# modules of finite type over a principal ideal domain

The modules of finite type over a principal ring(which, in this post, is considered the same as a principal ideal domain, just for simplify the long name) behaves very well, much like the vector spaces over a field. For example, we know that in linear algebra, any linear independent set of vectors of a vector space can be completed to be a basis of this space. In the theory of modules over a principal ring, we can’t expect that much, but we still have a similar result: we can simultaneously choose a set of elements that generate a free module of finite type and one of its submodule. We will try to prove this result in this post.

Recall that a principal ring $R$ is a ring that every ideal is principal. So as a consequence, $R$ is unique factorization domain. Hence, $R$ is a Noetherian ring(every ideal is finitely generated). And so, every module $M$ of finite type over $R$ is also Noetherian, since $M$ is a quotient of a free $R$-module of finite rank, while the later is a direct sum of Noetherian rings, hence is itself Noetherian, and so its quotient $M$, is Noetherian. But in general, $M$ is not Artinian, mainly because $R$ is integral, there always exists a descending sequence of infinite length of ideals $(a)\supset(a^2)\supset(a^3)...$.

But if pose a condition on $M$, then we can get what we want: if $M$ is torsion. Note that $M$ is of finite type, so we can write $M=\sum_{i=1}^n Rm_i$(the sum is a finite sum). And thus we can construct a projection from $R^n$ to $M$, by sending $(a_1,a_2,...,a_n)$ to $\sum_i a_im_i$. Now we denote $Ann_R(m)={a\in R| am=0}$. We see easily that this is an ideal of $R$, and thus is principal. For those generators of $M$, denote $Ann_R(m_i)=Rr_i(r_i\neq 0)$. So we have a surjection $R/Rr_1\times R/Rr_2\times...\times R/Rr_n\rightarrow M$. So if we can show each term in the product is Artinian, then the whole product is also Artinian, so is its quotient $M$. Note that if we have a descending sequence of ideals $R/Rr_i\supset Ra_1/Rr_i\supset Ra_2/Rr_i\supset...$. Then we must have that $a_1|a_2|...|a_k|a_{k+1}|...|r_i$. But since $r_i$ has only finite many factors non-equivalent, so the above sequence is stationary, hence we proved that $R/Rr_i$ is in fact Artinian, so is $M$. Here we use the fact crucial fact that $M$ is torsion, which guaranties that these $r_i$ are not zero elements.

So in summary we get the following result:

If $M$ is a torsion module of finite type over a principal ring, then $M$ is both Noetherian and Artinian.

So in fact we have seen that there are two important parts in a module, the torsion part and the torsion free part. In deed, if we denote by $T(M)$ the set of elements of $M$ which is torsion, then we see that this is a submodule of $M$. And continue to take a quotient, we get an exact sequence $0\rightarrow T(M)\rightarrow M\rightarrow M/T(M)\rightarrow 0$. Then there are three things to do, study the two non-trivial parts not in the center and see if we can recover $M$.

For the first step, we will study the structures of modules torsion-free. Before doing the real work, we can stop to have a look at the present question: does such a module over a principal ring resemble a vector space over a field? This is an interesting problem, and we will see that it is indeed the case in the sense that will be clarified.

In fact, we will prove that we can define an invariant for a free module over a commutative ring, this invariant will be called the rank of this free module. Note here we emphasis that the module is free, because otherwise it is hard to even to find a similar number to be a candidate for an invariant. Let’s first state the result:

Suppose that $R$ is a commutative ring, then for any injection of $R$-homomorphism $f:R^n\rightarrow R^m(n,m\in \mathbb{N})$, then we have that $n\leq m$

This result have a very nice proof using the method of Cayley-Hamilton theorem. The proof I found is here. Here I will just give the main idea of the proof(the proof itself is very short and simple). Suppose the opposite that $n>m$, then we can view $R^m$ as a submodule of $R^n$(the inclusion map being sending the $m$ coordinates of $R^m$ to the first $m$ coordinates of $R^n$), then we can say that $f$ is in fact a homomorphism from $R^n$ to itself, a transformation on $R^n$, which is still injective. So using the method in the proof of Cayley-Hamilton theorem, we see that $f$ satisfies its minimal polynomial, $f^k+a_{k-1}f^{k-1}+...+a_1f+a_0=0$. Using the fact that $f$ is injection, we have that $a_0\neq 0$. Noting that $a_0$ is an element of $R$, and when we apply both terms of this equation to the vector $(0,0,...,0,1)$. Using the new expanded definition of the $f$, we see that the left hand side is equal to $a_0$, while the right hand side is $0$. So we get a contradiction, which means that $n\leq m$.

