Chapter 1: Roots of Commutative Algebra

Exercises

A note: I use $(f_1, ...)$ to denote the submodule generated by $f_1,...$.

Exercise 1.1:

Let $M$ be a module over a commutative ring $R$. The following are equivalent:

1. The module $M$ is Noetherian (every submodule is finitely generated)
2. Every ascending chain of submodules of $M$ terminates
3. Every nonempty set $S$ of submodules of $M$ contains at least one maximal element; that is, at least one submodule which is contained in no other submodule in $S$.
4. Given a sequence $f_1, f_2, ...$ of elements of $M$, there is a natural number $m$ such that for each $n>m$ we may write $f_n = \sum_{i=1}^ma_if_i$ for $a_i\in R$.

Solution. We show $(1)\implies(2)\implies(3)\implies(4)\implies(1)$.

• Suppose (1) holds. If $0\subsetneq M_1\subsetneq M_2 \cdots$ is an ascending chain, then $\bigoplus_iM_i$ is a submodule of $M$ and thus, according to $(1)$, is finitely generated. This means that there is a finite set $g_i$ of generators, for $i < n$. If the sequence of $M_i$ does not terminate, for each $i$ there is an $m_i\in M_i$ such that $m_i\not\in M_j$ for $j< i$; in particular, this holds for $i = n+1$. But then $m_{n+1}$ cannot be generated by the $g_i$ by closure, which contradicts the finite generation of $\bigoplus_iM_i$. Thus $(1)\implies(2)$.
• Suppose (2) holds. Suppose there is a nonempty set $S$ with no maximal element, and say $M_1\in S$. Then there is an $M_2\in S$ satisfying $M_1\subsetneq M_2$, as otherwise $M_1$ would be maximal. We repeat this process to create a non-terminating chain, contradicting (2).
• Suppose (3) holds. Taking $S=\{(f_1), (f_1, f_2), (f_1, f_2, f_3), ...\}$ yields a set of submodules of $M$ which must have at least one element that is not a subset of any other element of $S$; but then every other element of $S$ is a submodule of this module, and so (since this module is generated by finitely many $f_i$) every $f_i$ can be written as a linear combination of finitely many of the $f_i$.
• Suppose (4) holds and $S$ is a submodule which is not finitely generated. Then there exists a sequence $g_i$ such that no finite subset will generate $S$. In particular, for any $n$, there is a $g_m$ such that for all $g_i$ with $i< n$, $g_m$ cannot be written as a linear combination of the $g_i$ with $i< n$.

QED.