LEC ALG4 5

A classification of modules over Euclidean domains

Let $R$ be a Noetherian ring, and $M$ be a finitely generated $R$-module with generating set $\{x_1,\dots ,x_n\}$. The kernel of the map $ϵ:R^n\to M$ given by Prp 373.14 is exactly all the $R$-linear relations among $\{x_1,\dots ,x_n\}$. Since $R^n$ is Noetherian by Prp 389.3, $\ker ϵ$ is finitely generated, say by $m$ elements. So, there exists a surjective map ${ϵ}_1:R^m\twoheadrightarrow \ker ϵ$. Let $\phi$ be the composite $\iota \circ {ϵ}_1$.

Observe that

$$\frac{R^n}{\mathrm{im}\phi }\cong \frac{R^n}{\ker ϵ}\cong \text{im}ϵ\cong M.$$

Thus, for Noetherian $R$ and a finitely generated $R$-module $M$, there exists a $R$-module homomorphism $\phi :R^m\to R^n$ such that $M\cong \text{coker}\phi$. Assuming the standard basis for $R^m$ and $R^n$, $\phi$ can be represented by a $n\times m$ matrix with entries in $R$, which we call a presentation matrix for $M$.

Definition 391.1(Definition).

An $R$-module $M$ is finitely presented if for some positive integers $m,n$ there is an exact sequence

$$R^n\stackrel{\phi }{\to }{R}^m\to M\to 0.$$

Such a sequence is called a presentation of $M$, and the matrix of $\phi$ is called a presentation matrix for $M$.

We have just shown that finitely generated modules over Noetherian rings are finitely presented.

For example, the $\mathbb{Z}$-module $\mathbb{Z}/5\mathbb{Z}$ is presented by the map $\mathbb{Z}\to \mathbb{Z}$ given by $1\mapsto 5$, or equivalently by the matrix $\left[\begin{array}{c}5\end{array}\right]$.

With the basis of $R^n$ being $\{x_1,\dots ,x_n\}$, the columns of the presentation matrix of $M$ form a complete set of relations among the generators of $M$ (a set $S$ of relations is said to be complete if every relation is a linear combination of $S$ with coefficients in $R$).

Let $A$ be the matrix of $\phi$. We can edit $\phi :R^m\to R^n$ in several ways while keeping $\text{coker}\phi$ unchanged. We can change the bases for $R^m$ or $R^n$:

$$\frac{R^n}{\text{im}\phi }\cong \frac{R^n}{{\sigma }^{-1}(\text{im}\phi )}.$$

Thus, if $R$ happens to be a euclidean domain, we can use Prp 405.3 to assume WLOG that $A$ is in Smith normal form. But this is not the whole story. Two different homomorphisms ${\phi }_1$, ${\phi }_2$ may have isomorphic cokernels, even if they act between different modules. Case in point: if $A$ has a column of zeroes, we can remove the column - the corresponding homomorphism ${\phi }^{\prime }:R^{m-1}\to R^n$ clearly has the same cokernel as $\phi$. Thus, we can assume WLOG that $A$ is of the form

$$\left[\begin{array}{ccc}{d}_1& & \\ & \ddots & \\ & & d_r\\ & 0_{k\times r}& \end{array}\right],$$

where $k⩾0$ and $d_1|\dots |d_r$. Finally, if $d_1$ is a unit, we have

$$\frac{R^n}{\text{im}\phi }=\frac{R^{n-1}\oplus R}{\text{im}{\phi }^{\prime }\oplus R}\cong \frac{R^{n-1}}{\text{im}{\phi }^{\prime }},$$

where ${\phi }^{\prime }:R^{m-1}\to R^{n-1}$ corresponds to the matrix obtained from $A$ by dropping column $1$ and row $1$. We can therefore assume that $d_1$ (and hence all $d_i$ for $1⩽i⩽r$) are nonunits. $\text{coker}\phi$ now takes the form

$$\text{coker}\phi \cong \frac{R^n}{d_{1}R\oplus \cdots \oplus d_{r}R}\cong \frac{R}{d_{1}R}\oplus \cdots \oplus \frac{R}{d_{r}R}\oplus R^k.$$

And that’s it! We’ve shown that when $R$ is a euclidean domain, any finitely generated $R$-module $M$ can be expressed as a direct sum of cyclic modules. In fact, this result is true over any PID, as we will prove in the coming lectures.

Prp 2 summarizes our observations on cokernel preserving matrix manipulations.

Proposition 391.2(Artin (2011) 14.5.7).