This is a very pleasant result, note that we have found an invariant of a free module over any commutative ring.

But what about the submodules of a free module? Are they similar to a free module? The following example shows that it is in general not the case.

The ring is polynomial ring of one variable over the integers, $R=\mathbb{Z}(X)$. Then consider one of its ideal(which is, of course, a module over $R$) $M=R<3,X^2>$. Suppose that there is a free module $R^n$of rank greater than $1$ and an injection $f:R^n\rightarrow M$. Consider the first two coordinate elements $e_1,e_2$, supposing that $f(e_1)=a_1, f(e_2)=a_2$. Then $f(a_2e_1-a_1e_2)=0$ since $f$ is an $R$-homomorphism. And obviously $a_2e_1-a_1e_2\neq 0$, thus contradicting the fact that $f$ is injective. So for $f$ to be injective, we must have $n=1$. But since the module $M$ has at least two generators, after examining all the possibilities, we conclude that $f$ can’t be surjective. Therefore, $M$ doesn’t have a free module structure.

But if we give the ring $R$ some more structures, things will change a lot. For example, when $R$ is a principal ideal domain(or equivalently, as put above, a principal ring), then any submodule of a free module of finite type is again a free module. That is

If $R$ is a principal ring, then any submodule of $R^n$ is a free module.

The proof proceeds as a recurrence on the rank $n$. For the beginning case, $n=1$, note that a submodule of $R$ is the same as an ideal of $R$, thus is a principal ideal, written as, say, $Rr$. Clearly we have that $Rr$ is isomorphic either to $R$, or to $0$, in both cases $Rr$ is a free module. For the induction part, suppose that $M$ is a submodule of $R^{n+1}$. Then consider the projection $p: R^{n+1}\rightarrow R$, sending an element of $R^{n+1}$ to its last component. Thus we  have a decomposition $R^{n+1}=ker(p)\bigoplus R$. And also we have a decomposition for the submodule $M$, that is to say, $M=(ker(p)\bigcap M)\bigoplus p(M)$. The decomposition is not so trivial as it first seems, and next we try to construct a map. Note that $p(M)$ is in fact an ideal of $R$, so we can write it as $Rr_0$. And we suppose that $p(m_0)=r_0$. The point in constructing this isomorphism is to start from the direct-sum-part, not the inverse direction. We define a map as $f: (ker(p)\bigcap M)\bigoplus p(M)\rightarrow M, (m,r)\mapsto m+rm_0$. That this map is $R$-isomorphism can be easily verified. So now we get a submodule $ker(p)\bigcap M$ of the free module $ker(p)=R^n$ of rank $n$, which satisfies the induction assumption, thus proving this result.

So now we have a very nice characterization of the submodule of a free module. We want to know if all modules torsion-free have a free module structure. Perhaps the first step is to show that any such module is a submodule of some free module. In fact we have an elegant result:

If $R$ is a principal ring, then any module of finite type over $R$ is a submodule of a free module of finite rank.

This fact is not hard to prove. In deed, if we denote the finite set of generators of $M$ as $G=\{m_1,m_2,...,m_r\}$. We can not expect that there is a direct map from $M=\sum_i Rm_i$ to $R^r$. The obstacle is that these generators are not ‘linearly independent’. So we have to find a subset of $G$ the elements of which are $R$-independent. The $R$-independence can be characterized as the existence of an injection from some $R^k$ to the submodule generated by these elements. And of course we have to consider the largest set of $R$-independent generators, which leads to the following consideration: we define the set of subsets of $G$ of which the elements are $R$-independent, that is to say $S=\{g\subset G|\exists f:R^{\#g}\rightarrow \sum_{m\in g}Rm, f \text{is injective}\}$. Using the inclusion relation as an order, we see there exists a maximal element $g$of $S$ with a morphism $f: R^{\#g}\rightarrow \sum_{m\in g}Rm:=M'$(note that this is in fact an isomorphism, since the elements in $g$ are $R$-independent). And for any generator $m_i\in G-g$, there exists a nonzero element $a_i\in R$ such that $a_im_i\in M'$. So we define $a=\prod_{m_i\in G-g}a_i$, which is not zero, since $R$ is integral. So now we can set $F:M\rightarrow M', m\mapsto am$. This homomorphism is injective since $M$ is torsion-free, and so we get an embedding $M\rightarrow R^n$ for some rank $n$.

Combining the above two results, we know that any module torsion-free of finite type is in fact a free module of finite rank.

So we know that the structures of modules torsion-free of finite type over a principal ring is rather simple, they are just $R^n$ for some integer $n$.

