# 数学代写|交换代数代写commutative algebra代考|Finitely Presented Ideals

#### Doug I. Jones

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## 数学代写|交换代数代写commutative algebra代考|Finitely Presented Ideals

Consider a ring $\mathbf{A}$ and a generator set $\left(a_1, \ldots, a_n\right)=(a)$ for a finitely generated ideal $\mathfrak{a}$ of $\mathbf{A}$. We are interested in the $\mathbf{A}$-module structure of $\mathfrak{a}$.

Trivial Syzygies
Among the syzygies between the $a_i$ ‘s there are what we call the trivial syzygies (or trivial relators if we see them as algebraic dependence relations over $\mathbf{k}$ when $\mathbf{A}$ is a k-algebra):
$$a_i a_j-a_j a_i=0 \text { for } i \neq j \text {. }$$
If $\mathfrak{a}$ is finitely presented, we can always take a presentation matrix of $\mathfrak{a}$ for the generator set $(a)$ in the form
$$W=\left[R_{a} \mid U\right]$$
where $R_{a}$ is “the” $n \times n(n-1) / 2$ matrix of trivial syzygies (the order of the columns is without importance). For example, for $n=4$
$$R_{a}=\left[\begin{array}{cccccc} a_2 & a_3 & 0 & a_4 & 0 & 0 \ -a_1 & 0 & a_3 & 0 & a_4 & 0 \ 0 & -a_1 & -a_2 & 0 & 0 & a_4 \ 0 & 0 & 0 & -a_1 & -a_2 & -a_3 \end{array}\right]$$

## 数学代写|交换代数代写commutative algebra代考|Regular Sequences

2.3 Definition A sequence $\left(a_1, \ldots, a_k\right)$ in a ring $\mathbf{A}$ is regular if each $a_i$ is regular in the ring $\mathbf{A} /\left\langle a_j ; j<i\right\rangle$.

Remark Here we have kept Bourbaki’s definition. Most authors also require that the ideal $\left\langle a_1, \ldots, a_k\right\rangle$ does not contain 1 .

As a first example, for every ring $\mathbf{k}$, the sequence $\left(X_1, \ldots, X_k\right)$ is regular in $\mathbf{k}\left[X_1, \ldots, X_k\right]$.

Our goal is to show that an ideal generated by a regular sequence is a finitely presented module.
We first establish a small lemma and a proposition.
Recall that a matrix $M=\left(m_{i j}\right) \in \mathbb{M}n(\mathbf{A})$ is said to be alternating if it is the matrix of an alternating bilinear form, i.e. $m{i i}=0$ and $m_{i j}+m_{j i}=0$ for $i, j \in \llbracket 1 . . n \rrbracket$.
The A-module of alternating matrices is free and of rank $\frac{n(n-1)}{2}$ and admits a natural basis. For example, for $n=3$,
$$\left[\begin{array}{rrr} 0 & a & b \ -a & 0 & c \ -b & -c & 0 \end{array}\right]=a\left[\begin{array}{rrr} 0 & 1 & 0 \ -1 & 0 & 0 \ 0 & 0 & 0 \end{array}\right]+b\left[\begin{array}{rrr} 0 & 0 & 1 \ 0 & 0 & 0 \ -1 & 0 & 0 \end{array}\right]+c\left[\begin{array}{rrr} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & -1 & 0 \end{array}\right] .$$

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