Thus $B X=0$ if and only if $X$ can be expressed as $G G^{\prime} Z$.

The above proposition is particularly important for systems of linear equations on $\mathbf{A}$ (i.e. when $M=\mathbf{A}$ ).

Comment The notion of a coherent ring is then fundamental from an algorithmic point of view in commutative algebra. Usually, this notion is hidden behind that of a Noetherian ring, ${ }^4$ and rarely put forward as we have here. In classical mathematics every Noetherian ring $\mathbf{A}$ is coherent because every submodule of $\mathbf{A}^n$ is finitely generated, and every finitely generated module is coherent for the same reason. Furthermore, we have the Hilbert theorem, which states that if $\mathbf{A}$ is Noetherian, every finitely generated A-algebra is also a Noetherian ring, whereas the same statement does not hold if one replaces “Noetherian” with “coherent.”

# 交换代数代考

## 数学代写|交换代数代写commutative algebra代考|A Fundamental Notion

$$L X=0 \text { with } L \in \mathbf{A}^{1 \times n} \text { and } X \in \mathbf{A}^{n \times 1}$$

$$\left{\begin{array}{c} \forall n \in \mathbb{N}, \forall L \in \mathbf{A}^{1 \times n}, \exists m \in \mathbb{N}, \exists G \in \mathbf{A}^{n \times m}, \forall X \in \mathbf{A}^{n \times 1}, \ L X=0 \Longleftrightarrow \exists Y \in \mathbf{A}^{m \times 1}, X=G Y . \end{array}\right.$$

$$\breve{V}: \mathbf{A}^n \longrightarrow M, \quad\left(x_1, \ldots, x_n\right) \longmapsto \sum_i x_i v_i$$

$$\left{\begin{array}{c} \forall n \in \mathbb{N}, \forall V \in M^{n \times 1}, \exists m \in \mathbb{N}, \exists G \in \mathbf{A}^{m \times n}, \forall X \in \mathbf{A}^{1 \times n}, \ X V=0 \Longleftrightarrow \exists Y \in \mathbf{A}^{1 \times m}, X=Y G \end{array}\right.$$

## 数学代写|交换代数代写commutative algebra代考|Let M be a coherent A-module

$$L X=0 \Longleftrightarrow \exists Y \in \mathbf{A}^{m \times 1}, X=G Y .$$

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