## Finite generation and the Noetherian Property.

Let $R$ be a ring and $M$ a left $R$ module. $M$ is finitely generated if there are finitely many elements $m_1,\ldots, m_k\in M$ so that, for any $m\in M$, there are $r_i\in R$, such that

\[m = \sum_{i=1}^{k} r_{i} m_{i}.\]Equivalently, $M$ is finitely generated if, for some $n\ge 0$ in $\Z$ there is a surjective $R$-module homomorphism

\[\pi: \oplus_{i=1}^{n} R =R^{n}\to M.\]The images of the basis elements of $R^{n}$ give the generating set $m_{i}$.

**Definition:** A module $M$ satisfies the *ascending chain condition* if any increasing sequence of submodules \(M_{1}\subset M_{2}\subset\cdots\subset M_{k}\subset\) eventually stabilizes, meaning that there is an $N$ so that $M_{i}=M_{j}$ for all $i,j\ge N$.

**Proposition:** The following are equivalent:

- $M$ satisfies the ascending chain condition.
- Every nonempty set of submodules of $M$ has a maximal element.
- Every submodule of $M$ is finitely generated.

A module that satisfies these equivalent conditions is called (left) *Noetherian* after Emmy Noether. A ring is Noetherian if it is Noetherian as a left module over itself. Since the submodules of a ring are the ideals, a ring is Noetherian if every ideal is finitely generated.

**Proof:** Suppose $M$ satisfies the ascending chain condition and let $\mathcal{M}$ be a nonempty collection of submodules of $M$. Every ascending chain in $\mathcal{M}$ has a maximal element (that’s basically what the chain condition says) and therefore by Zorn’s lemma there is a maximal element for $\mathcal{M}$. Now suppose $N$ is any submodule of $M$. Let $\mathcal{N}$ be the collection of finitely generated submodules of $N$. Since the zero module is in $\mathcal{N}$, it is nonempty, so it has a maximal element $N’\subset N$. Choose $x\in N$. Then $N’+Rx$ is a finitely generated submodule of $N$, and since $N’$ is maximal, we must have $N’+Rx=N’$. This means $x\in N’$. Therefore $N=N’$ so $N$ is finitely generated. Finally, if \(M_1\subset M_2\subset \cdots\) is an increasing chain of submodules, their union $M_{\infty}$ is a submodule which must be finitely generated by, say, $m_1,\ldots, m_n$. Then there is an integer $k$ such that $M_{k}$ that contains $m_1,\ldots, m_n$ so $M_{k}=M_{\infty}$ and the increasing chain stabilizes at $k$.

**Proposition:** Any principal ideal domain is Noetherian.

**Proof:** Any ideal is generated by one element.

**Proposition:** If $M$ is Noetherian, so is any quotient module of $M$.

**Proof:** Suppose $N=M/J$ where $J$ is a submodule of $M$. If $K\subset N$ is a submodule, then by the isomorphism theorem $K=K’/J$ for some $K’\subset M$ containing $J$. Since $M$ is Noetherian, $K’$ is finitely generated by, say $k_1,\ldots, k_r$ and then the corresponding $k_{i}+J$ generate $N$.

**Proposition:** If $R$ is Noetherian, so is $R^{n}$.

**Proof:** By induction on $n$. We know the result for $n=1$. Suppose it’s true for $R^{n-1}$. Let $M$ be a submodule of $R^{n}=R^{n-1}\oplus R$. Let $\pi: M\to R$ be the projection of $M$ onto the last component. Then $\pi(M)$ is an ideal of $R$, hence finitely generated. If $x_1,\ldots,x_k$ generate $\pi(M)$, then we know that each $x_i=\pi(m_i)$ for $m_i\in M$ and also, for any $m\in M$, we have \(\phi(m)=\sum r_i x_i =\phi(\sum r_{i}m_{i}).\) This means that any $m\in M$ can be written $m=m_{0}+\sum_{i=1}^{k} r_{i}m_{i}$ with $m_{0}\in \ker(\pi)\subset R^{n-1}$. Since $\ker(\pi)$ is finitely generated by induction, this shows that $M$ is finitely generated (by the finite set of generators for $\ker(M)$ together with $m_1,\ldots, m_k$)

**Remark 1:** The proof shows that if $N\subset M$ is finitely generated, and $M/N$ is finitely generated, then $M$ is finitely generated.

**Proposition:** If $R$ is Noetherian, and $M$ is a finitely generated $R$-module, then $M$ is Noetherian.

**Proof:** In this case $M$ is a quotient of $R^{n}$.

**Proposition:** If $R$ is a PID, and $M$ is a submodule of $R^{n}$, then $M$ can be generated by $n$ or fewer elements.

**Proof:** By induction. If $n=1$ this is true because $M$ is an ideal of $R$, hence generated by one element. If $M\subset R^{n}=R^{n-1}\oplus R$, and $\pi: M\to R$ is the projection on the last component, then $\pi(M)$ is generated by one elememt $m_{n}$ (which could be zero). Thus any element of $M$ is of the form $m_{0}+rm_{n}$ with $m_{0}\in\ker(\pi)\in R^{n-1}.$ By induction, $\ker(\pi)$ is generated by at most $n-1$ elements.

**Proposition:** (Hilbert) If $R$ is Noetherian, so is $R[x]$. This is called the Hilbert Basis Theorem; we won’t prove it. Paul Gordan is alleged to have said of this result “This isn’t mathematics, it is theology!” (quite possibly this is just a folk tale in mathematics. )

### Non-noetherian rings and modules

The polynomial ring in countably many variables is not Noetherian. The ring of continuous functions on $\R$ (or on $[-1,1]$) is not Noetherian because you can let $M_{i}$ be the space of continuous functions which vanish on $[-1/i,1/i]$ for $i=1,\ldots$. These are ideals in the ring and we have $M_{i}\subset M_{i+1}$ but the sequence doesn’t stabilize.

- There are finitely generated modules (over non-Noetherian rings) whose submodules are not finitely generated.
- There are modules over non-commutative rings that are left-Noetherian, but not right-Noetherian.