So our next step is to study the structures of torsion modules over a principal ring. Recall that we have shown that this kind of modules are, in fact, Noetherian and Artinian. So in some sense we can invoke the Krull-Schmidt theorem on decomposition of Noetherian or Artinian rings into indecomposable modules. By the way, looking back the above text, we see that the indecomposable modules over a principal ring include itself. But take the integer ring $R=\mathbb{Z}$ as an example. What are the indecomposable modules over $R$? It is easy to verify that $R/(Rp^n)(n\in\mathbb{N})$(we can do this by comparing the orders of the elements). We know that in a principal ring $R$, there are also prime elements, which are the generators of the prime ideals. So we expect that $R/(Rp^n)$ are also indecomposable modules over $R$(where $p$ is a prime element in $R$). This is what the following proposition says:

Suppose that $R$ is a principal ring, and $p$ is a prime element in $R$(possibly equal to $0$), then $R/(Rp^n)$(where $n$ is a non-negative integer) is an indecomposable module over $R$.

The problem then is reduced to how to show that a module is indecomposable. Note if the module $M$ is decomposable, that is, $M=M'\bigoplus M''$, then consider the endomorphisms on $M$. Each endomorphism $f$ on $M$ is, in fact, equivalent to four parts: the map $f_{11}$from $M'$ to $M'$, the map $f_{12}$ from $M'$ to $M''$, the map $f_{21}$ from $M''$ to $M'$, and the map $f_{22}$ from $M''$ to $M''$. Now consider two maps, the first is $F\in End(M)$ such that $F_{11}=id_{M'}, F_{12}=F_{21}=F_{22}=0$, the second $F\in End(M'')$ such that $G_{22}=id_{M''},G_{11}=G_{12}=G_{21}=0$. Consider the ideals $I_F,I_G$ of $End(M)$ generated by these two elements respectively. There are maximal ideals containing these two ideals, but these two maximal ideals can’t be identical, otherwise it would contain $F+G=id_{M}$, which is impossible. So through the opposite hypothesis, we get that the End-ring can’t be a local ring.

Now look at the particular module $M=R/(Rp^n)$. There is a series of identifications $End_R(M)=End_{M}(M)=M$. They are easy to verify. So we have to show that $M$ is a local ring(it indeed has a ring structure inherited from that of $R$) to show that $M$ is indecomposable. Note that an ideal of $M$ is always of the form $Rr/(Rp^n)$ for some element $r\in R$. Thus we have that $r|p^n$. So we conclude easily that there is only one maximal ideal in $M$, that is $Rp/(Rp^n)$. The above argument is valid for $p\neq0$. It remains to show that $R$ is indecomposable. If not, then we can write $R=Ra\bigoplus Rb(a,b\neq0)$. But this is impossible since $Ra\bigcap Rb=Rgcd(a,b)$, which is not zero.

So after a long journey, we get to the conclusion that $R/(Rp^n)$ are indecomposable modules over $R$.

But recall that in the beginning of this post, we have said that all torsion modules of finite type over a principal ring $R$ is in fact a finite product of modules like $R/(Ra)$. Then a simple application of the Chinese remainder theorem shows that $R/(Ra)$ is a finite product of modules of the form $R/(Rp^n)$ where $p$ is a prime element in $R$.

In fact one corollary of the above argument is that, any indecomposable torsion module of finite type over a principal ring $R$ is of the form $R/(Rp^n)$($p$ as above).

So all in all, we know the structures of the torsion part and the torsion-free part of a module over a principal ring. Now return to the exact sequence of a module $M$ of finite type: $0\rightarrow T(M)\rightarrow M\rightarrow M/T(M)\rightarrow0$. Note that $M/T(M)$ is torsion-free, thus is a free module over $R$ of finite rank. Therefore this sequence splits, that is $M=T(M)\bigoplus M/T(M)$. Explicitly, any module $M$ of finite type over a principal ring $R$ is of the form $M=R^n\bigoplus R/(Rp_i^{n_i})\bigoplus R/(Rp_2^{n_2})\bigoplus...\bigoplus R/(Rp_k^{n_k})$ where $p_1,p_2,...,p_k$ are prime elements of $R$, and they are not necessarily distinct. So after rearranging these prime terms, we can write it in another form $M=R^n\bigoplus R/R(a_1)\bigoplus R/(Ra_2) \bigoplus...\bigoplus R/(Ra_l)$ where $a_l|a_{l-1}|...|a_2|a_1$.

We see that the modules of finite type over a principal ring is semi-simple as well as totally decomposable. But the concepts of indecomposable module and simple module are still not the same.

