有限生成的ℤ-模的子模，是有限生成的?

hbghlyj

我看到了一个“有限生成模的子模不是有限生成的”的例子: MSE
但在解决以下问题时
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关于试卷第 3 页底部的题 1.b.ii. Show that $\overlineℤ$ is a sub-module of $ℂ$:

Suppose that $α, β \in \overline{ℤ}$. Then there are generators $p_1, \ldots, p_k \in ℤ[α]$ and $q_1, \ldots, q_m \in ℤ[β]$. But $ℤ[α+β]$ and $ℤ[α β]$ are both contained in the ℤ-module generated by $α^i β^j$ for $i, j \in \mathbb{N}_0$, which in turn is generated by the finite set $\left\{p_i q_j: 1 \leqslant i \leqslant k, 1 \leqslant j \leqslant m\right\}$. Hence $ℤ[α+β]$ and $ℤ[α β]$ are submodules of a finitely generated ℤ-module, and so themselves finitely generated. Finally, $ℤ[α]=ℤ[-α]$ and 1 is a root of $X-1$ and so $\overline{ℤ}$ is a ring by the subring test.

我总感觉这个证明过于轻松(比帖子的证明短呢!)
在红色部分，它使用了“有限生成的$ℤ$-模的子模是有限生成的”

Czhang271828

有限模生成模的子模当然不一定有限生成的. 考虑 $R=\mathbb Z[X_1,X_2,\ldots]$, 则 $R$ 作为 $R$-模是 $\{1\}$ 生成的模, 但是\[
R'=\bigoplus _{\deg f\geq 1}f\mathbb Z
\]不是有限生成 $R$-子模.

但 $\mathbb Z$ 是 Noetherian ring, 所以结论是对的. 一个更清晰的看法是把 $\mathbb Z$-模当作 Abelian group.

hbghlyj

Czhang271828 发表于 2023-4-19 07:31
有限模生成模的子模当然不一定有限生成的. 考虑 $R=\mathbb Z[X_1,X_2,\ldots]$, 则 $R$ 作为 $R$-模是 $\{ ...

在Finitely generated module找到了

Every homomorphic image of a finitely generated module is finitely generated. In general, submodules of finitely generated modules need not be finitely generated. As an example, consider the ring $R =\mathbf Z[X_1, X_2, ...]$ of all polynomials in countably many variables. $R$ itself is a finitely generated $R$-module (with {1} as generating set). Consider the submodule $K$ consisting of all those polynomials with zero constant term. Since every polynomial contains only finitely many terms whose coefficients are non-zero, the $R$-module $K$ is not finitely generated.

In general, a module is said to be Noetherian if every submodule is finitely generated...

Algebra3 (2022) ICL problem sheet 3

6. Let $R$ be a commutative ring. Show that $R$ is Noetherian if and only if every ideal $I ⊆ R$ is finitely generated.

Solution: $(\Leftarrow)$ : Suppose every ideal of $R$ is finitely generated. Given the chain $I_1 \subseteq I_2 \subseteq \cdots$, let:
$$
I=\bigcup_{i \geq 1} I_i
$$
This is an ideal (e.g. we proved this in lectures). We know $I$ is finitely generated, say $I=\left(r_1, \cdots, r_n\right)$, with $r_i \in I_{k_i}$. Let
$$
K=\max _{i=1, \cdots, n}\left\{k_i\right\}
$$
Then $r_1, \cdots, r_n \in I_K$. So $I_K=I$. So $I_K=I_{K+1}=I_{K+2}=\cdots$.
$(\Rightarrow)$ : To prove the other direction, suppose there is an ideal $I \triangleleft R$ that is not finitely generated. We pick $r_1 \in I$. Since $I$ is not finitely generated, we know $\left(r_1\right) \neq I$. So we can find some $r_2 \in I \backslash\left(r_1\right)$.

Again $\left(r_1, r_2\right) \neq I$. So we can find $r_3 \in I \backslash\left(r_1, r_2\right)$. We continue on, and then can find an infinite strictly ascending chain
$$
\left(r_1\right) \subseteq\left(r_1, r_2\right) \subseteq\left(r_1, r_2, r_3\right) \subseteq \cdots
$$
So $R$ is not Noetherian.