Let $A$ be an $n\times m$ presentation matrix for a finitely generated $R$-module $M$. The following matrices represent the same module $M$:

1. $Q^{-1}A$, with $Q\in G{L}_n(R)$;

2. $AP$, with $P\in G{L}_m(R)$;

3. A matrix obtained by deleting a column of zeroes from $A$;

4. If the $j$th column of $A$ is $e_i$, the matrix obtained from $A$ by deleting row $i$ and column $j$.

Proof.

Let $\phi :R^m\to R^n$ be the map that $A$ describes, interpreted using standard bases.

$(1)$ corresponds to changing the basis of $R^n$ to $Q$. Note that the map $\phi$ remains unchanged.

$(2)$ corresponds to changing the basis of $R^m$ to $P$. Again, $\phi$ is not impacted.

$(3)$ A column of zeros corresponds to a trivial relation, which can be omitted. More concretely, dropping a zero column gives us a map ${\phi }^{\prime }:R^{m-1}\to R^n$ with $\text{im}\phi =\text{im}{\phi }^{\prime }$, so $M=R^n/\text{im}\phi =R^n/\text{im}{\phi }^{\prime }$. Thus, the matrix of ${\phi }^{\prime }$ presents $M$.

$(4)$ A column of $A$ equal to $e_i$ corresponds to the relation $v_i=0$. The zero element is useless as a generator, and its appearance in any other relation is irrelevant. So we may delete $v_i$ from the generating set and from the relations.□

2296e1

---

Decomposition of finitely generated modules into torsion and free components

Our first step toward a classification theorem for modules over PIDs¹ (Thm 407.2) is showing that a finitely generated module over a PID is a direct sum of a free module and a ‘torsion’ module.

We’ve already classified finitely generated free modules over integral domains, so it makes sense to begin by examining how a module can fail to be free. There are several ways this can happen, with having torsion being one of them.

Definition 391.3(Definition).

Let $R$ be an integral domain, and $M$ an $R$-module.

1. A nonzero element $x\in M$ is called a torsion element if $\{x\}$ is linearly dependent, i.e, there exists nonzero $a\in R$ such that $ax=0$.
2. We say $M$ is torsion-free if $M$ does not contain any torsion elements.
3. We say $M$ is a torsion module if all elements of $M$ are torsion elements.
4. $\text{tor}(M):=\{x\in M:x\text{ is a torsion element}\}\cup \{0\}$. Note that $\text{tor}(M)$ is a submodule of $M$.

73f393

Note that free modules are torsion-free.

Lemma 391.4(Lemma).

Submodules and direct sums of torsion-free modules are torsion-free. Free modules over an integral domain are torsion-free.

Here’s another mechanism in which a module may fail to be free.

Example 391.5(Example).

Let $R=\mathbb{Z}[x]$, and let $I=(2,x)$. Then $I$ is not a free $R$-module. More generally, any nonprincipal ideal of an integral domain $R$ is a torsion-free module which is not free. Indeed, if $I$ were free, its rank would have to be $1$ at most, by Lem 373.8; thus one element would suffice to generate $I$, and $I$ would be principal.

Lemma 391.6(Lemma).

Let $M$ be a finitely generated $R$-module. Then, $M/\text{tor}(M)$ is torsion free (and finitely generated).

Proof.

If $x\in E$, let $\overline{x}$ denote its residue class mod $\text{tor}(M)$. Let $b\in R$, $b\ne 0$ be such that $b\overline{x}=0$. Then $bx\in \text{tor}(M)$, and hence there exists nonzero $c\in R$ such that $cbx=0$. Hence $x\in \text{tor}(M)$ and $\overline{x}=0$, thereby proving that $M/\text{tor}(M)$ is torsion free.□

4998de

Analogous to the case for abelian groups, submodules of free modules are free of rank less than or equal their parent.

Lemma 391.7(Lang (2002) 3.7.1).

Let $F$ be a free module over a PID, and $M$ a submodule. Then $M$ is free, and its dimension is less than or equal to the dimension of $F$.

84b3e5

Lemma 391.8(Isaacs (2009) 16.28).

Let $R$ be a PID, $M$ a finitely generated torsion free $R$-module. Then $M$ is free.

Proof.

Some general observations:

• By Prp 114.11, $R$ is a UFD. Therefore, we can talk of prime factorizations of elements of $R$.
• By Cor 389.4, $M$ is Noetherian.

We work by induction on the number $n$ of generators for $M$. If $n=0$, then $M=0$ and the result is trivially true. Assume $n>0$ and fix a generator $x$ of $M$. We consider singly generated submodules of the form $Ry$ for elements $y\in M$, and we write $\Lambda =\{yR:x\in yR\}$. By Prp 389.1.2, $\Lambda$ has a maximal element, say $N=Ry$. Note that $M/N$ is generated by $n-1$ elements.

