# Interlude: principal ideal rings

Recently I mentioned that Noetherian rings can be characterized either by the fact that their finitely generated modules don’t have non-finitely generated submodules, or by the a priori weaker condition that their ideals are all finitely generated. In other words, if you want a counterexample to the statement that a submodule of a finitely generated module $M$ is again finitely generated, it’s sufficient to consider the case $M = R$.

I also claimed that the function $\mu_R$ sending an $R$-module to its minimum number of generators is not generally increasing: for $R$-modules $N\subseteq M$ we don’t necessarily have $\mu_R(N)\leq \mu_R(M)$, even when $R$ is Noetherian. For example, over $R = \mathbb{C}[x,y]$, the ideal $I = (x,y)\subseteq R$ has $\mu_R(I)=2$, but $\mu_R(R) = 1$.

These two facts led me to ask the following question:

If we want a commutative ring $R$ to have the property that $\mu_R$ is increasing, is it again sufficient to look for counterexamples among the ideals of $R$? In other words, are the rings whose ideals can all be generated by single elements (the principal ideal rings) exactly the rings with this property?

Answer: Yes! For example, $\mathbb{Z}$ is a principal ideal ring (any ideal is generated by the gcd of its elements), so this tells us that any subgroup of an abelian group with $n$ generators also has at most $n$ generators. I imagine this can be proven directly from the structure theorem for finitely generated abelian groups, but dealing with arbitrary submodules of a direct sum sounds a little fiddly.

The key to the proof I’ll show below is to use exact sequences to build arbitrary finitely generated modules out of very simple ones, using this lemma reminiscent of the rank-nullity theorem for vector spaces:

Lemma: Let $0\to A\to B\to C\to 0$ be any exact sequence of $R$-modules. If $A$ and $C$ can be generated by $m$ and $n$ elements, respectively, then $B$ can be generated by $m+n$ elements. In other words, $\mu_R(B) \leq \mu_R(A) + \mu_R(C)$.

Proof of lemma: Let $\{a_1,\dots, a_m\}$ and $\{c_1,\dots,c_n\}$ be generating sets for $A$ and $C$. Let $\{b_1,\dots, b_n\}$ be arbitrary elements of $B$ mapping to the $c_i$ (which is possible since $B\to C$ is surjective). We show that $\{a_1,\dots,a_m,b_1,\dots,b_n\}$ is a generating set for $B$.

Let $b$ be an arbitrary element of $B$, and let $c$ be its image in $C$. Use our generating set $\{c_1,\dots,c_n\}$ for $C$ to write

$c = r_1c_1+\dots+r_nc_n$

for some $r_1,\dots,r_n\in R$. We don’t know immediately that $b$ equals $r_1b_1+\dots+r_nb_n$, but we know that their difference $b - (r_1b_1+\dots+r_nb_n)$ is sent to zero in $C$. Therefore this difference actually belongs to $A\subseteq B$, so call it $a$ and use our generating set $\{a_1,\dots,a_m\}$ to write

$a = s_1a_1+\dots+s_ma_m$

for some $s_1,\dots,s_m\in R$. Therefore

$b = (s_1a_1+\dots+s_ma_m)+(r_1b_1+\dots+r_nb_n)$

represents $b$ as a linear combination of $\{a_1,\dots, a_m,b_1,\dots,b_n\}$ as desired.

Now on to the proof that if $R$ is a principal ideal ring, then $\mu_R$ is increasing. There are two steps to the proof: reducing to the case of free modules, and inducting on the number of generators. Since these two steps can be done in either order, we present both possibilities.

Proof 1: Suppose $R$ is a principal ideal ring. We prove by induction on $n$ that if $M$ is an $R$-module with $\leq n$ generators, so is any submodule $M'$ of $M$.

Base cases: $n=0$ and $n=1$. For $n=0$, the only module generated by zero elements is the zero module, which has only itself as a submodule! For $n=1$, the single generator gives us a surjection of $R$-modules $R\to M$. Let $I\subseteq R$ be the inverse image of $M'$ under this surjection; since $R$ is a principal ideal ring, $I$ can be generated by one element, and the image of that element maps to the desired single generator of $M'$.

