交换代数习题

Please refer to [1, Section 3] .

$$ \DeclareMathOperator{\id}{id} \DeclareMathOperator{\img}{im} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\lcm}{lcm} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\norm}{\lVert}{\rVert} $$

Exericse 1

考虑 $A$-模映射

$$ A \to A/I_1 \times \cdots \times A/I_n, \quad a \mapsto (a + I_1, \cdots, a + I_n). $$

因为 $I_1 \cap \cdots \cap I_n = (0)$, 所以这个映射是单射. 每一个 $A/I_i$ 都是 Noetherian ring, 通过满射 $A \to A/I_i$ 拉回得到 $A/I_i$ 的 $A$-模结构, 容易看出每个 $A/I_i$ 都是 Noetherian $A$-模.

Exercise 2

Denote by $D$ the fibre product $A \times_C B$. Let $\alpha, \beta$ denote the map $D \to A, D \to B$ respectively. Since $f, g$ are surjective, so are $\alpha, \beta$. Note that $\ker(\alpha) \cap \ker(\beta) = (0) \in D$ and that $D/\ker(\alpha) \cong A, D/\ker(\beta) \cong B$.

Exercise 3

Assume $A$ is a local ring such that the maximal ideal $\km = (m)$ is principal. Note that $\km^n = (m^n)$.

Claim

For any $a \in A$, the principal ideal $(a)$ is strictly larger than $(ma)$.

Proof. Suppose not, then $ma \mid a$. There exists $k \in A$ such that $(1 - km)a = 0$. Note that $1 - km$ is a unit since $1 + \km \subseteq A^\times$. $\square$

Claim

Assume $\bigcap_n \km^n = (0)$. Then for any $a \in A$, it can be uniquely represented as $um^n$ for some $n \in \NN$ and $u \in A$ a unit.

Proof. 因为 $\bigcap_n \km^n = (0)$, 所以存在 $n \in \NN$ 使得 $a \in \km^n$, 但是 $a \notin \km^{n+1}$. Write $a = um^n$, then $u \notin \km$, so $u$ is a unit. $\square$

Let $I$ be an ideal of $A$. 设 $n_0$ 是集合

$$ \{n \in \NN: \text{There exists $a \in I$ such that $a = um^n$}\} $$

的最小值, then $I = (m^{n_0})$.

假若 $\bigcap_n \km^n \ne (0)$, then take $a \in \bigcap_n \km^n$ which is non-zero. Constuct an increasing sequence of ideals

$$ (a_0) \subset (a_1) \subset (a_2) \subset \cdots $$

where $a_0 = a$ and $ma_{n+1} = a_n$. This implies that $A$ is not Noetherian.

Exercise 4

If $II^{-1} = A$, then there exists $x_i in I, y_i \in I^{-1}$ such that $\sum_{i=1}^n x_iy_i = 1$. 显然 $\sum_{i=1}^n Ax_i \subseteq I$. 另一方面, 对任意 $x \in I$, 我们有 $x = \sum_{i=1}^n x_i(y_ix)$. 注意 $y_ix \in I^{-1}I \subseteq A$, 故 $I \subseteq \sum_{i=1}^n Ax_i$.

Exercise 5

Suppose $J$ is an invertible fractional ideal, then $J$ is finite generated, say by $\frac{a_1}{b_1},\cdots,\frac{a_n}{b_n}$. 现在环 $A$ 是 UFD, 我们可以不妨设 $\gcd(a_i, b_i) = 1$.

设 $u = \lcm(b_1,\cdots,b_n)$ and $v = \gcd(a_1,\cdots,a_n)$. Let $D$ be the fractional ideal generated by single element $\frac{u}{v}$. We claim that $D = J^{-1}$.

显然 $\frac{u}{v}\frac{a_i}{b_i} \in A$, 所以 $D \subseteq J^{-1}$. 另一方面, 设 $\frac{p}{q} \in J^{-1}$ with $\gcd(p,q) = 1$. Then by definition $\frac{p}{q}\frac{a_i}{b_i} \in A$ for all $i$, it follows that $q \mid a_i, b_i \mid p$ for all $i$. Hence $q \mid v, u \mid p$. 设 $v = kq, p = tu$, then $\frac{p}{q} = \frac{kp}{kq} = kt\frac{u}{v}$, 故 $J^{-1} \subseteq D$.

因为 $J^{-1} = (\frac{u}{v})$, 所以 $J = (\frac{v}{u})$. 也就是说, UFD 中的 invertible fractional ideal 都是 principal ideal.

Exercise 6

考虑 $I_n = \ker(\varphi^n)$, 则 $I_n$ 构成 $A$ 的一个理想升链, which must terminate. Replacing $\varphi$ by $\varphi^N$ for succificenly large $N$ wee see that we can assume $\ker\varphi = \ker\varphi^2$, which implies that $\ker\varphi \cap \img\varphi = \{0\}$. Since $\varphi$ is surjective, $\ker\varphi = \{0\}$.

Exercise 7

如果 $A$ 是 Noetherian 环, 那么 finite module 和 finitely presented 是一回事.

假如 $A$ 不是 Noetherian 环, 那么存在某个理想 $I$ 不是有限生成. Set $M = A/I$, then $M$ is generated by $1 + I$. 我们有 $A$-module 短正合列

$$ 0 \to I \to A \to M \to 0. $$

If $M$ is finitely presented, then $I$ is finite generated, contradiction.

References

1: H. Matsumura. Commutative Ring Theory.