Now the last to do is to prove the result mentioned in the beginning of this post, that is

Suppose that $R$ is a principal ring, and $M=R^n$ a free module over $R$, $N$ a submodule of $M$. Then we can find a basis $(e_1,e_2,...,e_n)$ of $M$ and a sequence of elements $(d_1,d_2,...,d_n)$ of $R$ such that $d_n|d_{n-1}|…|d_2|d_1$ such that $M=\sum_i Re_i, N=\sum_i Rd_ie_i$.

The proof utilize again recurrence on the rank $n$. For the starting step, it is very easy. For the induction step, suppose that $M=R^{n+1}$. We have to find a way to reduce to the $n$-case. Suppose that now we know the result, that is $M=\sum_i Re_i, N=\sum_i Rd_ie_i$. Then any morphism $f: M\rightarrow R$ must have that $f(n)=f(\sum_i r_id_ie_i)=d_nf(\sum_i r_id_i'e_i)(n\in N)$(where $d_i'd_n=d_i$). So the $d_n$ are, in some sense, independent of the choice of the coordinates, it can be characterized by the morphisms. This is a very important observation. Continue our imagination. Which map $f:M\rightarrow R$ can take the value $d_n$?Can we characterize this map? We can characterize this map using $d_n$, that is to say, $f$ is the morphism we are after if and only if $f(N)=Rd_n$. This is like an extreme point, in some sense the largest map. To give a precise definition, we consider $S=\{g(N)|g:M\rightarrow, \text{is a morphism}\}$. Since these $g(N)$ are, in fact ideals of $R$, so we can choose a maximal one, say, $f(N)=Rd_n\in S$. Suppose further that $f(r_n)=d_n(r_n\in N)$. We want to show for any other morphism $g:M\rightarrow R$, the value of $g$ at the point $r_n$ is ‘greater’ than $d_n$, that is, $d_n|g(r_n)$. For simplicity, we denote $g(r_n)=d$. Now consider the gcd of $d_n,d$, that is $d'=gcd(d_n,d)=ad_n+bd$(the last identity is due to Bezout’s lemma). Now the new morphism $g'=af+bg$ has an image of $N$ greater than that of $f$. Indeed, note that $g'(N)\supset Rd'\supset Rd_n$. But according to the maximality of $f$, we must have that these three ideals are equal, and thus $d_n|d'$, which implies that $d_n|d$. We note that $d_n$ can not be zero, which is easy to see. Next we want to divide $r_n$ by $d_n$. This can be done by considering those projections $p_i: M\rightarrow R$, projecting to the $i$-th component. The values of these maps at $r_n$ are all divisible by $d_n$. That is to say, if we write $r_n=\sum_i a_ie_i$ in the canonical coordinate form ,then it reads that $d_n|p_i(r_n)=a_i$, thus $r_n$ is divisible by $d_n$. Thus we can put $r_n=d_nr_n'$, and therefore $d_nf(r_n')=d_n$, which implies, due to the fact that $R$ is integral, that $f(r_n')=1$. So now we can define a map $F:M\rightarrow ker(f)\bigoplus f(M)=ker(f)\bigoplus Rr_n', m\mapsto (m-f(m)r_n',f(m)r_n')$. $F$ is well defined, and is a morphism. So, after a simple verification, we get a decomposition $M=ker(f)\bigoplus Rr_n'=M'\bigoplus Rr_n', N'=(ker(f)\bigcap N)\bigoplus Rr_n=N'\bigoplus Rr_n$. So $M'$ is a free module of rank $n$, and $N$ is one of its submodules, thus using the induction assumption, we have that there exist a basis $(e_0,...,e_{n-1})$ of $M'$ and $(d_0,...,d_{n-1})\in R^n(d_{n-1}|...|d_0)$ such that $M'=\sum_i Re_i, N'=\sum_i Rd_ie_i$. Now to show that $d_n|d_{n-1}$, we define a ‘smart’ morphism, $G: M=M'\bigoplus Rr_n'\rightarrow R, (\sum a_ie_i, rr_n')\mapsto a_{n-1}+r$. Restricting to $N$, we have that $G':N=N'\bigoplus Rr_n, (\sum a_id_ie_i, rr_n)\mapsto a_{n-1}d_{n-1}+rd_n$. One value that $G'$ can take is obviously $d_{n-1}$. What is more, $G$ is ‘greater’ than $f$ in that $G'(N)\supset Rd_n$, thus $d_{n-1}\in G'(N)=Rd_n$, which implies that $d_n|d_{n-1}$. Which completes the proof, and also this post.