Induction step: Suppose $n\geq 2$ and suppose $M$ can be generated by $\{m_1,m_2,\dots,m_n\}$. Let $M_1\subseteq M$ be the submodule generated by $m_1$ alone, and let $M_2$ be the quotient $M/M_1$, which can be generated by the classes of $m_2, \dots, m_n$. Therefore we have a short exact sequence

$0\to M_1\to M\to M_2\to 0$

with $M_1$ generated by a single element and $M_2$ generated by $n-1$ elements. Let $M'_1$ be the inverse image of $M'$ in $M_1$ (i.e. the intersection of submodules $M_1\cap M'$) and let $M'_2$ be the image of $M'$ in $M_2$. We get another short exact sequence

$0\to M'_1\to M' \to M'_2\to 0$

because $M'$ surjects onto $M'_2$, and the only elements sent to zero are those that also belong to $M_1$. Now since $M'_1\subseteq M_1$ and $M'_2\subseteq M_2$, by the base case and induction hypothesis, we know that $M'_1$ and $M'_2$ can be generated by $\leq 1$ and $\leq n-1$ elements, respectively. But then by the lemma about adding generating sets in short exact sequences, we find that $M'$ can be generated by $\leq n$ elements too! This completes the induction step, and the proof.

Proof 2: This time, we first prove by induction on $n$ that arbitrary submodules of $R^n$ can be generated by $\leq n$ elements, and then we generalize to arbitrary finitely generated $R$-modules.

Base cases: $n=0$ and $n=1$. Again, $R^0$ is the zero module, which is generated by zero elements and only has itself for a submodule. The $n=1$ case, that submodules of $R^1=R$ need at most one generator, is exactly our assumption that $R$ is a principal ideal ring.

Induction step: Let $n\geq 2$ and set up the short exact sequence

$0 \to R\to R^n\to R^{n-1}\to 0$

where the map $R\to R^n$ sends $r\mapsto (r,0,\dots,0)$ and the map $R^n\to R^{n-1}$ sends $(r_1,r_2,\dots,r_n)\mapsto (r_2,\dots,r_n)$. Let $K$ be any submodule of $R^n$, and as in Proof 1 let $K_1$ be its inverse image in $R$ and let $K_2$ be its image in $R^{n-1}$, so we get the short exact sequence

$0\to K_1\to K\to K_2\to 0$.

Then by the base case and induction hypothesis, $K_1$ and $K_2$ can be generated by $\leq 1$ and $\leq n-1$ elements, so $K$ can be generated by $\leq n$ elements as desired.

Now let $M$ be an arbitrary $R$-module with a generating set of size $n$, and let $M'$ be a submodule of $M$. The $n$ generators give us a surjection $R^n \to M$, and let $K$ be the inverse image of $M'$ in $R^n$. We have already shown that this $K$ can be generated by $n$ generators; their images then form a size-$n$ generating set for $M'$, as desired.

These proofs are inspired by the proof that over rings whose ideals are all finitely generated, submodules of arbitrary finitely generated modules are again finitely generated. That proof again shows by induction on $n$ that submodules of $R^n$ are finitely generated—the case $n=1$ is the hypothesis on ideals—and then generalizing straightforwardly to arbitrary finitely generated modules as quotients of finitely generated free modules.

So why use Noetherian rings, which only distinguish between finitely- and infinitely-generated modules, if we have a notion of principal ideal ring with much better behavior of generating set sizes? Simply put, it’s because Noetherian rings are common and principal ideal rings are not. The Hilbert Basis Theorem, that $R[x]$ is Noetherian if $R$ is, tells us in particular that all finitely generated algebras over fields are Noetherian, which allows us to do a lot of high-dimensional algebraic geometry without leaving the Noetherian setting. On the other hand, principal ideal rings only capture some one-dimensional behavior, and that’s just not general enough for most purposes.