We claim that $M/N$ is torsion free. In other words, we must show that if $z\in M$ is such that $za\in N$ for some $a\in R$, then $z\in N$. Let $z$ be such an element, with $za=yb\in N$ for some $a,b\in R$. If $a$ and $b$ have any common prime factor $\pi$, we can write $a=a_0\pi$, $b=b_0\pi$, and $(z{a}_0-y{b}_0)\pi =0$. Since $M$ is torsion free, we must have $z{a}_0=y{b}_0$. Therefore, we can assume WLOG that $a$ and $b$ are coprime. By Rmk 114.6, we can write $au+bv=1$ with $u,v\in R$.

Write $y=1\cdot y=(au+bv)y=auy+avz=a(yu+zv)$. Thus, we have $yR\subseteq (yu+zv)R$. Since $(yu+zv)R$ is clearly a member of $\Lambda$, we deduce $yR=(yu+zv)R$ by the maximality of $N$. It follows that $yu+zv\in N$. Since $y\in N$, this yields $zv\in N$. So, $z=1\cdot z=auz+bvz=buy+bvz\in N$, and $M/N$ is torsion free.

We have the short exact sequence

$$0\to N\hookrightarrow M\twoheadrightarrow M/N\to 0.$$

By the induction hypothesis, $M/N$ is free, and hence projective. By Cor 386.3 and Conrad (n.d.) 2.1, the sequence splits, and we have $M=N\oplus M/N$. Since $M$ is torsion free, $N$ is isomorphic to the free rank 1 $R$-module $R^1$. It follows that $M$ is free.□

Proof.

From Lang (2002) 3.7.3; Uses Lem 7.

Let $Y=\{y_1,\dots ,y_m\}$ be a generating set for $M$. Let $V=\{v_1,\dots ,v_n\}$ be a maximal linearly independent subset of $Y$. For $y\in Y$, there exist elements $a,b_1,\dots ,b_n\in R$, not all $0$, such that

$$ay+\sum _{i=1}^n{b}_i{v}_i=0.$$

Then $a\ne 0$, lest we contradict the linear independence of $V$. Hence $ay$ lies in the free submodule $R^{\oplus V}$ of $M$ generated by $V$. For each $j=1,\dots ,m$, we can find $a_j\in R$, $a_j\ne 0$, such that $a_j{y}_j\in R^{\oplus V}$. Let $a=a_1\dots a_m$ be the product. Then $aM\subseteq R^{\oplus V}$ and $a\ne 0$. Since $M$ is torsion free, the map $a\mapsto ax$ is an injective homomorphism, and $M$ is isomorphic to its image under this map. We conclude form Lem 7 that $M$ is free.□

fdca90

For example, $\mathbb{Q}$ is a non-free torsion free $\mathbb{Z}$ module. $\mathbb{Q}$ is not a finitely generated $\mathbb{Z}$ module.

Theorem 391.9(Theorem).

Let $R$ be a PID, and $M$ a finitely generated $R$-module. Then $M\cong \text{tor}(M)\oplus \frac{M}{\text{tor}(M)}$.

Proof.

We have the short exact sequence

$$0\to \text{tor}(M)\hookrightarrow M\twoheadrightarrow M/\text{tor}(M)\to 0.$$

Again, by Lem 6 and Lem 8, $M/\text{tor}(M)$ is free, and hence projective. By Cor 386.3 and Conrad (n.d.) 2.1, the sequence splits.□

1a540b

Thus, every finitely generated module is a direct sum of a finitely generated torsion module and a finitely generated free module. Free modules have a simple structure that we understand. It remains to study finitely generated torsion modules ($\text{tor}(M)$ must be finitely generated since it is a submodule of $M$, which is Noetherian).

Footnotes

1. Aluffi does a brilliant job of motivating why we’re interested in PIDs here using free resolutions - see Aluffi (2009, pp. 342–344), in particular VI.4.11. ↩

---

References

Aluffi, P. (2009). Algebra: Chapter 0. American Mathematical Society.

Artin, M. (2011). Algebra (2. ed). Pearson Education, Prentice Hall.

Conrad, K. (n.d.). SPLITTING OF SHORT EXACT SEQUENCES FOR MODULES. https://kconrad.math.uconn.edu/blurbs/linmultialg/splittingmodules.pdf

Isaacs, I. M. (2009). Algebra: A Graduate Course. American Mathematical Society. https://doi.org/10.1090/gsm/100

Lang, S. (2002). Algebra (Vol. 211). Springer New York. https://doi.org/10.1007/978-1-4613-0041-0